fundamentals of electrical & electronics engg

FUNDAMENTALS OF ELECTRICAL & ELECTRONICS ENGG.

1st Year (E-Contents)

Er. ANIL KUMAR

Head of Department ( ECE )

Department of Technical Education Haryana

1 FUNDAMENTAL OF ELECTRICAL & ELECTRONICS ENGG

DETAILED CONTENTS

UNIT – I Overview of DC Circuits (05 hrs)

1) Simple Problems on Series and Parallel combination of

Resistors and Capacitors with their Wattage consideration,

2) Application of Kirchhoff’s Current Law and Kirchhoff’s Voltage

Law to Simple Circuits.

3) Star – Delta Connections and their Conversion.

UNIT – II DC Circuit Theorems (05 hrs)

1) Thevenin’s theorem,

2) Norton’s theorem,

3) Application of Network Theorems in solving D.C. Circuit

Problems.

4) Superposition Nodal Analysis & Mesh Analysis,

5) Maximum Power Transfer Theorem.

UNIT – III Voltage and Current Sources (04 hrs)

1) Concept of Voltage Source, Symbol, Characteristics and

Graphical Representation of Ideal and Practical Sources.

2) Concept of Current Sources, Symbol, Characteristics and

Graphical Representation of Ideal and Practical Sources.


DETAILED CONTENTS

UNIT – IV Semiconductor Physics (06 hrs)

1) Review of Basic Atomic Structure and Energy Levels, Concept

of Insulators, Conductors and Semi Conductors,

2) Energy Level Diagram of Conductors, Insulators and Semi

Conductors;

3) Atomic Structure of Germanium (Ge) and Silicon (Si), Covalent

bonds

4) Concept of Intrinsic and Extrinsic Semi Conductor, Process of

Doping.

5) P and N type semiconductors and their conductivity, Effect of

Temperature on Conductivity of Intrinsic Semi Conductors.

6) Minority and Majority Charge Carriers.

UNIT – V Semiconductor Diode (08 hrs)

1) PN junction Diode, Mechanism of Current flow in PN junction,

Forward and Reverse biased PN junction, Potential Barrier,

Drift and Diffusion Currents, Depletion Layer,

2) Concept of junction Capacitance in Forward and Reverse

biased condition.

3) V-I Characteristics, Static and Dynamic Resistance and their

Value Calculation from the Characteristics.

4) Application of Diode as Half Wave, Full Wave and Bridge

Rectifiers. Peak Inverse Voltage, Rectification Efficiencies and

Ripple Factor calculations, Shunt Capacitor Filter, Series

Inductor Filter, LC and π Filters.

5) Types of Diodes, Characteristics and Applications of Zener

diodes , Zener and Avalanche Breakdown.


DETAILED CONTENTS

UNIT – VI Electro Magnetic Induction (06 hrs)

1) Concept of Electro-Magnetic Field produced by flow of Electric

Current, Magnetic Circuit, Concept of Magneto-Motive Force

(MMF), Flux, Reluctance, Permeability, Analogy between

Electric and Magnetic Circuit.

2) Faraday’s Laws of Electro-Magnetic Induction, Principles of

Self and Mutual Induction, Self and Mutually induced E.M.F,

Simple Numerical Problems.

3) Concept of Current Growth, Decay and Time Constant in an

Inductive (RL) Circuit.

4) Energy stored in an Inductor, Series and Parallel Combination

of Inductors.

UNIT – VII Batteries (05 hrs)

1) Basic idea of Primary and Secondary Cells

2) Construction, Working Principle and Applications of Lead-Acid,

Nickel-Cadmium and Silver-Oxide Batteries

3) Charging methods used for Lead-Acid Battery (Accumulator )

4) Care and Maintenance of Lead-Acid Battery

5) Series and Parallel connections of Batteries

6) General idea of Solar Cells, Solar Panels and their Applications

7) Introduction to Maintenance Free Batteries


DETAILED CONTENTS

UNIT – VIII AC Fundamentals (05 hrs)

1) Concept of Alternating Quantities

2) Difference between AC and DC

3) Concepts of: Cycle, Frequency, Time Period, Amplitude,

Instantaneous Value, Average Value, RMS. Value, Maximum

Value, Form Factor and Peak Factor.

4) Representation of Sinusoidal Quantities by Phasor Diagrams.

5) Equation of Sinusoidal Wave Form for an Alternating Quantity

and its Derivation.

6) Effect of Alternating Voltage applied to a Pure Resistance,

Pure Inductance and Pure Capacitance.

UNIT – IX AC Circuits (06 hrs)

1) Concept of Inductive and Capacitive Reactance

2) Alternating Voltage applied to Resistance and Inductance in

Series.

3) Alternating Voltage applied to Resistance and Capacitance in

Series.

4) Introduction to Series and Parallel Resonance and its

Conditions

5) Power in Pure Resistance, Inductance and Capacitance,

Power in combined RLC Circuits. Power Factor, Active and

Reactive Power and their Significance, Definition and

significance of Power Factor.

6) Definition of Conductance, Susceptance, Admittance,

Impedance and their Units


DETAILED CONTENTS

UNIT – X Introduction to Bipolar-Transistors (06 hrs)

1) Concept of a Bipolar Transistor, its structure, PNP and NPN

Transistors, their Symbols and mechanism of Current Flow;

Current relations in a Transistor; Concept of Leakage current;

2) CB, CE, CC Configurations of a Transistor; Input and Output

characteristics in CB and CE configurations; Input and Output

Dynamic Resistance in CB and CE Configurations; Current

Amplification Factors. Comparison of CB, CE and CC

Configurations;

3) Transistor as an Amplifier in CE Configuration; Concept of DC

load line and Calculation of Current Gain and Voltage Gain

using DC Load Line.

UNIT - XI Transistor Biasing Circuits (04 hrs)

1) Concept of Transistor Biasing and Selection of Operating

Point. Need for Stabilization of Operating Point.

2) Different Types of Biasing Circuits.

UNIT – XII Field Effect Transistors (05 hrs)

1) Construction, Operation and Characteristics of FETs and their

Applications.

2) Construction, Operation and Characteristics of a MOSFET in

Depletion and Enhancement Modes and its Applications.

3) CMOS - Advantages and Applications

4) Comparison of JFET, MOSFET and BJT.


DETAILED CONTENTS

UNIT – XIII Introduction to Electrical Machines (05 hrs)

1) Transformers : Principal of Operation, Construction Detail of

Single Phase Transformer, Turns ratio , Efficiency, Loses in a

Transformer.

2) DC Machine : Principal of Operation, Construction of DC Motor

and Generator,

3) Characteristics of Different Types of DC Machines , Starter .

4) AC machines: Principal and Working of Synchronous

Machines, Single Phase Induction Motor

Section

Percentage

of syllabus to

be covered

Units to

be

covered

Type of

assessment

Weight-

age of

Marks

Pass

Percentage

A 20% Unit 1 to

3 1st Internal

40%

40%(Combin

ed in internal

& final

assessment)

with

minimum

25% marks

in final

assessment)

B 20% Unit 4 , 5

2nd Internal

C 60% Unit 6 to

13 FINAL

60%


DETAILED CONTENTS

PRACTICALS

1) Operation and Use of measuring Instruments viz Voltmeter,

Ammeter, CRO, Wattmeter, Multi-meter and Other

accessories.

2) Measurement of Resistance of an Ammeter and a Voltmeter.

3) Verification of following Theorems:-

i. Thevenin’s theorem,

ii. Norton’s theorem,

4) Observation of Change in Resistance of a Bulb in Hot and Cold

conditions, using Voltmeter and Ammeter.

5) Verification of Krichhoff's Current and Voltage Laws in a DC

Circuit.

6) To find the Ratio of Inductance of a Coil having Air-Core and

Iron-Core respectively and to observe the effect of Introduction

of a Magnetic Core on Coil Inductance.

7) Charging and Testing of a Lead - Acid storage Battery.

8) Measurement of Power and Power Factor in a Single Phase R-

.L-.C. Circuit and Calculation of Active and Reactive Powers in

the Circuit.

9) Plotting of V-I Characteristics of a PN junction diode & Zener

diode.

10) Observe the output of waveform using;

i. Half-wave rectifier circuit using one diode

ii. Full-wave rectifier circuit using two diodes.

iii. Bridge-rectifier circuit using four diodes.

11) Plotting of the Wave Shape of Full Wave Rectifier with

i. Shunt Capacitor Filter.

ii. Series Inductor Filter.

12) Plotting of Input and Output Characteristics and Calculation of

Parameters of Transistors in CE configuration.

13) Plotting of Input and Output Characteristics and Calculation of

Parameters of Transistors in CB configuration.

14) Plotting of V-I Characteristics of a FET.

15) To determine the efficiency of Single Phase Transformer.

FUNDAMENTAL OF ELECTRICAL & ELECTRONICS ENGG Page 1 of 36

UNIT – I

OVERVIEW OF DC CIRCUITS


IMPOTANT DEFINITIONS:

1. Voltage : Voltage is the pressure ( Electromotive Force EMF ) from Power

Source of an Electrical Circuit's that pushes Charged Electrons (Current) through

a conducting closed Loop Circuit and enabling them to do work. It is measured

in volts (V).

Voltage is Potential Difference between the Two Points / Terminals of the

conductor.

2. Current: Electric Current is defined as the rate of flow of Negative Charges of

the Conductor. The conducting material consists a large number of free electrons

which move from one atom to the other at random.

In other words, the continuous flow of electrons in an Electric Circuit is

called an Electric Current. It is measured in Ampere ( A )

3. Resistance: It is the property of a substance which oppose the flow of Current

through it. It is measured in Ohms ( Ω ).

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UNIT – I


A Common Analogy of Voltage, Current, and Resistance, with

Water Tank

When describing Voltage, Current, and Resistance, a common analogy is

a Water Tank. In this analogy, Charge is represented by the Water amount,

Voltage is represented by the Water Pressure, and Current is represented by the

Water Flow. So for this analogy, remember:

i. Water = Charge (measured in Coulombs)

ii. Pressure = Voltage (measured in Volts)

iii. Flow of Water = Current (measured in Amperes,)

iv. Pipe Width = Resistance ( measured in Ohms )

4. Electric Power: Electric Power is defined as the rate at which Electrical

Energy is consumed in an electrical circuit. The SI unit of power is the Watt,

which is one joule per second.

Electric Power (P) can be calculated as Energy consumption (E) divided

by the time consumed (t):

P = E / t, with P in Watts, E in Joules and t in Seconds

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UNIT – I


5. Electric Energy: Electrical Energy is the Energy generated by the movement

of Electrons from one Point to another Point. Electrical energy is the work done

by Electric Charge.

If Current I ampere flows through a Conductor of Potential Difference V

volts across it, for time t Second,

Electric Energy ( E ) = Vx I x t = VIt = P. t Joules

6. Ohm’s Law: It is State that the Electric Current passing through a Conductor is

directly proportional to the Potential Difference across it, under physical conditions

remains same.

I α V

I = G. V Where G is Conductance of Conductor, G = 1 /

R

I = V / R , V = I. R , R = V / I

3

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https://www.electrical4u.com/voltage-or-electric-potential-difference/

https://www.collinsdictionary.com/dictionary/english/electric

https://www.collinsdictionary.com/dictionary/english/current

https://www.collinsdictionary.com/dictionary/english/passing

https://www.collinsdictionary.com/dictionary/english/directly

https://www.collinsdictionary.com/dictionary/english/proportional

https://www.collinsdictionary.com/dictionary/english/potential

https://www.collinsdictionary.com/dictionary/english/difference


UNIT – I


Factors that affect resistance of Conductor:

There are Four different factors which affect the value of Resistance:

1) Type of material used for Conductor. The resistance of a

conductor depends on the material used for makeup the Conductor.

Copper wire has less resistance than Steel wire of the same size.

2) Length of the Conductor. Resistance of Conductor also depends

upon the length of Conductor. Longer wires have more Resistance than

short wires and vice versa.

3) Cross-sectional Area (Thickness) of the Conductor.

Resistance of a Conductor depends upon the cross-sectional area of

Conductor. Thick wires have less Resistance than thin wires and vice-

versa.

4) Temperature of the Conductor. Electrical resistance also

depends on temperature. Resistance decreases with an increase of

Temperature and vice versa.

Relation of resistance with Length:

The resistance R of the wire is directly proportional to the length ( L ) of the wire :

R α L …..(1)

It means, if we double the length of the wire, its resistance will also be

doubled, and if its length is halved, its resistance would become one half.

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UNIT – I


Relation of resistance with area:

The resistance R of a wire is inversely proportional to the Area of Cross-section

(A) of the wire as:

R α 1/A ……(2)

It means that a thick wire would have smaller Resistance than a thin wire.

After combining the equations (1) and (2) we get;

R α L/A

R = ρ L / A ….(3)

Where ρ is the Constant of proportionality, known as specific resistance

(Resistivity). Its value depends upon the nature of conductor i.e copper, iron, tin,

and silver would each have different values of ρ. From equation (3) we have;

ρ = R A / L ….(4)

Resistivity: Resistivity is the resistance per Unit Length of a Substance. The

unit of ρ is ohm-meter (Ωm).

SERIES COMBINATION OF RESISTERS

Resistors are said to be connected in Series when they are connected together in

a single line resulting in a Common ( same ) Current flowing through them.

Or

Resistors are said to be in Series whenever the Current flows through the

Resistors sequentially.

As the Resistors are connected together in series the same current passes

through each Resistor in the chain and the Total Resistance, RT of the Circuit

must be equal to the sum of all the individual Resistors

RT = REQ = R1 + R2 + R3 = 1 kΩ + 2 kΩ + 6 kΩ = 9kΩ

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UNIT – I


The value of all above three individual Resistors is equivalent to one single

Resistor of value of 9kΩ.

Characteristics of Series Combination:

1. If there is only one path for the flow of current in a circuit then the

combination of resistances is called Series Combination.

2. In Series Combination, the current flowing through each Resistor is equal.

3. In Series Combination, Potential difference ( Voltage ) across each Resistor

is different depending upon the value of Resistance.

4. In Series Combination, Total (Equivalent) Resistance of Circuit is equal to

the Sum of individual Resistances.

Equivalent Resistance In Series Combination:

Consider Three Resistances R1, R2, & R3 connected in Series combination with

a Power Supply of Voltage.

Potential difference of each Resistor is V1, V2, & V3 respectively.

Let electric current I is passing through the circuit.

Then V = V1 + V2 + V3

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UNIT – I


According to Ohm’s law V = IR

I.RT = I.R1 + I.R2 + I.R3

IRT = I ( R1 + R2 + R3 )

IRT / I = R1 + R2 + R3

RT = R1 + R2 + R3

This shows that in Series combination Total (equivalent) Resistance of Circuit is

Sum of Individual Resistance and always greater than Individual Resistances.

RT = R1 + R2 + R3 + ………… Rn

Disadvantages of Series Combination:

If One Component is fused, then the Other Components of Circuit will not

function.

PARALLEL COMBINATION OF RESISTERS

Resistors are said to be connected in Parallel when one end of all the resisters

are connected together and other end of all the resisters also connected together

through a continuous wire of negligible resistance.

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UNIT – I


The voltage drop across all of the Resistors connected in Parallel is the same.

Then, Resistors in Parallel have a Common Voltage across them and this is

true for all Parallel connected Elements. Since there are multiple paths for the

supply Current to flow through, the current may not be the same through all the

branches in the Parallel Network.

Characteristics of Parallel Combination:

1. If there are more than one path for the flow of current in a circuit then the

combination of resistances is called Parallel Combination.

2. In Parallel combination current through each Resistor is different.

3. Potential difference across each Resistor is same.

4. Equivalent Resistance of Circuit is always less than either of the

Resistances included in the Circuit.

Equivalent Resistance In Parallel Combination:

Consider Three Resistances R1 , R2 & R3 connected in Parallel Combination with

a Power Supply of Voltage V.

Now

I = I1 + I2 + I3

According to Ohm’s law I = V / R

Therefore, V / Re = V / R1 + V / R2 + V / R3

V / Re = V (1 / R1 + 1 / R2 + 1 / R3 )

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UNIT – I


V / Re. V = 1 / R1 + 1 / R2 + 1 / R3

OR

This shows that in Parallel Combination , Equivalent Resistance of Circuit is

always less than Individual Resistances.

Advantages of Parallel Combination:

In Parallel Combination of Resistors, if one Component of Circuit (Resistor) is

damaged then rest of the Component of the Circuit will perform their work without

any disturbance. It is due to the presence of more than paths for the flow of

electric current.

Numerical No. – 1 Calculate the Equivalent Resistance for the below

Circuit which consists of 7 resistors and the supply voltage is 5 V. Also

calculate the Current supplied to the circuit.

9


UNIT – I


Solution:

The resistors R6 and R7 are in series combination. If the equivalent resistance of

R6 and R7 in series is Ra, then

Ra = R6 + R7 = 2+2 = 4Ω

The resulting circuit is reduced to the one shown below.

In the above circuit the resistors Ra and R5 are in parallel combination. Hence the

equivalent resistance of Ra and R5 is

Rb = (Ra X R5) / ( Ra + R5 ) = (4 X 4) / (4 + 4) = 2Ω.

Then the Simplified Circuit is shown is below.

In this Circuit the Resistors R4 and Rb are in Series Combination.

Rc = R4 + Rb = 10 + 2 = 12 Ω.

Now we can replace the resistors R4 and Rb with resistor Rc as shown below.

10


UNIT – I


In the above circuit again the resistors R2 and R3 are in series combination. If

Rd is the equivalent resistance of R2 and R3 then

Rd = R2 + R3 = 4 + 8 = 12 Ω.

The equivalent circuit is

Here resistors Rc and Rd are in parallel combination. Let Rp be the equivalent

resistance of Rc and Rd in parallel. Then

Rp = (Rc X Rd) / (Rc + Rd) = (12 X 12) / (12 + 12) = 6 Ω.

The Resulting Circuit is

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UNIT – I


Here, the resistors R1 and Rp are in series combination. Let REQ be the

equivalent resistance of this combination. Then

REQ = R1 + Rp = 4 + 6 = 10 Ω.

This is the equivalent resistance of the circuit. Hence the given circuit can be

finally redrawn as

The Current in the Circuit can be calculated from Ohm’s law

I = V / REQ = 5 / 10 = 0.5 A

Numerical No. – 2 Calculate the Equivalent Resistance for the below

Circuit which consists of 10 resistors and the supply voltage is 6 V. Also

calculate the Current supplied to the circuit.

12


UNIT – I


Solution:

Here the resistors R9 and R10 are in Series combination. Let RA is the equivalent

resistance of this combination.

Therefore, RA = R9 + R10 = 3 + 3 = 6 Ω.

The Circuit after replacing R9 and R10 with RA is

In this Circuit, the Resistors R8 and RA are in Parallel Combination. Then the

Equivalent Resistance of R8 and RA is

RB = ( R8 X RA ) / ( R8 + RA ) = (6 X 6 ) / ( 6 + 6 ) = 3 Ω.

Now Replacing R8 and RA with RB, we get the following Circuit.

In this circuit, the resistors R7 and RB are in Series Combination.

RC = R7 + RB = 9 + 3 = 12 Ω.

The equivalent circuit after replacing R7 and RB with RC is

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UNIT – I


It is clear that the Resistors R6 and Rc are in Parallel Combination. If RD is the

Equivalent Resistance of this combination, then

RD = (R6 X Rc) / (R6 + Rc) = (12 X 12) / (12 + 12) = 6 Ω.

The circuit with RD replacing R6 and Rc is

Now the Resistors R4 and RD are in Series Combination. If RE is the Equivalent

Resistance of R4 and RD then

RE = R4 + RD = 6 + 6 = 12 Ω.

The resulting reduced Circuit after replacing R4 and RD with RE is

In this Circuit, the Resistors R5 and RE are in Parallel Combination.

Let RF be the Equivalent Resistance of R5 and RE in Parallel.

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UNIT – I


Then, RF = (R5 X RE) / (R5 + RE) = (12 X 12) / (12 + 12) = 6 Ω.

The simplified circuit is as shown below.

Here resistors R2 and R3 are in series. If RG is equivalent of this combination,

then

RG = R2 + R3 = 4 + 2 = 6 Ω.

After replacing R2 and R3 with RG, the Simplified Circuit will be as shown below:

The resistors RF and RG are in Parallel.

Let RT be the Equivalent of this Combination.

Then RT = (RF X RG) / (RF + RG) = (6 X 6) / (6 + 6) = 3 Ω.

Now the resistors R1 and RT are in Series. If REQ is the Total Circuit Equivalent

Resistance, then REQ = R1 + RT = 3 + 3 = 6 Ω.

Finally the above Circuit is as follows

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UNIT – I


Now the Total Current in the Circuit can be calculated using Ohm’s law

I = V1 / REQ = 6 / 6 = 1 A.

SERIES COMBINATION OF CAPACITORS

Capacitors are said to be connected in Series when they are connected together

in a single line resulting in a Common ( same ) Charging Current flowing

through them.

Characteristics of Series Combination of Capacitors:

1. When Capacitors are connected in Series, the magnitude of charge Q on

each Capacitor is same.

Q = Q1 = Q2 = Q3

2. When Capacitors are connected in Series, The Potential difference across

each Capacitors (C1 , C2 and C3) is different i.e., Vc1 , Vc2. and VC3

respectively.

3. The Total Voltage of the Battery connected Series combinations of

capacitors has been divided among the various Capacitors.

Hence

VAB = VC1 + VC2 + VC3

= Q / C1 + Q / C2 + Q / C3

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UNIT – I


= Q [ 1 / C1 + 1 / C2 + 1 / C3 ]

VAB / Q = [ 1 / C1 + 1 / C2 + 1 / C3 ]

1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3

4. The Calculation of Total Series Capacitance is analogous to the calculation

of Total Resistance of Parallel Resistors.

5. When Capacitors are connected in Series, the Total Capacitance is less than

the Smallest Capacitance Value because the effective Plate separation

increases.

PARALLEL COMBINATION OF CAPACITORS

Capacitors are said to be connected in Parallel, when both of its terminals are

connected to each terminal of another Capacitor.

Characteristics of Parallel Combination of Capacitors:

1. When Capacitors are connected in Parallel, the magnitude of charge Q on

each Capacitor is different and Total Charge Q will be:

Q = Q1 + Q2 + Q3

2. When Capacitors are connected in Parallel, The Potential difference across

each Capacitors (C1 , C2 and C3) is Same

i.e. VAB = Vc1 = Vc2 = VC3

3. The Total Charge Q Supplied by the Battery connected Parallel combinations

of capacitors has been divided among the various Capacitors.

Hence

Q = Q1 + Q2 + Q3

CT x VAB = C1 x VAB + C2 x VAB + C3 x VAB

CT x VAB = VAB [ C1 + C2 + C3 ]

CT = C1 + C2 + C3

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https://physicsabout.com/capacitor/


UNIT – I


4. The Calculation of Total Parallel Capacitance is analogous to the calculation

of Total Resistance of Series Resistors.

5. When Capacitors are connected in Parallel, the Total Capacitance CT is being

the Sum of all the individual Capacitance’s

CT = C1 + C2 + C3 = 0.1 uF + 0.2uF + 0.3uF = 0.6uF

6. When Capacitors are connected in Parallel, The Total Capacitance CT is

always be greater than the value of the Largest Individual Capacitor.

Problem No: 3 When three Capacitors, C1 = 16 μF, C2 = 8 μF, C3 = 8 μF,

are connected in Series and Parallel as shown in figure below. Determine

the Equivalent Capacitance that will have the same effect as the

combination.

Solution :

We know that:

Capacitor C1 = 16 μF



Equivalent capacitance (C) = ?

Capacitor C2 and C3 connected in Parallel.

Equivalent capacitance : CP = C2 + C3 = 8 + 8 = 16 μF

18

https://physicsabout.com/capacitor/

https://physics.gurumuda.net/capacitors-in-series-and-parallel-problems-and-solutions.htm

https://physics.gurumuda.net/electric-charge-stored-in-capacitor-problems-and-solutions.htm


UNIT – I


Now, Capacitor C1 and Cp connected in Series.

Equivalent Capacitance : 1 / C = 1 / C1 + 1 / CP = 1 / 16 + 1 / 16

1 / C = 2 / 16 = 1 / 8

C = 8 μF Ans.

Problem No: 4. When Five Capacitors, C1 = 5 μF, C2 = 4 μF, C3 = 6 μF, C4 = 5 μF,

C5 = 10 μF, are connected in series and parallel as shown in figure. Determine the

equivalent Capacitance that will have the same effect as the combination.

Solution :

We know






Equivalent Capacitance (C) = ?

Capacitor C2 and C3 are connected in Parallel.

Equivalent Capacitance : CP = C2 + C3

CP = 4 + 6 = 10 μF

Now, Capacitors C1, CP, C4 and are connected in Series.

Equivalent Capacitance : 1 / CS = 1 / C1 + 1 / CP + 1 / C4

1 / CS = 1 / 5 + 1 / 10 + 1 / 5

1 / CS = 5 / 10 = 1 / 2

CS = 2 μF

Now, Capacitors CS and C5, are connected in Parallel.

Equivalent Capacitance (C) = CS + C5 = 2 μF + 10 μF = 12 μF

Equivalent Capacitance (C) = 12 μF Ans.

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https://physics.gurumuda.net/kirchhoffs-first-rule-problems-and-solutions.htm


UNIT – I


KIRCHHOFF’S LAWS

In 1845, German Physicist Gustav Kirchhoff was described relationship of two

Quantities in Current and Potential Difference (Voltage) inside a Circuit. This

relationship or rule is called as Kirchhoff’s Circuit Law.

Kirchhoff’s Circuit Law consist two Laws:

1) Kirchhoff’s Current Law - Which is related with current flowing,

inside a closed circuit and called as KCL

2) Kirchhoff’s Voltage Law -Which is to deal with the voltage sources of

the circuit, known as Kirchhoff’s voltage law or KVL.

Kirchhoff’s Current Law : According Kirchhoff’s Current Law (KCL), the

Algebraic Sum of Currents in an Electrical Circuit meeting at a Junction ( Node )

is Zero.

Here, the Three Currents entering the Node (Incoming Currents), I1, I2, I3 are all

Positive in value and the Two Currents leaving the node (Outgoing

Currents), I4 and I5 are Negative in value. Then this means:

I1 + I2 + I3 – I4 – I5 = 0

I1 + I2 + I3 = I4 + I5

OR

According Kirchhoff’s Current Law (KCL), at any Junction ( Node ) of the

Conductors Network, Incoming Current is equal to Outgoing Current.

Σ IIN = Σ IOUT

20


UNIT – I


Kirchhoff’s Voltage Law: According Kirchhoff’s Voltage Law (KVL), in any

Closed Loop of the Circuit, the Algebraic Sum of all the Voltage Drops is equal to

Zero.

In other words the Algebraic Sum of All the Potential differences around the loop

must be equal to zero as: ΣV = 0.

The term “Algebraic Sum” means to take into account the Polarities and

Signs of the Sources and Voltage drops around the loop.

So when applying Kirchhoff’s Voltage Law to a Specific Circuit

Component, it is important that we need a special attention to the Algebraic

signs, (+ and -) of the Voltage drops across the components and the EMF’s of

Sources otherwise our calculations may be wrong.

When, the flow of Current through the Resistor is from point A to point B, that is

from Positive terminal to a Negative terminal, the Potential difference across the

resistance will be ( - IR ) Voltage drop across it.

When, the flow of Current is in the opposite direction from point B to

point A, that is from Negative terminal to Positive terminal, the Potential

difference across the resistance will be ( +IR ) Voltage drop across it.

By Appling KVL on the above Circuit,

VS + ( - I x R1 ) + ( - I x R2 ) = 0

So, VS = IxR1 + IxR2

21


UNIT – I


VS = I ( R1 + R2 )

VS = I x RT ( Where RT = R1 + R2 )

I = _VS_ = ____VS____ RT R1 + R2

VR1 = I x R1 = VS x [ __R1__ ] R1 + R2

VR2 = I x R2 = VS x [ __R2__ ] R1 + R2

Numerical No. – 5; Find the Currents flowing in all branches of the

Circuit given below by using Kirchhoff’s Laws ( KVL & KCL ).

Solution: IT is the Total Current flowing around the Circuit by the 12V DC Supply

Voltage. At point A, I1 is equal to IT, thus there will be an I1xR voltage drop

across resistor R1.

Now, the Circuit has two Branches, 3 Nodes (B, C and D) and 2

Independent Loops, thus the IxR Voltage drops around the two Loops will be:

Loop ABC ⇒ 12 = 4I1 + 6I2

Loop ABD ⇒ 12 = 4I1 + 12I3

Since Kirchhoff’s Current Law States that at node B, I1 = I2 + I3, we can therefore

substitute Current I1 for (I2 + I3) in both of the following loop equations and then

simplify.

Kirchhoff’s Loop Equations

22


UNIT – I


We now have two simultaneous equations that relate to the currents flowing

around the circuit.

12 = 10I2 + 4I3 ………………………................eq. (1 )

12 = 4I2 + 16I3 ………………………………...eq ( 2 )

By Multiplying equation no. 1 by 4 & equation no-2 by 1 and now subtracting

equation 4 from equation 3 and by reducing both equations, the value of I2 will

be calculated.

( 12 = 10I2 + 4I3 ) x ( 4 ) ⇒ 48 = 40I2 + 16I3 ………..eq (3)

( 12 = 4I2 + 16I3 )( x1 ) ⇒ 12 = 4I2 + 16I3…………eq (4)

Eq. No 3 – Eq. No 4 ⇒ 36 = 36I2 + 0

I2 = 1.0 Amps

Now Multiplying the equation-1 by 4 and equation no-2 by 10. Again by

subtracting equation ( 6 ) from equation ( 5) , and by reducing both equations to

give us the values of I3

12 = 10I2 + 4I3 ( x4 ) ⇒ 48 = 40I2 + 16I3 ……… eq (5 )

12 = 4I2 + 16I3 ( x10 ) ⇒ 120 = 40I2 + 160I3 ……… eq (6 )

Eq. No ( 5 ) – Eq. No ( 6 ) ⇒ 72 = 0 + 144I3

Thus I3 = 0.5 Amps

According to KCL: I1 = I2 + I3

I1 = 1.0 Amp + 0.5 Amp = 1.5 Amps

Thus I1 = IT = 1.5 Amps, I2 = 1.0 Amps and I3 = 0.5 Amps

Numerical - 6; Three Resistor of values: 10 ohms, 20 ohms and 30 ohms,

respectively are connected in series across a 12 volt DC Power Supply as

shown in figure:

Calculate: a) Total Resistance,

b) Circuit Current,

23


UNIT – I


c) Current through each Resistor,

d) Voltage Drop across each Resistor,

e) Verify that Kirchhoff’s Voltage Law, KVL is also true.

Solution:

a) Total Resistance (RT)

RT = R1 + R2 + R3 = 10Ω + 20Ω + 30Ω = 60Ω

Then the total circuit resistance RT = 60Ω

b) Circuit Current (I)

Thus the total Circuit Current ( I ) = 0.2 Amperes or 200mA

c) Current Through Each Resistor

The Resistors are connected together in Series, therefore same current

will flow through each registers

Thus: IR1 = IR2 = IR3 = I = 0.2 Amperes

d) Voltage Drop Across Each Resistor

VR1 = I x R1 = 0.2 x 10 = 2 volts

VR2 = I x R2 = 0.2 x 20 = 4 volts

VR3 = I x R3 = 0.2 x 30 = 6 volts

e) Verify Kirchhoff’s Voltage Law

Thus Kirchhoff’s voltage law is also true as the individual voltage drops around

the closed loop add up to the total.

Kirchhoff’s Voltage Law, KVL is Kirchhoff’s Second Law and states that the

Algebraic Sum of all the Voltage drops in a Closed Loop Circuit is always Zero.

ΣV = 0

24


UNIT – I


Numrical-7; Find the Current flowing in the 40Ω Resistor ( R3 ) by using

Kirchhoff’s Voltage Law and Kirchhoff’s Current Law as shown in figure

given below :

Solution: The Circuit has 3 Branches, 2 Nodes (A and B) and 2 Loops.

Using Kirchhoffs Current Law, KCL the equations are given as:

At node A : I1 + I2 = I3

At node B : I3 = I1 + I2

Using Kirchhoffs Voltage Law, KVL the equations are given as:

Loop 1 is given as : 10 = R1 I1 + R3 I3 = 10I1 + 40I3

Loop 2 is given as : 20 = R2 I2 + R3 I3 = 20I2 + 40I3

Loop 3 is given as : 10 – 20 = 10I1 – 20I2

As I3 is the sum of I1 + I2 we can rewrite the equations as;

10 = 10I1 + 40(I1 + I2) = 50I1 + 40I2 ……………………eq- 1

20 = 20I2 + 40(I1 + I2) = 40I1 + 60I2 …………………….eq -2

By Multiplying equation no.-1 by 4 & equation no-2 by 5 and subtract

equation no-4 from equation no-3 and by reducing both equations, the

value of I2 will be calculated.

40 = 200 I1 + 160 I2 ……………………eq- 3

100 = 200 I1 + 300 I2 ……………………eq -4

Eq. No 3 – Eq. No 4 ⇒ - 60 = 0 + -140 I2

I2 = + 0.429 Amps

25


UNIT – I


Now By Multiplying equation no.-1 by 3 & equation no-2 by 2 and subtract

equation no-6 from equation no-5 and by reducing both equations, the

value of I1 will be calculated.

30 = 150 I1 + 120 I2 ……………………eq- 5

40 = 80 I1 + 120 I2 ……………………eq -6

Eq. No 5 – Eq. No 5 ⇒ - 10 = 70 I1 + 0

I1 = -0.143 Amps

As : I3 = I1 + I2

I3 = -0.143 + 0.429

I3 = 0.286 Amps

The Current flowing in Resistor R3 is I3 = 0.286 Amps

Voltage across Resistor R3 is VR3 = I3 x R3 = 0.286 x 40

VR3 = 11.44 volts

Applications of Kirchhoff’s Laws:

1. Kirchhoff’s Laws can be used to determine the values of unknown values like

Current, Voltage, Current as well as the direction of the flowing values in the

circuit.

2. Kirchhoff’s Laws can be applied on any Electric Circuit ( excluding High

Frequency Circuits and fluctuating Magnetic Field linking the closed Loop )

and useful to find the unknown values in Complex Circuits and Networks.

3. Kirchhoff’s Laws also used in Nodal and Mesh analysis to find the values of

Current and Voltage.

4. Current through each independent loop is carried by applying KVL (each

loop) and current in any element of a circuit by counting all the current

(Applicable in Loop Current Method).

5. Current through each branch is carried by applying KCL (each junction)and

KVL in each loop of a circuit (Applicable in Loop Current Method).

6. Kirchhoff’s Laws are useful in understanding the Transfer of Energy through

an Electric Circuit.

26


UNIT – I


STAR – DELTA CONNECTIONS AND THEIR CONVERSION

Star Connection: When the terminals of the three Branches are connected to

a common point. The Network formed is known as Star Connection.

A star connection has a Common or a Star Point to which all the Three

terminals are connected forming a Star Shape as shown below.

Delta Connection: When the three Branches of the Network are

connected in such a way that it forms a closed Loop known as Delta

Connection

In Delta Connection, All the three Terminals are connected together

forming a Closed Loop. In this configuration, there is no Common or Neutral

Point, and it is used for Power Transmission for Short Distances. The connection

diagram is shown below.

27


UNIT – I


Delta To Star Conversion: The replacement of Delta (mesh) Connection

by its equivalent Star Connection is known as Delta – Star Conversion.

The Two Connections are equivalent or identical to each other if the

Impedance is measured between any pair of Lines will be the same irrespective

of whether the delta is connected between the lines or its equivalent star is

connected between that lines.

Consider a Delta System that’s three Corner Points are A, B and C as shown in

the above figure. Electrical resistance of the Branch between Points A and B, B

and C and C and A are R1, R2 and R3 respectively.

The resistance between the points A and B will be,

Now, one Star System is connected to these points A, B, and C as shown in the

figure. Three arms RA, RB and RC of the Star System are connected with A, B and

C respectively. Now if we measure the resistance value between points A and B,

we will get,

Since the Two Systems are identical, Resistance measured between terminals A

and B in both systems must be equal.

Similarly, Resistance between points B and C being equal in the two systems,

28

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UNIT – I


Similarly, Resistance between points C and A being equal in the two systems,

Adding Equations (I), (II) and (III) We will get,

Subtracting equations (I), (II) and (III) from equation (IV) we get,

The relation of Delta – Star Conversion can be expressed as follows.

The Equivalent Star Resistance connected to a given terminal, is equal to the

Product of the Two Delta Resistances connected to the same Terminal divided by

the Sum of the Delta connected Resistances.

If the Delta connected System has Same Resistance ( R1 = R2 = R3 = R )

at its Three Sides then equivalent Star Resistance RSTARwill be,

RA = RB = RC = RSTAR = ____R x R____ = R2 = _R_ R + R + R 3R 3

Star To Delta Conversion

29



UNIT – I


In order to Convert of the Resistances of Delta Network in to Resistances of Star

Network.

We know, In Star Connection the value of RA, RB, RC are given below:

For finding the value of Resistance in Delta Connection, We just multiply each set

of two equations and then add.

that is by doing (V) × (VI) + (VI) × (VII) + (VII) × (V)

We get,

30


UNIT – I


Now dividing equation (VIII) by equations (V), (VI) and equations (VII) separately

we get,

By using above relations, we can find Resistances of Delta Network from

Resistances of Star Network. In this way, we can convert Star Network into Delta

Network.

Numerical – 8; Find the equivalent resistance between A & B in the

network given below:

Solution:-

For the given Network, we can easily determine the value of equivalent

Resistance i.e, RAB through Star-Delta conversion.

We have

31

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UNIT – I


Above Network can be represent as below:-

Now, This Network can be converted from Delta to its equivalents Star

configuration as shown in the figure given below:-

For the value of new star connected resistance are finding through direct formula

of delta to star conversion, as shown below

32

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https://2.bp.blogspot.com/-TZ9knbWxk6Q/UzsLW8ipsBI/AAAAAAAAAeI/F5oNXgCoReM/s1600/Copy+of+star-delta-solved.jpg

https://4.bp.blogspot.com/-DfUTqNjxfL0/UzsQmDvKykI/AAAAAAAAAe4/Is7Ucp2TNuM/s1600/Copy+of+Copy+of+star-delta-solved.jpg

https://2.bp.blogspot.com/-Ve5glIjUW3o/UzsPyIhj3GI/AAAAAAAAAew/YS9yaIr9uVw/s1600/Copy+of+DELA+TO+STAR+FORMULA.png


UNIT – I


So, RAB = Requivalent = R1 + R2 + R3 = 4Ω + 3.88Ω + 1.77Ω = 9.65

Ω Answer

33

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UNIT – I


Difference Star Connection and Delta Connection

Sr. No.

BASIS Star Connection Delta Connection

1. Basic Definition

The terminals of the Three

Branches are connected to a

common point. The Network

formed is known as Star

Connection

The Three Branches of the

Network are connected in

such a way that it forms a

Closed Loop known as Delta

Connection

2. Connection of Terminals

The Starting and the

Finishing Point that is the

Similar ends of the Three

Coils are connected together

The End of Each Coil is

connected to the Starting

Point of the Other Coil that

means the opposite terminals

of the Coils are connected

together.

3. Neutral Point Neutral (Star Point) exists in

the Star Connection.

Neutral Point does not exist in

the Delta Connection.

4.

Relation between line and Phase Current

Line Current is equal to the

Phase Current.

Line Current is equal to root

three times of the Phase

Current.

5.

Relation between Line and Phase Voltage

Line Voltage is equal to root

three times of the Phase

Voltage

Line Voltage is equal to the

Phase Voltage.

6. Speed

The Speed of the Star

connected motors is slow as

they receive 1/√3 of the

Voltage.

The Speed of the Delta

connected motors is high

because each Phase gets the

total of the Line Voltage.

7. Phase Voltage Phase voltage is low as 1/√3

times of the Line Voltage.

Phase Voltage is equal to the

Line Voltage.

8. Number of Turns

Requires less number of

turns

Requires large number of

turns.

9. Insulation Level

Insulation required is low. High insulation is required.

10 Network Type Mainly used in the Power

Transmission Networks.

Used in the Power Distribution

Networks.

11. Received voltage

In Star Connection each

winding receive 230 volts

In Delta Connection each

winding receives 414 volts.

12. Type of System

Both Three Phase four wire

and Three Phase three wire

system can be derived in star

connection.

Three Phase four wire system

can be derived from the Delta

connection.

34


UNIT – I


Fill IN THE BLANKS:

1. Resistivity of a wire depends on ………………….

2. A circuit contains two un-equal resistances in parallel, potential difference

across each is …………….

3. A circuit contains two un-equal resistances in parallel, Current flowing

through branches is …………….

4. The conductivity of material is measured in ……………… .

5. The ………………. is the reciprocal of Conductivity of that material.

6. A Circuit containing Two un-equal Resistance connected in Series, the

voltage drop across low value Resistance is …………… than High Value

Resistance.

7. The unit of resistance is ……………………

8. ………………. is the reciprocal of Resistance.

9. When Resistances are connected in Series, …………….. Current flows

through all Resistances.

10. The unit of conductance is………………...

11. Total capacitance will be ……………., when three capacitors, C1, C2 and C3

are connected in parallel

12. When capacitors are connected in Parallel, the total Capacitance is always

…………………… the individual capacitance values.

13. When Resistors are connected in Parallel, the equivalent Resistance is

always …………………… the individual Resistance values.

14. When Capacitors are connected in Parallel, the effective Plate Area will

…………….…

15. When Capacitors are connected in Series, the equivalent Capacitance is

…………… each individual Capacitance.

Answers : 1) Material 2) Same 3) Different 4) Ω / m

5) Resistivity 6) Lesser 7) ohm (Ω). 8) Conductance

9) Same 10) mho 11) C1 + C2 + C3 12) Greater than

13) Less Than 14) Increases 15) Less Than

35


UNIT – I


Fill IN THE BLANKS:

16. The resistance of the wire is Inversely proportional to its ………………..of

the wire/conductor.

17. Resistance of a wire is directly proportional to its …………….

18. In a Series Circuit, the total resistance is …………. the largest resistance in

the circuit.

19. In a Parallel Circuit, the total resistance is ……………. the smallest

resistance in the circuit.

20. In parallel connection of resisters, the voltage across each resistor

is…………..

21. Resistors are connected end to end in ……………. combination.

22. ……………….. is measured in mho .

23. Kirchhoff’s current law is applied at ……………. of the circuit.

24. The sum of the voltages over any closed loop is equal to ………. .

25. KVL is applied in ………….. Loop circuit..

26. Delta connection is also known as ……………. Connection.

27. …………….. connection is also known as Y-Connection.

Answers :

16) Cross sectional area 17) Length 18) Greater than

19) Smaller than 20) Same 21) Series

22) Conductance 23) Junction 24) Zero

25) Closed 26) Mesh 27) Star

36


UNIT – II

DC CIRCUIT THEORMS

DC CIRCUIT THEORMS

THEVENIN’S THEOREM

Thevenin’s Theorem states that “Any linear circuit containing several voltages and

resistances can be replaced by just one single voltage Source in series with a

single resistance connected across the load“.

In other words, it is possible to simplify any Electrical Circuit to an its

Equivalent Two-Terminal Circuit with just a Single Constant Voltage Source VTHin

Series with a Resistance RTH connected to the Load.

Basic Procedure for Solving a Circuit using Thevenin’s

Theorem is as follows:

1. Remove the Load Resistor RL.

2. Find the Open Loop Voltage (VTh ) across the both terminals, where Load

was connected

3. Replacing All Sources by their Internal Resistances. If Sources are Ideal

then Short Circuit the Voltage Sources and Open Circuit the Current Sources.

4. Find the Equivalent Resistance at the Load Terminal known as Thevenin

Resistance RTh.

5. Re-Connect Open Loop Thevenin Voltage (VTh ), Thevenin Resistance

RTh and Load Resistor RL in series as shown in Figure:

37


UNIT – II

DC CIRCUIT THEORMS

6. Find the Current flowing through the Load Resistor RL.

Load Current IL is given as

Where,

VTH is Thevenin’s Equivalent Voltage.

RTH is Thevenin’s Equivalent Resistance

RL is Load Resistance

Explanation of Thevenin’s Theorem

The Thevenin’s Statement is explained with the help of a Circuit shown below;

Let us Consider a Simple DC Circuit as shown in the figure above, Where we have

to find the Load Current IL by the Thevenin’s Theorem.

In order to find the Equivalent Voltage Source, RL is removed from the

Circuit as shown in the figure below;

38


UNIT – II

DC CIRCUIT THEORMS

Replacing All Sources by their Internal Resistance. If Sources are Ideal then

Short Circuit the Voltage Sources and Open the Current Sources. Find the

Equivalent Resistance at the Load Terminals AB known as Thevenin Resistance

RTh .

Now, Re-Connect VTh and RTh in Series and this will be the Thevenin’s Equivalent

Circuit as shown in Figure below:

39


UNIT – II

DC CIRCUIT THEORMS

Connect Load Resistance RL across the Open Terminals A-B of the above Circuit

as shown in Figure below:

Numerical – 1; Find VTH, RTH and the Load Current IL flowing through and

Load Voltage across the Load Resistor RL as Shown in figure below by

using Thevenin’s Theorem.

Solution:-

Step - 1: Open the 5 KΩ Load Resistor from the Circuit as shown in figure below:

40


UNIT – II

DC CIRCUIT THEORMS

Step - 2.: Calculate the open circuit voltage Thevenin Voltage (VTH) across the

Terminal AB. Load Resistor RL already removed from above Circuit and became

an open circuit as shown in figure above.. Now Calculate the Thevenin’s Voltage

(VTH). Since 48 V Supply connected in Series with both 12kΩ and 4kΩ resistors

(Total Resistance = 16Ω) So 3mA (48/16 = 3 mA) Current flows in

both 12kΩ and 4kΩ resistors as these are connected in Series Circuit and

the 8kΩ resistor is as open Circuit and hence no Current will flow through it.

So VTH = VAB = (3mA x 4kΩ) + (0mA x 8kΩ)

VTH = 12 V + 0 V = 12 V

VTH = 12 V

Step – 3 : Replacing Voltage Sources by their Internal Resistance and The

Voltage Sources is Ideal, so Short Circuited the Voltage Source. Now, find the

Equivalent Resistance at the Load Terminals AB known as Thevenin Resistance

RTH .

41


UNIT – II

DC CIRCUIT THEORMS

Step – 4 : Calculate Open Circuit Resistance across the terminal AB i. e.

Thevenin Resistance (RTH)

Removed 48V DC Source to zero as Equivalent i.e. 48V DC source has

been replaced with a short. Resistor 8kΩ Resistor is in series with a Parallel

connection of 4kΩ Resistor and 12k Ω Resistor. i.e.:

RTH = 8 kΩ + ( 4k Ω || 12 kΩ )

RTH = 8kΩ + [ ( 4 kΩ x 12 kΩ ) / (4 kΩ + 12 kΩ ) ]

RTH = 8kΩ + 3kΩ

RTH = 11kΩ

Step - 5: Connect the RTH in Series with Voltage Source VTH and re-connect the

Load Resistor as shown in figure below.

This is Thevenin Equivalent Circuit with Load Resistor RL .

Thevenin’s equivalent circuit

42


UNIT – II

DC CIRCUIT THEORMS

Step 6: Now apply the Last Step i.e Ohm’s Law . Calculate the Total Load

Current & Load Voltage:

IL = VTH / ( RTH + RL )

= 12V / ( 11 kΩ + 5kΩ )

= 12V / 16 kΩ

IL = 0.75 mA

VL = IL x RL

VL = 0.75mA x 5kΩ

VL = 3.75 V

Numerical – 2: Find VTH, RTH and the Load Current IL flowing through and

Load Voltage VL across the Load Resistor RL of 40 Ω across A-B terminal as

Shown in figure below by using Thevenin’s Theorem.

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UNIT – II

DC CIRCUIT THEORMS

Solution:

Step 1 – Calculate Thevenin Resistance RTH

First remove the 40 Ω Load Resistor connecting terminals A and B, along with all

Voltage Sources with short Circuit as there is no internal Resistance in both

Voltage sources. .

To calculate the total Thevenin Resistance, we can use the following process:

Step 2 – Calculate Thevenin Voltage VTH

By using Ohm’s Law to Calculate the Total Current flowing through the Circuit :

Since these resistors are wired in series, they will share same 0.33 amps. We can

use these resistor values and our current to calculate the voltage drop, which is:

44


UNIT – II

DC CIRCUIT THEORMS

Step - 3 Calculate Load Current

Connect the RTH in Series with Voltage Source VTH and re-connect the Load

Resistor as shown in figure below.

This is Thevenin Equivalent Circuit with Load Resistor RL .

By Using Ohm’s Law , Calculate Total Current flowing across the Load Resistor:

Step -4 : Calculate Load Voltage VL

VL = IL x RL

VL = 0. 286 A x 40 kΩ

VL = 11.44 V

Numerical - 3 Find the Current in 3Ω Resistor of the Circuit as shown in

figure given below using Thevenin’s theorem.

45


UNIT – II

DC CIRCUIT THEORMS

Solution: The Thevenin’s theorem has four steps.

Step - 1 : Calculat Thevenin’s voltage VTh

To find Thevenin’s voltage VTh , remove 3Ω Resistor leaving other parts of the

circuit as it is and Calculate the Voltage across the Open Circuited terminals A-B.

To find voltage across the Open Circuited terminals A-B lets assume potential at

point C to be zero. Then Potential at point B is equal to 10V .

In left side loop, by Kirchoff’s voltage law

20 - 2I1 - 5I1 - I1 - 5 = 0

46


UNIT – II

DC CIRCUIT THEORMS

Step - 2 :- Calculate RTh

Remove 3Ω Resistor and replace all independent voltage sources by short circuits

(as internal resistance of ideal voltage source is zero) and all independent current

sources by open circuits (as internal resistance of ideal current source is infinite) .

Leave dependent Voltage and Current Sources as it is and obtain a circuit

with only resistances as shown.

Now find the Equivalent Resistance of this circuit looking through the open

circuited terminals a and b. This equivalent resistance will be the RTh .

Step-3 : Find Thevenin’s Equivalent Circuit.

Connect VTh , Rth and 3Ω resistor in series, and get Thevenin’s Equivalent Circuit

as shown in figure

47


UNIT – II

DC CIRCUIT THEORMS

Step-4 : Find Current through Load Resistor

Current through 3Ω resistor can be easily calculated as

NORTON’S THEOREM

Norton’s Theorem states that Current flowing through any Resistance between

Two Terminals of a Linear Circuit can be determined by replacing entire Linear

Circuit in to its Equivalent Circuit consisting with just a Single Current Source and

Parallel Resistance connected to a Load.

The Norton’s Theorems reduce the Entire Linear Networks in to its Equivalent

circuit having one Current Source, Parallel Resistance and Load.

Norton’s theorem is the converse of Thevenin’s Theorem. It consists of

Equivalent Current Source instead of an Equivalent Voltage Source as in

Thevenin’s Theorem. The determination of Internal Resistance of the Source

network is Identical in both the theorems. In Equivalent Circuit, Current Source is

placed in Parallel to Internal Resistance in Norton’s Theorem whereas In

Thevenin’s Theorem Equivalent Voltage Source is placed in Series with Internal

Resistance.

Norton’s Theorem is a method to reduce a Network to an Equivalent Circuit

composed of a Single Current Source, Parallel Resistance, and Parallel Load.

48


UNIT – II

DC CIRCUIT THEORMS

Basic Procedure for Solving a Circuit using Norton’s Theorem is as

follows:

1. Removing the Load Resistor (RL) from the original Circuit

2. Find the Norton Current Source (INorton) by shorting the both terminals

where the load resistor was connected and Calculate Total Current flowing

through these terminals.

3. Find the Norton Resistance by removing all Power Sources in the original

Circuit (voltage sources shorted and current sources open) and Calculating

total Resistance between the open connection Points.

4. Draw the Norton Equivalent Circuit, with the Norton Current Source in

Parallel with the Norton resistance. The Load Resistor RL re-connect

between the two open Points of the Equivalent Circuit as shown in figure.

5. Determine Load Current (IRL) and Load Voltage (VRL) across the Load

Resistor following the rules for Parallel Circuits.

IRL = INorton x RNorton VRL = IRL x RL RNorton + RL

Explanation of Norton’s Theorem

Step – 1: Identify Load Resistance

The first step is to Identify the Load Resistance and remove it from the original

circuit as shown in Figure below:

49


UNIT – II

DC CIRCUIT THEORMS

Step-2: Find Norton Current For Current Source

Now, Find the Norton Current (for the current source in the Norton Equivalent

Circuit), by shorting the connection between the Load Points and determine the

resultant Current.

Note This step is exactly opposite the respective Step in Thevenin’s Theorem,

where the load resistor with open circuit:

With Zero Voltage dropped between the Load Resistor connection Points, the

current through R1 is only a function of Battery (B1) Voltage and R1 resistance

IR1 = VB1 / R1 = 28 V / 4Ω = 7 Amps

Similarly, the Current through R3 is now only a function of Battery (B2) voltage and

R3 resistance

IR2 = VB2 / R3 = 7 V / 1Ω = 7 Amps

Total Current through the Short between the Load connection Points is the sum of

these two currents

ISORT = IR1 + IR2 = 7 Amps + 7 Amps = 14 Amps.

Norton Source Current (INorton) = 14 Amps in Equivalent Circuit:

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UNIT – II

DC CIRCUIT THEORMS

Note The arrow notation for Current Source indicates the direction of flow of

current

Step – 3: Find Norton Resistance

To Calculate Norton Resistance (RNorton),: Take original circuit (with the load

resistor still removed), remove the Power sources voltage sources replaced with

Short and current sources replaced with Open), and Calculate Equivalent

Resistance :

Note Exact same as for Calculating Thevenin resistance (RThevenin)

RNorton = R1 // R3

RNorton = R1 x R3 = 4 x 1 = 4 / 5 = 0.8 Ω R1 + R3 4 + 1

Step – 4: Norton equivalent circuit looks like this:

51


UNIT – II

DC CIRCUIT THEORMS

Step – 4: Determine Load Current & Load Voltage

Re-connect original Load Resistance of 2 Ω as shown above, The Norton Circuit

as a Simple Parallel Arrangement:

IRL = INorton x RNorton = 14 x 0.8 = 11.2 / 2.8 = 4 Amps RNorton + RL 0.8 + 2

VRL = IRL x RL = 4 x 2 = 8 V

Numerical - 4 Find RN, IN, the current flowing through Load IL and Voltage

across the Load Resistor VL of the Figure as shown below by using

Norton’s Theorem.

Solution:-

Step-1: Identify the Load Resistance RL of 1.5 Ω and remove it from the original

circuit .

Step 2. Find Norton Current For Current Source. This is Norton Current (IN).

Short the terminal A-B in place of 1.5Ω Load resistor as shown above. The AB

terminals shorted to determine the Norton current, IN. The 6Ω and 3Ω are then in

Parallel and this Parallel combination of 6Ω and 3Ω are then in series with 2Ω.

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UNIT – II

DC CIRCUIT THEORMS

So the Total Resistance of the circuit to the Source is:-

Total Resistance (RT) = 2Ω + (6Ω || 3Ω)

RT = 2Ω + [(3Ω x 6Ω) / (3Ω + 6Ω)]

RT = 2Ω + 2Ω = 4Ω.

RT = 4Ω

IT = V / RT = 12V / 4Ω = 3A

Now ISC = IN

(By Applying Current Divider Rule)…

ISC = IN = 3A x [(6Ω / (3Ω + 6Ω)] = 2A.

ISC = IN = 2A.

Step – 3: Find Norton Resistance:

To Calculate Norton Resistance (RN),: Take original circuit (with the load resistor

still removed), remove the Power sources voltage sources replaced with Short and

current sources replaced with Open), and Calculate Equivalent Resistance :

53


UNIT – II

DC CIRCUIT THEORMS

Reduced the 12V DC source to zero is Equivalent to replace it with a short as

shown in figure above. The 3Ω Resistor is in Series with a Parallel combination of

6Ω Resistor and 2Ω Resistor. i.e.:

RN = 3Ω + (6Ω || 2Ω)

RN = 3Ω + [ ( 6Ω x 2Ω ) / ( 6Ω + 2Ω ) ]

RN = 3Ω + 1.5Ω

RN = 4.5Ω

Step 4: Norton equivalent circuit looks like this. Connect the RN in Parallel with

Current Source IN

Step 5: Determine Load Current & Load Voltage.

Now, Re-connect the Load Resistor as shown in fig below i.e. Norton Equivalent

circuit with Load Resistor

54


UNIT – II

DC CIRCUIT THEORMS

Calculate the Load Current IL and Load Voltage VL by Ohm’s Law

Load Current

IL = IN x [RN / ( RN + RL ) ]

= 2A x (4.5Ω / 4.5Ω + 1.5Ω ) = 2 x ( 4.5 / 6) = 2 x 3 / 4 = 1.5 A

IL = 1. 5A

And Load Voltage across Load Resistor VL = IL x RL = 1.5A x 1.5Ω

VL = 2.25V

Numerical – 5: Find Norton’s Equivalent Circuit of the circuit given below.

Step 1: Find Norton Current For Current Source. This is Norton Current (IN).

Now, Find the Norton Current (for the current source in the Norton Equivalent

Circuit), by shorting the connection between the Load Points (a - b) and determine

the resultant Current.

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UNIT – II

DC CIRCUIT THEORMS

Using loop analysis; i1 = 2 A

20 i2 − 4 i1 − 12 = 0

20 i2 − 4 x 2 − 12 = 0

20 i2 − 20 = 0

20 i2 = 20

i2 = 1 A

So;

IN = Isc = i2 = 1A

Step - 2 Find Norton Resistance:

To Calculate Norton Resistance (RN),: Take original circuit (with the load resistor

still removed), remove the Power sources voltage sources replaced with Short and

current sources replaced with Open), and Calculate Equivalent Resistance :

Reduced the 12V DC source to zero is Equivalent to replace it with a short and

reduced 2A Current source with Open Circuit as shown in figure above. The

Series combination of Resisters ( 8 + 4 + 8 = 20Ω ) is Parallel with of 5Ω

i.e.: RN = 5 || (8+4+8)

RN = 5 x 20 / ( 5 + 20 ) = 100 / 25

RN = 4 Ω

Step 3: Norton equivalent circuit looks like this. Connect the RN in Parallel with

Current Source IN.

56


UNIT – II

DC CIRCUIT THEORMS

SUPERPOSITION THEOREM

Superposition Theorem States that in any Linear Network having more than one

Sources ( Voltage or Current Source ), then Resultant Current flowing through any

branch is Algebraic sum of Current obtained from each Source considered

separately and all other sources are replaced by their internal resistance.

Or

The Superposition Theorem States that in any Linear Bilateral Network that

consisting of Two or more Independent Sources, Current flowing through (or

Voltage drop across) an element is the algebraic Sum of the Currents flowing

through (Voltages across) that element caused by each Independent Source acting

alone with all other Sources are replaced by their internal resistances.

I1 = I1' - I1''

I2 = I2'' - I2'

I3 = I3' + I3''

As long as the Linearity exists between the Source and Contribution, the Total

contribution due to various Sources acting simultaneously is equal to the algebraic

Sum of Individual contributions due to Individual Source acting at a time.

Basic Procedure for Solving a Circuit using Superposition

Theorem is as follows:

1. Consider the Various Independent Sources in a given Circuit.

2. Select and Retain one of the Independent Sources and replace all other

Sources with their internal Resistances ( Replace the Current Sources with

open Circuits and Voltage Sources with short Circuits.

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UNIT – II

DC CIRCUIT THEORMS

3. Find out the desired Voltage/Currents due to the one Source acting alone

using various circuit reduction techniques.

4. Repeat the steps 2 to 3 for each independent Source in the given Circuit.

5. Algebraically add all the Voltages/Currents that are obtained from each

individual Source (Consider the Voltage signs and Current directions while

adding).

Numerical - 6 Find the Voltage drop across the Resistance of 10Ω in a

Simple DC Circuit given below by applying the Superposition.

Solution :

Let us consider the above Simple DC circuit and by apply the Superposition

Theorem for finding the voltage across the resistance 10 Ohms.

Step-1: Consider that in a given Circuit there are Two independent Sources as

One Voltage Source and other One Current Sources.

Step-2: First, one Source at a time that means , only Voltage Source is acting in

the Circuit and the Current Source is replaced with open Circuited as shown in

figure.

Consider VL1 is the Voltage across the Load terminals with voltage source acting

alone, then

VL1 = Vs × RL / (RL + R1)

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UNIT – II

DC CIRCUIT THEORMS

= 20 × 10 / (10 + 20)

= 6. 66 Volts

Step-3 Retain the Current Source alone and replace the Voltage Source with its

internal Resistance (zero) so it becomes a short circuited as shown in figure.

Consider that VL2 is the voltage across the load terminals when current source

acting alone.

Then

VL2 = IL × RL

IL = I × R1 / (R1 + RL)

= 1 × 20 / (20 +30)

= 0.4 Amps

VL2 = 0.4 × 10

= 4 Volts

Step-4 : According the Superposition Theorem, the Voltage across the Load is

the sum of VL1 and VL2

VL = VL1 + VL2

= 6.66 + 4

VL = 10.66 Volts

Numerical - 7 Find the Current flowing through 20 Ω Resistor of the

circuit given below using Superposition Theorem.

59


UNIT – II

DC CIRCUIT THEORMS

Solution:

Step - 1 : Let us find the Current flowing through 20 Ω Resistor by considering

only 20 V voltage Source. In this case, we can eliminate the 4 A Current Source

by making open Circuit of it. The modified Circuit diagram is shown below:

There is only one Principal Node except Ground in the above circuit. So, we can

use Nodal analysis method. The Node Voltage V1 is labeled in the figure below.

Here, V1 is the Voltage from Node 1 with respect to ground.

The nodal equation at Node 1 is

V1−20 + __V1__ + ___V1 ___ = 0 5 10 ( 10+20)

⇒ 6V1−120+3V1+V1 = 0 30

⇒ 10V1 = 120

⇒ V1 = 12V

The Current Flowing through 20 Ω Resistor can be found by doing the following

simplification.

I1 = V1 / (10+20)

Substitute the value of V1 in the above equation.

60


UNIT – II

DC CIRCUIT THEORMS

I1 = 12 / (10+20 ) = 12 / 30 = 0.4 A

Therefore, the current flowing through 20 Ω resistor is 0.4 A, when only 20 V

voltage source is considered.

Step 2 : Let us find the Current flowing through 20 Ω Resistor by considering

only 4 A Current Source. In this case, we can eliminate the 20 V Voltage Source

by making short-circuit of it. The modified circuit diagram is shown in the following

figure.

In the above Circuit, there are three Resistors to the left of terminals A & B. We

can replace these Resistors with a Single equivalent Resistor. Here, 5 Ω & 10 Ω

resistors are connected in parallel and the entire combination is in series with 10 Ω

resistor.

The equivalent Resistance to the left of terminals A & B will be

RAB = [ 5×10] +10 = 10 + 10 = 40 Ω 5+10 3 3

The simplified circuit diagram is shown in the following figure.

The Current flowing through 20 Ω resistor can be by using current division

principle.

61


UNIT – II

DC CIRCUIT THEORMS

I2 = IS [ R1 ]

R1+R2 Substitute IS = 4A, R1 = 40 Ω and R2 = 20 Ω

3 in the above equation.

I2 = 4 [ 40 / 3 ] = 4 [ 40 / 100 ] = 1.6 A 40/3+20

Therefore, the current flowing through 20 Ω resistor is 1.6 A, when only 4 A

current source is considered.

Step 3: The Current flowing through 20 Ω Resistor of the given Circuit by doing

the addition of Two Currents that got in Step 1 and Step 2.

Mathematically, it can be written as

I = I1 + I2

Substitute, the values of I1 and I2 in the above equation.

I = 0.4 + 1.6 =2A

Therefore,

the Current flowing through 20 Ω Resistor of given circuit is 2 A.

Numerical - 8 Find the Current flowing through 4 Ω Resistor of the circuit

given below using Superposition Theorem.

Solution: Consider the Above Circuit to determine the Current I through the 4 Ω

Resistor using Superposition Theorem.

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UNIT – II

DC CIRCUIT THEORMS

Consider I1, I2 and I3 are the Currents due to Sources 12 V, 20 V and 4 A Sources

respectively. Then, based Superposition Theorem

I = I1 + I2 + I3.

Step -1: Only with 12V Voltage Source :

Consider the below Circuit where only 12V Source is retained in the Circuit and

other Sources are replaced by their internal Resistances.

By combining the Resistance 6 Ω with 10 Ω we get 16 Ω resistance which is

parallel with 6 Ω Resistance. Then this combination produce, 16 × 6 / (16 + 6) =

4.36 Ω. Therefore the Equivalent Circuit will be as shown in figure.

Then the current through 4 Ω Resistance,

I1 = 12 / 8.36

I1 = 1.43 A

Step – 2 : Only with 20 V Voltage Source:

Retain only 20 V Voltage Source and replace other Sources with their internal

Resistance, then the circuit becomes as shown below.

63


UNIT – II

DC CIRCUIT THEORMS

Apply the Mesh analysis to the Loop a,

22 Ia – 6 Ib + 20 = 0

22 Ia – 6 Ib = - 20 ……………….(1)

For Loop b,

10 Ib – 6 Ia = 0

Ia = 10 Ib / 6

Substituting Ib in Equation 1

22 ( 10 Ib / 6 ) – 6 Ib = - 20

Ib = – 0.65

Therefore, I2 = Ib = - 0.65

Step - 3: Only with 4A Current Source

Consider the below Circuit where only Current Source is retained and other

Sources are replaced with their internal Resistances.

By applying Nodal analysis at Node - 2

4 = ( V2 / 10 ) + ( V2 – V1 ) / 6 ………………..(2)

At Node - 1,

( V1 / 6 ) + ( V1 / 4 ) = ( V2 – V1 ) / 6

2V1 + 3V1 = V2 - V1

12 6 6 5 V1 + V1 = V2 12 6 6

V2 = 3.496 V1

Substituting V2 in Equation - 2,

4 = ( 3.496 V1 / 10 ) + (3.496 V1 – V1 ) / 6

V1 = 0.766 Volts.

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UNIT – II

DC CIRCUIT THEORMS

Therefore I3 = V1 /4

= 0. 766/4

I3 = 0.19 Amps. Thus, as per Superposition Theorem, I = I1 + I2 + I3

= 1.43 – 0.65 + 0.19 I = 0.97 Amps.

MAXIMUM POWER TRANSFER THEOREM

Maximum Power Transfer Theorem stated as – When a Resistive Load, being

connected to a DC Network, it receives Maximum Power When the Load

Resistance is equal to the Internal Resistance of Source Network.

RL = RS

The Maximum Power Transfer Theorem is used to find the Load Resistance for

which there would be the Maximum amount of Power Transfer from the Source to

the Load.

As far as the Load Resistor RL is concerned, any Complex “one-port”

Network consisting of multiple resistive Circuit elements and Energy Sources can

be replaced by one Single Equivalent Resistance Rs and one Single Equivalent

Voltage Vs. Rs is the Source Resistance value looking back into the Circuit

and Vs is the Open Circuit Voltage at the terminals.

However, when a Load Resistance RL is connected across the Output

terminals of the Power Source, the impedance of the Load will vary from an Open-

Circuit state to a Short-Circuit, The resulting Power being absorbed by the Load

becoming dependent on the impedance of the actual Power Source. For

transferring Maximum Power to the Load, Impedance Matching of Source with

Load is required .

65


UNIT – II

DC CIRCUIT THEORMS

Transformer Impedance Matching

The maximum power transfer can be obtained even if the output Impedance is not

the same as the load impedance. This can be done using a suitable “turns ratio” on

the Transformer with the corresponding ratio of load impedance, ZLOAD to output

impedance, ZOUT matches that of the ratio of the transformers primary turns to

secondary turns as a resistance on one side of the transformer becomes a

different value on the other.

If the load impedance, ZLOAD is purely resistive and the source impedance is

purely resistive, ZOUT then the equation for finding the maximum power transfer is

given as:

Where: NP is the number of Primary turns on Transformer

NS the number of Secondary turns on Transformer

Then by varying the value of the transformers turns ratio the output impedance can

be “matched” to the source impedance to achieve maximum power transfer.

Numerical- 9: Determine the Maximum Power that can be delivered to

the Variable Resistor R in the Figure as shown below:

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UNIT – II

DC CIRCUIT THEORMS

Solution: (a) Find VTh : Open circuit voltage

From the circuit, Vab = VTh = 40 - 10 = 30

(b) RTh: Let’s apply Input Resistance Method:

Then Rab = (10 // 20 ) + ( 25 // 5 )

= 10 x 20 + 25 x 5 = 200 + 125 10+ 20 25 + 5 30 30 Rab = Rth = 6.67 + 4.16 = 10.83 Ω

(c) Thevenin Circuit:

67


UNIT – II

DC CIRCUIT THEORMS

Numerical – 10: If an 8Ω Loudspeaker is to be connected to an Amplifier

with an output impedance of 1000Ω, calculate the turns ratio of the matching

transformer required to provide maximum power transfer of the audio signal.

Solution: Assume the amplifier source impedance is Z1, the load impedance

is Z2 with the turns ratio given as N.

Generally, Small High Frequency Audio Transformers used in Low Power Amplifier

Circuits are nearly always regarded as ideal for simplicity, so any losses can be

ignored.

68


UNIT – II

DC CIRCUIT THEORMS

Fill IN THE BLANKS:

1. The Thevenin voltage is the ……………. circuit voltage.

2. Thevenin resistance is found by …………… all voltage sources and …………..

all current sources.

3. Thevenin’s theorem is true for …………….. Circuits.

4. In Thevenin’s theorem, Vth is found across the …… terminals of the Network.

5. In Superposition Theorem, when the effect of one Voltage Source to be

considered, all the other Voltage Sources are …………..

6. In Superposition Theorem, when the effect of one Current Source to be

considered, all the other Voltage Sources are………….. .

7. In Superposition Theorem, when the effect of one Current Source to be

considered, all the other Current Sources are ……………… .

8. In Superposition Theorem, when the effect of one Voltage Source to be

considered, all the other Current Sources are …………….

9. Superposition theorem is valid for ………….. Circuit.

10. The Norton Current ( IN or ISC ) is the ……………. circuit current.

11. Norton’s theorem is true for …………….. Circuits.

12. Thevenin’s Theorem is also known as the ……………. of Norton’s Theorem.

13. Norton’s Theorem is also known as the …………… of Thevenin’s Theorem.

14. The Norton Current ( IN or ISC ) is found across the ………. terminals of the

Network.

15. The maximum power drawn from source depends on Value of ……..

Resistance.

16. The maximum power is delivered to a circuit when source resistance is

……….. load resistance.

17. The Thevenin’s Voltage ( VTH ) is the ……………. Circuit Voltage.

Answers :

1) Open 2) Shorting, Opening 3) Linear 4) Output

5) Shorted 6) Shorted 7) Opened 8) Opened

9) Linear 10) Short 11) Linear 12) Dual or Converse

13) Dual or Converse 14) Output 15) Load 16) Equal to

17) Open

69


UNIT – III

VOLTAGE AND CURRENT SOURCES

VOLTAGE AND CURRENT SOURCES ELECTRICAL SOURCE

An Electrical Source is a Device which converts Mechanical, Chemical, Thermal

or Some other Form of Energy into Electrical Energy.

In other words, the Source is an Active Network Element meant for

generating Electrical Energy. The Various Types of Sources available in the

Electrical Network are Voltage Source and Current Sources. A Voltage Source

has a forcing function of EMF whereas the Current Source has a forcing function

of Current.

IDEAL VOLTAGE SOURCE

A Ideal Voltage Source is a Two-terminal Device whose Voltage at any instant of

time is Constant and is Independent of the Current drawn from it, is called

an Ideal Voltage Source and have Zero internal Resistance.

Or

An Imaginary Voltage Source, which can provide a Constant Voltage to

load ranging from zero to infinity. Such voltage Source is having Zero

Internal Resistance in Series with the Source and is called Ideal Voltage

Source.

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UNIT – III


The Figure shows the Symbol and Characteristics of an ideal Voltage Source:

Practically it is not possible to build a Voltage Source with no internal Resistance

and Constant Voltage for that long range of the Load.

PRACTICAL VOLTAGE SOURCE

Voltage Sources having some amount of internal Resistances in Series with

Source are known as Practical Voltage Source. Due to this internal Resistance;

voltage drop takes place, and it causes the Terminal Voltage to reduce.

The Figure shows the Symbol and Characteristics of an Practical Voltage Source:

The Smaller is the internal Resistance (r) of a Voltage Source, the more Closer it

is to an Ideal Voltage Source.

IDEAL CURRENT SOURCE

An Ideal Current Source is a Two-Terminal Device which Supplies the same

(Constant ) Current to any Load Resistance connected across its terminals.

Or

An Imaginary Current Source, which can provide a Constant Current to the load

ranging from zero to infinity. Such Current Source is having Infinity

Internal Resistance in Parallel with the Source and is called Ideal Current

Source.

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UNIT – III


It is important to keep in mind that the Current supplied by the Current

Source is independent of the Voltage of Source terminals. It has Infinite

Resistance.

The Figure shows the Symbol and Characteristics of an ideal Current Source:

Practically it is not possible to build a Current Source with infinity internal

Resistance and Constant Current for that long range of the Load.

PRACTICAL CURRENT SOURCE

A Current Sources having Very High amount of internal Resistances (in place of

Infinity Resistance) in Parallel with Source are known as Practical Current

Source. Due to this internal Resistance; some Current may flow in the Source ,

and it causes the Source Current to be decreased.

The Figure shows the Symbol and Characteristics of an Practical Current Source:

The Larger is the internal Resistance (r) of a Current Source, the more Closer it is

to an Ideal Current Source.

72


UNIT – III


CONCEPT OF SOURCE TRANSFORMATIONS ( CONVERSION )

Source Transformation ( Conversion ) Methods are used for Circuit simplification

to modify the complex circuits by Transforming Independent Current Sources into

Independent Voltage Sources and Vice-Versa.

Consider both Practical Voltage and Current Sources with Load Resistance of RL.

Let us see how the circuit behaves for Resistance change at the Load.

i. If the Load Resistance, RL = 0 in Practical Voltage Source Circuit, then the

Load acts as a Short Circuit and hence the Short Circuit Current flows

through the Load.

So the VL is zero

VL = IL x RL

IL would be, IL = VS / RS

ii. Similarly for RL = 0 in Practical Current Source Circuit, Load also behaves

as Short Circuit as it prefers the Current flow through Non-Resistance Path.

This Load Current is equal to the Source Current is which is equal to the

value of VS / RS in Practical Voltage Source Circuit.

Therefore,

IL = IS = VS / RS

VS = RS x IS when RL = 0………………..……(1)

iii. If the Load Resistance RL is infinity, Both Circuits behaves as an Open

Circuit. Therefore Load Current is Zero in Both Circuits and the Voltage drop

across the Resistance RS in Practical Current Source Circuit is

VS = IS x RS

73


UNIT – III


iv. Similarly, the Voltage across the RS in a Practical Voltage Source Circuit is

equal to the Vs which is equal to the Practical Current Source Circuit.

VS = IS x RS when RL is infinity ………………(2)

Hence,

From equations 1 and 2, we get

VS = RS x IS

VS = RS x IS

By observing above two Equations, if the Internal Resistance of the Two Sources

is same then the two Sources are Electrically Equivalent.

These Two Sources are equivalent and can produce the same values of IL

and VL when connected to same Load Resistance. Hence, these Equivalent

Sources can produce identical values of Short Circuited Current and Open Circuit

Voltage when Zero Load Resistance and Infinity Resistances respectively.

Therefore, by interchanging the internal resistors we can transform their

properties from Current Source to a Voltage Source and Vice-versa.

CONVERSION OF VOLTAGE SOURCE TO CURRENT SOURCE

A Voltage Source can be converted or transformed into a Current Source by

interchanging a Series Resistor to Parallel as shown in figure.

Steps: i. Find the Internal Resistance of the Voltage Source and keep this Resistor in

Parallel with a Current Source.

ii. Determine the Current flow provided by Current Source by applying Ohm’s

Law.

IS = VS / RS

In the above figure, a Voltage Source with a Resistance RS is converted

(transformed) into an Equivalent Current Source with a Parallel Resistor RS.

74


UNIT – III


Numerical - 1 : A Voltage Source Circuit with a voltage of 20 V and a

Internal Resistance of 5 ohms shown below. Convert this Voltage Circuit

into the Current Source by placing a Resistor of the same value with a

Current Source.

Solution:

Consider the above Voltage Source Circuit with a Voltage of 20 V and a Internal

Resistance of 5 ohms.

This Circuit is converted into the Current Source by placing a Resistor of

the same value with a current source.

This current source value can be determined by,

IS = VS / RS

= 20 / 5

= 4 Amps

The Equivalent Current source with a Current of 4A and Parallel Resistor of 5

ohms is shown below.

75


UNIT – III


CONVERSION OF CURRENT SOURCE TO VOLTAGE SOURCE

The Current Source can be converted (transformed) into a Voltage Source by

interchanging Parallel Resistor in Series as shown in figure below:

Steps:

i. Find the Parallel Resistance of the Constant Current Source and place in

Series with a Voltage Source.

ii. Determine the Open Circuit Voltage Value of the Voltage Source by applying

Ohm’s law.

Vs = Is x Rs

In the above figure, a Current Source with a Resistance Rs in Parallel with the

Current Source is converted (transformed) into an Equivalent Voltage Source with

a Series Resistor Rs.

Numerical- 2: A Current Source of 10A with a Parallel Resistance of 3

ohms as shown in Figure below. Convert this Current Source into its

Equivalent Voltage Source.

76


UNIT – III


Solution:

Consider the above Current Source to be converted in its Equivalent

Voltage Source with a resister of same value connected in Series with Voltage

Source.

To Calculate the value of Voltage in Voltage Source apply the simple

Ohm’s Law, then,

VS = IS x RS

VS = 10 x 3

= 30 Volts.

Therefore, Equivalent Voltage Source of this conversion consists a Voltage

Source 30 V with a Series Resistance 3 ohms.

Calculation of Current and Voltage in Practical Voltage Source

with Load Resistance RL:

77


UNIT – III


Consider the above Figure, where a Voltage Source VS with Series Resistance RS

is connected to a Load Resister RL.

Total Resistance RT = RS + RL ( As Both Resistance are connected in

Series)

Current Flowing in the Circuit IL = VS / (RS + RL)

IL = VS / RT

Now, the Voltage across Load Resistance VL is given by:

VL = RL x IL

VL = RL x VS / RT

Calculation of Current and Voltage in Practical Current Source

with Load Resistance RL:

Consider the below Figure, where a Current Source IS with Parallel Resistance RS

is connected to a Load Resister RL.

Current Flowing in the Circuit IL = IS x RS___ RS + RL

Now, the Voltage across Load Resistance VL is given by:

VL = RL x IL

Numerical - 3 : A Voltage Source Circuit with a voltage of 20 V and a

Internal Resistance of 5 ohms connected with Load Resistance RL of 15

ohms shown below .

78


UNIT – III


i. Calculate the Current flowing in the Load ( IL )

ii. Voltage drop across the Load ( VL )

Solution :

i. Consider the above Figure, where a Voltage Source VS = 20V with Series

Resistance RS = 5 Ω is connected to a Load Resister RL = 15 Ω

Total Resistance RT = RS + RL ( As Both Resistance are connected in

Series)

RT = 5 Ω + 15 Ω = 20 Ω

Current Flowing in the Circuit IL = VS / (RS + RL)

IL = VS / RT = 20 / 20 = 1 Amp.

IL = 1 Amp

ii. Now, the Voltage across Load Resistance VL is given by:

VL = RL x IL

VL = RL x VS / RT = 15 x 1/ 20 = 3 / 4 = 0.75 Amp

VL = 0.75 Amp

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UNIT – III


Numerical - 4 : A Current Source Circuit with a Current of 10 Amp having a

Internal Resistance of RS of 5 ohms connected with Load Resistance RL of

15 ohms shown below .

i. Calculate the Current flowing in the Load ( IL )

ii. Voltage drop across the Load ( VL )

Solution:

i. Consider the above Figure, where a Current Source IS = 10 Amps with

Parallel Resistance RS = 5 Ω is connected to a Load Resister RL = 15 Ω

Current Flowing in the Circuit IL = IS x RS___ RS + RL IL = 10 x 5 = 50 / 20 = 2.5 Amps 5 + 15 IL = 2.5 Amps

ii. Now, the Voltage across Load Resistance VL is given by:

VL = RL x IL 15 x 2.5 = 37.5 Volts

VL = 37.5 Volts

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UNIT – III


Fill IN THE BLANKS:

1. An ideal current source is one whose internal resistance is …………...

2. An ideal voltage source is that which has ………… internal resistance.

3. A Practical constant voltage source should have …………. internal resistance.

4. A Practical constant current source should have …………internal resistance.

5. In Constant Current Source, the internal resister is connected in ………. with

the Source.

6. In Constant Voltage Source, the internal resister is connected in ………. with

the Source.

7. Output of Constant Current Source is ………… .

8. Output of Constant Voltage Source is …………..

ANSWERS:

1) Infinite 2) Zero 3) Minimum 4) Maximum

5) Parallel 6) Series 7) Constant 8) Constant.

81


UNIT – IV

SEMICONDUCTOR PHYSICS


BASIC ATOMIC STRUCTURE

An atom is the smallest unit of a substance that retains all of the chemical

properties of an element. Atoms combine to form molecules, which interact to

form solids, gases, or liquids. For example, water is composed of hydrogen and

oxygen atoms that have combined to form water molecules.

Atomic Particles: Atoms consist of three basic particles:

i. Protons: Positively charged subatomic particle forming part of the

nucleus of an atom and determining the atomic number of an element.

Proton has approximately the mass, about 1.67 × 10-24 grams.

ii. Electrons: Negatively charged subatomic particle forming an outer part

of an atom. Electrons are much smaller in mass than protons, weighing

only 9.11 × 10-28 grams.

iii. Neutrons: A subatomic particle forming part of the nucleus of an atom.

It has no charge. Neutron has approximately the mass, about 1.67 × 10-

24 grams. It is equal in mass to a proton.

The nucleus (center) of the atom contains the protons (Positively Charged) and

the neutrons (no charge). The outermost regions of the atom are called electron

shells (Orbits) and contain the electrons (Negatively Charged). Atoms have

different properties based on the arrangement and number of their basic particles.

Atomic Number: Atomic number is the number of protons in an element. while

the mass number is the number of protons plus the number of neutrons.

Mass Number: Mass Number of an atom is sum of the number of protons and

the number of neutrons in that atom.

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ATOMIC STRUCTURE : BOHR MODELS In 1913, Neils Bohr, developed a new model of the atom. He proposed that

electrons are arranged in concentric circular orbits around the nucleus. This

model is patterned on the solar system and is known as the planetary model. The

Bohr model can be summarized by the following principles:

1. In an atom, electrons (negatively charged) revolve around the positively charged

nucleus in a definite circular path called as orbits or shells.

2. Each orbit or shell has a fixed energy and these circular orbits are known as

orbital shells.

3. The orbits n = 1, 2, 3, 4… are assigned as K, L, M, N…. shells and the number of

Electrons lies in an orbit depends upon the relation 2 n2.

Therefore,

1st orbit (energy level) is represented as K shell and it can hold up to 2

electrons.

2nd orbit (energy level) is represented as L shell and it can hold up to 8

electrons.

3rd orbit (energy level) is represented as M shell and it can contain up to 18

electrons.

4th orbit (energy level) is represented as N Shell and it can contain maximum

32 electrons.

The orbits continue to increase in a similar manner.

4. However, the last orbit cannot more contain more than 8 Electrons and second

last orbit cannot contain more than 18 Electrons.

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5. The electrons in an atom move from a lower energy level to a higher energy level

by gaining the required energy and an electron moves from a higher energy level

to lower energy level by losing energy.

ATOMIC STRUCTURE OF SILICON AND GERMANIUM ATOM :

Silicon Atom:

Atomic No. = 14 No. of Protons = 14 No. of Electron = 14

1st Orbit ( K ) has 2 Electrons

2nd Orbit ( L ) has 8 Electrons

3rd Orbit ( M ) has 4 Electrons

Germanium Atom:

Atomic No. = 32 No. of Protons = 32 No. of Electron = 32

1st Orbit ( K ) has 2 Electrons

2nd Orbit ( L ) has 8 Electrons

3rd Orbit ( M ) has 18 Electrons

4th Orbit ( N ) has 4 Electrons

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CONCEPT OF INSULATORS, CONDUCTORS AND SEMI-CONDUCTORS

Insulators : Insulators are the materials or substances which don’t allow the

current to flow through them.

In general, they are solid in nature. They do not allow the flow of heat also.

The property which makes insulators different from conductors is its resistivity.

Wood, cloth, glass, mica, and quartz are some good examples of insulators. Also,

insulators are protectors. They give protection against heat, sound and of course

passage of electricity. Insulators don’t have any free electrons. It is the main reason

why they don’t conduct electricity.

Conductors: Conductors are the materials or substances which allow electricity

to flow through them.

The Conductor have free electrons which allow current to pass through them

easily. Conductors also allow the transmission of heat or light from one source to

another. Metals, humans, earth, and animals are all conductors. This is the reason

we get electric shocks! Moreover, the Metals are a good conductors.

Semi-Conductor : Semi-Conductors are the materials or substances which

conductivity lies between Conductors and Insulators.

At 0o temperature, Semi-Conductor behaves as Insulator and when

temperature increases, the conductivity of semiconductor also increases. Good

examples of semiconductor materials are germanium, selenium, and silicon.

Silicon is used in most semiconductors for computer and electronic components,

as is it considered to be the best semiconductor material.

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ENERGY LEVEL DIAGRAM OF CONDUCTORS, INSULATORS AND

SEMI-CONDUCTORS

Energy Level Diagram: It is the diagram which represent the energy levels of

Electrons revolving in the orbits from the Nucleus. Electrons revolves in an orbit

have same level of energy. The Electrons in innermost Orbit have lowest energy

level and electrons at higher levels have more energy.

The above figure shows the Energy level diagram of a substance. The 1st orbit (K)

has lowest energy level and last orbit has highest energy level of electrons.

Energy Band : Energy band is the range of energies of electrons revolving in

any one orbit of the atom. Each orbit has its energy band and range of energy

band depends upon the electrons energies in that orbit. The last Energy band in

which electrons revolving is called as Valance band and there is extra band after

valance in the atomic structure is called conduction band.

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Forbidden Energy Gap: It is gap of energy between the valance band and

conduction band of an atom. In other words, the Energy required by an electron to

jump from valance band to conduction band. It is measured in electron Volts (eV).

Energy Level / Band Diagram of Insulators : Figure shows the Energy

Level diagram of an Insulator.

Normally, in insulators the valence band is fully occupied with electrons due to

sharing of outer most orbit electrons with the neighboring atoms. Where as

conduction band is empty, I.e, no electrons are present in conduction band.

The forbidden gap between the valence band and conduction band is

very large in insulators. The energy gap of insulator is approximately equal to or

greater than 10 electron volts (eV).

The electrons in valence band cannot move because they are locked up

between the atoms. In order move the valence band electrons in to conduction

band large amount of external energy is applied which is equal to the forbidden

gap. But in insulators, this is practically impossible to move the valence band

electrons in to conduction band.

Rubber, wood, diamond, plastic are some examples of insulators.

Insulators such as plastics are used for coating of electrical wires. These

insulators prevent the flow of electricity to unwanted points and protect us from

electric shocks.

Energy Level / Band Diagram of Conductor : Figure shows the Energy

Level diagram of an Conductor.

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In a conductor, valence band and conduction band overlap each other as shown

in above figure. Therefore, there is no forbidden gap in a conductor.

A small amount of applied external energy provides enough energy for the

valence band electrons to move in to conduction band. Therefore, more number of

valence band electrons can easily moves in to the conduction band.

When valence band electrons moves to conduction band they becomes

free electrons. The electrons present in the conduction band are not attached to

the nucleus of a atom. In conductors, large number of electrons are present in

conduction band at room temperature, i.e, conduction band is almost full with

electrons. Whereas valence band is partially occupied with electrons. The

electrons present in the conduction band moves freely by carrying the electric

current from one point to other.

Energy Level / Band Diagram of Semi-Conductor : Figure shows the

Energy Level diagram of an Semi-Conductor.

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UNIT – IV


In semiconductors, the forbidden gap between valence band and conduction band

is very small. It has a forbidden gap of about 1 electron volt (eV).

At low temperature, the valence band is completely occupied with electrons

and conduction band is empty because the electrons in the valence band does not

have enough energy to move in to conduction band. Therefore, semiconductor

behaves as an insulator at low temperature.

However, at room temperature some of the electrons in valence band gains

enough energy in the form of heat and moves in to conduction band. When the

temperature is goes on increasing, the number of valence band electrons moving

in to conduction band is also increases. This shows that electrical conductivity of

the semiconductor increases with increase in temperature. i.e. a semiconductor

has negative temperature co-efficient of resistance.

The resistance of semiconductor decreases with increase in temperature.

COVALENT BOND:

A covalent bond is a chemical bond in which pairs of electrons are shared between

two atoms. The covalent bond is also called a molecular bond. The forces of

attraction or repulsion between two atoms, when they share electron pair or bonding

pair, is called as Covalent Bonding.

Covalent bonding occurs between non-metal elements when pairs of

electrons are shared by atoms. When elements share their electrons, they do not

become positive or negative, since they are neither gaining nor loosing electrons.

Thus, no ions are formed by covalent bonding.

Covalent Bonding in Silicon ( Si ) : The outermost shell of atom is capable

to hold up to eight electrons. The atom which has eight electrons in the outermost

orbit is said to be completely filled and most stable. But the outermost orbit of

silicon has only four electrons. Silicon atom needs four more electrons to become

most stable. Silicon atom forms four covalent bonds with the four neighboring

atoms. In covalent bonding each valence electron is shared by two atoms.

When silicon atoms comes close to each other, each valence electron of

atom is shared with the neighboring atom and each valence electron of

neighboring atom is shared with this atom. Likewise each atom will share four

valence electrons with the four neighboring atoms and four neighboring atoms will

share each valence electron with this atom. Therefore, total eight electrons are

shared.

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The outermost shell of silicon is completely filled and valence electrons are tightly

bound to the nucleus of atom because of sharing electrons with neighboring

atoms. In intrinsic semiconductors, free electrons are not present at absolute zero

temperature. Therefore intrinsic semiconductor behaves as perfect insulator.

Covalent Bonding in Germanium (Ge): The outermost orbit of germanium

has only four electrons. Germanium atom needs four more electrons to become

most stable. Germanium atom forms four covalent bonds with the four neighboring

atoms. In covalent bonding each valence electron is shared by two atoms.

When germanium atoms comes close to each other each valence electron

of atom is shared with the neighboring atom and each valence electron of

neighboring atom is shared with this atom. Each atom will share four valence

electrons with the four neighboring atoms and four neighboring atoms will share

each valence electron with this atom. Therefore, total eight electrons are shared.

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The outermost shell of germanium is completely filled and valence electrons are

tightly bound to the nucleus of atom because of sharing electrons with neighboring

atoms. In intrinsic semiconductors, free electrons are not present at absolute zero

temperature. Therefore intrinsic semiconductor behaves as perfect insulator.

INTRINSIC AND EXTRINSIC SEMI-CONDUCTORS

The semiconductor is divided into two types. One is Intrinsic Semiconductor and

other is an Extrinsic semiconductor. The pure form of the semiconductor is

known as the intrinsic semiconductor and the semiconductor in which impurities is

added for making it conductive is known as the extrinsic semiconductor. The

conductivity of the intrinsic semiconductor become zero at room temperature while

the extrinsic semiconductor is very little conductive at room temperature. The

detailed explanation of the two types of the semiconductor is given below.

Intrinsic Semiconductor An extremely pure semiconductor is called as

Intrinsic Semiconductor.

An Intrinsic (Pure) Semiconductor, also called an undoped semiconductor is a

pure semiconductor without any significant dopant species present. The number

of charge carriers is therefore determined by the properties of the material itself

instead of the amount of impurities. In intrinsic semiconductors the number of

excited electrons and the number of holes are equal: N ( Electrons ) = P ( Holes ).

This may even be the case after doping the semiconductor, though only if it is

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doped with both donors and acceptors equally. In this case, N ( Electrons ) = P

(Holes) still holds, and the semiconductor remains intrinsic, though doped.

Effect of Temperature on Conductivity of Intrinsic Semi

Conductors

The valence band of Intrinsic Semiconductor is completely filled and the

conduction band is completely empty at 0o C temperature and hence there is no

conduction at 0o C temperature. When the temperature is raised and some heat

energy is supplied to it, some of the valence electrons are lifted to conduction

band leaving behind holes in the valence band as shown below.

The electrons reaching at the conduction band move randomly. The holes created

in the crystal also free to move anywhere in valence band. This behavior of the

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semiconductor shows that they have a negative temperature coefficient of

resistance. This means that with the increase in temperature, the resistivity of the

material decreases and the conductivity increases.

An intrinsic semiconductor is capable to conduct a little current even at

room temperature, but it is not useful for the preparation of various electronic

devices. Thus, to make it conductive a small amount of suitable impurity is added

to the material.

Extrinsic Semiconductor : A Semiconductor to which an impurity at controlled

rate is added to make it conductive, is known as an extrinsic Semiconductor.

Doping : The process by which an impurity is added to a semiconductor is

known as Doping. The amount and type of impurity which is to be added to a

material has to be closely controlled during the preparation of extrinsic

semiconductor. Generally, one impurity atom is added to a 108 atoms of a

semiconductor.

The purpose of adding impurity in the semiconductor crystal is to increase

the number of free electrons or holes to make it conductive. If a Pentavalent

impurity, having five valence electrons is added to a pure semiconductor a large

number of free electrons will exist.

If a trivalent impurity having three valence electrons is added, a large

number of holes will exist in the semiconductor.

Depending upon the type of impurity added the extrinsic semiconductor may be

classified as N -Type Semiconductor and P-Type Semiconductor.

P - Type Semiconductor : The extrinsic P-Type Semiconductor is formed

when a trivalent impurity is added to a pure semiconductor in a small amount,

and as a result, a large number of holes are created in it. A large number of holes

are provided in the semiconductor material by the addition of trivalent impurities

like Aluminium, Gallium and Indium. Such type of impurities which produces P-

Type semiconductor are known as an Acceptor Impurities because each atom of

them create one hole which can accept one electron.

A trivalent impurity like Aluminium ( Al ), having three valence electrons is

added to Silicon crystal in a small amount. Each atom of the impurity fits in the

Silicon crystal in such a way that its three valence electrons form covalent bonds

with the three surrounding Silicon atoms as shown in the figure below.

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UNIT – IV


In the fourth covalent bonds, only the Silicon atom contributes one valence

electron, while Aluminium atom has no valence bonds. Hence, the fourth covalent

bond is incomplete, having one electron short. This missing electron is known as

a Hole. Thus, each Aluminium atom provides one hole in the Silicon crystal.

As an extremely small amount of Aluminium impurity has a large number of

atoms, therefore, it provides millions of holes in the semiconductor. The holes are

mainly responsible for the conduction in the P-type semiconductor to take place.

Hence, in this case, charge carriers are holes rather than electrons.

Energy Band Diagram of P-Type Semiconductor : A large number of

holes or vacant space in the covalent bond is created in the crystal with the

addition of the trivalent impurity in valance band. A small quantity of free electrons

is also available in the conduction band. These electrons are jumped when

thermal energy at room temperature is imparted to the Silicon crystal forming

electron-hole pairs.

On the other hand, trivalent impurity is added to P type semiconductor.

This is called an acceptor.

The energy band diagram of a P-Type Semiconductor is shown below.

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UNIT – IV


As shown in above figure, the energy level of the acceptor is close to the

valence band. Since there are no electrons here, electrons in the valence band

are excited and jumped in the conduction band. As a result, holes are formed in

the valence band, which contributes to the conductivity.

As it is clear from the above figure that there exists a very small energy

difference between valence band and the acceptor energy level. Thus, electrons

easily drift to acceptor energy level creating a vacancy of electrons in the valance

band. Hence, producing holes in the valence band.

Conduction Through P - Type Semiconductor : In P - type semiconductor

large number of holes are created by trivalent impurity. When a potential

difference is applied across this type of semiconductor, the current flows the

Semiconductor due to majority carrier ( Holes ) as shown in the figure below.

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UNIT – IV


The holes are available in the valence band are directed towards the Negative

terminal. As the current flow through the crystal is by holes, which are carrier of

positive charge, therefore, this type of conductivity is known as Positive or P-

Type conductivity. In a P type conductivity the valence electrons move from one

covalent to another.

N- Type Semiconductor : When a small amount of Pentavalent

impurity (Having Five Electrons in the Valance Band) is added to a pure

semiconductor providing a large number of free electrons in it, the extrinsic

semiconductor thus formed is known as N - Type Semiconductor. The

conduction in the N-type semiconductor is because of the free electrons denoted

by the Pentavalent impurity atoms.

The addition of Pentavalent impurities such as Phosphorous, Arsenic and

antimony provides a large number of free electrons in the semiconductor crystal.

Such impurities which produce N-type semiconductors are known as Donor

Impurities. Electrons are mainly responsible for the conduction in the N-type

semiconductor to take place. Hence, in this case, charge carriers are Electrons

rather than Holes.

When a few Pentavalent impurities such as Phosphorous whose have five

valence electrons, is added to Silicon crystal. Each atom of the impurity fits in four

Silicon atoms as shown in the figure above. Hence, each Phosphorous atom

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UNIT – IV


provides one free electron in the Silicon crystal. Since an extremely small amount

of Phosphorous impurity have a large number of atoms; it provides millions of free

electrons for conduction.

Energy Diagram of N - Type Semiconductor

The Energy diagram of the N - Type semiconductor is shown in the figure below.

Here, from the figure, it is clear that the existence of the Donor Energy level is

near the conduction band. So there is small energy difference exists between

donor energy level and the conduction band. So, less energy is needed by the

electrons to reach the conduction band.

So, a large number of free electrons are available in the conduction band

because of the addition of the Pentavalent impurity. These Free Electrons are

mainly responsible for the conduction in the N-type semiconductor to take place.

Hence, in this case, charge carriers are Electrons.

Conduction Through N - Type Semiconductor: In the N - Type

semiconductor, a large number of free electrons are available in the conduction

band which are donated by the impurity atoms. When the Potential difference is

applied to this type of Semiconductor, the current flows through this

semiconductor is due to majority carrier (Electrons) The figure below shows the

conduction process of an N - Type Semiconductor.

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UNIT – IV


When a potential difference is applied across this type of semiconductor, the free

electrons are directed towards the positive terminals. It carries an electric current.

As the flow of current through the crystal is constituted by free electrons which are

carriers of negative charge, therefore, this type of conductivity is known

as Negative or N-type conductivity.

The electron-hole pairs are formed at room temperature. These holes

which are available in small quantity in valence band also consists of a small

amount of current. For practical purposes, this current is neglected.

Majority and Minority Carriers : In an N-type semiconductor, the electrons

are the majority carriers whereas, the holes are the minority carriers. In the P-

type semiconductor material, the holes are the majority carriers, whereas, the

electrons are the minority carriers as shown in the figures below.

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UNIT – IV


When a small amount of Pentavalent impurity is added to a pure semiconductor,

it provides a large number of free electrons in the crystal forming the N-type

semiconductor. Some of the covalent bonds break even at the room temperature,

releasing a small number of electron-hole pairs.

Thus, an N-type semiconductor contains a large number of free electrons

and a few numbers of holes. This means the electron provided by Pentavalent

impurity added and a share of electron-hole pairs. Therefore, in N-type

semiconductor, the most of the current conduction is due to the free electrons

available in the semiconductor.

Similarly, in the P-type semiconductor, the holes are in the majority as

compared to electrons, and the current conduction in P-Type takes place due to

Holes available as majority carrier in the Semiconductor.

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Fill IN THE BLANKS:

1. …………………. are those Substances which allow electricity to flow through them.

2. Insulators are those substances which ………. … electricity to flow through them.

3. ……………are Positively charged subatomic particle forming part of the

nucleus of an atom.

4. ……… are Negatively charged subatomic particle forming an outer part of an

atom.

5. The Atoms in a silicon crystal are held together by …………. bonds.

6. An electron can move to another atom's orbit only while in the ………. Band.

7. Majority carriers in N -Type Semiconductor are …………...

8. A Semiconductor is formed by ………….. bonds.

9. The random motion of holes and free electrons due to thermal agitation is

called ……………. .

10. Process of adding impurities to a pure semiconductor is known as …………...

11. A doped semiconductor is known as ………………….. Semi-Conductor.

12. The junction break down voltage for Silicon diode is ……… Volt.

13. The junction break down voltage for Germanium diode is ………… Volt.

14. The most commonly used semiconductor is ……………. Semiconductor.

15. The leakage current across a P-N junction is due to …………. Carriers.

16. Majority carrier in P -Type Semiconductor are…………….

17. With forward bias to a P-N junction , the width of depletion layer……………….

18. A Semiconductor has …………….. temperature coefficient of resistance.

19. The leakage current in a P-N junction is of the order of ............................ .

20. The electrons in the outermost orbit are called …………… electrons.

ANSWERS:

1) Conductors 2) Do not allow 3) Protons 4) Electrons

5) Covalent 6) Conduction 7) Electrons 8) Covalent

9) Diffusion 10) Doping 11) Extrinsic 12) 0.7

13) 0.3 14) Silicon 15) Minority 16) Holes

17) Decreases 18) Negative 19) µA 20) Valance

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Fill IN THE BLANKS:

21. A Semiconductor has generally …………….. Valence Electrons.

22. The larger the orbit number, the …………. is the energy of the electron.

23. The merging of a free electrons and a hole is called ………………. Bonds.

24. A Pure Germanium Crystal is an …………… Semiconductor.

25. When a Pure Semiconductor is heated, its resistance ……………..

26. At 00 C temperature, an intrinsic silicon crystal acts as …………….

27. In an intrinsic semiconductor, the number of free electrons equal to the

numbers of …………. .

28. As the temperature of a semiconductor increases, its conductivity ………….

29. .............................. impurity has Five Valence electrons.

30. A ……………. impurity has Three Valence electrons.

31. Addition of ………………impurity to a semiconductor creates many free

Electrons.

32. Addition of ………………. impurity to a semiconductor creates many Holes.

33. When a Pentavalent impurity is added to a pure semiconductor, it becomes

………………… Semi-Conductor.

34. In P- Type Semiconductor, the minority carriers are ……………….

35. When a ……………….. impurity is added to a pure semiconductor, it becomes

P-Type Semi-Conductor.

36. In a N - Type semiconductor, current conduction is due to ……………… .

37. In a P - Type semiconductor, current conduction is due to ………………. .

38. In ……………… Semiconductor, the minority carriers are Holes.

39. A Hole in a Semiconductor is as incomplete part of an ………………. bond.

40. As the doping level to a pure semiconductor increases, the resistance of the

semiconductor ……………….

ANSWERS:

21) Four 22) Greater 23) Recombination 24) Intrinsic

25) Decreases 26) Insulator 27) Holes 28) Increases

29) Pentavalent 30) Trivalent 31) Pentavalent 32) Trivalent

33) N-Type 34) Electrons 35) Trivalent 36) Free Electrons

37) Holes 38) N- Type 39) Electron pair 40) Decreases

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UNIT – V

SEMI-CONDUCTOR DIODE

SEMI - CONDUCTOR DIODE

PN JUNCTION DIODE

A P-N junction diode is two-terminals (Anode and Cathode) semiconductor device,

which allows the electric current in only one direction while blocks the electric

current in opposite or reverse direction.

If the diode is forward biased, it allows the electric current flow. On the other hand,

if the diode is reverse biased, it blocks the electric current flow. P-N junction

semiconductor diode is also called as p-n junction semiconductor device.

In N-Type Semiconductors, free electrons are the majority charge carriers

whereas in P-Type Semiconductors, holes are the majority charge carriers. When

the P-type Semiconductor is joined with the N-Type Semiconductor, a P-N

junction is formed. The P-N junction, which is formed when the P-Type and N-

Type semiconductors are joined, is called as P-N junction diode.

The P-N junction diode is made from the semiconductor materials such as

silicon, germanium, and gallium arsenide. For designing the diodes, Silicon is

more preferred over Germanium. The P-N junction diodes made from Silicon

semiconductors works at higher temperature when compared with the p-n junction

diodes made from Germanium semiconductors.

IMPORTANT TERMS

Forward Biased P-N Junction: When Anode ( P- Type region ) of Diode is

connected with the positive terminal of the battery and Cathode ( N-Type region )

with the negative terminal of the Battery, then the junction is said to be forward

biased.

When a diode is connected in a Forward Bias condition, a negative voltage

is applied to the N-type material and a positive voltage is applied to the P-type

material. If this external voltage becomes greater than the value of the potential

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https://www.physics-and-radio-electronics.com/electronic-devices-and-circuits/introduction/free-electrons.html

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barrier, approx. 0.7 volts for silicon and 0.3 volts for germanium, the potential

barriers opposition will be overcome and current will start to flow.

Reverse Biased PN Junction Diode : When Anode ( P- Type region ) of

Diode is connected with the Negative terminal of the battery and Cathode ( N-

Type region ) with the Positive terminal of the Battery, then the junction is said to

be Reverse biased.

The positive voltage applied to the N-type material attracts electrons

towards the positive electrode and away from the junction, while the holes in the

P-type end are also attracted away from the junction towards the negative

electrode.

The Net result is that the depletion layer grows wider due to a lack of electrons

and holes and presents a high impedance path, almost an insulator. The result is

that a high potential barrier is created thus preventing current from flowing through

the semiconductor material.

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UNIT – V


Depletion Layer: Depletion region or depletion layer is a region in a P-N

junction diode where no mobile charge carriers are present. Depletion layer acts

like a barrier that opposes the flow of electrons from n-side and holes from p-side.

When P Type and N – Type Semiconductors are joined together and form a

junction between them. The Opposite charges diffuse and form a depletion Zone

in the middle making a potential barrier consecutively.

Under forward bias condition, the barrier reduces and hence the energy

required to cross the depletion region minimizes, therefore forward saturation

current flows. For ideal diode, the forward biased equivalent to Zero resistance

(Short Circuit ).

Under reverse biased condition, the barrier increases and hence the diode

act as an open circuit negligible leakage reverse saturation current flows. For ideal

diode, the reverse biased equivalent to Infinity resistance ( Open Circuit ).

Potential barrier: The potential barrier in the PN- Junction diode is the barrier

in which the charge requires additional force for crossing the region.

In other words, the barrier in which the charge carrier stopped by the

obstructive force is known as the potential barrier.

When P and N-type semiconductor material are placed together, the depletion

layer is created on both P and N side region. The free electrons from N-side cross

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the region and start combining with the holes, leaving behind the immobile positive

donor ions. Similarly, the holes of the P-region combine with the electrons of the

N-region and leaving behind the negative acceptor ions.

The process is continued until the P and N-region have enough charge

carrier for opposing the electrons and holes respectively. The immobile ions

(negative acceptor ions and positive donor ions) are concentrated between the N

and P-region and create the electric field which acts as a barrier between the

flows of charges. The region is created because of the depleted ions, and hence it

is called the depletion region.

The depletion region acts as a barrier and opposes the flow of charge

carrier. The value of barrier potential lies between 0.3 – 0.7V depends on the type

of material used.

Diffusion Current: The diffusion current can be defined as the flow of charge

carriers within a semiconductor travels from a higher concentration region to a

lower concentration region.

In a semiconductor material, the charge carriers have the tendency to move from

the region of higher concentration to that of lower concentration of the same type

of charge carriers. Thus the movement of charge carriers takes place resulting in

a current called diffusion current.

Diffusion: Diffusion occurs when substance (Charge Carrier) move from an area

of High Concentration to low Concentration. This does not require any energy.

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There is a potential barrier set up across the depletion layer due to immobile charges

which set up an electric field across the junction. This electric field is directed from

positive charge to negative charges across the junction. This electric field exerts a

force on electrons in p-side to move towards n-side and exerts a force on holes in n-

region to move towards p-side.

Thus, a drift current begins to flow due to drifting of holes and electrons across the

junction. The flow of drift current is opposite to diffusion current.

When drift current is equal to diffusion current, the net current becomes zero.

Drift Current : Drift current can be defined as the charge carrier’s moves in a

semiconductor because of the electric field. There are two kinds of charge carriers

in a semiconductor like holes and electrons. Once the voltage is applied to a

semiconductor, then electrons move toward the +Ve terminal of a battery whereas

the holes travel toward the –Ve terminal of a battery.

Here, holes are Positively charged carriers whereas the electrons are Negatively

charged carriers. Therefore, the electrons attract by the +Ve terminal of a

Battery whereas the holes attract by the -Ve terminal of a Battery.

Conclusion: The flow of current in a semiconductor are of two types

namely, drift and diffusion current. The current produced due to the movement of

charge carriers by external applied voltage is known as drift current. Whereas,

the current produced due to the change in concentrations is called diffusion

current.

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Voltage – Current ( V - I ) Characteristics of Diode

The Volt-Ampere or V-I characteristics of a P-N junction diode is basically the

curve between voltage across the junction and the circuit current.

The characteristics can be explained under three conditions namely

i. Forward bias

ii. Reverse bias.

Forward Bias: When Anode ( P- Type region ) of Diode is connected with the

positive terminal of the battery and Cathode ( N-Type region ) with the negative

terminal of the Battery, then the junction is said to be forward biased.

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When a diode is connected in a Forward Bias condition, a negative voltage is

applied to the N-type material and a positive voltage is applied to the P-type

material. If this external voltage becomes greater than the value of the potential

barrier, approx. 0.7 volts for silicon and 0.3 volts for germanium, the potential

barriers opposition will be overcome and current will start to flow.

From now onwards, the current increases with the increase in forward voltage.

Thus a rising curve OB is obtained with forward bias as shown in above figure.

From the forward characteristics, it is seen that at first (i.e region OA ), the

current increase very slowly and curve is non-linear. It is because the external

applied voltage is used to overcome the potential barrier.

However, once the external applied voltage exceeds the potential barrier

voltage, the p-n junction behaves like an ordinary conductor. Therefore, current

rises very sharply with increase in voltage (region AB). The curve is almost linear.

Reverse Bias: When Anode ( P- Type region ) of Diode is connected with the

Negative terminal of the battery and Cathode ( N-Type region ) with the Positive

terminal of the Battery, then the junction is said to be Reverse biased.

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With reverse bias to the p-n junction i.e. p-type connected to negative terminal and

n-type connected to positive terminal, potential barrier at the junction is increased.

Therefore, the junction resistance becomes very high and practically no current

flows through the circuit.

Reverse Bias Region : The reverse bias region exists between zero current

and breakdown.

In this region from O to C of the above figure , a small reverse current flows

through the diode. This reverse current is caused by the thermally produced

minority carriers. This reverse current is very small in order of μA that it cannot

even notice and it is considered almost zero.

In N-type and P-type semiconductors, very small number of minority charge

carriers is present. Hence, when a small voltage applied on the diode pushes all

the minority carriers towards the junction and a small reverse current is caused.

But, further increase in the external voltage does not increase the electric current.

This electric current is called reverse saturation current.

In other words, the voltage or point at which the electric current reaches its

maximum level and further increase in voltage does not increase the electric

current is called reverse saturation current.

The reverse saturation current depends on the temperature. If temperature

increases the generation of minority charge carriers increases. Hence, the reverse

current increases with the increase in temperature.

However, the reverse saturation current is independent of the external

reverse voltage. Hence, the reverse saturation current remains constant with the

increase in voltage.

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Reverse Breakdown Region: If the voltage applied on the diode is

increased continuously from point C towards D, the kinetic energy of minority

carriers may become high enough to knock out electrons from the semiconductor

atom. At this stage breakdown of the junction occurs.

At this point, a process called Avalanche Breakdown occurs in the semiconductor

depletion layer and the diode starts conducting heavily in the reverse direction, a

sudden rise of reverse current and a sudden fall of the resistance of barrier region

occurs. This may destroy the junction permanently and hence diode may also

destroyed.

STATIC AND DYNAMIC REISTANCE AND THEIR VALUES

CALCULATION FROM THE CHARACTERISTICS

Resistance of a Diode: In practice, no diode is an Ideal diode, this means

neither it acts as a perfect conductor when forward biased nor it acts as an

insulator when it is reverse biased.

In other words an actual diode offers a very small resistance (not zero)

when forward biased and is called a Forward Resistance. Whereas, it offers a

very high resistance (not infinite) when reverse biased and is called as a Reverse

Resistance.

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Forward Resistance : Under the Forward biased condition, the opposition

offered by a diode to the forward current is known as Forward Resistance. The

forward current flowing through a diode may be constant, i.e., direct current or

changing i.e., alternating current.

The forward resistance is classified as:

1. Static Forward Resistance

2. Dynamic Forward Resistance

Static or DC Forward Resistance ( RF ) : The opposition offered by a diode

to the direct current flowing forward bias condition is known as its DC forward

resistance or Static Resistance. It is measured by taking the ratio of DC voltage

across the diode to the DC current flowing through it.

The forward characteristic of a diode is shown below.

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It is clear from the graph that for the operating point P, the forward voltage is OA

and the corresponding forward current is OB. Therefore, the static forward

resistance of the diode is given as

Dynamic or AC Forward Resistance ( rf ) : The opposition offered by a

diode to the changing current flow I forward bias condition is known as its AC

Forward Resistance.

It is measured by a ratio of change in voltage across the diode to the

resulting change in current through it. From the above figure, it is clear that for an

operating point P the AC forward resistance is determined by varying the forward

voltage ( CE ) on both the sides of the operating point equally and measuring the

corresponding forward current ( DF ).

The Dynamic or AC Forward Resistance is represented as shown below.

The value of the forward resistance of a crystal diode is very small, ranging from 1

to 25 Ohms.

Reverse Resistance ( RR ) : Under the Reverse biasing condition, the

opposition offered by the diode to the reverse current is known as Reverse

Resistance. Ideally, the reverse resistance of a diode is considered to be infinite.

However, in actual practice the reverse resistance is not infinite because diode

conducts a small leakage current (due to minority carriers) when reverse biased.

The value of reverse resistance is very large as compared to forward

resistance. The ratio of reverse to forward resistance is 1 00 000 : 1 for silicon

diodes, whereas it is 40 000 : 1 for germanium diode.

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JUNCTION CAPACITANCE IN FORWARD AND REVERSE BIASED

CONDITIONS

Capacitor: We know that capacitors store electric charge in the form of electric

field. This charge storage is done by using two electrically conducting plates

(placed close to each other) separated by an insulating material called dielectric.

The conducting plates or electrodes of capacitor are good conductors of

electricity. Therefore, they easily allow electric current through them. On the other

hand, dielectric material is poor conductor of electricity. Therefore, it does not

allow electric current through it. However, it efficiently allows electric field.

When voltage is applied to the capacitor, charge carriers starts flowing through the

conducting wire. When these charge carriers reach the electrodes of the

capacitor, they experience a strong opposition from the dielectric or insulating

material. As a result, a large number of charge carriers are trapped at the

electrodes of the capacitor. These charge carriers cannot move between the

plates. However, they exerts electric field between the plates. The charge carriers

which are trapped near the dielectric material will stores electric charge. The

ability of the material to store electric charge is called capacitance.

In a basic capacitor, the capacitance is directly proportional to the size of

electrodes or plates and inversely proportional to the distance between two plates.

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Junction Capacitance in Reverse Biased Conditions: Just like the

capacitors, a reverse biased P-N junction diode also stores electric charge at

the depletion region. The depletion region is made of immobile positive and

negative ions.

In a reverse biased P-N junction diode, P-type and N-type regions have low

resistance. Hence, P-type and N-type regions act like the electrodes or conducting

plates of the capacitor. The depletion region of the P-N junction diode has high

resistance. Hence, the depletion region acts like the dielectric or insulating

material. Thus, P-N junction diode can be considered as a parallel plate capacitor.

In depletion region, the electric charges (Positive and Negative ions) do not

move from one place to another place. However, they exert electric field or electric

force. Therefore, charge is stored at the depletion region in the form of electric

field. The ability of a material to store electric charge is called capacitance. Thus,

there exists a capacitance at the depletion region.

The capacitance at the depletion region changes with the change in applied

voltage. When reverse bias voltage applied to the P-N junction diode is increased,

a large number of holes (majority carriers) from P-side and electrons (majority

carriers) from N-side are moved away from the P-N junction. As a result, the width

of depletion region increases whereas the size of P-type and N-type regions

(plates) decreases.

We know that capacitance means the ability to store electric charge. The P-

N junction diode with narrow depletion width and large P-type and N-type regions

will store large amount of electric charge whereas the P-N junction diode with wide

depletion width and small P-type and N-type regions will store only a small amount

of electric charge. Therefore, the capacitance of the reverse bias P-N junction

diode decreases when voltage increases.

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Junction Capacitance in Forward Biased Conditions: In a forward

biased diode, the junction capacitance exist. However, the junction capacitance is

very small. Hence, junction capacitance is neglected in forward biased diode.

The amount of capacitance changed with increase in voltage is called

junction capacitance. The junction capacitance is also known as depletion region

capacitance or barrier capacitance. Junction capacitance is denoted as CJ.

The change of capacitance at the depletion region can be defined as the

change in electric charge per change in voltage.

CJ = dQ / dV

Where,

CJ = Junction capacitance

dQ = Change in electric charge

dV = Change in voltage

The junction capacitance can be mathematically written as,

CJ = ε A / W

Where,

ε = Permittivity of the semiconductor

A = Area of plates or P-type and N-type regions

W = Width of depletion region

RECTIFIER

A Rectifier is an electrical device that is made of one or more than one diodes

which converts the alternating current (AC) into direct current (DC). It is used

for rectification of the signal.

Rectification is the process of conversion of the alternating current (which

periodically changes direction) into direct current (flow in a single direction).

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Based on the type of rectification circuit, the rectifiers are classified into

followings categories.

1. Half Wave Rectifier

2. Full Wave Rectifier

i. Full Wave Centre Tapped Rectifier

ii. Full Wave Bridge Rectifier

HALF WAVE RECTIFIER

A Half Wave Rectifier is a circuit, which converts an AC voltage into a pulsating

DC voltage using one half cycles of the applied AC voltage. It uses only one diode

which conducts during one-half cycle and switched off during the other half cycle

of the applied AC voltage.

When AC supply is applied at the input and positive half cycle appears across the

load, whereas the negative half cycle is suppressed, the rectifier is known as Half

Wave Rectifier. This can be done by using the semiconductor PN – junction diode.

The diode allows the current to flow only in one direction. Thus, convert the AC

voltage into DC voltage.

In half wave rectification, only one diode is used. It is connected in the circuit as

shown below.

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The AC supply to be rectified is generally given through a transformer. The

transformer is used to step down or step up the main supply voltage as per the

requirement. It also isolates the rectifier from power lines and thus reduces the

risk of electric shock.

Operation of Half Wave Rectifier: When AC supply is switched ON the

alternating voltage (Vin) shown in the figure above appears across the terminal AB

at the secondary winding.

During the positive half cycle, the terminal A is positive with respect to B and the

diode is forward biased. Therefore, it conducts and current flows through the load

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resistor RL. The output voltage waveform is shown in the above figure with respect

to Input signal. In the output, only positive half cycle of the Input will appear across

the load and negative half cycle will be suppressed.

Peak Inverse Voltage of Half Wave Rectifier

Peak Inverse Voltage (PIV) is the maximum voltage that the diode can withstand

during reverse bias condition. If a voltage is applied more than the PIV, the diode

will be destroyed.

Peak Inverse Voltage (PIV) of Half Wave Rectifier = Vm

Output DC Voltage ( Average Value of DC Output Voltage ) of Half

Wave Rectifier:

Negative half cycles are absent in the output wave form of a half wave rectifier.

So, in order to find the average value of the rectifier, the area under the positive

half cycle has divided by the total base length.

The area under the positive half cycle is the integral of sinusoidal wave equation

from the limits 0 to π. The total base length is the difference of limits of a complete

cycle (2π – 0 = 2π), which includes the base length of both the positive and

negative cycles.

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The average output voltage of a half wave rectifier can be derived as,

Average voltage, VDC = Vm /2π 0∫π sinωt dωt

= Vm/2π [ – cosωt]0π = Vm/2π [- cosπ + cos0]

= Vm/2π [1+1] = 2Vm/2π = Vm/π

The average voltage equation for a half wave rectifier is VDC = Vm / π.

Similarly, The average Current equation for a half wave rectifier is IDC = Im / π

RMS value of Half Wave Rectifier

RMS (Root mean square) value is the square root of the mean value of the

squared values. The RMS value of an alternating current is the equivalent DC

value of an alternating or varying electrical quantity. RMS value of an AC current

produces the same amount of heat when an equal value of DC current flows

through the same resistance.

RMS value of a signal = √ Area under the curve squared / base length.

For a function f(x) the RMS value for an interval [a, b] = √ (1/b-a) a ∫b f2(x) dx.

To derive the RMS value of half wave rectifier, we need to calculate the Voltage

across the load. If the instantaneous load Voltage is equal to VL = Vmsinωt, then

the average of load Voltage (VDC) is to be determine as follows:

In a half wave rectifier, the Negative Half Cycle will be removed from the output.

So, the total base length (2π) should be taken from the interval 0 to 2π.

The RMS voltage, VRMS = √ Vm2/2π 0∫π sin2ωt dωt

= √ Vm2/2π 0 ∫π(1 – cos2ωt) / 2 ) dωt = √ Vm

2/4π [ωt – sin2ωt / 2]0π

= √ Vm2/4π [ π – (sinπ) / 2 – (0 – (sin0) / 2)] = √ Vm

2/4π ( π ) = √ Vm2/ 4

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Therefore the RMS voltage, VRMS = Vm / 2

Similarly, RMS Current, IRMS = Im / 2

Ripple Factor of Half Wave Rectifier: ‘Ripple’ is the unwanted AC

component remaining when converting the AC voltage waveform into a DC

waveform.

Even though we try out best to remove all AC components, there is still

some small amount left on the output side which pulsates the DC waveform. This

undesirable AC component is called ‘ripple’.

To quantify how well the half-wave rectifier can convert the AC voltage into

DC voltage, we use what is known as the ripple factor (represented by γ or r).

The ripple factor is the ratio between the RMS value of the AC Component (on the

input side) and the average value of DC voltage (on the output side) of the

rectifier.

Putting the value of rms value of Input AC Current and average DC Output

Current, the value of ripple factor can easily be calculated from the above formula .

RMS Current for Half Wave rectifier , IRMS = Im / 2 = 0.5 Im

Average Current for a half wave rectifier is IDC = Im / π. = 0.318 Im

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Or

The ripple factor of half wave rectifier is equal to 1.21 (i.e. r = 1.21).

Ripple factor a Rectifier should be minimum as possible. So that capacitors

and inductors are used as filters to reduce the ripples in the circuit.

Efficiency of Half Wave Rectifier: Rectifier efficiency (η) is the ratio of the

output DC power to the input AC power.

The formula for the efficiency is equal to:

Pdc = I2dc RL = V2dc / RL = Im / π 2 = Vm / π 2

Pac = I2RMS RL = V2RMS / RL = Im / 2 2 = Im / 2 2

The efficiency of a half wave rectifier is equal to 40.6% (i.e. ηmax = 40.6%)

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Applications of Half Wave Rectifier

Half wave rectifiers are not as commonly used as full-wave rectifiers. Despite this,

they still have some uses:

1. For rectification applications

2. For signal demodulation applications

Advantages of Half Wave Rectifier

The main advantage of half-wave rectifiers is in their simplicity. As they don’t

require as many components, they are simpler and cheaper to setup and

construct.

The main advantages of half-wave rectifiers are:

1. The Circuit is very Simple (lower number of components)

2. The of half wave Rectifier is Cheaper in cost.

Disadvantages of Half Wave Rectifier

1. Half Wave Rectifier only allow a half-cycle of of the Input Signal and the

other half-cycle is wasted. This leads to power loss.

2. They produces a low output voltage.

3. The output current we obtain is not purely DC, and it still contains a lot of

ripple (i.e. it has a high ripple factor)

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FULL WAVE CENTRE TAPPED RECTIFIER

A Full Wave Centre Tapped Rectifier is a circuit, which converts an AC voltage

into a pulsating DC voltage using both half cycles of the applied AC voltage. It

uses two diodes of which one conducts during one-half cycle while the other

conducts during the other half cycle of the applied AC voltage.

The Full Wave Center Tapped Rectifier employs a transformer with the

secondary winding AB tapped at the centre point C. It converts the AC input

voltage into DC voltage. The two diode D1, and D2 are connected in the circuit as

shown in the circuit diagram below.

Each diode uses one-half cycle of the input AC voltage. The diode D1 utilises the

AC voltage appearing across the upper half (AC) of the secondary winding for

rectification. The diode D2 uses the lower half (CB) of the secondary winding.

Operation of the Full Wave Center Tapped Rectifier

When AC supply is applied at the input of the Transformer, Vin appears across the

terminals AB of the secondary winding of the transformer.

During the Positive Half Cycle of the secondary voltage, the terminal A is Positive,

and terminal B is Negative. Thus, the diode D1 becomes forward biased, and

diode D2 becomes reversed biased. When the Diode D1 is conducting, the

current (i) flows through the diode D1 load resistor RL (from M to L) and the upper

half of the secondary winding as shown in the circuit diagram marked by the arrow

heads Line.

During the Negative Half Cycle, the terminal B is positive and end A becomes

negative. This makes the diode D2 forward biased, and diode D1 reverse biased.

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So, the diode D2 conducts while the diode D1 does not conduct. The current (i)

flows through the diode D2 load resistor RL (from M to L) and the lower half of the

secondary winding as shown by the dotted arrow arrows line.

The current flowing through the load resistor RL is in the same direction

(i.e., from M to L) during both the Positive as well as the Negative half cycle of the

input. Hence, the DC output voltage (Vout = i RL) is obtained across the load

resistor.

The Wave form of Full Wave Centre Tapped Rectifiers with the input

voltage and the output voltage developed across the load are shown in the figure

below.

Peak Inverse Voltage of Full Wave Center Tapped Rectifier

The circuit diagram is shown below shows the instant when the secondary voltage

attains its maximum positive value.

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At this instant, Vm developed in the upper half of the secondary winding of the

transformer will forward bias the diode D1. This diode conducts, and the current

flows through RL, developing Vm voltage across it.

The diode D2 at this instant is reverse biased, and the voltage was coming

across it is the sum of the maximum value of voltage developed by the lower half

of the secondary winding and the voltage developed across the load. Hence, the

peak inverse voltage across the diode D2 is 2Vm.

Output DC Voltage ( Average Value of DC Output Voltage ) of Full

Wave Centre Tapped Rectifier:

In a Full Wave Rectifier, the Negative polarity of the wave will be converted to

positive polarity. So the average value can be found by taking the average of one

positive half cycle. The total base length ( π ) should be taken from the interval 0

to 2π as shown in figure below.

Derivation for average voltage of a full wave rectifier,

The Average Voltage, VDC = Vm/π 0∫π sinωt dωt

= Vm/π [ – cosωt]0π = Vm/π [- cosπ + cos0]

= Vm/π [1+1] = 2Vm/π

Average voltage equation for a full wave rectifier is VDC = 2Vm/π.

So during calculations, the average voltage can be obtained by substituting the

value of maximum voltage in the equation for VDC.

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RMS value of Full Wave Centre Tapped Rectifier








To derive the RMS value of Full Wave Rectifier, we need to calculate the Voltage




positive polarity. So the RMS value can be found by taking the RMS of one



Derivation for RMS voltage of a full wave rectifier is as under:

The RMS voltage, VRMS = √ Vm2/π 0∫π sin2ωt dωt

= √ Vm2/π 0 ∫π(1 – cos2ωt) / 2 ) dωt = √ Vm

2/2π [ωt – sin2ωt / 2]0π

= √ Vm2/2π [ π – (sinπ) / 2 – (0 – (sin0) / 2)] = √ Vm

2/2π ( π ) = √ Vm2/ 2

RMS voltage of Full Wave Rectifier, VRMS = Vm/ √2

Similarly, RMS Current of Full Wave Rectifier, IRMS = Im/ √2

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Ripple Factor of Full Wave Center Tapped Rectifier

The ripple factor for a Full Wave Rectifier is given by

Average Output Voltage for a full wave rectifier is VDC = 2Vm/π.

RMS voltage for a full wave rectifier is VRMS = Vm / √2

The ripple factor of Full wave rectifier is equal to 0.482 (i.e. r = 0.482).

Ripple factor Full wave rectifier is less than Half Wave rectifier. Ripple

factor a Rectifier should be minimum as possible. So that capacitors and

inductors are used as filters to reduce the ripples in the circuit.

Efficiency of Center Tapped Full Wave Rectifier

Efficiency, (η) is the ratio of dc output power to ac input power

Pdc = I2dc RL = V2dc / RL = 2 Im / π 2 = 2 Vm / π 2

Pac = I2RMS RL = V2RMS / RL = Im / √ 2 2 = Im / √ 2 2

The maximum efficiency of a Full Wave Rectifier is 81.2%.

The Efficiency of Full Wave Rectifier is double of Half Wave Rectifier.

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Advantages of Center Tapped Full Wave Rectifier

1. The ripple factor of Centre Tapped Full Wave Rectifier is much less than

that of half wave rectifier.

2. The rectification efficiency of Centre Tapped Full Wave Rectifier is twice

than a half wave rectifier. For a full wave rectifier, maximum possible value

of rectification efficiency is 81.2 % while that half wave rectifier is 40.6 %.

3. The DC output voltage and DC load current values of Centre Tapped Full

Wave Rectifier are twice than a half wave rectifier.

Disdvantages of Center Tapped Full Wave Rectifier

1. It is more expensive to manufacture a center tapped transformer which

produces equal voltage on each half of the secondary windings.

2. The output voltage is half of the secondary voltage, as each diode utilizes

only one half of the transformer secondary voltage.

3. The PIV (peak inverse voltage) of a diode used twice that of the diode

used in the half wave rectifier, so the diodes used must have high PIV.

Applications of Center Tapped Full Wave Rectifier

1. Full Wave Rectifiers are used in Car alternator.

2. Full Wave Rectifiers are used in any cell phone charger.

3. Full Wave Rectifiers are used in Laptop/tablet charger.

4. Full Wave Rectifiers are used in Power bank.

5. Full Wave Rectifiers are used in any other switching supply: alarm,

charger, Bluetooth device charger, lan / router supply etc

6. These are used in Audio power supply in pre amp and power amplifier

7. These are used in Any video device.

8. These are used in Lead battery charger.

FULL WAVE BRIDGE RECTIFIER

A Full Wave Bridge Rectifier is a circuit, which converts an AC voltage into a

pulsating DC voltage using both half cycles of the applied AC voltage. It uses four

diodes of which two conduct during one-half cycle while the other two conduct

during the other half cycle of the applied AC voltage.

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In Full Wave Bridge Rectifier, an ordinary transformer is used in place of a

center tapped transformer.

The circuit forms a bridge connecting the four diodes D1, D2, D3, and D4.

The circuit diagram of Full Wave Bridge Rectifier is shown below.

The AC supply which is to be rectified is applied diagonally to the opposite

terminals of the bridge. Whereas, the load resistor RL is connected across the

remaining two diagonals of the opposite terminals of the bridge.

Operation of Full Wave Bridge Rectifier

When an AC supply is switched ON, the alternating voltage Vin appears across the

terminals AB of the secondary winding of the transformer which needs

rectification. During the Positive half cycle of the secondary voltage, the terminal A

becomes Positive, and terminal B becomes Negative as shown in figure below:

The diodes D1 and D3 are forward biased and the diodes D2 and D4 are reversed

biased. Therefore, diode D1 and D3 conduct and diode D2 and D4 are switched off.

The current (I) flows from terminal A of the Transformer Secondary, through diode

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D1, load resistor RL (from M to N), diode D3 , the terminal B and terminal A of

transformer secondary.

During the negative half cycle, the terminal A becomes Negative and terminal B

Positive as shown in the figure below.

From the above Circuit, it is observed that the diode D2 and D4 are forward biased

and the diodes D1 and D3 are reverse biased. Therefore, diode D2 and D4 conduct

while diodes D1 and D3 switched OFF. Thus, current (I) flows from terminal B of

the Transformer Secondary through the diode D2, load resistor RL (from M to N),

diode D4 , the terminal A and B transformer Secondary.

The current flows through the load resistor RL in the same direction (M to

N) during both the half cycles. Hence, a DC output voltage Vout is obtained across

the load resistor.

The waveform of the full wave bridge rectifier is shown below.

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Peak Inverse Voltage of Full Wave Bridge Rectifier

When the secondary voltage attains its maximum Positive value and the terminal

A is Positive, and B is Negative as shown in the circuit diagram below:

At this instant diode, D1 and D3 are forward biased and conducts current.

Therefore, terminal M attains the same voltage as that A’ or A, whereas the

terminal N attains the same voltage as that of B’ or B. Hence the diode D2 and

D4 are reversed biased and the peak inverse voltage across both of them is Vm.

Therefore,

Output DC Voltage ( Average Value of DC Output Voltage ) of Full

Wave Bridge Rectifier:

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positive polarity. So the average value can be found by taking the average of one



Derivation for average voltage of a full wave rectifier,

The Average Voltage, VDC = Vm/π 0∫π sinωt dωt

= Vm/π [ – cosωt]0π = Vm/π [- cosπ + cos0]

= Vm/π [1+1] = 2Vm/π

Average voltage equation for a full wave rectifier is VDC = 2Vm/π.

So during calculations, the average voltage can be obtained by substituting the

value of maximum voltage in the equation for VDC.

RMS value of Full Wave Bridge Rectifier








To derive the RMS value of Full Wave Rectifier, we need to calculate the Voltage




positive polarity. So the RMS value can be found by taking the RMS of one



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Derivation for RMS voltage of a full wave rectifier is as under:

The RMS voltage, VRMS = √ Vm2/π 0∫π sin2ωt dωt

= √ Vm2/π 0 ∫π(1 – cos2ωt) / 2 ) dωt = √ Vm

2/2π [ωt – sin2ωt / 2]0π

= √ Vm2/2π [ π – (sinπ) / 2 – (0 – (sin0) / 2)] = √ Vm

2/2π ( π ) = √ Vm2/ 2

RMS voltage of Full Wave Rectifier, VRMS = Vm/ √2

Similarly, RMS Current of Full Wave Rectifier, IRMS = Im/ √2

Ripple Factor of Full Wave Bridge Rectifier

The ripple factor for a Full Wave Rectifier is given by

Average Output Voltage for a full wave rectifier is VDC = 2Vm/π.

RMS voltage for a full wave rectifier is VRMS = Vm / √2

The ripple factor of Full wave rectifier is equal to 0.482 (i.e. r = 0.482).

Ripple factor Full wave rectifier is less than Half Wave rectifier. Ripple

factor a Rectifier should be minimum as possible. So that capacitors and

inductors are used as filters to reduce the ripples in the circuit.

Efficiency of Center Tapped Full Wave Rectifier

Efficiency, (η) is the ratio of dc output power to ac input power

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Pdc = I2dc RL = V2dc / RL = 2 Im / π 2 = 2 Vm / π 2

Pac = I2RMS RL = V2RMS / RL = Im / √ 2 2 = Im / √ 2 2

The maximum efficiency of a Full Wave Rectifier is 81.2%.

The Efficiency of Full Wave Rectifier is double of Half Wave Rectifier.

Advantages of Full Wave Bridge Rectifier

1. The center tap transformer is eliminated in bridge Rectifier.

2. The output is double to that of the center tapped full wave rectifier for the

same secondary voltage.

3. The peak inverse voltage across each diode is one-half of the center tap

circuit of the diode.

Disadvantages of Full Wave Bridge Rectifier

1. It needs four diodes.

2. The circuit is not suitable when a small voltage is required to be rectified. It is

because, in this case, the two diodes are connected in series and offer double

voltage drop due to their internal resistance.

Applications of Bridge Full Wave Rectifier

1. Full Wave Rectifiers are used in Car alternator.

2. Full Wave Rectifiers are used in any cell phone charger.

3. Full Wave Rectifiers are used in Laptop/tablet charger.

4. Full Wave Rectifiers are used in Power bank.

5. Full Wave Rectifiers are used in any other switching supply: alarm, charger,

Bluetooth device charger, lan / router supply etc

6. These are used in Audio power supply in pre amp and power amplifier

7. These are used in Any video device.

8. These are used in Lead battery charger.

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FILTERS

The devices which converts the pulsating DC in to pure DC is called filter. As the

name specifies it filters the oscillations in the signal and provides a pure DC at the

output. The electronic reactive elements like capacitor and inductors are used to

do this work.

Series Inducter Filter (L) : The property of the inductor is that it opposes any

sudden change that occurs in a circuit an provides a smoothed output. In the case

of AC, there is change in the magnitude of current with time. So the inductor offers

some impendence (opposing force) for AC ((XL = 2π f L) and offers shot circuit for

DC ( Zero frequency Signal ). So by connecting inductor in series with the supply

blocks AC and allows DC to pass through it.

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The waveform of series inductor filter are shown above. It can be seen that

waveforms without filter consist of AC ripples while the waveform with filter is

regulated.

Capacitor Filter : The quality of the capacitor is to stores the electrical energy

for short time and discharges it. By controlling the charging and discharging rate of

the capacitor the pure DC can be obtained from the pulsating DC. The Capacitor

offers infinity impendence (For Zero Frequency Signal ) for DC ((XL = 1 / 2π f C)

and offers some resistance for AC Signal and this resistance also decreases with

increase in frequency.

In simple the Capacitor allows AC and blocks DC, so the Capacitor can

connect parallel to the output of Rectifier Circuit ( Power Supply ) so that the AC is

filtered out to the ground and DC will reach the load.

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When Input Signal rises upward, the Capacitor starts charging and stores energy

in the form of the electrostatic field. The Capacitor will charge to its peak value

because the charging time constant is almost zero because Capacitor is directly

connected to the Rectifier Circuit. During decreases period of the input signal, the

Capacitor will discharge through the load resistor, which have resistance. The

Capacitor will discharging slowly and in this way, the capacitor will maintain

constant output voltage and provide the regulated output.

The shunt capacitor filters use the property of capacitor which blocks DC

and provides low resistance to AC. Thus, AC ripples can bypass through the

capacitor.

If the value of Capacitance of the Capacitor is high, then it will offer very

low impedance to AC and extremely high impedance to DC. Thus, the AC ripples

in the DC output voltage gets bypassed through parallel capacitor circuit, and DC

voltage is obtained across the load resistor.

The ripple factor of series inductor filter is directly proportional to the load

resistance it means as the load resistance increases, ripple factor also starts

increasing. And in the case of shunt capacitor, the ripple factor is inversely

proportional to the value of load resistance. It implies that in shunt capacitor filter

the ripple factor decreases with increase in load resistance and increases with the

decrease in load resistance.

Thus, for better performance, we need a filter circuit in which ripple factor is

low and do not vary with the variation in load resistance. This can be achieved by

using the combination of series inductor filter and shunt capacitor filter. The

voltage stabilization property of shunt capacitor filter and current smoothing

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property of series inductor filter is utilized for the formation of choke filter or L-

section filter.

The combination of series inductor filter and shunt capacitor filter is

generally used for most of the applications. The combination results in two types:

1. L-section filter or Choke Filter

2. Pi Filter.

LC Filter ( L – Section or Choke Fiter ) : When the pulsating DC signal

from the output of the rectifier circuit is feed into choke filter, the AC ripples

present in the output DC voltage gets filtered by choke coil. The inductor has the

property to block AC and pass DC. This is because DC resistance of an inductor

is low and AC impedance of inductor coil is high. Thus, the AC ripples get blocked

by inductor coil.

Although the inductor efficiently removes AC ripples, a small percentage of AC

ripples is still present in the filtered signal. These ripples are then removed by the

capacitor connected in parallel to the load resistor. Now, the DC output signal is

free from AC components, and this pure DC can be used in any application.

If the inductor of high inductive reactance (XL), greater than the capacitive

reactance at ripple frequency is used than filtering efficiency gets improved.

Waveform of Choke Filter or L-section Filte

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Pi or π filter : Pi ( π ) Filter consists of a shunt capacitor at the input side,

and it is followed by an L-section filter. The output from the rectifier is directly

given across capacitor. The pulsating DC output voltage is filtered first by the

capacitor connected at the input side and then by choke coil and then by another

shunt capacitor.

The construction arrangement of all the components resembles the shape of

Greek letter Pi (π). Thus it is called Pi filter. Besides, the capacitor is present at

the input side. Thus, it is also called capacitor input filter.

The output voltage coming from rectifier also consist of AC components. Thus it is

a crucial need to remove these AC ripples to improve the performance of the

device. The output from the rectifier is directly applied to the input capacitor. The

capacitor provides a low impedance to AC ripples present in the output voltage

and high resistance to DC voltage. Therefore, most of the AC ripples get

bypassed through the capacitor in input stage only.

The residual AC components which are still present in filtered DC signal

gets filtered when they pass through the inductor coil and through the capacitor

connected parallel across the load. In this way, the efficiency of filtering increases

multiple times.

Waveform of Pi or π filter

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DIFFERENT TYPES OF DIODES

Light Emitting Diode : The term LED Stands for "Light-Emitting Diode." An

LED is an electronic device that emits light when an electrical current is passed

through it. It is one of the most standard types of the diode. When the diode is

connected in forwarding bias, the current flows through the junction and generates

the light.

Photo Diode: A special type of PN junction diode that generates current when

exposed to light is known as Photodiode. It is also known as Photo-Detector or

Photo-Sensor. It operates in reverse biased mode and converts light energy

into electrical energy.

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Schottky Diode: Schottky diode is a metal-semiconductor junction diode that

has less forward voltage drop than the P-N junction diode and can be used in

high-speed switching applications.

In a normal P-N junction diode, a P-type semiconductor and an N-type

semiconductor are used to form the P-N junction. When a P-type semiconductor is

joined with an N-type semiconductor, a junction is formed between the P-type and

N-type semiconductor. This junction is known as P-N junction.

In Schottky Diode, metals such as aluminum or platinum replace the P-type

semiconductor. When aluminum or platinum Metal is joined with N-type

semiconductor, a junction is formed between the Metal and N-type semiconductor.

This junction is known as a metal-semiconductor junction or M-S junction. A

metal-semiconductor junction formed between a metal and N-type semiconductor

creates a barrier or depletion layer known as a Schottky barrier.

Schottky diode can switch on and off much faster than the P-N junction

diode. Also, the schottky diode produces less unwanted noise than P-N junction

diode. These two characteristics of the schottky diode make it very useful in high-

speed switching power circuits.

Varactor Diode : Varactor Diode is a P-N junction diode whose Capacitance is

varied by varying the reverse voltage. The term varactor is originated from a

variable capacitor. Varactor diode operates only in reverse bias. The varactor

diode acts like a variable capacitor under reverse bias.

It is also sometimes referred to as Varicap Diode, Tuning Diode, Variable

Reactance Diode, or Variable Capacitance Diode.

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The symbol of the varactor diode is similar to that of the PN-junction diode.

The diode has two terminals namely anode and cathode. The one end of a symbol

consists the diode, and their other end has two parallel lines that represent the

conductive plates of the capacitor.

Tunnel Diode: A Tunnel diode is a heavily doped P-N junction diode in which

the electric current decreases as the voltage increases. In Tunnel Diode, electric

current is caused by “Tunneling”. The tunnel diode is used as a very fast

switching device in computers. It is also used in high-frequency oscillators and

amplifiers.

The tunnel diode is a heavily doped PN-junction diode. The concentration

of impurity in the normal PN-junction diode is about 1 part in 108. But, In the tunnel

diode, the concentration of the impurity is about 1 part in 103. Because of the

heavy doping, the diode conducts current both in the forward as well as in the

reverse direction. It is a fast switching device; thereby it is used in high-frequency

oscillators, computers and amplifiers.

Zener Doide: A Zener Diode is a special type of Diode which is designed to

operate in the Zener Breakdown Region. The Zener Diode consists of a special,

heavily doped P-N junction, designed to conduct in the reverse direction when a

certain specified voltage is reached.

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Zener diodes acts like normal P-N junction diodes under forward biased condition.

When forward biased voltage is applied to the Zener Diode, It allows electric

current in forward direction like a normal diode. But, It also allows electric current

in the reverse direction, if the applied reverse voltage is greater than the Zener

Voltage.

Zener diode is always connected in reverse direction because it is

specifically designed to work in reverse direction.

Zener diode is heavily doped than the normal P-N junction diode. Hence, it

has very thin depletion region. Therefore, Zener Diodes allow more electric

current than the normal P-N junction diodes.

V-I Characteristics of Zener Diode:

Zener diode works in the reversed biasing conditions. In reversed biased mode, its

Anode is connected with the Negative terminal and Cathode with the Positive

terminal of Battery / DC Supply.

The figure shows the reverse biasing connection of Zener diode with DC Supply or

Battery.

The reversing biasing effect of the Zener Diode is as shown in the curve between

the Voltage and Current.

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1. When reverse voltage is applied to the Zener diode, Initially the small amount

of the leakage current flows in the diode, till that point the applied voltage is

less than the Zener voltage.

2. When the value of applied voltage approaches the Zener voltage, a large

amount of the reversed current flows in the diode and curve suddenly

changes its state from the OFF to ON.

3. Due to the instant increase in the current value, the breakdown happens in

the diode that is called the Zener breakdown. But, Zener diode manifests a

restrained breakdown that does harm the component.

The quantity of the Zener breakdown voltage fluctuates according to the doping

level of the diode. If the doping level of the diode is larger then breakdown occurs

at lesser voltage. If doping is less then breakdown happens at the higher value of

the revered supplied voltage.

Usually, the value of the Zener voltage for the diodes is (1.8) volts to (400) volts.

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AVALANCHE BREAKDOWN AND ZENER BREAKDOWN

Avalanche Breakdown: Avalanche breakdown occurs in a P-N junction

diode which is moderately doped and has a thick junction (means its depletion

layer width is high). Avalanche breakdown usually occurs when we apply a high

reverse voltage across the diode. So as we increase the applied reverse voltage,

the electric field across junction will keep increasing.

The collision increases the electron-hole pair. As the electron-hole induces

in the high electric field, they are quickly separated and collide with the other

atoms of the crystals. The process is continuous, and the electric field becomes

so much higher than the reverse current starts flowing in the PN junction. The

process is known as the Avalanche breakdown. After the breakdown, the junction

cannot regain its original position because the diode is completely burnt off.

Zener Breakdown : The Zener breakdown usually occurs in heavily doped &

thin junction (means depletion layer width is very small) diodes ( Zener Diodes ).

Zener breakdown does not result in damage of diode. Since current is only due to

drifting of electrons, there is a limit to the increase in current as well.

When a high reverse voltage is applied across the Zener diode and as

increasing this applied reverse voltage, an electric field across junction will keep

increasing, the electrons start moving across the junction and electron hole

recombination occurs. So a net current is developed and it increases rapidly with

increase in electric field.

Thus, as long as the current in the diode is limited the Zener diode will not

destroy the junction. But avalanche breakdown destroys the junction.

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Differences Between Avalanche & Zener Breakdown

Sr. No.

Avalanche Breakdown Zener Breakdown

1.

The breakdown which occurs

because of the collision of the

electrons inside the PN-junction is

called avalanche breakdown.

The Zener breakdown occurs when

the heavy electric field is applied

across the PN- junction.

2.

The avalanche breakdown occurs in

the thick region.

The Zener breakdown occurs in

the thin region.

3.

After avalanche breakdown, the

junction of the diode will not regain

its original position.

After Zener breakdown, the

junction regains its original

position.

4.

The existence of the electric field is

less on the avalanche breakdown

as compared to the Zener

breakdown. Because the

mechanism of avalanche

breakdown occurs in the moderate

doped region.

The existence of the electric field is

more on the Zener breakdown as

compared to the avalanche

breakdown. Because the mechanism

of Zener breakdown occurs in the

heavily doped region.

5.

The avalanche breakdown

produces the pairs of electrons and

holes because of the thermal

effects.

The Zener breakdown produces the

electrons.

6.

The avalanche breakdown occurs in

low doping material.

The Zener breakdown occurs in high

doping material.

7.

The avalanche breakdown voltage

causes because of high reverse

potential because it is lightly doped.

Zener breakdown is because of low

reverse potential as it is highly doped.

8.

The avalanche breakdown voltage

is directly proportional to the

temperature.

The Zener breakdown voltage is

inversely proportional to the

temperature.

9.

In avalanche breakdown, the

mechanism of ionisation occurs

because of collision of electrons.

In the Zener breakdown, the

mechanism of ionisation occurs

because of the electric field.

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Advantages of Zener Diode:

There are some advantages of the Zener diode over the general diode that make

it effective to operate in the high voltage conditions.

1. The Power consumption capability of Zener Diode is higher than the

Normal Diode.

2. Efficiency of Zener Diode is very high.

3. Zener Diode is available in a smaller size.

4. It is a less expensive diode.

Applications of Zener Diode

1. Zener Diode is commonly used as a voltage references device.

2. It used in the voltage regulators.

3. It used for switching purposes.

4. Zener diode is an important part of the clamp and clipping circuits.

5. It used in many security circuits.

6. It also used in electronics devices like mobile laptops, computer, etc.

Numercal -1 : Half wave Rectifier having the Input AC Supply 220 V 50 Hz

and with a Transformer having coil Turn ratio 20: 1.

Calculate :

i. Output DC Voltage

ii. PIV of Doide.

Solution:

(i) Given: Input AC Supply 220 V 50 Hz, VRMS = 220 V , f = 50 Hz

Transformer having coil Turn ratio 20: 1

Maximum Input Voltage ( Vm ) at Primary of Transformer = √2 VRMS

= √2 x 220

= 311 V

Maximum Input Voltage (Vm) at Secondary of Transformer = 311/ 20

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= 15.55 V

Output DC Voltage ( VDC ) = Vm / π = 15.55 / π = 15.55 / 3.14 = 4.95 V

VDC = 4.95 V

(ii) Peak Inverse of Diode in Half Wave Rectifier = Vm

PIV of Diode = 15.55 V

Numerical -2 : A Full Wave Centre Tapped Rectifier having a Input voltage 20 Sin

314t at each Half of Secondary Winding of Transformer. A Load Resistor of 500 Ω

is connected to the Rectifier.

Calculate:

i. Im ii. IDC iii. IRMS

iv. Ripple Factor v. Rectification Efficiency vi. PIV of Diode

Solution:

(i) Vm = 20 Volts

Im = Vm / RL ( Considering Diode Forward Resistance Negligible )

Im = 20 / 500 = 40 mA

(ii) IDC = 2 Im / π = 2 x 40 / π = 80 / 3.14 = 25.47 mA

IDC = 25.47 mA

(iii) lRMS = Im / √2 = 40 mA / √2 = 28.28 mA

lRMS = 28.28 mA

(iv) Ripple Factor ( R. F.)

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Ripple Factor = 0.482

(v) Rectification Efficiency

Pdc = I2dc RL

Pac = I2RMS RL

Rectification efficiency = I2dc RL / I2RMS RL = I2dc / I2RMS

= ( 25.47 mA )2 / ( 28.28 mA )2

= 0.812 = 81.2 %

Rectification efficiency = 81.2 %

(vi) Peak Inverse of Diode in Full Wave Centre Tapped Rectifier = 2 Vm

= 2 x 20 = 40 V

PIV of Diode = 40 V

Numerical -3 : A Full Wave Bridge Rectifier having a Input voltage 20 Sin 314t at

Secondary Winding of the Transformer. A Load Resistor of 500 Ω is connected to

the Rectifier.

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Calculate: i. Im ii. IDC iii. IRMS

iv. Ripple Factor v. Rectification Efficiency vi. PIV of Diode

Solution:

(i) Vm = 20 Volts

Im = Vm / RL ( Considering Diode Forward Resistance Negligible )

Im = 20 / 500 = 40 mA

(ii) IDC = 2 Im / π = 2 x 40 / π = 80 / 3.14 = 25.47 mA

IDC = 25.47 mA

(iii) lRMS = Im / √2 = 40 mA / √2 = 28.28 mA

lRMS = 28.28 mA

(iv) Ripple Factor ( R. F.)

Ripple Factor = 0.482

(v) Rectification Efficiency

Pdc = I2dc RL

Pac = I2RMS RL

Rectification efficiency = I2dc RL / I2RMS RL = I2dc / I2RMS

= ( 25.47 mA )2 / ( 28.28 mA )2

= 0.812 = 81.2 %

Rectification efficiency = 81.2 %

(vi) Peak Inverse of Diode in Full Wave Bridge Rectifier = Vm = 20 V

PIV of Diode = 20 V

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Fill IN THE BLANKS:

1. The Diode is ……………… Component.

2. A crystal / Semi-Conductor diode has ………………. P-N junction.

3. An ideal diode has ……………. forward biased resistance.

4. The knee voltage of a crystal / semiconductor diode is approximately equal to

……………… of diode.

5. An ideal diode has …………. reverse biased resistance.

6. When the graph between current through and voltage across a device is a

straight line, the device is referred to as …………. Device.

7. A semiconductor diode is a ………….. device.

8. If the doping level of a semi-conductor diode is increased, the breakdown

voltage will …………… .

9. The forward voltage drop across a silicon diode is about ……….. Volts.

10. If the doping level in a semiconductor diode is increased, the width of

depletion layer is …………… .

11. An ideal Semiconductor diode is one which behaves as a perfect …………

when forward biased.

12. A crystal / semiconductor diode utilises …………characteristic for rectification.

13. If the PIV rating of a diode is exceeded, the diode is …………….

14. The leakage current in a Semi-Conductor Diode is due to ……….. carriers.

15. If the temperature of a Semiconductor Diode increases, then leakage current

will………….. .

16. The small amount of ac signal present on the output of a filtering network for a

dc power supply is known as…………. .

17. Special diodes designed to conduct in reverse direction are called …………..

18. Rectification efficiency of Full Wave bridge rectifier is …………….

ANSWERS:

1) Unidirectional 2) One 3) Zero 4) Barrier Potential

5) Infinity 6) Linear 7) Non-linear 8) Decreased

9) 0.7 10) Increased 11) Conductor 12) Forward

13) Destroyed 14) Minority 15) Increased 16) Ripples

17) Zener Diodes. 18) 81.2%.

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Fill IN THE BLANKS:

19. A Full Wave Bridge Rectifier having ………… number of diodes.

20. Peak Inverse Voltage of Full wave Bridge Rectifier is ………….. .

21. The ripple factor of DC Power Supply is measures of its filters ………… .

22. The maximum efficiency of a Full Wave rectifier is ………..

23. The process of making DC from AC is known as ………….

24. Ripple factor of Full Wave Bridge Rectifier is …………..

25. Peak Inverse Voltage of Half Wave Rectifier is …………….. .

26. Ripple factor of Half Wave Rectifier is …………...

27. A Half Wave Rectifier having ………. number of diode.

28. A single-phase full wave rectifier is a ………… pulse rectifier.

29. Rectification Efficiency of Half Wave Rectifier is………….. .

30. RMS value (lRMS ) of Half Wave Rectifier is ………..

31. Peak Inverse Voltage of Full wave Centre Tapped Rectifier is ……………..

32. A Full Wave Centre Tapped ( mid-point type) rectifier requires ……. number

of diodes.

33. The PIV rating of each diode in a bridge rectifier is …….. that of the equivalent

centre-tap rectifier.

34. Zener diode are used as…………...

35. A zener diode utilizes ………… characteristics for its operation.

36. Filters are used in rectifiers to ……………. ripples at the output.

37. A Zener diode has ………… P-N junction.

38. The doping level in a Zener diode is ……….than that of a crystal diode.

39. A series resistance is connected in Zener diode circuit to ……… Zener diode.

40. A Zener diode has ………….. breakdown voltage.

ANSWERS:

19) Four 20) Vm 21) Efficiency 22) 81.2%

23) Rectification 24) 0.482 25) Vm 26) 1.21

27) One 28) Two 29) 40.6% 30) Im / √2

31) 2 Vm 32) Two 33) Half 34) Voltage Regulator

35) Reverse. 36) Minimize 37) One 38) More

39) Protect 40) Sharp

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ELECTRO - MAGNETIC INDUCTION

ELECTRO – MAGNETIC INDUCTION

CONCEPT OF ELECTRO - MAGNETIC FIELD PRODUCED FLOW

ELECTRIC CURRENT

Magnet : An object which is capable of producing magnetic field and attracting

unlike poles and repelling like poles.

Magnetic Field: “A magnetic field is the area around a magnet, magnetic

object, or an electric charge in which magnetic force is exerted.”

If a magnetic compass is placed on a table, it is found that its needle rests in

geographic north-south direction. But when it is placed near a magnet, the needle

swings and then rests in some other direction. As the compass is placed at

different positions around a magnet, the direction in which the needle rests,

changes such that its one end always points towards the nearer pole of the

magnet. This behaviour of the compass needle is due to the influence of the

magnet near it. So, the region in which the compass gets influenced is called the

magnetic field of the magnet.

Types of Magnets :There are three types of magnets and are as follows:

1. Permanent Magnet : Permanent Magnets are those magnets that are

commonly used. They are known as permanent magnets because they do not

lose their magnetic property once they are magnetized.

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2. Temporary Magnet : Temporary magnets can be magnetized in the

presence of a magnetic field. When the magnetic field is removed, these

materials lose their magnetic property. Iron nails and paperclips are examples

of the temporary magnet..

3. Electromagnets: Electromagnets consist of a coil of wire wrapped around

the metal core made from iron. When this material is exposed to an electric

current, the magnetic field is generated making the material behave like a

magnet. The strength of the magnetic field can be controlled by controlling the

electric current.

An advantage of the electromagnet over a permanent magnet because

controlling the electric current also controls the magnetic field, i.e., the

strength of electric field controls the strength of magnetic field. In fact, the

poles of an electromagnet can even be reversed by reversing the flow of

electricity.

Magnetic Field Produced by Electric Current: An electric current on a

long straight wire produces a magnetic field whose field lines are made up of

circles with center on the wire. This magnetic field may be detected by placing a

magnetic compass close to the wire as shown in the figure below.

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The direction of the magnetic field ( B ) may be determined by the right hand rule:

imagine grasping the wire using the RIGHT hand with the thumb in the direction of

the current. The direction of the magnetic field is in the same direction along which

the four fingers curl.

In order to produce a stronger magnetic field using electric currents, several

loops are grouped together to form a solenoid.

A solenoid not only produce a strong magnetic field but also a uniform one

with a North and a South pole similar to magnets. Solenoids have many

applications. The magnetic filed produced by solenoids may be controlled by

controlling the current in the solenoid. The current in the solenoid may be switched

on or off and also by increasing or decreasing the electric current in the solenoid,

we can control the intensity of the magnetic field produced.

The following Points may be noted for the Magnetic Effect of Electric Current:

1. When the current flowing through the conductor is higher, the Magnetic

field is stronger and vice-versa.

2. The magnetic field near the conductor id stronger and become weaker

as move away from the conductor.

3. The Magnetic lines of force around the conductor may be either

clockwise or anti-clockwise, depending upon the direction of current.

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IMPORTANT TERMS:

Magnetic Flux: The number of magnetic lines of forces set up in a magnetic

circuit is called Magnetic Flux.

It is analogous to Current as in Electric Circuit. Its SI unit is Weber (Wb)

and denoted by φ. The magnetic flux measures through flux meter.

Properties of Magnetic Flux

1. They always form a closed loop.

2. They always start from the north pole and ends in the south pole.

3. They never intersect each other.

4. Magnetic lines of forces that are parallel to each other and are in the

same direction repel each other.

Reluctance : The opposition offered by a magnetic circuit to the magnetic flux

is known as Reluctance.

It is analogous to Resistance as in Electric Circuit. Its Unit is AT / Wb

(ampere-turns / Weber). It is denoted by S.

Where, l – the length of the conductor

μo – Permeability of vacuum which is equal to 4π Χ107 Henry /metre.

μr – Relative permeability of the material.

A – cross-section area of the conductor.

The reluctance of the magnetic circuit is directly proportional to the length of the

conductor and inversely proportional to the cross-section area of the conductor.

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The reluctance in the DC field is defined as the ratio of the magnetic motive

force and to the magnetic flux of the same circuit. The reluctance in the DC field is

expressed as

Where, S – Reluctance in ampere-turns per weber.

F – Magnetic Motive Force

Φ – magnetic flux

Permeance. It is the ability of the Magnetic Circuit to allow magnetic flux

through it and reciprocal of the magnetic reluctance is known as the

Magnetic Permeance. It is analogous to conductance in Electric Circuit and its SI

Unit is Henry . It is given by the expression

Permeability: The magnetic permeability is defined as the property of the

material to allow the magnetic flux to pass through it. It is analogous to

conductivity in Electric Circuit and its SI unit is Henry per meter.

The magnetic permeability of the material is directly proportional to the

number of lines passing through it.

Consider the soft iron ring is placed inside the magnetic field shown above.

The most of the magnetic line of force passes through the soft iron ring because

the ring provides the easy path to the magnetic lines. This shows that the

magnetic permeability of the iron is much more than the air or the permeability of

air is very poor.

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Figure shows the magnetic lines ( Flux ) passing through air. The

permeability of the air or vacuum is represented by μ0 which is equal to 4π×17-

7 H/m. The permeability of air or vacuum is very poor.

The Permeability of the material is equal to the ratio of the field intensity to

the flux density of the material. It is expressed by the formula shown below.

Where, B – Magnetic Flux Density

H – Magnetic Field Intensity

The permeability of any medium is represented μ.

Where μo – Permeability of vacuum which is equal to 4π Χ107 Henry /metre.

μr – Relative permeability of the material.

Relative Permeability (μr ) –The relative permeability of the material is the

ratio of the permeability of any medium to the permeability of air or vacuum. It is

expressed as

Magneto Motive Force ( MMF ) : The magnetic pressure, which sets up the

magnetic flux in a magnetic circuit is called Magneto Motive Force. It is analogous

to EMF ( Voltage ) in Electric Circuit. The SI unit of MMF is Ampere-turn (AT).

The MMF for the inductive coil shown in the figure below is expressed as

MMF = N. I

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Where, N – numbers of turns of the inductive coil

I – current

The strength of the MMF is equivalent to the product of the current around the

turns and the number of turns of the coil.

The MMF is also known as the Magnetic Potential. It is the property of a

material to give rise to the magnetic field. The Magneto Motive Force is the

product of the magnetic flux and the magnetic reluctance. The reluctance is the

opposition offers by the magnetic field to set up the magnetic flux on it.

The MMF regarding reluctance and magnetic flux is given as

Where S – Reluctance

Φ – Magnetic Flux

The Magneto Motive Force can measure regarding magnetic field intensity and

the length of the substance. The magnetic field strength is the force act on the unit

pole placed on the magnetic field. MMF regarding field intensity is expressed as

Where H is the magnetic field strength,

l is the length of the substance.

Magnetic Circuit: The closed path followed by magnetic lines of forces is

called the magnetic circuit. In the magnetic circuit, magnetic flux or magnetic

lines of force starts from a point and ends at the same point after completing its

path.

Consider a solenoid having N turns wound on an iron core. The magnetic

flux of ø Weber sets up in the core when the current of I ampere is passed through

a solenoid.

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Let, l = mean length of the magnetic circuit

A = cross-sectional area of the core

µr = relative permeability of the core

Now the Flux Density ( B ) is the Magnetic flux per unit area in the core material

Magnetic Field Intensity ( Magnetising force ) in the core

According to work law, the work done in moving a unit pole once round the

magnetic circuit is equal to the ampere-turns enclosed by the magnetic circuit.

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The above equation explains the following points:

1. The Magnetic Flux is directly proportional to the number of turns (N) and

current (I). It shows that the Magnetic flux increase if the number of turns or

current increases and decreases when either of these two quantity

decreases. NI is the magneto motive force (MMF).

2. The Magnetic Flux is inversely proportional to l /a µ0 µr. Where ( l / a µ0 µr) is

known as Reluctance. When reluctance is lower, the magnetic flux will be

higher and vice- verse.

Analogy Between Magnetic Circuit and Electric Circuit

Sr. No.

Magnetic Circuit Electric Circuit

1. The closed path for magnetic

flux is called magnetic circuit.

The closed path for electric current is

called electric circuit.

2. In Magnetic Circuit, the unit of

flux is Weber.

In Electric Circuit, the unit of current is

Ampere.

3. In Magnetic Circuit, Magneto

motive force (MMF) is the

driving force and it is measured

in Ampere-turns (AT).

In Electric Circuit, Electro motive force

(EMF) is the driving force and it is

measured in volts (V).

4. In Magnetic Circuit, Reluctance

opposes the flow of magnetic

flux. It is represented by

S = l / aµ

and measured in (AT/wb)

In Electric Circuit, Resistance opposes

the flow of current. It is represented

by,

R = ρ. l / a

and measured in (Ώ).

5. In the magnetic circuit,

Permeance = 1 / Reluctance

In the electric circuit,

Conductance = 1 / Resistance.

6. In the magnetic circuit, there is

existence of Permeability.

In Electric Circuit, there is existence of

Conductivity.

7. In the magnetic circuit, there is

existence of Reluctivity.

In Electric Circuit, there is existence of

Resistivity.

8. For magnetic flux, there is no

perfect insulator. It can set up

even in non-magnetic materials

like air, rubber, glass, etc.

For electric circuit, there are a large

number of perfect insulators like

glass, air, rubber, PVC etc which do

not allow it to flow through them.

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9. The reluctance (S) of a magnetic

circuit is not constant rather it

varies with the value of Magnetic

Flux Density (B).

The resistance (R) of an electric

circuit is almost constant as its value

depends upon the value of ρ. The

value of ρ and R can change slightly

if the change in temperature takes

place.

10. In Magnetic Circuit, Magnetic

lines of flux start from North Pole

and ends at the South Pole.

In Electric Circuit, Electric current

starts from the positive charge and

ends on the negative charge.

FARADAY’S LAWS OF ELECTRO-MAGNETIC INDUCTION

Faraday’s Laws of Electromagnetic Induction consists of two laws. The first law

describes the induction of emf in a conductor and the second law quantifies the

emf produced in the conductor.

Faraday’s First Laws of Electromagnetic Induction: Faraday’s First

law of electromagnetic induction states that “Whenever there is change in

magnetic flux associated with a coil, an EMF is induced in that Coil.”.

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Faraday’s Second Law of Electromagnetic Induction: Faraday’s Second

law of electromagnetic induction states that “ The magnitude of induced emf in a

coil is directly proportional to the rate of change magnetic flux associated with that

coil”.

ε = - N Δ Φ Δt

Where,

ε is the electromotive force

Φ is the magnetic flux

N is the number of turns

The Negative sign indicates that the direction of the induced emf and change in

direction of magnetic fields have opposite signs.

Numerical -1: A coil of 500 turns of wire is wounded on a magnetic core of

reluctance 5000 AT/Wb. If the of 2 mA flowing in the coil is reversed in 20

mSec. Calculate the average EMF induced in the Coil.

Solution: Given N = 500, Current ( I ) = 2 mA, Reluctance = 5000 AT/ Wb

MMF = N x I = 500 x 2 mA = 1 AT

Flux in Coil = MMF / Reluctance = 1 / 5000 = 0.2 mWb

When the current 2 mAmp in the coil is reversed, the flux will also reversed

ε = - N Δ Φ = - N d Φ Δt dt

N = 500 Turns

d Φ = 0.2 – ( -0.2 ) = 0.4 mWb

dt = 20 m Sec = 20 x 10-3 Sec

e = 500 x 0.4 x 10-3 / 20 x 10-3 = 10 V

Numerical -2 : A Coil of 200 Turns is linked with a flux of 30 mWb. If this

flux is reversed in the time of 5 mSec. Calculate the Average EMF induced in

the that coil.

Solution : Given, N = 200 Turns

Change in Flux d Φ = 30 – ( -30 ) = 60 m Wb. = 60 x 10-3 Wb

Time required to change , dt = 5 m Sec = 5 x 10-3 Sec

ε = - N Δ Φ = - N d Φ Δt dt ε = 200 x 60 x 10-3 / 5 x 10-3 = 2400 V

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Self-inductance : Self-inductance of the coil is defined as the property of the

coil due to which it opposes the change of current flowing through it. Inductance is

attained by a coil due to the self-induced emf induced in the coil itself by changing

the current flowing through it.

If the current in the coil is increasing, the self-induced emf produced in the

coil will oppose the rise of current, that means the direction of the induced emf is

opposite to the applied voltage.

If the current in the coil is decreasing, the emf induced in the coil is in such

a direction as to oppose the fall of current; this means that the direction of the self-

induced emf is same as that of the applied voltage. Self-inductance does not

prevent the change of current, but it delays the change of current flowing through

it.

This property of the coil only opposes the changing current (alternating

current) and does not affect the steady current that is (Direct Current) when flows

through it. The unit of inductance is Henry (H).

Where,

N – Number of turns in the coil

Φ – Magnetic Flux

I – Current flowing through the Coil

Self-Induced EMF: Self-induced emf is the e.m.f induced in the coil due to the

change of flux produced by linking it with its own turns.

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Consider a coil having N number of turns as shown in the above figure. When the

switch S is closed and current I flows through the coil, it produces flux (φ) linking

with its own turns. If the current flowing through the coil is changed by changing

the value of variable resistance (R), the flux linking with it, changes and hence emf

is induced in the coil. This induced emf is called Self Induced emf.

The direction of this induced emf is such that it opposes its very own cause

which produces it, that means it opposes the change of current in the coil. This

effect is because of Lenz’s Law.

Since the rate of change of flux linking with the coil depends upon the rate

of current in the coil.

The magnitude of self-induced emf is directly proportional to the rate of change of

current in the coil. L is constant of proportionality and called as Self Inductance or

the Coefficient of Self Inductance or Inductance of the coil.

Mutual Inductance : Mutual Inductance between the two coils is defined as

the property of the coil due to which it opposes the change of current in the other

coil. When the current in the neighbouring coil changes, the flux sets up in the coil

and because of this, changing flux emf is induced in the coil called Mutually

Induced emf and the phenomenon is known as Mutual Inductance.

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Two coils namely coil A and coil B are placed nearer to each other. When the

switch S is closed, and the current flows in the coil, it sets up the flux φ in the coil

A and emf is induced in the coil and if the value of the current is changed by

varying the value of the resistance (R), the flux linking with the coil B also changes

because of this changing current.

Thus this phenomenon of the linking flux of the coil A with the other coil, B is

called Mutual Inductance.

The value of Mutual Inductance (M) depends upon the following factors

1. Number of turns in the neighboring coil .

2. Cross-sectional area.

3. Closeness of the two coils.

Mutually Induced EMF : The emf induced in a coil due to the change of flux

produced by another neighbouring coil linking to it, is called Mutually Induced

emf.

Consider Coil B is having N2 number of turns and is placed near another

coil A having N1 number of turns, as shown in the figure below:

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When the switch (S) is closed in the circuit shown above, current I1 flows through

the coil A, and it produces the fluxφ1. Most of the flux says φ12 links with the other

coil B.

If the current flowing through the coil A is changed by changing the value of

variable resistor R, it changes flux linking with the other coil B and hence emf is

induced in the coil. This induced emf is called Mutually Induced emf.

The direction of the induced emf is such that it opposes the cause which

produces it, that means it opposes the change of current in the first coil. This

effect of opposition caused by its own reason of production is called Lenz’s Law.

A galvanometer (G) is connected to coil B for measuring the induced emf.

Since the rate of change of flux linking with the coil, B depends upon the rate of

change of current in the coil A.

The rate of change of current in coil A is directly proportional to the magnitude of

the mutually induced emf. M is called the Constant of Proportionality and is also

called as Mutual Inductance or Coefficient of Mutual Inductance.

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Numerical – 3: A Coil is wound with 400 Turns and A current of 2 Amp

flowing through this coil induced flux of 100 mWb. Calculate its Self

Inductance.

Solution: Given, N = 400 Turns, Flux (φ ) = 100 m Wb, I = 2 Amp.

L = N x φ / I = 400 x 100 x10-3 / 2 = 40,000 x10-3 / 2 = 20 Henry

L = 20 H

Numerical – 4 : If the current flowing through a coil having an Inductance

of 2 H is reduced from 4 Amp to 1 Amp in 100 m Sec, Calculate the average

value of EMF induced in the Coil.

Solution : Given, L = 2 H

Chane in Current ( di ) = 1 – 4 = - 3 Amp

Time required to change this Current = 100 mSec = 100 x 10-3 Sec

eL = - L di / dt = - 2 x ( - 3 ) / 100 x 10-3 = 60 V

Numerical - 5 : A Insulating material ring has a mean diameter of 200 mm

and a cross-sectional area 300 mm2. It is wound with 2000 turns of wires. A

second coil of 1000 turns wound on the top of the first. Assuming that all the

flux produced by the first coil links with the second, Calculate the Mutual

Inductance.

Solution: Given : No. of Turns on First Coil ( A ) N1 = 2000

No. of Turns on Second Coil ( B ) N2 = 1000

Diameter of Ring ( D ) = 200 mm

Cross-section Area of the Ring ( a ) = 300 mm2

Relative Permeability of Insulating martial ( µr ) = 1.0

Relative Permeability of Air ( µ0 ) = 4 π x 10-7 H /m

Length ( l ) of Ring = 2 π r = π D = 200 x 10-3 π = 0.2 π

Mutual Inductance ( M ) = N1 N2 x µ0 µr a l

M = 2000 x 1000 x 4 π x 10-7 x 1.0 x 300 x 10-3

0.2 π

M = 2000 x 1000 x 4 x 3.14 x 10-7 x 1.0 x 300 x 10-3

0.2 x 3.14

M = 1.2 H

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Numerical - 6 : The Self Inductance of 1200 Turns coils is 0.5 H. If 80% of

the Flux is linking with a second coil of 1000 Turns.

Calculate: ( i ) The Mutual Inductance

( ii ) EMF induced in the Second coil when the current in the

First Coil changing at the rate 200 Amp per second.

Solution : Given Number of Turns in the First coil ( A ) N1 = 1200 Turns

Number of Turns in the Second Coil ( B ) N2 = 1000 Turns

Self Inductance of First coil L1 = 0.5 H

Linking Flux with second Coil φ2 = 0.8 φ1

Rate of Change of Current in First Coil = di1/dt = 200 Amp/sec

( i ) We know that, L = N x φ / I

L1 = N1 . φ1 / I1 or L1 / N1 = φ1 / I1

0.5 / 1200 = φ1 / I1

Mutual Inductance M = N2 x φ2 / I2

φ2 = 0.8 φ1

M = 1000 x 0.8 x 0.5 / 1200 = 0.33 H = 330 mH

Mutual Inductance ( M ) = 330 mH

( ii ) EMF induced in the Second Coil ( B ) = M . di1 /dt

= 0.33 x 200

= 66.66 Volts

Emf2 = 66.66 Volts

CONCEPT OF CURRENT GROWTH, DECAY AND TIME CONSTANT

IN AN INDUCTIVE ( RL ) CIRCUIT

Growth of Current : Figure below shows a circuit containing resistance R and

inductance L connected in series combination through a battery of constant emf E

through a switch S1 and S2.

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Suppose in the beginning, the switch S1 is closed as shown in figure above.

So, Battery E, inductance L and resistance R are now connected in series.

Because of self induced emf current will not immediately reach its steady value

but grows at a rate depending on inductance and resistance of the circuit as

shown in figure below;

Let at any instant I be the current in the circuit increasing from 0 to a maximum

value at a rate of increase dI/dt

If we put t= τL = L / R

Hence, the time in which the current in the circuit increases from zero to 63% of

the maximum value of Imax is called the Growth Time Constant of the RL Circuit.

Decay of Current : When the switch S2 is closed as shown in the figure

below, the R - L circuit is again closed and battery is cut off. In this condition the

current in the circuit begins to decay.

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Because of self induced emf current will not immediately decayed, but decreases

at a rate depending on inductance and resistance of the circuit as shown in figure

below;

Let at any instant I be the current in the circuit decreasing from maximum value at

a rate of dI/dt to initial value 0.

If in the above equation put t = τL = L/R

then

I = Imax e-1 = .37 Imax

Hence the time in which the current decrease from the maximum value to 37% of

the maximum value Imax is called the Decay Time Constant of the RL Circuit.

ENERGY STORED IN AN INDUCTOR

An inductor is a passive electronic component which is capable

of storing electrical energy in the form of magnetic energy. Basically, it uses a

conductor that is wound into a coil, and when electricity flows into the coil from the

left to the right, this will generate a magnetic field in the clockwise direction.

When a battery of voltage ( V ) is connected to Inductor ( L ) with internal

Resistance ( R ) as shown in figure below:

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Now, the battery establishes the current ( I ) in an inductor has to do work against

the opposing induced emf. The energy supplied by the battery is stored in the

inductor.

Let I be the instantaneous current in the circuit then applying Kirchoff's

voltage law, we get

V = I R + L dI /dt

Multiplying by idt on both sides

V x I dt = I2 x R dt + L x I dI

Where I2R dt is Energy dissipated in the resistor of the coil in the form of

Heat

L x I dI represents the energy utilized by the Inductance of the Coil for

setting up the magnetic field.

Energy ( E ) = L x I dI

The total energy stored when the current rises from 0 to I is found by integration.

This energy is actually stored in the magnetic field generated by the current

flowing through the inductor and measured in Joules.

In a pure inductor, the energy is stored without loss, and is returned to the

rest of the circuit when the current through the inductor is ramped down, and its

associated magnetic field collapses.

SERIES AND PARALLEL COMBINATION OF INDUCTOR

Series Combination of Inductors: Inductors are said to be connected in

Series when they are connected together in a single line resulting in a Common

(same ) Current flowing through them.

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Or

Inductors are said to be in Series whenever the Current flows through the

Inductors sequentially.

As the Inductors are connected together in series, the same current passes

through each Inductor in the chain and the Total Inductance, LT of the Circuit must

be equal to the sum of all the individual Inductances;

LT = L1 + L2 + L3

Characteristics of Series Combination:

1. If there is only one path for the flow of current in a circuit then the

combination of Inductances is called Series Combination.

2. In Series Combination, the current flowing through each Inductor is equal.

3. In Series Combination, Potential difference ( Voltage ) across each Inductor

is different depending upon the value of Inductance..

4. In Series Combination, Total (Equivalent) Inductance of Circuit is equal to

the Sum of individual Inductances.

Equivalent Resistance In Series Combination

Consider Three Inductors L1, L2, & L3 connected in Series combination with a

Power Supply of Voltage V.

Potential difference of each Resistor is V1, V2, & V3 respectively.

Let electric current I is passing through the circuit.

Then V = V1 + V2 + V3

The self-induced emf across an inductor is given as: V = L di/dt.

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So,

By dividing by di/dt on both sides,

LT = L1 + L2 + L3

Mutually Connected Inductors in Series

When inductors are connected together in series so that the magnetic field of one

links with the other, the effect of mutual inductance either increases or decreases

the total inductance depending upon the amount of magnetic coupling.

The effect of this mutual inductance depends upon the distance apart of the coils

and their orientation to each other.

Mutually connected series inductors can be classed as either “Aiding” or

“Opposing” the total inductance.

1. If the magnetic flux produced by the current flows through the coils in the

same direction then the coils are said to be Cumulatively Coupled.

2. If the current flows through the coils in opposite directions then the coils are

said to be Differentially Coupled.

Cumulatively Coupled Series Inductors

When Two Inductors L1 & L2 are connected together in series so that the

magnetic field of one links with other in additive mode as shown in figure above.

When the current flowing between points A and D through the two

cumulatively coupled coils is in the same direction, the equation above for the

voltage drops across each of the coils needs to be modified to take into account

the interaction between the two coils due to the effect of mutual inductance. The

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self inductance of each individual coil, L1 and L2 respectively will be the same as

before but with the addition of M denoting the mutual inductance.

Then, The total EMF induced into the cumulatively coupled coils is given as:

Where : 2M represents the influence of coil L1 on L2 and likewise coil L2 on L1.

By dividing by di/dt on both sides,

LT = L1 + L2 + 2M

Differentially Coupled Series Inductors

When Two Inductors L1 & L2 are connected together in series so that the

magnetic field of one links with the other in subtractive mode as shown in figure

above.

The emf that is induced into coil -1 by the effect of the mutual inductance of

coil - 2 is in opposition to the self-induced emf in coil -1 as now the same current

passes through each coil in opposite directions. To take account of this cancelling

effect a minus sign is used with M when the magnetic field of the two coils are

differentially connected giving the final equation for calculating the total inductance

of a circuit when the inductors are differentially connected as:

LT = L1 + L2 - 2M

Numerical – 7: Three inductors of 10mH, 40mH and 50mH are connected

together in a series combination with no mutual inductance between them.

Calculate the total inductance of the series combination.

Solution:

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Numerical - 8: Two inductors of 10mH respectively are connected together

in a series combination so that their magnetic fields aid each other giving

cumulative coupling. Their mutual inductance is given as 5mH. Calculate the

total inductance of the series combination.

Solution:

Numerical – 9 : Two coils connected in series have a self-inductance of

20mH and 60mH respectively. The total inductance of the combination was

found to be 100mH. Determine the amount of mutual inductance that exists

between the two coils assuming that they are aiding each other.

Solution:

Parallel Combination of Inductors

Inductors are said to be connected in Parallel when one end of all the Inductors

are connected together and other end of all the Inductors also connected together

through a continuous wire.

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The voltage drop across all of the inductors in parallel will be the same.

Then, Inductors in Parallel have a Common Voltage across them and the voltage

across the inductors is given as:

VL1 = VL2 = VL3 = VAB

Characteristics of Parallel Combination:

1. If there are more than one path for the flow of current in a circuit then the

combination of Inductors is called Parallel Combination.

2. In Parallel combination current through each Inductor is different.

3. Potential difference across each Inductor is same.

4. Equivalent Inductance of Circuit is always less than either of the Inductances

included in the Circuit.

Equivalent Inductance In Parallel Combination

Consider Three Inductances L1 , L2 & L3 connected in Parallel Combination with a

Power Supply of Voltage V.

Now

IT = I1 + I2 + I3

By substituting di/dt in the above equation with V/L gives:

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Mutually Coupled Inductors in Parallel

When inductors are connected together in parallel so that the magnetic field

of one links with the other, the effect of mutual inductance either increases or

decreases the total inductance depending upon the amount of magnetic coupling

that exists between the coils. The effect of this mutual inductance depends upon

the distance apart of the coils and their orientation to each other.

Mutually connected inductors in parallel can be classed as either “aiding” or

“opposing”.

1. The total inductance with parallel aiding connected coils increasing

the total equivalent inductance.

2. Parallel opposing coils decreasing the total equivalent inductance.

Parallel Aiding Inductors:

Mutual coupled parallel coils are connected in an aiding configuration by the use

of polarity dots or polarity markers as shown below.

The voltage across the two parallel aiding inductors above must be equal since

they are in parallel so the two currents, i1 and i2 must vary so that the voltage

across them stays the same. Then the total inductance, LT for Two Parallel Aiding

Inductors is given as:

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Parallel Opposing Inductors

Mutual coupled parallel coils are connected in an opposing configuration by the

use of polarity dots or polarity markers as shown below.

Then the total inductance, LT for two parallel opposing inductors is given as:

Numerical – 10: Three inductors of 60mH, 120mH and 75mH respectively,

are connected together in a parallel combination with no mutual inductance

between them. Calculate the total inductance of the Parallel Combination.

Solution:

Numerical – 11: Two inductors whose self-inductances are of 75mH and

55mH respectively are connected together in parallel aiding. Their mutual

inductance is given as 22.5mH. Calculate the total inductance of the parallel

combination.

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Solution:

Numerical - 12: Calculate the equivalent inductance of the following

inductive circuit.

Solution:

Calculate the first inductor branch LA, (Inductor L5 in parallel with inductors L6

and L7

Calculate the second inductor branch LB, (Inductor L3 in parallel with inductors L4

and LA

Calculate the equivalent circuit inductance LEQ, (Inductor L1 in parallel with

inductors L2 and LB)

Then, the equivalent inductance for the above circuit was found to be: 15 mH.

LEQ = 15 mH

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Fill IN THE BLANKS:

1. An ………… can be induced by Change in the magnetic field, area or angle

between the coils.

2. The total number of magnetic field lines passing through an area is known as

………………… .

3. EMF stand for ………………. .

4. According to Faraday’s law induced emf is equal to rate of change of

……………..

5. According to Lenz law ……………….oppose the cause due to which they are

produced.

6. Magneto Motive Force is equal to …………….. x number of turns.

7. ………………….. substances have relative permeability Less than 1.

8. ………………… substances have relative permeability Greater than 1.

9. Reciprocal of reluctance is ……………. .

10. Reluctance is directly proportional to the ………….. of the material.

11. Reluctance is Inversely proportional to the ………….. the material.

12. When the length of the material increases, reluctance will ……………..

13. When the area of cross section of the material increases, reluctance will

………….. .

14. Unit of reluctance is…………….. .

15. The electrical equivalent of reluctance is ………………. .

16. As the magnetic field strength increases, reluctance …………… .

17. As the magnetic flux density increases, the reluctance …………….. .

18. Unit for inductive reactance is …………. .

19. Unit for Inductance is ………….. .

20. As the number of turns in the coil increases, the inductance of the coil will

…………….

ANSWERS:

1) E.M.F. 2) Magnetic flux 3) Electromotive force 4) Magnetic flux

5) Induced emf 6) Current 7) Diamagnetic 8) Paramagnetic

9) Permeance 10) Length 11) Area of cross section 12) Increases

13) Decreases 14) A / Wb 15) Resistance 16) Increases

17) Decreases. 18) Ohm 19) Henry ( H ) 20) Increases

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Fill IN THE BLANKS:

21. Reluctance in a magnetic material is a property by virtue of which it opposes

the creation of ……………….

22. The unit of magnetic flux is …………….

23. Inductance opposes the change in circuit …………… .

24. Magnetic field line do not ………….. each other.

25. Magnetic field strength ……………. on increasing the current through the

wire.

26. Magnetic field strength …………… as the distance from the wire increases.

27. The Unit of Flux Density is ……………… .

28. The Energy stored by Inductor is given by the expression ……………… .

29. When two Inductor L1 & L2 are connected in Series, and mutual magnetized

with each other in the same direction, the total Inductance is given by

…………………….

30. When two Inductor L1 & L2 are connected in Series, and mutual magnetized

with each other in the opposite direction, the total Inductance is given by

……………………………

31. ……………… is a unit of Magnetic flux density.

32. A permeable substance is one through which the magnetic lines of force can

pass …………….. .

33. Air gap has ………………. reluctance as compared to iron or steel path.

34. One Telsa is equal to ……………….. .

35. The magnetic reluctance of a material …………… with increasing cross

sectional area of material.

ANSWERS:

21) Magnetic flux 22) Weber 23) Current 24) Intersect

25) Increases 26) Decreases 27) Wb / m2 28) ½ LI2

29) LT = L1 + L2 + 2M 30) LT = L1 + L2 - 2M

31) Tesla 32) Very easily 33) Lower 34) 1 Wb/m2

35) Decreases

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BATTERIES

BASIC IDEA OF PRIMARY AND SECONDARY CELLS

A Battery is a combination of two or more electrochemical cells that store electrical

charges / energy and generate direct current ( DC ) through the conversion of

Chemical Energy in to Electrical Energy, when connected to an electrical circuit.

A cell consists of two electrodes with an electrolyte placed between them.

The negative electrode is known as the cathode, while the positive electrode is

known as the anode, but both are in contact with the same electrolyte where

chemical reactions take place which create the electrical charge. The electrolyte

between them can either be a liquid or a solid.

Type of Cells : There are two types of Cells

1. Primary Cells

2. Secondary cells

PRIMARY CELL

A primary cell or battery is the one that cannot be recharged after one use, and are

discarded following discharge. These cell are not chargeable because the electrode

reaction occurs only once and after the use over a period of time the Cells / batteries

become dead and cannot be reused.

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Primary cells have high density and get discharged slowly. Since there is no fluid

inside these cells they are also known as dry cells. The internal resistance is high

and the chemical reaction is irreversible. Its initial cost is cheap and also primary

cells are easy to use.

The batteries used for children’s toys, radios and similar consumer

electronics products are some examples of primary cell.

SECONDARY CELL

A secondary cell or battery is one that can be electrically recharged after its complete

discharge. In the secondary cells, the reactions can be reversed by an external

electrical energy source. Therefore, these cells can be recharged by passing

electric current and used again and again. These are also celled storage cells.

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Secondary cells have low energy density and are made of molten salts and wet

cells. The internal resistance is low and the chemical reaction is reversible. Its

initial cost is high and is a little complicated to use when compared to the primary

cell.

Examples of secondary cells are, lead storage battery and nickel –

cadmium storage cell.

DIFFERENCE BETWEEN PRIMARY CELL AND SECONDARY CELL

Sr. No.

Primary Cell Secondary cell

1. In Primary cell, Chemical

reaction is irreversible.

In Secondary cell, Chemical reaction is

reversible.

2.

In Primary Cell, Chemical

energy is converted into

electrical energy only.

In Secondary cell, Chemical Energy is

converted into Electrical Energy during

Energy Supply ( Discharging ) &

Electrical Energy is converted into

Chemical Energy during Charging.

3. It cannot be recharged It can be recharged

4. Internal resistance of Primary

Cell is high.

Internal resistance of Secondary cell is

low.

5. The Primary Cell can supply

weak currents only.

The Secondary Cell can supply weak

currents / high current.

6. It is Light in weight and cheaper

in cost.

It is Heavy in weight and costly.

7.

Example of Primary Cells:

Simple voltaic cells, Dry cells

Example of secondary cells : Lead

storage battery and nickel – cadmium

storage cell / Batteries.

LEAD ACID BATTERY

The battery which uses sponge lead and lead peroxide for the conversion of the

chemical energy into electrical power, such type of battery is called a lead acid

battery. The lead acid battery is most commonly used in the Power Stations and

Substations because it has higher cell voltage and lower cost.

Lead-acid batteries can be classified as secondary batteries. The chemical

reactions that occur in secondary cells are reversible. The reactants that generate

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an electric current in these batteries (via chemical reactions) can be regenerated

by passing current through the battery (recharging).

The chemical process of extracting current from a secondary battery

(forward reaction) is called discharging. The method of regenerating active

material is called charging.

Construction of Lead Acid Battery : The various parts of the lead acid

battery are shown below. The container and the plates are the main part of the

lead acid battery. The container stores chemical energy which is converted into

electrical energy by the help of the plates.

1. Container / Protective casing – The container of the lead acid battery

is made of glass, lead lined wood, the hard rubber of bituminous compound,

ceramic materials or moulded plastics and are seated at the top to avoid the

discharge of electrolyte. At the bottom of the container, there are four ribs, on

two of them rest the positive plate and the others support the negative plates. It

is also called as Protective casing.

2. Plates / Electrodes – The Plates of the lead-acid cell are of diverse

design and they all consist some form of a grid which is made up of lead and

the active material. The grid is essential for conducting the electric current and

for distributing the current equally on the active material. If the current is not

uniformly distributed, then the active material will loosen and fall out.

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The grids are made up of an alloy of lead and antimony. These are usually

made with the transverse rib that crosses the places at a right angle or

diagonally. The grid for the positive and negative plates are of the same

design, but the grids for the negative plates are made lighter because they are

not as essential for the uniform conduction of the current.

3. Active Material – The material in a cell which takes active participation in a

chemical reaction during charging or discharging is called the active material of

the cell. The active elements of the lead acid are :

i. Lead Peroxide (PbO2) – It forms the Positive Active Material. The

PbO2 are dark chocolate broom in colour.

ii. Sponge lead – Its form the Negative Active Material. It is grey in colour.

iii. Dilute Sulfuric Acid (H2SO4) – It is used as an electrolyte. It contains

31% of sulfuric acid.

The lead peroxide and sponge lead, which form the Negative and

Positive Active Materials have the little mechanical strength and therefore can

be used alone.

4. Separators / Cell Dividers – The separators are thin sheets of non-

conducting material made up of chemically treated leadwood, porous rubbers,

or mats of glass fibre and are placed between the cells to insulate them from

each other. Separators are grooved vertically on one side and are smooth on

the other side.

5. Battery Terminals – A Battery has two terminals the Positive and the

Negative. The Positive terminal with a diameter of 17.5 mm at the top is slightly

larger than the Negative terminal which is 16 mm in diameter.

Working Principle of Lead Acid Battery: When the sulfuric acid dissolves,

its molecules break up into positive hydrogen ions (2H+) and sulphate negative

ions (SO4—) and move freely. If the two electrodes are immersed in solutions and

connected to DC supply then the hydrogen ions being positively charged and

moved towards the electrodes and connected to the negative terminal of the

supply. The SO4— ions being negatively charged moved towards the electrodes

connected to the positive terminal of the supply main (i.e., anode).

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Each hydrogen ion takes one electron from the cathode, and each sulphates ions

takes the two negative ions from the anodes and react with water and form sulfuric

and hydrogen acid.

The oxygen, which produced from the above equation react with lead oxide

and form lead peroxide (PbO2.) Thus, during charging the lead cathode remain as

lead, but lead anode gets converted into lead peroxide, chocolate in colour.

If the DC source of supply is disconnected and if the voltmeter connects

between the electrodes, it will show the potential difference between them. If wire

connects the electrodes, then current will flow from the positive plate to the

negative plate through external circuit i.e. the cell is capable of supplying electrical

energy.

Chemical Action During Discharging

When the cell is full discharge, then the anode is of lead peroxide (PbO2)

and a cathode is of metallic sponge lead (Pb). When the electrodes are connected

through a resistance, the cell discharge and electrons flow in a direction opposite

to that during charging.

The hydrogen ions move to the anode and reaching the anodes receive

one electron from the anode and become hydrogen atom. The hydrogen atom

comes in contacts with a PbO2, so it attacks and forms lead sulphate (PbSO4),

whitish in colour and water according to the chemical equation.

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The each sulphate ion (SO4—) moves towards the cathode and reaching there

gives up two electrons becomes radical SO4, attack the metallic lead cathode and

form lead sulphate whitish in colour according to the chemical equation.

Chemical Action During Recharging

For recharging, the anode and cathode are connected to the positive and

the negative terminal of the DC supply mains. The molecules of the sulfuric acid

break up into ions of 2H+ and SO4—. The hydrogen ions being positively charged

moved towards the cathodes and receive two electrons from there and form a

hydrogen atom. The hydrogen atom reacts with lead sulphate cathode forming

lead and sulfuric acid according to the chemical equation.

SO4— ion moves to the anode, gives up its two additional electrons becomes

radical SO4, react with the lead sulphate anode and form leads peroxide and lead

sulphuric acid according to the chemical equation.

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The charging and discharging are represented by a single reversible

equation given below.

The equation should read downward for discharge and upward for recharge.

Advantages

1. It is simple to manufacture.

2. It has low cost as compared to other batteries.

3. It has low internal resistance.

4. It is reliable battery and its technology is well-understood.

5. Lead Acid Battery is durable and provides dependable service.

6. This battery can deliver very high currents.

7. Overcharging Tolerance capability of this type of battery is high.

8. A range of size and capabilities are available of this type of batteries.

Disadvantages: 1. It has low energy density.

2. It is heavy in weight and bulky

3. It cannot be stored in a discharged condition.

4. Lead content and electrolyte make the battery environmentally unfriendly.

5. Thermal runaway can occur if improperly charged.

6. It is not suitable for fast charging.

Applications:

1. It is used in automotive and traction applications.

2. It is used as standby / back-up/ Emergency power for electrical

Installations.

3. It is used in UPS.

4. It is used in high current drain applications.

5. Lead Acid Sealed battery available for use in portable equipment.

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NIKEL - CADMIUM BATTERY

A Nickel-Cadmium Battery (Ni-Cd or Ni-Cad) is a rechargeable battery used for

portable computers, drills, camcorders and other small battery-operated devices

requiring an even power discharge. Ni-Cds Battery use electrodes made of nickel

oxide hydroxide, metallic cadmium and an alkaline electrolyte of potassium

hydroxide.

During discharge, Ni-Cd Batteries transform chemical energy into electric

energy. During recharge, Ni-Cds retransform electric energy into chemical

energy.

Construction

Nickel cadmium, NiCd cells consist of four main elements:

1. Anode: This comprises a net that is cadmium plated.

2. Cathode: The cathode electrode of the NiCd cell is a mesh that is nickel-

plated.

3. Separator: The separator is used to ensure that the anode and cathode do

not physically touch and cause a short circuit.

4. Electrolyte: The electrolyte serves to carry the ions and charge carriers

between the anode and cathode. It can be either potassium hydroxide, KOH

or Sodium Hydroxide. Of the tow potassium hydroxide conducts better, but

sodium hydroxide does not leak as much.

In the discharged state the Positive Active Material is consists of Nickel

Hydroxide Ni(OH)2 and the Negative Material is Cadmium Hydroxide. During

charging these convert to NiO OH and cadmium together with some water.

Although the separator does not take place in the reaction it serves to insulate

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between the plates. An electrolyte is also needed. Potassium hydroxide is used for

this. It does not participate in the reaction, but enables electron transfer to take

place between the two plates.

Chemical reaction on both Electrodes during charging and

discharging

During Charging

When a charging current is applied to a NiCad battery, the negative plates lose

oxygen and begin forming metallic cadmium. The active material of the positive

plates, nickel-hydroxide, becomes more highly oxidized. This process continues

while the charging current is applied or until all the oxygen is removed from the

negative plates and only cadmium remains.

Toward the end of the charging cycle the cells emit gas. This will also occur

if the cells are overcharged. This gas is caused by decomposition of the water in

the electrolyte into hydrogen at the negative plates and oxygen at the positive

plates. The voltage used during charging, as well as the temperature, determines

when gassing will occur. To completely charge a NiCad battery, some gassing,

however slight, must take place; thus some water will be used.

During Discharging

The chemical action is reversed during discharge. The positive plates slowly give

up oxygen, which is regained by the negative plates. This process results in the

conversion of the chemical energy into electrical energy. During discharge the

plates absorb a quantity of the electrolyte.

On recharge the level of the electrolyte rises and at full charge the

electrolyte will be at its highest level. Therefore, water should be added only when

the battery is fully charged.

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Advantages:

1. A nickel-cadmium battery has a greater recharge cycle life than a lead-acid

battery. Nickel-cadmium batteries can be discharged and recharged more

times than lead-acid batteries before battery cell failure occur.

2. A Nickel-Cadmium Battery can be stored in a fully discharged condition

without any detrimental effects.

3. Nickel-Cadmium Batteries perform repeatedly throughout normal duty

cycles with no loss of active material or electrolyte, therefore, requiring no

maintenance.

4. The Nickel-Cadmium battery maintains available capacity over a wide

temperature range while the lead-acid battery capacity decreases

dramatically as temperature decreases.

5. The Nickel Cadmium can be charged very fast.

6. The Nickel Cadmium battery is lighter in weight and more compact than

lead-acid batteries. So, NiCd Battery is preferable when size and weight

are key factors, such as in airplanes.

Disadvantages:

1. The Initial cost Nickel Cadmium battery is very high as compared to lead

Acid battery.

2. It has low energy density.

3. Nickel Cadmium battery is Environmentally unfriendly — as it contains toxic

metals. Some countries are limiting the use of the NiCd battery.

4. It had high self discharge and need to charge after storage.

Applications:

1. These batteries can be used in Emergency Light.

2. These batteries are used in power tools.

3. These batteries are used in Toys.

4. These batteries are used in Digital Camera.

5. These batteries are used in UPS.

6. These are used in electric razors.

7. These batteries are used in Commercial and Industrial portable products.

8. These batteries are used in medical instruments.

9. These batteries are used in motorized Equipments.

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SILER OXIDE BATTERY

A silver-oxide battery is a primary cell with a very high Energy-to-Weight Ratio.

These cells are available in small sizes as button cells, where the amount of silver

used is minimal and not a significant contributor to the product cost.

Maxell is the first company in Japan to successfully market button-type

silver oxide batteries. Based on many years of experience and know-how in

various fields, the Silver Oxide Batteries are suitable for electronic devices such

as quartz watches, where high-energy density per unit volume and a stable

operating voltage are required. Various product lineups are available to meet the

growing need for supplying power to various types of watches, ranging from large

to compact, thin models.

Principle of operation and Reaction

The button-type silver oxide battery uses silver oxide (Ag2O) as its Positive Active

Material and Zinc (Zn) as its Negative Active Material. Potassium hydroxide (KOH)

or sodium hydroxide (NaOH) is used as an electrolyte.

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Battery discharge reactions are follows

With open circuit voltage of approx. 1.55V, the battery voltage is extremely stable

and impedance remains low during discharge.

Advantages:

1. Silver Oxide battery has high energy density, it means high capacity per unit

weight.

2. The Size of Silver Oxide Cell / Battery is very small.

3. The cost of Silver Oxide Cell / battery is Low.

4. Its operating life is very long. A tiny button cell will keep a watch running 24

hours per day for 3 to 5years.

5. The use of antioxidant, high performance separators in the battery, it has

improved storage characteristics.

6. It has excellent stable discharge characteristics and it gives constant voltage

output during the discharging.

Disadvantages:

1. This Cell / battery can deliver very low current.

2. It has poor low temperature performance.

3. It has limited cycle life.

4. It cannot recharged.

Applications:

1. It is used in Electronic Watches.

2. It is used in calculators.

3. It is used in Film Camera.

4. It is used in medical Instruments.

5. Due to small size, these are used under water and aerospace applications.

6. It is used in onboard Microcomputers.

7. It is used in sensors.

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LEAD - ACID BATTERY CHARGING METHODS

The lead-acid battery stores chemical energy and this energy is converted into

electrical energy whenever requires. The conversion of energy from chemical to

electrical is known as the charging. When the electric power changes into

chemical energy then it is known as discharging of the battery. During the

charging process, the current passes inside the battery because of chemical

changes. The lead-acid battery mainly uses two types of charging methods;

1. Constant Voltage Charging

2. Constant Current Charging

Constant voltage Charging : It is the most common method of charging the

lead acid battery. It reduces the charging time and increases the capacity up to

20%. But this method reduces the efficiency by approximately 10%.

In this method, the charging voltage is kept constant throughout the

charging process. The charging current is high in the beginning when the battery

is in the discharge condition. The current is gradually dropping off as the battery

picks up charge resulting in increase back emf.

The advantages of charging at constant voltage are that it allows cells with

different capacities and at the different degree of discharge to be charges. The

large charging current at the beginning of the charge is of relatively short duration

and will not harm the cell.

At the end of the charge, the charging current drops to almost zero

because the voltage of the battery becomes nearly equal to the voltage of the

supply circuit.

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Constant Current Charging : In this method of charging, the batteries are

connected in series so as to form groups and each group charges from the DC

supply mains through loading rheostats. The number of charging in each group

depends on the charging circuit voltage which should not be less than the 2.7 V

per cell.

The charging current is kept constant throughout the charging period by

reducing the resistance in the circuit as the battery voltage goes up. In order of

avoiding excessive gassing or overheating, the charging may be carried out in two

steps. An initial charging of approximately higher current and a finishing rate of

low current.

In this method, the charge current is approximately one-eighth of its ampere

ratings. The excess voltage of the supply circuit is absorbed in the

series resistance. The groups of the battery to be charged should be so connected

that the series resistance consumes as little energy as possible.

The current carrying capacity of series resistance should be greater than or

equal to the required charging current otherwise, the resistance will overheat and

burn out.

The group of batteries which is to be selected should have the same

capacity. If the battery has a different capacity, then they will have to be set

according to the least capacity.

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LEAD - ACID BATTERY CARE AND MAINTENANCE RULES

1. Watering is the most neglected maintenance feature of flooded lead-acid

batteries. As overcharging decreases water, we need to check it frequently.

Less water creates oxidation in plates and decreases the lifespan of the

battery. Add distilled or ionized water when needed.

2. Check for the vents, they need to be perfected with rubber caps, often the

rubber caps sticks with the holes too tightly.

3. Recharge lead-acid batteries after each use. A long period without recharging

provides sulfating in the plates.

4. Do not freeze the battery or charge it more than 49-degree centigrade. In cold

ambient batteries need to be fully charged as fully charge batteries safer than

the empty batteries in respect of freezing.

5. Do not deep discharge the battery less than 1.7V per cell.

6. To store a lead acid battery, it needs to be completely charged then the

electrolyte needs to be drained. Then the battery will become dry and can be

stored for a long time period.

SERIES AND PARALLEL CONNECTIONS OF BATTERY

Series and Parallel connections of batteries are done to increase total voltage and

increasing Ah capacity. Series connections are done to increase total

voltage. Parallel connections are made to increase total Ah ( Ampere Hours ) of

battery bank.

Series Connection of Batteries

When the Positive (+) terminal of one Battery is connected to Negative (-)

terminal of the second battery and Positive terminal the second Battery to

Negative terminal of the third Battery and so on, until the desired voltage is

reached. The final voltage is the sum of all battery voltages added together

while the final amp-hours remains unchanged, then the batteries configuration is

called as Series connection of Batteries.

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In Series connection of Batteries, current is same in each wire or section while

voltage is different i.e. voltages are additive ;

VT = V1 + V2 + V3….Vn

VT = 12 Volts + 12 Volts +12 Volts +12 Volts = 48 Volts ( Voltage Additive )

I = 150 AH ( Current Same )

Parallel Connection of Batteries

When the Positive (+) terminals of all batteries are connected together, or to a

common conductor, and all Negative terminals (-) are connected in the same

manner. The final voltage remains unchanged while the capacity of the

assembly ( AH ) is the sum of the Ampere Hours Capacities of the individual

batteries of this connection.

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In Parallel connection of batteries, Voltage is same across each section while

Current is different i.e. Currents are additive;

I1 + I2 + I3….In

IT = 120 AH + 120 AH = 240 AH ( Current Capacity is Additive )

V = 12 Volts ( Voltage Same )

Series-Parallel Connection of Batteries

When combination of batteries connected in series up-to the desired level of

voltage and then these series connected batteries are connected in parallel up-to

the desired level of current capacity , this configuration of batteries is called

Series-Parallel connection of Batteries.

In other words, It is neither series, nor parallel circuit, but known as Series-

Parallel Circuit. Some of the components are in series and other are in parallel or

complex circuit of series and parallel connected devices and batteries.

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Battery - 1 ( 12 V, 150 AH ) and battery - 2 ( 12 V, 150 AH ) are connected in

Series which increases the voltage rating of the batteries. In Series connection of

Batteries, current is same in each wire or section while voltage is different i.e.

voltages are additive ;

VT1 = V1 + V2

VT1 = 12 Volts + 12 Volts = 24 Volts ( Voltage Additive )


Similarly, Battery - 3 ( 12 V, 150 AH ) and battery - 4 ( 12 V, 150 AH ) are

connected in Series which increases the voltage rating of the batteries. In Series

connection of Batteries, current is same in each wire or section while voltage is

different i.e. voltages are additive ;

VT2 = V3 + V4

VT1 = 12 Volts + 12 Volts = 24 Volts ( Voltage Additive )


As these two combinations of batteries are connected in Parallel, which increases

the current capacity ( AH ) of the both sets of batteries. In Parallel connection of

batteries, Voltage is same across each section while Current is different i.e.

Currents are additive;

IT = 150 AH + 150 AH = 300 AH ( Current Capacity is Additive )

VT = VT-1 = VT-2 = 24 Volts ( Voltage Same )

The Overall Voltage Rating and Current Capacities of the Batteries in above

Series-Parallel Connections will be 24 Volts and 300 AH respectively.

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Review of Series, Parallel Connections

1. Connecting Batteries in Series increases Voltage, but does not increase

overall Amp-Hour Capacity.

2. All Batteries in a Series Connection must have the same Amp-Hour Rating.

3. Connecting Batteries in Parallel increases total current capacity by decreasing

total resistance, and it also increases overall Amp-Hour Capacity.

4. All batteries in a parallel Connection must have the same Voltage Rating.

GENERAL IDEA OF SOLAR CELLS, SOLAR PANELS AND THEIR

APPLICATIONS

Solar Cell: A Solar Cell, or Photovoltaic ( PV ) Cell, is an electrical device that

converts the energy of light directly into electricity by the photovoltaic effect, which

is a physical and chemical phenomenon. It is a form of Photoelectric Cell, defined

as a device whose electrical characteristics, such as current, voltage, or

resistance, vary when exposed to light. Individual solar cell devices can be

combined to form modules are known as Solar Panels.

The common single junction silicon solar cell can produce a maximum open-circuit

voltage of approximately 0.5 to 0.6 volts. Solar cells are described as being

photovoltaic, irrespective of whether the source is sunlight or an artificial light. In

addition to producing energy, they can be used as a Photo Detector.

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Solar Cell System

Solar cells produce direct current (DC), therefore these are only used for DC

equipments. If alternating current (AC) is needed for AC equipments or backup

energy is needed, solar photovoltaic systems require other components in addition

to solar modules. These components are specially designed to integrate into solar

PV system, that is to say these are Renewable Energy products or energy

conservation products and one or more of components may be included

depending on type of application. The components of solar photovoltaic

system are

1. Solar Module / Pannel / Array: It is the essential component of any solar

PV system that converts sunlight directly into DC electricity.

2. Solar Charge Controller: It regulates voltage and current from solar

arrays, charges the battery, prevents battery from overcharging and also

performs controlled over discharges.

3. Battery : It stores current electricity that produces from solar arrays for using

when sunlight is not visible, nighttime or other purposes.

4. Inverter: It is a critical component of any solar PV system that converts DC

power output of solar arrays into AC for AC appliances.

.

5. Lightning protection : It prevents electrical equipments from damages

caused by lightning or induction of high voltage surge. It is required for the large

size and critical solar PV systems, which include the efficient grounding.

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UNIT – VII

BATTERIES

Advantages of Solar Cell System

1. There is No pollution associated with it. It means this system is Pollution free.

2. This system has long life span about more 30 years .

3. It is easy to install and transportable.

4. It requires minimal maintenance.

5. It has stable efficiency.

6. With the modular characteristic, it can be constructed any sizes as required.

Dis-advantages of Solar Cell System

1. This System require high cost of Installation.

2. This System has low efficiency.

3. During the cloudy day and at night , this system cannot produced energy.

Applications of Solar Cell System

1. This solar system is used at home for Indoor and Outdoor Lighting

System, Electrical equipments, Electric Gate Opener, Security System,

Ventilators, Water Pump, Water Filter and Emergency Light etc.

2. This solar system is used for Lighting Systems at Bus stop lighting,

telephone booth lighting, billboard lighting, parking lot lighting, indoor and

outdoor lighting and street lighting, etc.

3. This solar system is used for Water Pumping for Consumption, public

utility, livestock watering, agriculture, gardening and farming, mining and

irrigation, etc.

4. This solar system is used for Battery charging System like Emergency

power system, battery charging center for rural village and power supply for

household use and lighting in remote area, etc.

5. This solar system is used for agriculture purpose like Water Pumping,

agricultural products fumigator, thrashing machines and water sprayer, etc.

6. This solar system is used for Cattle like Water Pumping, oxygen filling

system for fish-farming and insect trapped lighting, etc.

7. This solar system is used in Health centers for Refrigerator and cool box

for keeping medicines and vaccines and medical equipment, etc.

8. This solar system is used in Communication system like Air navigational

aid, air warning light, lighthouse, beacon navigation aid, illuminated road

sign, railway crossing sign, street lighting and emergency telephone, etc.

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BATTERIES

9. This solar system is used in remote areas like Hill, island, forest and

remote area that the utility grids are not available, etc.

10. This solar system is used in Satellite, international space station and

spacecraft, etc.

INTRODUCTION TO MAINTENANCE FREE BATTERIES

Maintenance Free" means that the manufacturer didn't provide any means of

maintaining the water/acid level in the battery, which means that if a battery boils

dry you can only replace it instead of refilling it yourself with water or acid,

whichever is appropriate (probably water).

Maintenance-free batteries should never be topped up, therefore there are no

filler caps on top. The filler cap is replaced by an over-pressure valve that is

normally closed. Any gas that forms ends up being recombined in the cell as

water. This way there is always sufficient electrolyte in the battery. Good quality

maintenance-free batteries have the advantage of being guaranteed for life!

Naturally the user never forgets to carry out maintenance. Maintenance-free

batteries have to be charged up with a charger that is suitable for this type of

battery.

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BATTERIES

Advantages

1. Maintenance Free batteries come sealed for life from the factory, and

do not require maintenance of the electrolyte levels.

2. The most significant benefit of the sealed design is that no service attention

is required.

3. Maintenance Free batteries come sealed for life from the factory, that will

not produce any harmful gases when charging.

4. Easy to place, and unnecessary to consider the ventilation problem of

resettlement sites.

5. The charging speed of Maintenance Free battery is much faster than that of

an ordinary battery.

6. The performance of the maintenance-free battery is much better than that

an ordinary battery.

7. The maintenance-free battery is shockproof, high temperature resistant,

small in size, less self-discharged, and has a relatively long service life.

Dis-advantages

1. The maintenance-free battery is relatively more expensive than that an

ordinary battery.

2. The maintenance-free battery cannot be repaired.

3. The maintenance-free battery has low reserve capacity.

4. In Maintenance free battery, Jumpstarting is not possible in case of total

discharge.

Applications

1. The maintenance-free batteries are very popular for use in Electric Vehicles

because of its compact size.

2. These batteries are used in UPS System.

3. These Batteries are used in Solar and Power Applications.

4. These batteries are used in Inverters at home for Indoor Lighting System,

where ventilation is not proper .

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BATTERIES

Fill IN THE BLANKS:

1. A ……………. Cell is the one that cannot be recharged after one use.

2. The capacity of a battery is expressed in terms of …………………… rating.

3. The storage battery generally used in electric power station is ………………

battery.

4. The Electrode for a battery must be a good conductor of …………… .

5. Cells are connected in series in order to Increase the ……………….

6. Five 2 V cells are connected in parallel. The output voltage is ………….

7. Self Life of a small dry cell is ………….. than that of large dry cell.

8. The current in a chemical cell is a movement of Positive and negative

………...

9. The output voltage of a charger is ………….. than the battery voltage.

10. Under charging ………………. specific gravity of the electrolyte.

11. A typical output of a solar cell is …………….. V.

12. Satellite power requirement is provided through …………… cells.

13. Number of cells connected in …………. to provide a High current carrying

capacity.

14. Specific Gravity of a electrolyte in a single cell or a battery is always

………….. 1.0.

15. The emf of the dry cell is about ………. V.

16. In cell, the ………. flows in outer circuit from Positive terminal to Negative

terminal.

17. In cell, the ………………… flows in outer circuit from Negative terminal to

Positive terminal.

18. The colour of a Positive Plate of a Lead-Acid Battery is ………….. .

19. A Negative Plate of a Lead-Acid Battery is ………. in colour.

20. In Secondary cell, Chemical reaction is…………. .

ANSWERS:

1) Primary 2) Ampere Hour 3) Lead-acid 4) Electricity

5) Voltage rating 6) 2 V 7) Less 8) Ions

9) Higher 10) Reduces 11) 0.5 to 0.6 12) Solar

13) Parallel 14) Greater than 15) 1.5 16) Current

17) Electrons. 18) Brown 19) Grey 20) Reversible

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Fill IN THE BLANKS:

21. Level of electrolyte in a cell should be …………. the level of plates.

22. When two batteries are connected in parallel, it should be ensured that both

have …………. e.m.f.

23. In constant voltage charging method, the charging current from discharged to

fully charged condition ……………… .

24. The common impurity in the electrolyte of lead-acid battery is …………… .

25. The active material of the Positive Plates of silver-zinc batteries is …………….

26. When a lead-acid battery is in fully charged condition, the color of its Positive

Plate is …………. .

27. In a lead-acid cell, lead is called as ……….. Active Material.

28. The specific gravity of electrolyte is measured by ……………….

29. In a lead-acid battery the energy is stored in the form of ………… energy.

30. During charging the specific gravity of the electrolyte of a lead-acid battery

……………..

31. In constant- ……………. charging method, the supply voltage from discharged

to fully charged condition increases.

32. The current flow through electrolyte is due to the movement of ………. .

33. The substances of the cell which take active part in chemical combination and

hence produce electricity during charging or discharging are known as …….…

34. Electrolyte used in a lead-acid cell is …………….. .

35. The lead-acid cell should never be discharged beyond ……….. V.

36. …………………. of electrolyte indicates the state of charge of the battery.

37. During …………….. , Chemical Energy is converted into Electrical Energy in

the Secondary Cell.

38. During Charging, Electrical Energy is converted into ………… Energy in the

Secondary Cell.

ANSWERS:

21) Above 22) Same 23) Decreases 24) Iron

25) Silver Oxide 26) Dark Brown 27) Negative 28) Hydrometer

29) Chemical 30) Increases 31) Current 32) Ions

33) Active materials34) Dilute H2SO4 35) 1.8 36) Specific Gravity

37) Discharging 38) Chemical

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AC FUNDAMENTALS

CONCEPT OF ALTERNATING QUANTITIES

Alternating Quantity: An Alternating Quantity is one which acts in alternate

Positive and Negative directions, whose magnitude undergoes a definite series of

changes in definite intervals of time and in which the sequence of changes while

Negative is identical with the sequence of changes while Positive.

Alternating Current ( A. C. ) : The current that changes its magnitude and

polarity at regular intervals of time is called an Alternating Current.

When the resistive load RL is connected across the Alternating Source

shown in the figure below, the current ( I ) flows through it. The alternating current

flows in one direction and in the opposite direction when the polarity is reversed.

The wave shape of the source voltage and the current flow through the

circuit Load RL is shown in the figure below.

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AC FUNDAMENTALS

The above graph represents the manner in which an alternating current changes

with respect to time is known as wave shape or waveform. Usually, the alternating

value is taken along the Y-axis and the time taken to the X- axis.

The alternating current varies in a different manner as shown in the figure

below. Accordingly, their wave shapes are named in different ways, such as

irregular wave, a triangular wave, square wave, periodic wave, sawtooth wave, the

sine wave.

An Alternating Current which varies according to the Sine of angle θ is known as

sinusoidal alternating current. The alternating current is generated in the power

station because of the following reasons.

1. The Alternating Current produces low iron and copper losses in AC rotating

machine and transformer because it improves the efficiency of AC

machines.

2. The Alternating Current offer less interference to the nearby communication

system (telephone lines etc.).

3. The AC produce the least disturbance in the electrical circuits.

4. The major advantage of using the Alternating Current instead of Direct

Current is that the alternating current is easily transformed from higher

voltage level to lower voltage level.

The Alternating Supply is always used for Domestic and Industrial applications.

Direct Current ( D. C ) :

When the Electric Current flows only in one direction in the Circuit, such type of

current is called Direct Current. The magnitude of the Direct Current always

remains constant and the frequency of the Current is Zero. It is used in cell

phones, electric vehicles, welding, electronic equipment, etc.

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UNIT – VIII

AC FUNDAMENTALS

The circuit and graphical representation of the Direct current is shown in the

above figures.

DIFFERENCE BETWEEN AC AND DC

Sr. No.

Alternating Current ( AC ) Direct Current ( DC )

1.

The current which changes its

direction at a regular interval of time

such type of current is called

Alternating Current.

Direct Current is unidirectional or

flows only in one direction.

2.

The frequency of the Alternating

Current is 50 to 60 hertz depends

on the country standard.

The frequency of the Direct Current

always remains zero.

3. The Power Factor of the Alternating

Current lies between Zero to One.

The Power Factor of the Direct

Current always remains One.

4.

The Alternating Current is

generated by the alternator ( AC

Generator).

The Direct Current is generated by

the DC Generator, Battery and Cells.

5.

The load of the Alternating Current

is Capacitive, Inductive or

Resistive.

The load of the Direct Current is

always Resistive in nature.

6.

The Alternating Current can be

graphically represented through

different irregular wave shape such

as Triangular Wave, Square Wave,

Periodic Wave, the Saw-tooth

Wave, Sine Wave, etc.

The Direct Current is graphically

represented by the straight line.

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AC FUNDAMENTALS

7.

The Alternating Current transmits

over a long distance with some

losses.

The Direct Current transmits over

very long distances with negligible

losses.

8.

The Alternating Current is

converted into Direct Current with

the help of Rectifier.

The Direct Current is converted into

Alternating Current with the help of

the Inverter.

9.

The Alternating Current is used in

Industries, Factories, and for the

Household purposes.

The Direct Current is mainly used in

Electronic Equipment, Flash Lighting,

Hybrid Vehicles etc.

10.

The Alternating Current is less

dangerous than the Direct Current.

In Alternating Current, the

magnitude of the current becomes

high and low at regular interval of

time. When the human body is

getting shocked, the Alternating

Current enter and exit from the

body at regular time interval

The Direct Current is more

dangerous than the Alternating

Current.

In Direct Current, the magnitude of

the current remains the same. When

the human body is getting shocked,

the Direct Current, afflict the body

continuously

IMPORTANT TERMS:

Waveform: “The graph between an alternating quantity (voltage or current) and

time is called waveform”. Generally, Alternating Quantity is depicted along the Y-

axis and time along the X-axis. The Below Figure shows the waveform of a

sinusoidal voltage.

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AC FUNDAMENTALS

Instantaneous value: The value of an Alternating Quantity at any instant is

called instantaneous value. The instantaneous values of alternating voltages and

current are represented by ‘e’ and ‘I’ respectively.

Alternation : When an Alternating Quantity goes through one half cycle

(complete set of Positive Half Cycle or Negative Half Cycle) it completes an

Alternation.

Cycle : When an Alternating Quantity goes through a complete set of Positive

and –Negative values, it is said to have completed one Cycle.

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UNIT – VIII

AC FUNDAMENTALS

Periodic Time or Time Period ( T ) : The Time taken in seconds by an

Alternating Quantity to complete one Cycle is known as Periodic Time OR Time

Period. It is denoted by T.

Frequency : The number of Cycles completed per second by an Alternating

Quantity is known as Frequency. It is denoted by ‘f’. The frequency is expressed in

Hz. Frequency is the reciprocal of the Time Period, ( ƒ = 1 / T )

In India, the standard frequency for Power Supply is 50 Hz. It means that

Alternating Voltage or Current completes 50 cycles in one second.

Amplitude: – This is the magnitude or intensity of the signal waveform

measured in volts or amps.

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Peak Value or Maximum Value : The maximum value attained by an

alternating quantity during one cycle is called its Peak value. It is also known as

the Maximum Value or Maximum Amplitude or crest value.

The Sinusoidal Alternating Quantity obtains its Peak value at 90 degrees as

shown in the figure below. The peak values of alternating voltage and current is

represented by Em and Im respectively.

Average Value : The average of all the instantaneous values of an Alternating

Voltage and Currents over one complete cycle is called Average Value.

If we consider symmetrical waves like sinusoidal current or voltage

waveform, the Positive Half Cycle will be exactly equal to the Negative Half Cycle.

Therefore, the Average Value over a complete cycle will be zero.

So, the only Positive Half Cycle is considered to determine the Average

Value of Alternating Quantities of Sinusoidal Waves.

1. Average Voltage Graphical Method : The Positive Half of the

waveform is divided up into any number of “n” equal portions or mid-

ordinates. The width of each mid-ordinate will therefore be no degrees

(or t seconds) and the height of each mid-ordinate will be equal to the

instantaneous value of the waveform at that point along the X-axis of the

waveform.

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AC FUNDAMENTALS

Each mid-ordinate value of the voltage waveform is added to the next and

the summed total, V1 to V12 is divided by the number of mid-ordinates used

to give us the “Average Voltage”. Then the average voltage (VAV) is the

mean sum of mid-ordinates of the voltage waveform and is given as:

The Average Voltage is therefore calculated as:

So as before lets assume again that an alternating voltage of 20 volts peak

varies over one half cycle as follows:

Voltage 6.2 V 11.8 V 16.2 V 19.0 V 20.0 V 19.0 V 16.2 V 11.8 V 6.2 V 0.0 V

Angle 18o 36o 54o 72o 90o 108o 126o 144o 162o 180o

The Average voltage value is therefore calculated as:

The Average Voltage value for one half-cycle using the graphical method is

given as: 12.64 Volts.

2. Average Voltage Analytical Method: The average voltage of a

periodic waveform whose two halves are exactly similar, either sinusoidal

or non-sinusoidal, will be zero over one complete cycle. The average value

is obtained by adding the instantaneous values of voltage over one half

cycle only. But in the case of an non-symmetrical or complex wave, the

average voltage (or current) must be taken over the whole periodic cycle

mathematically.

The average value can be taken mathematically by taking the

approximation of the area under the curve at various intervals to the

distance or length of the base and this can be done using rectangles as

shown.

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UNIT – VIII

AC FUNDAMENTALS

The mathematical area under the positive half cycle of the periodic wave

with a period of T using integration is given as:

Where: 0 and π are the limits of integration, since, the average value

of voltage over one half a cycle is to be determined. The area under the

positive (or negative) half cycle can determine the average value of the

positive (or negative) region of a sinusoidal waveform by integrating the

sinusoidal quantity over half a cycle and dividing by half the period.

The average voltage (VAV) of a sinusoidal waveform is determined by

multiplying the peak voltage (Maximum Voltage Vm) value by the

constant 0.637, which is 2 / π . The average voltage, which can also be

referred to as the mean value, depends on the magnitude of the waveform

and is not a function of either the frequency or the phase angle.

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AC FUNDAMENTALS

R.M.S Value: The term “RMS” stands for “Root-Mean-Squared”. It is defined as

the “amount of AC power that produces the same heating effect as an equivalent

DC power”. The RMS value is the square root of the mean (average) value of the

squared of the instantaneous values. The symbols used for defining an RMS

value are VRMS or IRMS.

The term RMS, ONLY refers to time-varying sinusoidal voltages, currents

or complex waveforms which magnitude changes over time and is not used in DC

circuit analysis which magnitude is always constant. When comparing the

equivalent RMS voltage value of an Alternating Sinusoidal Signal that supplies the

same Electrical Power to a given load as an equivalent DC Signal, the RMS value

is called the “Effective Value” and is generally presented as: Veff or Ieff.

The RMS voltage of a Sinusoid or Complex Waveform can be determined by two

basic methods.

1. RMS Voltage Graphical Method: The method of calculation is the

same for both halves of an AC waveform, consider only the positive half cycle.

The Positive Half Cycle of the waveform is divided up into any number of “n”

equal portions or mid-ordinates and the more mid-ordinates that are drawn

along the waveform, the more accurate will be the final result. The width of

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UNIT – VIII

AC FUNDAMENTALS

each mid-ordinate will therefore be no degrees and the height of each mid-

ordinate will be equal to the instantaneous value of the waveform at that time

along the X-axis of the waveform.

Each mid-ordinate value of a waveform (the voltage waveform in this case) is

multiplied by itself (squared) and added to the next. This method gives us the

“square” or Squared part of the RMS voltage expression. Next this squared

value is divided by the number of mid-ordinates used to give us the Mean part

of the RMS voltage expression, and in above example, the number of mid-

ordinates are twelve (12). Finally, the square root of the previous result is

found to give us the Root part of the RMS voltage.

An RMS voltage (VRMS) as being “the square root of the mean of

the square of the mid-ordinates of the voltage waveform” and this is given as:

The RMS voltage will be calculated as:

So lets assume that an Alternating Voltage has a peak voltage (Vm) of 20 volts

and by taking 10 mid-ordinate values is found to vary over one half cycle as

follows:

Voltage 6.2 V 11.8 V 16.2 V 19.0 V 20.0 V 19.0 V 16.2 V 11.8 V 6.2 V 0.0 V

Angle 18o 36o 54o 72o 90o 108o 126o 144o 162o 180o

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AC FUNDAMENTALS

The RMS voltage is therefore calculated as:

The RMS Voltage value using the Graphical Method is given as: 14.14 Volts.

2. RMS Voltage Analytical Method: This is another method for finding

the RMS value of Alternating Quantity which are sinusoidal in nature. The

equation of a Sinusoidal Alternating Voltage is as under:

V = Vm Sin Ǿ

Effective Value or RMS Value,

Vrms = Area of First Half of Squared Wave

Vrms = √ Vm2π / 2π

Vrms = Vm / √2 = 0.707 Vm

Vrms = 0.707 Vm or Vm / √2

Similarly Irms = 0.707 Im or Im / √2

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AC FUNDAMENTALS

FORM FACTOR : The ratio of the root mean square value to the average value

of an alternating quantity (current or voltage) is called Form Factor.

The average of all the instantaneous values of current and voltage over one

complete cycle is known as the average value of the alternating quantities.

Mathematically, it is expressed as:

Ir.m.s and Vr.m.s are the roots mean square values of the current and the voltage

respectively, and Iav and Vav are the average values of the alternating current and

the voltage respectively.

For the current varying Sinusoidally, the Form Factor is given as:

The value of Form Factor is 1.11

PEAK FACTOR : The ratio of maximum value to the R.M.S value of an

alternating quantity is called as Peak factor.

The alternating quantities can be voltage or current. The maximum value is

the peak value or the crest value or the amplitude of the voltage or current.

The root mean square value is the amount of heat produced by the

alternating current will be same when the direct supply of current is passed

through the same resistance in the same given time.

Mathematically it is expressed as:

Where, Im and Vm are the maximum value of the current and the voltage

respectively, and Ir.m.s and Vr.m.s are the roots mean square value of the alternating

current and the voltage respectively.

For the current varying sinusoidally, the peak factor is given as:

The value of Peak Factor is 1.4142

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AC FUNDAMENTALS

REPRESENTATION OF SINUSOIDAL QUANTITIES BY PHASOR

DIAGRAM

Phasor Diagrams: Phasor Diagrams are a graphical way of representing the

magnitude and directional relationship between two or more Alternating

Quantities.

Sinusoidal waveforms of the same frequency can have a Phase Difference

between themselves which represents the angular difference of the two sinusoidal

waveforms. Also the terms “Lead” and “Lag” as well as “In-phase” and “Out-of-

phase” are commonly used to indicate the relationship between them.

Phaser: A phasor is a vector that has an arrow head at one end which signifies

partly the maximum value of the vector quantity ( V or I ) and partly the end of the

vector that rotates.

Generally, Vectors are assumed to pivot at one end around a fixed zero

point known as the “Point of Origin” while the arrowed end representing the

Quantity, freely rotates in an anti-clockwise direction at an angular velocity, ( ω )

of one full revolution for every cycle.

This anti-clockwise rotation of the vector is considered to be a Positive

Rotation. Similarly, a clockwise rotation is considered to be a Negative Rotation.

The phase of an Alternating Quantity at any instant in time can be

represented by a Phasor Diagram. So, The phasor diagrams can be thought of as

“functions of time”. A complete sine wave can be constructed by a single vector

rotating at an angular velocity of ω = 2πƒ, where ƒ is the frequency of the

waveform. Then a Phasor is a quantity that has both “Magnitude” and “Direction”.

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UNIT – VIII

AC FUNDAMENTALS

Phase Difference: When two Alternating Quantities of the same frequency

have their values Zero and Maximum at different instants, they are said to have a

Phase Difference.

The mathematical expression to define these two sinusoidal quantities will be

written as:

The current, i is lagging the voltage, v by angle Φ and in above example this

is 30o. So the difference between the two phasors representing the two sinusoidal

quantities is angle Φ and the resulting phasor diagram will be;

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UNIT – VIII

AC FUNDAMENTALS

EFFECT OF ALTERNATING VOLTAGE APPLIED TO A PURE

RESISTANCE, PURE INDUCTANCE, PURE CAPACITANCE.

Pure Resistive AC Circuit : The circuit containing only a pure resistance of R

ohms in the AC circuit is known as Pure Resistive AC Circuit.

The presence of inductance and capacitance does not exist in a purely

resistive circuit. The alternating current and voltage both move forward as well as

backwards in both the direction of the circuit. Hence, the alternating current and

voltage follows a shape of the Sine wave or known as the sinusoidal waveform.

In the purely resistive circuit, the power is dissipated by the resistors and

the phase of the voltage and current remains same i.e., both the voltage and

current reach their maximum value at the same time. The resistor is the passive

component which neither produce nor consume electric power. It converts

the Electrical Energy into Heat.

Let the alternating voltage applied across the circuit be given by the

equation

Then the instantaneous value of current flowing through the resistor shown

in the figure below will be:

The value of current will be maximum when ωt= 90° or sinωt = 1

Putting the value of sinωt in the above equation

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AC FUNDAMENTALS

Phase Angle and Waveform of Resistive Circuit

From the above Voltage and Current equations, it is clear that there is no phase

difference between the applied voltage and the current flowing through a Purely

Resistive Circuit, i.e. phase angle between voltage and current is Zero. Hence, in

an AC circuit containing Pure Resistance, the current is in phase with the voltage

as shown in the waveform figure below.

The above waveforms indicate the curve for current, voltage and power

respectively. From the phasor diagram, it is clear that the current and voltage are

in phase with each other that means the value of current and voltage attains its

peak at the same instant of time, and the power curve is always positive for all the

values of current and voltage.

Power in Pure Resistive Circuit

As in DC Supply Circuit, the product of Voltage and Current is known as the

Power in the Circuit. Similarly, the power is the same in the AC circuit also, the

only difference is that in the AC circuit the instantaneous value of voltage and

current is taken into consideration.

Therefore, the instantaneous Power in a Purely Resistive Circuit is given by the

equation shown below:

Instantaneous Power, P= V x I

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UNIT – VIII

AC FUNDAMENTALS

The average power consumed in the circuit over a complete cycle is given by

As the valve of cosωt is Zero.

So, putting the value of cosωt in equation (4) the value of power will be given by

Where,

P – Average Power

Vr.m.s – Root Mean Square value of the Voltage

Ir.m.s – Root Mean Square value of the Current

Hence, the Power in a Purely Resistive Circuit is given by:

Power, P= V x I

The Voltage and the Current in the Purely Resistive Circuit are in phase

with each other having no phase difference with phase angle zero. The

Alternating Quantity reaches their Peak Value at the interval of the same time

period that is the rise and fall of the Voltage and Current occurs at the same time.

Pure inductive AC Circuit

The circuit which contains only Inductance (L) and not any other quantities like

Resistance (R) and Capacitance (C) in the circuit is called a Pure inductive

circuit. In this type of circuit, Current lags behind Voltage by an angle of 900.

The Inductor is a type of coil which reserves electrical energy in the

magnetic field when the current flow through it. The inductor is made up of wire

which is wound in the form of a coil. When the current flowing through inductor

changes then time-varying magnetic field causes emf which obstruct the flow of

current. The inductance is measured in Henry. The opposition of flow of current is

known as the inductive reactance.

The Circuit containing Pure Inductance is shown below:

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UNIT – VIII

AC FUNDAMENTALS

Let the alternating voltage applied to the circuit is given by the equation:

As a result, an alternating current I flows through the Inductance which induces an

emf in it. The equation is shown below:

The emf which is Induced in the circuit is equal and opposite to the applied

voltage. Hence, the equation becomes,

Putting the value of e in equation (2) we will get the equation as

Integrating both sides of the equation (3), we will get

where, XL = ω L is the opposition offered to the flow of alternating current by a

pure inductance and is called inductive reactance.

The value of current will be maximum when sin (ωt – π/2) = 1

Therefore,

Substituting this value in Im from the equation (5) and putting it in equation (4)

The current in the Pure Inductive AC Circuit lags the voltage by 90 degrees.

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UNIT – VIII

AC FUNDAMENTALS

Phasor Diagram and Power Curve of Inductive Circuit

The Waveform, Power Curve and Phasor Diagram of a Purely Inductive Circuit is

shown below

The voltage, current and power waveform are shown in above figure ( I ) . When

the values of Voltage and Current are at its peak as a Positive value, the Power is

also Positive and similarly, when the Voltage and Current give Negative waveform

the Power will also become Negative. This is because of the Phase difference

between Voltage and Current.

When the value of current is at its maximum or peak value of the voltage at

that instance of time will be zero, and therefore, the voltage and current are out of

phase with each other by an angle of 90 degrees.

The phasor diagram is also shown above on the right-hand side of the

waveform where current (Im) lag voltage (Vm) by an angle of π/2.

Power in Pure Inductive Circuit

Instantaneous Power in the Inductive Circuit is given by

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Hence, the Average Power consumed in a Purely Inductive Circuit is Zero.

The Average Power in one alteration, i.e., in a half cycle is zero, as the

Negative and Positive loop is under Power Curve is the same.

In the Purely Inductive Circuit, during the first quarter cycle, the power

supplied by the source, is stored in the magnetic field set up around the coil. In the

next quarter cycle, the magnetic field diminishes and the power that was stored in

the first quarter cycle is returned to the source.

This process continues in every cycle, and thus, no power is consumed in

the circuit.

Pure Capacitor AC Circuit

The Circuit containing only a Pure Capacitor of Capacitance C farads is known as

a Pure Capacitor Circuit. The capacitors stores electrical power in the electric

field, their effect is known as the capacitance. It is also called the condenser.

The Capacitor consists of two conductive plates which are separated by the

dielectric medium. The dielectric material is made up of glass, paper, mica, oxide

layers, etc. It stores energy in electrical form. The capacitor works as a storage

device, and it gets charged when the supply in ON and gets discharged when the

supply is OFF. If it is connected to the direct supply, it gets charged equal to the

value of the applied voltage.

When the Voltage is applied across the Capacitor, then the electric field is

developed across the plates of the capacitor and no current flow between them. If

the variable voltage source is applied across the capacitor plates then the ongoing

current flows through the source due to the charging and discharging of the

capacitor. In pure AC Capacitor Circuit, the Current leads the Voltage by an angle

of 90 degrees.

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Let the Alternating Voltage applied to the circuit is given by the equation:

Charge of the Capacitor at any instant of time is given as:

Current flowing through the circuit is given by the equation:

Putting the value of Q from the equation (2) in above equation;

Now, putting the value of V from the equation (1) in the equation (3)

Where Xc = 1/ωC is the opposition offered to the flow of alternating current by a

pure capacitor and is called Capacitive Reactance.

The value of current will be maximum when sin(ωt + π/2) = 1. Therefore,

the value of maximum current Im will be given as:

Substituting the value of Im in the equation (4):

In the Pure Capacitor Circuit, the Current flowing through the Capacitor leads the

Voltage by an angle of 90 degrees.

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Phasor Diagram and Power Curve

The Phasor Diagram and the Waveform of Voltage, Current and Power of pure

Capacitive AC Circuit are shown below:

The Waveform show Current, Voltage, and Power curve. When the voltage is

increased, the capacitor gets charged and reaches or attains its maximum value

and, therefore, a Positive half cycle is obtained. Further when the voltage level

decreases the capacitor gets discharged, and the Negative half cycle is formed.

When the voltage attains its maximum value, the value of the current is zero that

means there is no flow of current at that time.

When the value of voltage is decreased and reaches a value π, the value

of Voltage starts getting Negative, and the Current attains its peak value. As a

result, the Capacitor starts discharging. This cycle of charging and discharging of

the capacitor continues.

The values of voltage and current are not maximised at the same time

because of the phase difference as they are out of phase with each other by an

angle of 90 degrees.

The phasor diagram is also shown in the waveform indicating that the

current (Im) leads the voltage (Vm) by an angle of π/2.

Power in Pure Capacitor Circuit

Instantaneous Power is given by P = V x I

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Hence, from the above equation, it is clear that the average power in the

capacitive circuit is zero. The average power in a half cycle is zero as the Positive

and Negative loop area in the waveform shown are same.

In the first quarter cycle, the Power which is supplied by the source is

stored in the electric field set up between the capacitor plates. In the another or

next quarter cycle, the electric field diminishes, and thus the Power stored in the

field is returned to the source. This process is repeated continuously and,

therefore, no power is consumed by the capacitor circuit.

Numerical - 1 : An Alternating Voltage is expressed by the expression V = 300

Sin 314 t. Determine

i. Maximum value of Voltage

ii. RMS Value of Voltage

iii. Average value of Voltage

iv. Form Factor

v. Peak Factor

vi. Frequency of Voltage

vii. Time Period

viii. Instantaneous Voltage at 300

Solution : Given V = Vm Sin ώt V = 300 Sin 314 t

i. Maximum Voltage ( Vm ) = 300 Volts

ii. RMS Value of Voltage VRMS = Vm / √2 = 300 / √2 = 212 Volts

VRMS = 212 Volts

iii. Average Value of Voltage ( Vav ) = 2 Vm / π = 2 x 300 / 3.14 = 191 Volts

Value of Voltage ( Vav ) = 191 Volts

iv. Form Factor = VRMS / Vav = 212 / 191 = 1.11

Form Factor = 1.11

v. Peak Factor = Vm / VRMS = 300 / 212 = 1.41

Peak Factor = 1.41

vi. ώ = 314 , 2π f = 314

f = 314 / 2π = 50 Hz

f = 50 Hz

vii. Time Period ( T ) = 1 / f = 1 / 50 = 0.02 Sec = 20 m Sec

viii. V = Vm Sin ώt = Vm Sin Ǿ = 300 Sin 300 = 300 x 0.5 = 150 Volts

V = 150 Volts

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Numerical - 2 : An Alternating Current is expressed by the expression I = 50 Sin

314 t. Determine

i. Maximum value of Current

ii. RMS Value of Current

iii. Average value of Current

iv. Form Factor

v. Peak Factor

vi. Frequency of Current

vii. Time Period

viii. Instantaneous Current at 300

Solution : Given I = Im Sin ώt I = 50 Sin 314 t

i. Maximum Current ( Im ) = 50 Amperes

ii. RMS Value of Current IRMS = Im / √2 = 50 / √2 = 35.35 Amperes

IRMS = 35.35 Amperes

iii. Average Value of Current ( Iav ) = 2 Im / π = 2 x 50 / 3.14 = 31.85 Amperes

Average Value of Current ( Iav ) = 31.85 Amperes

iv. Form Factor = IRMS / Iav = 35.35 / 31.85 = 1.11

Form Factor = 1.11

v. Peak Factor = Im / IRMS = 50 / 35.35 = 1.41

Peak Factor = 1.41

vi. ώ = 314 , 2π f = 314

f = 314 / 2π = 50 Hz

f = 50 Hz

vii. Time Period ( T ) = 1 / f = 1 / 50 = 0.02 Sec = 20 m Sec

viii. I = Im Sin ώt = I m Sin Ǿ = 50 Sin 300 = 50 x 0.5 = 25 Amperes

Instantaneous Current at 300 ( I ) = 25 Amperes

Numerical - 3 : A Circuit Consists of a pure resistance of 100 Ω and connected to

AC Power Supply having a voltage V = 300 Sin 314 t . Determine:

i. Current flowing in Resistance

ii. Power Consumed in Resistance

iii. Instantaneous Current Expression

Solution : Given Resistance R = 100 Ω , V = 300 Sin 314 t

i. Current flowing in Resistance ( I ) = V / R = VRMS / R

VRMS = Vm / √2 = 300 / √2 = 212 Volts

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I = VRMS / R = 212 / 100 = 2.12 Amps

I = 2.12 Amps

ii. Power Consumed in resistance P = V x I = 212 x 2.12 = 449.44 Watts

P = 449.44 Watts

iii. Instantaneous Current Expression I = Im Sin ώt

IRMS = Im / √2 Im = IRMS x √2 = 2.12 x √2 = 3 Amperes

I = 3 Sin 314 t

Numerical - 4 : A Circuit Consists of a Capacitor having Capacitance of 100 µF

and connected to AC Power Supply having a voltage V = 300 Sin 314 t

Determine: i. Current ii. Instantaneous Current Expression

Solution : Given Capacitance ( C ) = 100 µF , V = 300 Sin 314 t

i. Current flowing in Capacitance ( I ) = V / XC = VRMS / XC

VRMS = Vm / √2 = 300 / √2 = 212 Volts

ώ = 2 π f

XC = 1/ 2 π f C = 1 / 314 x 100 x10-6 = 31.85 Ω

I = VRMS / XC = 212 / 31.85 = 6.66 Amps

I = 6.66 Amps

ii. Instantaneous Current Expression I = Im Sin ώt

IRMS = Im / √2 Im = IRMS x √2 = 6.66 x √2 = 9.4 Amperes

I = 9.4 Sin 314 t + 900

Numerical - 5 : A Circuit Consists of a Inductor having Inductance of 100 mH

and connected to AC Power Supply having a voltage V = 300 Sin 314 t

Determine: i. Current ii. Instantaneous Current Expression

Solution : Given Inductance ( L ) = 100 mH , V = 300 Sin 314 t

i. Current flowing in Inductance ( I ) = V / XL = VRMS / XL

VRMS = Vm / √2 = 300 / √2 = 212 Volts

ώ = 2 π f

XL = 2 π f L = 314 x 100 x10-3 = 31.4 Ω

I = VRMS / XL = 212 / 31.4 = 6.75 Amps,

I = 6.75 Amps

ii. Instantaneous Current Expression I = Im Sin ώt

IRMS = Im / √2 Im = IRMS x √2 = 6.75 x √2 = 9.55 Amperes

I = 9.55 Sin 314 t - 900

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Fill IN THE BLANKS:

1. The frequency of the Direct Current always remains …………… Hz.

2. Form factor for a sine wave is ………….

3. All the rules and laws of D.C. circuit also apply to A.C. circuit containing only

…………..

4. In an A.C. circuit, Power is dissipated in ……………. only.

5. Power factor of Capacitance will be …………...

6. The voltage of domestic supply is 220 V. This Voltage represents ……….

value of the AC Supply.

7. The Power consumed in a Circuit element will be least when the phase

difference between the Current and Voltage is ………….. .

8. Form Factor is the ratio of ……………… to Average Value.

9. Power factor of Inductance will be ……………… .

10. In a Pure ………….. Circuit , the Voltage lags behind the Current by 90°.

11. The frequency of domestic power supply in India is ………….. Hz.

12. In a Pure Inductive Circuit, if the supply frequency is reduced to 1/2, the

current will ……………. .

13. Two waves of the same frequency have opposite phase when the phase

angle between them is …………….

14. In a Pure …………… Circuit , the Current lags behind the Voltage by 90°.

15. The Power Factor of a D.C. circuit is always ………….. .

16. In a Pure …………. Circuit, if the supply frequency is reduced to ½ , the

current will be reduced by ½ .

17. The Time Period of a wave is time required to complete ………….. .

18. The Time period of a sine wave is 0.02 seconds. Its frequency is ………. Hz.

19. In a Purely Inductive Circuit, Actual power is ………… .

20. The Peak Value of a Sine wave is 200 V. Its average value is ……….. V.

ANSWERS:

1) Zero 2) 1.11 3) Resistance 4) Resistance

5) Zero 6) RMS 7) 90o 8) RMS Value

9) Zero 10) Capacitive 11) 50 12) Doubled

13) 180o 14) Inductive 15) Unity 16) Capacitive

17) One cycle. 18) 50 19) Zero 20) 127.4

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Fill IN THE BLANKS:

21. RMS stands for ……………… .

22. RMS current is also known as the …………… current.

23. The RMS Value and Average Value of Square wave is …………….. .

24. …………… is the ratio of RMS Value to Average Value.

25. The value of the form factor for sinusoidal current ………….. .

26. ……………. of an electrical circuit is equal to R / Z .

27. Inductance of coil increases with the increase in supply …………….. .

28. Time constant of an Inductive Circuit increases with increase of ……………. .

29. Time constant of an Inductive Circuit increases with decrease of …………….

30. Time constant of a Capacitive Circuit increase with increase of ………….. .

31. Time constant of a capacitive circuit increase with increase of …………. .

32. The ratio of active power to apparent power is known as ………….. .

33. The Peak factor of Sinusoidal Wave is ……………

34. Average value of current over a half cycle is ………… Im.

35. Average value of current over a full cycle is.

36. The expression of ω is equal to ……………. .

37. Ammeters and Voltmeters are calibrated to read the …………….. of AC

Signal.

38. The RMS value is 0.707 times the ……………. of AC Signal.

39. The RMS Value of a Sinusoidal AC Quantity will be equal to its …………

Value.

40. Sinusoidal Wave repeat itself after an interval of ………… Electrical .

ANSWERS:

21) Root Mean Square22) Effective 23) Same 24) Form Factor

25) π/2 26) Power factor 27) Frequency 28) Inductance

29) Resistance 30) Capacitance 31) Resistance 32) Power Factor

33) 1.414 34) 0.67 35) Zero 36) 2πf

37) RMS value 38) Peak value 39) Effective 40) 360o

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AC CIRCUITS

CONCEPT OF INDUCTIVE AND CAPACITIVE REACTANCE

Reactance: Reactance is the opposition offered by Capacitor and Inductor in a

circuit to the flow of AC current in the circuit. It is quite similar to resistance but

reactance varies with the frequency of the ac voltage source. It is measured in

ohms.

Inductive Reactance (XL) : Inductive reactance is the opposition offered by

the Inductor to the flow of AC current in AC Circuit.

Inductive Reactance is directly proportional to the Inductance and the

Signal Frequency. It is represented by XL and measured in ohms (Ω). Inductive

reactance is mostly low for lower frequencies and high for higher frequencies. It is

Zero for steady DC current.

Inductive Reactance, XL = ώ L = 2 π f L

Capacitive Reactance: Capacitive Reactance is the opposition offered by a

Capacitor to the flow of AC Current in the AC Circuit.

A Capacitor opposes the changes in the potential difference or the voltage

across its plates. Capacitive reactance is said to be inversely proportional to the

Capacitance and the Signal Frequency. It is normally represented by (Xc) and

measured in ohm (Ω).

Capacitive Reactance, Xc = 1 / ώ C = 1 / 2 π f C

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ALTERNATING VOLTAGE APPLIED TO RESISTANCE AND

INDUCTANCE IN SERIES ( R - L SERIES AC CIRCUIT )

The Current flowing through a purely inductive coil lags the voltage by 90o and a

purely inductive coil has no resistance and therefore, no I2R losses. But in the real

world, it is impossible to have a purely AC Inductance only.

All Electrical Coils, Relays, Solenoids and Transformers will have a certain

amount of resistance no matter how small associated with the coil turns of wire

being used. This is because copper wire has resistivity. The Inductive coil as

being one that has a resistance ( R ) in series with an inductance ( L ) producing

an “Impure Inductance”.

If the coil has some “Internal” resistance, then to represent the total

impedance of the coil as a resistance in series with an inductance and in an AC

circuit that contains both inductance (L) and resistance (R). The voltage (V)

across the combination will be the phasor sum of the two component

voltages, VR and VL. This means then that the current flowing through the coil will

still lag the voltage, but by an amount less than 90o depending upon the values

of VR and VL, the phasor sum. The new angle between the voltage and the current

waveforms gives their phase difference which is the phase angle of the circuit

given the Greek symbol phi, Φ.

Consider the circuit below a resistance R is connected in series with a

Inductance L.

In the RL series circuit above, the current is common to both Resistance and

Inductance, while the voltage across these two component voltages, VR and VL

are different. The resulting voltage of these two components can be found by

drawing a vector diagram. To be able to produce the vector diagram a reference

or common component must be found and in a series AC circuit the current is the

reference source as the same current flows through the resistance and the

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inductance. The individual vector diagrams for a pure resistance and a pure

inductance are given as:

The voltage and current in a resistive circuit are both in phase and therefore

vector VR is drawn superimposed to scale onto the current vector. The current

lags the voltage in an AC inductance (pure) circuit therefore vector VL is drawn

90o in front of the current and to the same scale as VR as shown.

From the Above Vector Diagram, the line OB is the horizontal current reference

and line OA is the voltage across the resistive component which is in-phase with

the current. Line OC shows the inductive voltage which is 90o in leads the current

therefore the current lags the purely inductive voltage by 90o. Line OD gives the

resulting supply voltage. Then:

V is the RMS value of the applied AC Voltage.

I is the RMS value of the series AC Current.

VR = I.R is the voltage drop across the resistance which is in-phase

with the current.

VL = I.XL is the voltage drop across the inductance which leads the

current by 90o.

As the current lags the voltage in a pure inductance by exactly 90o the resultant

phasor diagram drawn from the individual voltage drops VR and VL represents a

right angled voltage triangle shown above as OAD.

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As VR = I.R and VL = I.XL the applied voltage will be the vector sum of the two as

follows:

The quantity √ R2 + XL2 represents the impedance ( Z ) of the circuit.

Impedance of an AC Inductance

Impedance, Z is the “TOTAL” opposition to current flowing in an AC circuit that

contains both Resistance, ( the real part ) and Reactance ( the imaginary part ).

Impedance also has the units of Ohms, Ω. Impedance depends upon the

frequency, ω of the circuit as this affects the circuits reactive components and in a

series circuit all the resistive and reactive impedance’s add together. Impedance

can also be represented by a complex number, Z = R + jXL.

Phase Angle

If two or more inductive coils are connected together in series or then the

total resistance for the resistive elements would be equal to: R1 + R2 + R3 etc,

giving a total resistive value for the circuit.

Similarly, the total reactance for the inductive elements would be equal to:

XL1 + XL2 + XL3 etc, giving a total reactance value for the circuit. Then, an

impedance value, Z comprising of a single resistance in series with a single

reactance, Z2 = R2 + XL2.

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ALTERNATING VOLTAGE APPLIED TO RESISTANCE AND

CAPACITANCE IN SERIES ( R - C SERIES AC CIRCUIT )

The current flowing through a pure capacitance leads the voltage by 90o. It is

impossible to have a pure AC Capacitance, as all capacitors will have a certain

amount of internal resistance across their plates giving rise to a leakage current.

Then capacitor as being one that has a resistance (R) in series with a

capacitance (C) producing an “Impure Capacitor”.

If the capacitor has some “Internal Resistance” then to represent the total

impedance of the capacitor as a resistance in series with a capacitance. In an AC

circuit that contains both capacitance (C) and resistance (R), the voltage phasor

(V) across the combination will be equal to the phasor sum of the two component

voltages, VR and VC. This means then that the current flowing into the capacitor

will still lead the voltage, but by an amount less than 90o depending upon the

values of R and C giving a phasor sum with the corresponding phase angle by phi

(Φ).

Consider the series RC circuit below where a resistance (R) is connected in

series with a pure capacitance (C).

In the RC Series circuit as above, the current flowing through the circuit is

common (Same) to both the resistance and capacitance, while the voltage is

made up of the two component voltages, VR and VC. The resulting voltage of

these two components can be found by vectors VR and VC are 90o out-of-phase,

these can be added vectorially by constructing a vector diagram.

In a series AC circuit, the current is common and can therefore be used as

the reference source because the same current flows through resistance and

capacitance. The individual vector diagrams for a pure resistance and a pure

capacitance are given as:

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Both the voltage and current vectors for Resistance are in phase with each other

and therefore the voltage vector VR is drawn superimposed to scale onto the

current vector. The current leads the voltage in a pure capacitance circuit,

therefore the voltage vector VC is 90o lagging the current vector and to the same

scale as VR as shown.

In the vector diagram above, line OB represents the horizontal current reference

and line OA is the voltage across the resistive component which is in-phase with

the current. Line OC shows the capacitive voltage which is 90o behind ( lagging )

the current therefore the current leads the purely capacitive voltage by 90o.

Line OD gives the resulting supply voltage. Then:

V is the RMS value of the applied AC Voltage.

I is the RMS value of the series AC Current.

VR = I.R is the voltage drop across the resistance which is in-phase

with the current.

VL = I.XC is the voltage drop across the Capacitor which lags the

current by 90o.

As the current leads the voltage in a pure capacitance by 90o , the resultant

phasor diagram from the individual voltage drops VR and VC represents a right

angled voltage triangle shown above as OAD.

As VR = I.R and VC = I.XC the applied voltage will be the vector sum of the

two as follows.

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The quantity √ R2 + XL2 represents the Impedance Z of the circuit.

Impedance of Capacitance

Impedance (Z) which has the units of Ohms (Ω) is the “TOTAL” opposition to

current flowing in an AC Circuit that contains both Resistance, ( the real part ) and

Reactance ( the imaginary part ). A purely resistive impedance will have a phase

angle of 0o while a purely capacitive impedance will have a phase angle of -90o.

However when resistors and capacitors are connected together in same

circuit, the total impedance will have a phase angle between 0o and 90o

depending upon the value of the components used. Then the impedance of our

simple R-C Circuit shown above can be found by using the impedance triangle.

Phase Angle

If two or more Capacitors are connected together in series or then the total

resistance for the resistive elements would be equal to: R1 + R2 + R3 etc, giving a

total resistive value for the circuit.

Similarly, the total reactance for the capacitive elements would be equal

to: XC1 + XC2 + XC3 etc, giving a total reactance value for the circuit. Then, an

impedance value, Z comprising of a single resistance in series with a single

reactance, Z2 = R2 + XC2.

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INTRODUCTION TO SERIES AND PARALLEL RESONANCE AND

ITS CONDITIONS

According to the type of connection to the main circuit, the Resonance Circuit

are of Two Types :

1. Series Resonant Circuit

2. Parallel Resonant Circuit

1. Series Resonance Circuit : The Inductor and Capacitor are connected in

series make a Series Resonant (Tuned) Circuit. When this circuit is connected to

AC voltage source of constant amplitude and varying frequency from Zero to

Infinity, at a frequency resonance will occur for which impedance of circuit will be

minimum and draw maximum current.

When Resonance occurs in a Series Circuit, the Supply voltage causes the

voltages across L and C to be equal and opposite in phase.

In a Series RLC Circuit, at frequency when inductive reactance of the

Inductor becomes equal in value to the capacitive reactance of the Capacitor or In

other words, XL = XC resonant occurs. The frequency point at which this resonant

occurs is called the Resonant Frequency point, ( ƒr ) of the circuit.

Series Resonance Circuits are one of the most important circuits used in

many electrical and electronic circuits such as in AC mains filters, noise filters and

also in radio and television tuning circuits producing a very selective tuning circuit

for the receiving of the different frequency channels.

Simple Series RLC Circuit :

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When AC Signal is connected in series of RLC Circuit as above, the Inductive

reactance (XL) causes total current to lag the applied voltage, whereas Capacitive

reactance ( XC ) cause the current to lead. Thus the effect of XL and XC are

opposite. Therefore the net reactance and Impedance in series circuit can be

calculated as :

Series Resonance Frequency:

The values of these resistances ( reactance ) depend upon the frequency of the

Input Sigal. At a higher frequency XL is high and at a low frequency XC is high.

Then there must be a frequency point were the value of XL is the same as the

value of XC and there is. When the curve for inductive reactance placed on the

curve for capacitive reactance , the point of intersection will give the result of the

series resonance frequency point, i. e. ( ƒr or ωr ) as shown below.

where: ƒr is resonant frequency in Hertz,

L is Inductance in Henries

C is Capacitance in Farads.

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Electrical resonance occurs in an AC Circuit, when the two reactances which are

opposite and equal cancel each other out as XL = XC and the point on the graph

at which this happens, where these two reactance curves cross each other.

In a series resonant circuit, the resonant frequency, ƒr point can be

calculated as follows.

XL = XC

2π fr L = 1/ 2π fr C

fr = ___1___

2π √LC

Impedance in a Series Resonance Circuit:

At resonance, the two reactances cancel each other, which makes a Series LC

Circuit act as a short circuit. But only the resistance R opposes to the current flow

in a Series Resonance Circuit.

In complex form, the resonant frequency is the frequency at which the total

impedance of a series RLC circuit becomes purely “real”, that is no imaginary

impedance’s exist. This is because at resonance they are cancelled out.

So the total impedance of the series circuit becomes just the value of the

resistance and therefore: Z = R.

Therefore, at resonance the impedance of the series circuit is at its

minimum value and equal only to the resistance, R of the circuit.

Series RLC Circuit Voltage at Resonance :

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The voltage across a series combination is the phasor sum of VR, VL and VC. At

resonance the two reactances are equal and opposite so cancel each other. Then

two voltages representing VL and VC must also be opposite and equal in value

thereby cancelling each other. Because of pure ( Resistance ) components the

phasor voltages are drawn at +90o and -90o respectively.

So that, in a Series Resonance Circuit as VL = -VC the resulting reactive

voltages are zero and all the supply voltage is dropped across the resistor only.

Therefore, VR = Vsupply

Due to this reason, Series Resonance Circuits are known also as Voltage

Resonance Circuits.

2. Parallel Resonance (Tuned) Circuit : The Inductor and Capacitor are

connected in Parallel make a Parallel Resonant (Tuned) Circuit. When this parallel

circuit is connected to AC voltage source of constant amplitude and varying

frequency from Zero to Infinity, at a frequency resonance will occur for which

impedance of circuit will be maximum and draw minimum current.

Parallel Resonance means when the circuit current is in phase with the applied

voltage of an AC circuit containing an Inductor and a Capacitor connected

together in parallel.

Parallel Resonance circuit diagram shown below:

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Consider an Inductor of L Henry having some resistance of R ohms connected in

parallel with a capacitor of capacitance C farads. A supply voltage of V volts is

connected across these elements. The circuit current Ir will only be in phase with

the supply voltage when the following condition given below in the equation is

satisfied.

In Parallel Resonant Circuit, resistance R actually represents resistance of

the coil. The Total current Ir is divided into two parts, one of which current IC flows

through the capacitor whereas the current IL flows through the Inductor should be

lags by 90o. As Inductor has small resistance R, therefore current does not lags by

90o but less than 90o let us say that phase lags by ǿ as shown in Phasor diagram.

Phasor Diagram:

Frequency at Resonance Condition in Parallel resonance Circuit

The value of inductive reactance XL = 2πfL and capacitive reactance XC = 1/2πfC

can be changed by changing the supply frequency. As the frequency increases

the value of XL and consequently the value of ZL increases. As a result, there is a

decrease in the magnitude of current IL and this IL current lags behind voltage V.

On the other hand, the value of capacitive reactance decreases and

consequently the value of IC increases.

At some frequency, fr called resonance frequency.

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If R is very small as compared to L, then Resonant frequency will be

Impedance in a Parallel Resonance Circuit

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In Parallel Circuits at resonance frequency fr, impedance is very high so that the

circuits is having very low limiting the circuits current. Unlike the series resonance

circuit, the resistor in a parallel resonance circuit has a damping effect on the

circuits bandwidth making the circuit less selective.

At the resonant frequency, ƒr the current drawn from the supply must be

“in-phase” with the applied voltage as effectively there is only the resistance

present in the parallel circuit, so the power factor becomes one or unity, ( θ = 0o ).

Also as the impedance of a parallel circuit changes with frequency, this

makes the circuit impedance “dynamic” with the current at resonance being in-

phase with the voltage since the impedance of the circuit acts as a resistance.

Then we have seen that the impedance of a parallel circuit at resonance is

equivalent to the value of the resistance and this value must, therefore represent

the maximum dynamic impedance (Zd) of the circuit as shown.

Comparison Between Series and Parallel Resonant Circuits

Sr. No.

Specifications Series Resonant

Circuits

Parallel Resonant Circuits

1. Impedance at

resonance Minimum ( Zr = R ) Maximum ( Zr = L / CR )

2. Current at

resonance Maximum ( Ir = V / R ) Minimum ( Ir = V / Zr

3. Effective

impedance R L/CR

4.

Resonant

frequency

5. It magnifies

Voltage Current

6. It is known as Acceptor circuit Rejector circuit

7. Power Loss (Ir2R) is High as Ir is high (Ir2R) is Low as Ir is Low

8. Power Factor Unity Unity

9. Quality Factor Q = XL / R ( Same ) Q = XL / R ( Same )

10. Frequency

Response Same Same

11. Bandwidth Same Same

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Power in Pure Resistive Circuit

As in DC Supply Circuit, the product of Voltage and Current is known as the

Power in the Circuit. Similarly, the power is the same in the AC circuit also, the

only difference is that in the AC circuit the instantaneous value of voltage and

current is taken into consideration.

Therefore, the instantaneous Power in a Purely Resistive Circuit is given by the

equation shown below:

Instantaneous Power, P= V x I

The average power consumed in the circuit over a complete cycle is given by

As the valve of cosωt is Zero.

So, putting the value of cosωt in equation (4) the value of power will be given by

Where,

P – Average Power

Vr.m.s – Root Mean Square value of the Voltage

Ir.m.s – Root Mean Square value of the Current

Hence, the Power in a Purely Resistive Circuit is given by:

Power, P= V x I

The Voltage and the Current in the Purely Resistive Circuit are in phase

with each other having no phase difference with phase angle zero. The

Alternating Quantity reaches their Peak Value at the interval of the same time

period that is the rise and fall of the Voltage and Current occurs at the same time.

Power in Pure Inductive Circuit

Instantaneous power in the inductive circuit is given by

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Hence, the Average Power consumed in a Purely Inductive Circuit is Zero.

The average power in one alteration, i.e., in a half cycle is zero, as the

negative and positive loop is under power curve is the same.

In the Purely Inductive Circuit, during the first quarter cycle, the power

supplied by the source, is stored in the magnetic field set up around the coil. In the

next quarter cycle, the magnetic field diminishes and the power that was stored in

the first quarter cycle is returned to the source. This process continues in every

cycle, and thus, no power is consumed in the circuit.

Power in Pure Capacitive Circuit

Instantaneous power in the Capacitive Circuit is given by

P = V x I

Hence, from the above equation, it is clear that the Average Power in the

Capacitive Circuit is Zero.

The Average Power in a half cycle is zero as the positive and negative loop

area in the waveform shown are same.

In the Capacitive Circuit, during first quarter cycle, the power which is

supplied by the source is stored in the electric field set up between the capacitor

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plates. In the another or next quarter cycle, the electric field diminishes, and thus

the power stored in the field is returned to the source. This process is repeated

continuously and, therefore, no power is consumed by the capacitor circuit.

Power in RLC Circuit

When XL > XC, the Phase Angle ϕ is Positive. The Circuit behaves as RL Series

circuit in which the current lags behind the applied voltage. When XL < XC, the

Phase Angle ϕ is Negative, and the Circuit acts as a Series RC circuit in which the

current leads the voltage.

Let us consider the Circuit behaves as Inductive circuit and the alternating voltage

applied across the circuit is given by the equation:

The equation of current I is given as:

The product of Voltage and Current is defined as Power.

Putting the value of V and I from the equation (1) and (2) in the equation (3)

The average power consumed in the circuit over one complete cycle is given by

the equation shown below:

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Where cosϕ is called the power factor of the circuit.

The Power Factor is defined as the ratio of Resistance to the Impedance of an AC

Circuit. Putting the value of V and cosϕ from the equation (4) the value of power

will be:

From equation (5) it is clear that the inductor does not consume any power in the

circuit.

When the RLC Circuit behaves as RL Series circuit (XL > XC), the

inductor does not consume any power in the circuit, only Resistance

consume Power in the circuit. .

Similarly, When the RLC Circuit behaves as RC Series circuit (XL < XC) ,

the Capacitor does not consume any power in the circuit, only Resistance

consumes the Power in the Circuit.

Similarly, When RLC Circuit behave as Resonance Circuit (XL = XC), the

phase angle ϕ is zero, as a result, the circuit behaves like a purely

resistive circuit. In this stage of circuit, the current and voltage are in phase

with each other. Resistance consumes the Power in the Circuit. The value

of the power factor is unity.

Active Power: The Power which is actually consumed or utilised in an AC

Circuit is called True power or Active power or Real power. It is measured in

kilowatt (KW) or MW. It is the actual outcomes of the electrical system which runs

the electric circuits or load.

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Reactive Power: The Power which flows back and forth that means it moves in

both the directions in the circuit or reacts upon itself, is called Reactive Power.

The Reactive Power is measured in Kilo Volt-Ampere reactive (KVAR) or MVAR.

Apparent Power: The product of root mean square (RMS) value of voltage and

current is known as Apparent Power. This Power is measured in KVA or MVA.

It has been seen that power is consumed only in resistance. A pure

inductor and a pure capacitor do not consume any power since in a half cycle

whatever power is received from the source by these components, the same

power is returned to the source. This power which returns and flows in both the

direction in the circuit, is called Reactive power. This reactive power does not

perform any useful work in the circuit.

In a purely resistive circuit, the current is in phase with the applied voltage,

whereas in a purely inductive and capacitive circuit the current is 90 degrees out

of phase, i.e., if the inductive load is connected in the circuit the current lags

voltage by 90 degrees and if the capacitive load is connected the current leads the

voltage by 90 degrees.

Hence, it is cleared that the current in phase with the voltage produces true

or active power, whereas, the current 90 degrees out of phase with the voltage

contributes to reactive power in the circuit.

Therefore,

True Power = Voltage x Current in phase with the voltage

Reactive Power = Voltage x Current out of phase with the voltage

The phasor diagram for an inductive circuit is shown below:

Taking voltage V as reference, the current I lags behind the voltage V by an angle

ϕ. The current I is divided into two components:

i. I Cos ϕ in phase with the voltage V

ii. I Sin ϕ which is 90 degrees out of phase with the voltage V

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Therefore, the following expression shown below gives the Active, Reactive and

Apparent power respectively.

i. Active power P = V x I cosϕ = V I cosϕ

ii. Reactive power Pr or Q = V x I sinϕ = V I sinϕ

iii. Apparent power Pa or S = V x I = V I

Power Triangle : Power Triangle is the representation of a right angle triangle

showing the relation between Active Power, Reactive Power and Apparent Power.

When each component of the current that is the active component (Icosϕ)

or the reactive component (Isinϕ) is multiplied by the voltage V, a power triangle is

obtained shown in the figure below:

The following point shows the relationship between the quantities and is explained

by graphical representation called Power Triangle shown above.

i. When an active component of current is multiplied by the circuit voltage V, it

results in active power. This power produces torque in the motor, heat in the

heater, etc. This power is measured by the wattmeter.

ii. When the reactive component of the current is multiplied by the circuit

voltage, it gives reactive power. This power determines the Power Factor,

and it flows back and forth in the circuit.

iii. When the circuit current is multiplied by the circuit voltage, it results in

apparent power.

iv. From the power triangle shown above the power factor may be determined

by taking the ratio of true power to the apparent power.

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As we know simply power means the product of voltage and current but in AC

circuit except for Pure Resistive Circuit, there is usually a phase difference

between Voltage and Current and thus V x I does not give real or true power in the

circuit.

Power Factor : Power factor (PF) is the ratio of Active Power working power,

measured in kilowatts (kW), to apparent power.

The Active Power is also known as working power, measured in kilowatts

(kW). Apparent power, also known as demand, measured in kilovolt amperes

(kVA). It is the measure of the amount of power used to run machinery and

equipment during a certain period. It is found by multiplying (kVA = V x A).

PF expresses the ratio of true power used in a circuit to the apparent power

delivered to the circuit. A 96% power factor demonstrates more efficiency than a

75% power factor. PF below 95% is considered inefficient in many regions.

Conductance ( G ) : Conductance is the ability of a Substance or a Circuit to

allow electric Direct Current to pass through it. Conductance is the reciprocal of

resistance (R), as measured in ohms. Conductance is measured in mhos, which

is ohms spelled backwards.

Admittance ( Y ) : Admittance is the ability of a Substance or a Circuit to allow

Alternating Current to pass through it, when AC voltages (its AC conductance) is

applied. It is equal to the reciprocal of the impedance of the circuit, just

as conductance is equal to the reciprocal of resistance, and is similarly

measured in mhos.

Impedance ( Z ) : Impedance is the ability of Substance or a Circuit to oppose

the flow of Alternating Current through it.

Impedance is measured in ohms. It is just as resistance of a circuit to

oppose direct current through it (also measured in ohms) is generally not the

same as its impedance, due to the effects of capacitance and induction in and

among the components of the AC Circuit.

Susceptance ( B ) : Susceptance (B) is ability a Capacitor or Inductor which

allow to pass AC Current through these components, when AC Voltage is Applied.

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It is the imaginary part of admittance, where the real part is conductance.

Inductive Susceptance is assigned Negative Imaginary number values, and

Capacitive Susceptance is assigned Positive imaginary number values. In

Susceptance is measured in Siemens or Mho. It is reciprocal of Reactance and is

equal to 1 / X.

Numerical – 1: A coil of inductance 150mH and zero resistance is

connected across a 100V, 50Hz supply.

Calculate the inductive reactance of coil and the current flowing through it.

Solution: Given: L= 150mH R = 0 Ω V = 100 V f = 50 Hz

Numerical – 2: A solenoid coil has a resistance of 30 Ohms and an

inductance of 0.5H. If the current flowing through the coil is 4 Amps.

Calculate:

i. Voltage of the supply if frequency is 50Hz.

ii. Phase angle between Voltage and Current.

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Solution: L= 0.5 H R = 30 Ω I = $ Amps f = 50 Hz

i. Supply Voltage

ii. Phase angle between Voltage and Current

Numerical - 3: A single-phase sinusoidal AC supply voltage defined

as: V(t) = 240 sin(314t – 20o) is connected to a pure AC capacitance of 200uF.

Determine:

i. Current flowing into Capacitor

ii. Draw Resulting Phasor Diagram

Solution: Given: V(t) = 240 sin(314t – 20o) C = 200uF

The voltage across the Capacitor will be same as supply voltage.

So, VC = 240 ∠-20o Volts

Capacitive Reactance ( XC ) = 1/( ω.200uF ).

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ii. Draw Resulting Phasor Diagram: With the current leading the voltage by

90o in an AC capacitance circuit the Phasor Diagram

Numerical – 4: A capacitor which has an internal resistance of 10Ω and a

capacitance value of 100uF is connected to a supply voltage given as V(t) =

100 sin (314t).

Calculate:

i. Current flowing into Capacitor.

ii. Construct a Voltage Triangle showing the individual voltage drops.

Solution: Given : R = 10Ω, C = 100uF, V(t) = 100 sin (314t).

The Capacitive Reactance and Circuit Impedance:


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ii. Voltage Triangle

The Phase Angle between the Current and Voltage is

Individual Voltage drops in the circuit

Resultant Voltage Triangle

Numerical- 6: For the given circuit diagram as below:

I. Calculate RLC series circuit impedance

II. Current

III. Voltage across each component

IV. Power Factor.

V. Also draw the Phasor diagram of current and voltage, impedance

triangle and voltage triangle.

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Solution:

i. First of all Calculate the total impedance with the following formula

Resistance: R = 12 Ω

Inductive Reactance: XL = ωL = 2 π f L = 2 × π × 50 × 0.15 = 47.1 Ω

Capacitive Reactance XC = 1/ ωL = 1 / 2 π f L = 1 / 2 π x 50 x 100 x10-6

= 31.83 Ω

The Total Impedance Z = √R2 + ( XL2 - XC

2) = √ (12)2 + ( 47.1 – 31.83 )2

Z = √ 144 + 234 = 19.4 Ω

ii. Current I = V / Z = 100 / 19.4 = 5.14 Amps.

iii. Voltage across each component

Voltage across Resister VR = I x R = 5.14 x 12 = 61.7 Volts

Voltage across Capacitor VC = I x XC = 5.14 x 31.8 = 163.5 Volts

Voltage across Inductor VL = I x XL = 5.14 x 47.13 = 242.2 Volts

iv. Power Factor:

P. F = R / Z = 12 / 19.4 = 0.619 ( Lagging )

As from the above calculation, it is observed that inductive reactance is larger

than capacitive, so the power factor is considered lagging.

P. F. = Cos ǿ = 0.619

ǿ = Cos-1 x 0.619 = 51.8o

v i) Phasor diagram of Current and Voltage,

ii) Impedance Triangle

iii) Voltage Triangle

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Numerical – 7: For a Series RLC Circuit, R = 25 Ohm, C = 50 pF , L = 10 mH

and source voltage = 100Volts.

Find ( i ) resonant frequency ( ii ) Current.

Solution :

i. Given R = 25 Ohm, C = 50 pF , L = 10 mH and

Source Voltage = 100Volts.

fr = ____1____ = ____1________________ = _____1______

2 π √ LC 2 π √ 10 x10-3 x 50 x 10-12 2 π x7.707 x 10-7

fr = 225 KHz

ii. At resonant Z = R

I = VS / R = 100 / 25 = 5 Amps.

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Fill IN THE BLANKS:

1. In case of Inductive circuit, Inductive reactance (XL) is …….. Proportional to

Inductance (L)

2. In inductive circuit, when Inductive Reactance (XL) increases, circuit current

………..

3. In case of Capacitive circuit, Capacitive reactance (XC) is ……….. Proportional

to Frequency.

4. In case of Capacitive circuit, Capacitive reactance (XC) is ………… Proportional

to the Capacitance (C).

5. In a Capacitive circuit, when Capacitive reactance………… , then the circuit

power factor Decreases.

6. A parallel AC circuit in resonance will have a ………….. impedance.

7. The unit of frequency is ……………… .

8. Unit of reactive power is ………….. .

9. Unit of inductive reactance( XL ) is …………….

10. Resonance is occurs in AC circuit when Reactances are equal and ………….. .

11. In a Series RLC circuit, the phase difference between the current in the

capacitor and the current in the inductor is …………. .

12. In a series RLC circuit, the phase difference between the current in the circuit

and the voltage across the resistor is …………… .

13. In a series RLC circuit, the phase difference between the current in the circuit

and the voltage across the capacitor is ……………… .

14. In a parallel circuit, current in each impedance is…………….. .

15. A Series Resonant Circuit magnifies …………… .

16. The Quality Factor of Coil is given by …………… .

17. Product of RMS values of current and voltage is called as …………….. .

18. Admittance is reciprocal of …………… .

19. Resonance occurs in a LC Circuit, when XL is ……………… XC.

ANSWERS:

1) Directly 2) Decreases 3) Inversely 4) Inversely

5) Increases 6) High 7) Hertz 8) VAR

9) Ohm 10) Opposite 11) 0o 12) 0o

13) 90o 14) Different 15) Voltage 16) XL / R

17) Apparent power. 18) Impedance 19) Equal to

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INTRODUCTION TO BIPOLAR - TRANSISTORS


CONCEPT OF BIPOLAR TRANSISTOR

A Bipolar Junction Transistor is a Three terminal Semiconductor Device consisting

of Two P-N Junctions which is able to amplify or magnify a signal.

The three terminals of the BJT are Base, Collector and Emitter. Two Junctions of

BJT are J1 and J2. A signal of small amplitude is applied to the base and available

in the amplified form at the collector of the transistor. This is the amplification

provided by the BJT. It require an external source of DC power supply to carry out

the amplification process.

1. Emitter – The Heavily Doped Section of Transistor that supplies the large

section of majority charge carrier is called emitter. The emitter is always

connected in forward biased with respect to the base so that it supplies the

majority charge carrier to the base. The emitter-base junction injects a large

amount of majority charge carrier into the base because it is heavily doped and

moderate in size.

2. Collector – The Moderately Doped Section which collects the major portion of

the majority charge carrier supplied by the emitter is called a collector. The

collector-base junction is always in reverse bias. Its main function is to remove

the majority charges from its junction with the base. The collector section of the

transistor is moderately doped, but larger in size so that it can collect most of

the charge carrier supplied by the emitter.

3. Base – The middle section of the transistor is known as the base. The base

forms two circuits, the input circuit with the emitter and the output circuit with

the collector. The emitter-base circuit is in forward biased and offered the low

resistance to the circuit. The collector-base junction is in reverse bias and

offers the higher resistance to the circuit. The base of the transistor is lightly

doped and very thin due to which it offers the majority charge carrier to the

base.

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The Transistor is a Current Controlled Semiconductor Device which transfers a

weak signal from low resistance circuit to high resistance circuit. The

words trans mean transfer property and istor mean resistance property

offered to the junctions. In other words, it is a switching device which regulates

and amplify the electrical signal likes voltage or current.

The Bipolar Transistors have the ability to operate within Four Different Regions:

i. Active Region: When Junction J1 is forward biased and J2 is

reverse biased , the Transistor work in Active Region and operates as

an Amplifier and Ic = β x IB

ii. Saturation Region: When Junction J1 is forward biased and J2 is

also forward biased , the Transistor work in Saturation Region. It

operates as an Switch at “Fully-ON” Position & allow Electric Current

through it and IC = I(saturation).

iii. Cut-off Region: When Junction J1 is reversed biased and J2 is also

reversed biased , the Transistor work in Cut-off Region. It operates as

an Switch at “Fully-OFF” Position & does not allow Electric Current

through it and IC = 0.

iv. Inverted Region: When Junction J1 reversed biased and Junction J2

is forward biased, the Transistor work in Inverted Region. As the

collector is lightly doped as compared to the emitter junction it does not

supply the majority charge carrier to the base. Hence poor transistor

action is achieved and not useful for any application.

TYPES OF BI-POLAR TRANSISTORS

There are Two Types of Transistor, namely NPN transistor and PNP transistor.

1. PNP Transistor: The Transistor in which one N-Type Semiconductor is placed

(sandwiched) between -Two P-Type Semiconductors is known as PNP

Transistor.

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It is a current controlled device. The small amount of base current controlled both

the emitter and collector current. The hole is the majority carriers of the PNP

transistors which constitute the current in it. The current inside the transistor is

constituted because of the changing position of holes and in the leads of the

transistor it is because of the flow of the electrons. The PNP transistor turns on

when a small current flows through the base. The direction of current in PNP

transistor is from the emitter to collector.

Construction of PNP Transistor

The construction of PNP transistor is shown in the figure below. The PNP

transistor has two crystal diodes connected back to back. The left side of the

diode in known as the emitter-base diode and the right side of the diode is known

as the collector-base diode.

The PNP transistor has three terminals, namely emitter, collector and base. The

middle section of the PNP transistor is lightly doped, and it is the most important

factor of the working of the transistor. The emitter is moderately doped, and the

collector is heavily doped.

Working of PNP Transistor

PNP transistor works when the emitter-base junction is forward biased while

collector-base junction is reverse biased. The emitter terminal is formed by P-type

semiconductor thus, for forward biasing the P-type terminal should be connected

with Positive terminal and N-type with Negative terminal.

The circuit diagram of the PNP transistor is shown in the figure below.

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Similarly, in order to reverse bias collector-base junction, the P-type is connected

with Negative terminal while the N-type is connected with Positive terminal.

After applying biasing the depletion region formed at the emitter-base

junction will be narrow while the depletion region formed at the collector-base

junction will be wide. This is because emitter-base junction is forward biased and

in forward biased the depletion layer in narrow. Therefore, due to reverse biasing

of collector-base junction the depletion width is broad.

Concept of Effective base width

The holes are the majority charge carrier in P-type semiconductor and electrons

are the majority charge carriers in N-type electrons. The hole in emitter region will

be repelled by the positive terminal of the battery and will be attracted by electrons

present in the N-region. Thus, the effective base width between both the junctions

will be reduced.

Effective of Doping and Size of Emitter

The emitter is heavily doped than base and collector and area of the emitter is

also more than the base but less than the collector. The base is lightly doped so it

has fewer electrons.

As a consequence of which these few electrons will combine with holes emitted

from emitter region due to repulsion from the positive terminal of the battery. But

only a few holes will combine with electrons present in the base region due to the

small size of the base and light doping.

The majority of electrons are remaining that has not combined with electrons of

base terminal. These electrons will flow towards collector. They will further move

towards the end of collector region because they are attracted by negative

terminal of the battery through which collector is connected.

In this due to movement of holes, electric current flows in the circuit. Some of the

holes also away from base constituting the base current in the circuit.

The direction of current flowing in the emitter will be towards emitter while the

direction of current flowing in base and collector will be outwards. The equation of

current in PNP transistor is given below.

IE = IB + IC

2. NPN Transistor: The Transistor in which one P-Type Semiconductor is placed

between two N-Type Semiconductors is known as NPN Transistor.

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The NPN transistor amplifies the weak signal enter into the base and produces

strong amplify signals at the collector end. In NPN transistor, the direction

of movement of an electron is from the emitter to collector region due to which

the current constitutes in the transistor. Such type of Transistors are mostly used

in the circuit because their majority charge carriers are electrons which have high

mobility as compared to holes.

Construction of NPN Transistor: The NPN transistor has two diodes

connected back to back. The diode on the left side is called an emitter-base diode,

and the diodes on the left side are called collector-base diode. These names are

given as per the name of the terminals.

The NPN transistor has three terminals, namely emitter, collector and base. The

middle section of the NPN transistor is lightly doped, and it is the most important

factor of the working of the transistor. The emitter is moderately doped, and the

collector is heavily doped.

Working of NPN Transistor:

The base-emitter junction should be forward biased, and the collector-base

junction should be reversed biased. Therefore, the N-terminal of emitter-base

junction is connected to the negative terminal of VBE, and the P-terminal of the

battery is connected to the positive terminal of the VBE.

The circuit diagram of the NPN transistor is shown in the figure below.

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To reverse bias the collector-base junction, the N-terminal is connected to the

Positive terminal of the VCB and the P-terminal is connected to the Negative

terminal of the battery VCE. This will make the wide depletion layer at the collector-

base junction and narrow depletion layer at emitter-base junction.

When forward biased is applied to the emitter-base junction, the electrons

in N-region will repel from the Negative terminal of the battery and will move

towards the base region. The base region is very small as compared to emitter

and collector region. Besides, the doping intensity of base is lowest. Thus, it

consists of fewer holes.

Due to few holes in the base region, only a few electrons will recombine

with holes. The other electrons which have not recombined yet will move towards

collector region. This will constitute current in the circuit. The size of the collector

is large so that it can collect more charge carriers and can dissipate heat.

The current in NPN transistor is due to electrons because electrons are the

majority charge carriers in NPN transistor. The emitter current in NPN transistor is

equal to sum of base and collector current. Mathematically it can be written as:-

IE = IB + IC

TRANSISTOR LEAKAGE CURRENT: It is the current flowing in the

transistor due to the minority charge carriers. It flows in the same direction as the

current due to the majority charge carriers.

Let us consider a common base configuration. In this case the base emitter

junction is forward biased and base collector junction is reverse biased. When the

supply at the emitter base junction is open circuited, there is only reverse biasing

in the base collector junction. Therefore, this sets up a small amount of current

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called the leakage current. It is highly temperature dependent because, it depends

upon number of minority charge carriers which are thermally generated and in turn

depends upon temperature. The leakage current is present even when the base

emitter junction is supplied with a source.

It can be classified as-

(i) Collector-to-Base Leakage Current (lCBO)

(ii) Collector-to-Emitter Leakage Current (ICEO)

(iii) Emitter-to-Base Leakage Current (IEBO).

i. Collector-to-Base Leakage Current (lCBO): If the Emitter is open

circuited and the Collector-Base junction is reversed biased, a small Collector

Current flowing is called as the Collector-to-Base leakage current (ICBO).

In the symbol ICBO, the subscript CB shows a collector-base current

while the subscript O indicates that the current in the third electrode (viz., the

emitter, E) is zero. In transistor biasing circuits, the current ICBO has high

importance.

When the emitter current is zero, the transistor is said to be off and in this

condition the leakage current continues to flow. For the common base

configuration of the transistor with the emitter-base junction forward biased

and the collector-base junction reverse biased, the part of the emitter current

which reaches the collector is IC – ICBO.

ii. Collector-to-Emitter Leakage Current (lCEO): If the Base is open

circuited and the Collector-Emitter junction is reversed biased, a small

Collector Current flowing is called as the Collector-to-Emitter leakage current

(ICEO).

In the symbol ICEO, the subscript CE shows a collector-Emitter current while

the subscript O indicates that the current in the third electrode (viz., the Base,

B) is zero. In transistor biasing circuits, the current ICEO has high importance.

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When the Base current is zero, the transistor is said to be off and in this

condition the leakage current continues to flow. For the common Emitter

configuration of the transistor with the Base- Emitter junction forward biased

and the collector-Emitter junction reverse biased, the part of the emitter

current which reaches the collector is IC – ICEO.

iii. Emitter-to-Base Leakage Current (IEBO):

If the Collector is open circuited and the Emitter – Base junction is reversed

biased, a small Emitter Current flowing is called as the Emitter-to-Base

leakage current (IEBO).

In the symbol IEBO, the subscript EB shows a Emitter-Base current

while the subscript O indicates that the current in the third electrode (viz., the

Collector, C) is zero. In transistor biasing circuits, the current IEBO has high

importance.

When the Collector current is zero, the transistor is said to be off and in this

condition the leakage current continues to flow. For the common Emitter

configuration of the transistor with the Base- Emitter junction forward biased

and the collector-Emitter junction reverse biased, the part of the emitter

current which reaches the collector is IC – IEBO.

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TRANSISTOR CONFIGURATIONS

A Transistor have three terminals and one terminal is common for both Input and

Output is called common terminal or reference terminal. Therefore transistor have

three type of Configurations.

1. Common Base Configuration

2. Common Emitter Configuration

3. Common Collector Configuration

1. Common Base Configuration : The common base circuit arrangement for

NPN and PNP transistor is an arrangement in which base is common for both

Input and Output. In Common Base Connection, the input is connected between

emitter and base while the output is taken across collector and base.

Current Amplification factor (α) : The ratio of output current to input current

is known as a current amplification factor.

In the Common Base Configuration, the collector current IC is the output

current, and the emitter current IE is the input current. Thus, the ratio of change in

emitter current to the collector at constant collector-base voltage is known as a

current amplification factor of a transistor in common base configuration. It is

represented by α (alpha). The value of current amplification factor is less than

unity.

Where ΔIC is the change in the collector and ΔIE is changed in emitter current at

constant VCB.

We know that

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The value of current amplification factor is less than unity. The value of the

amplification factor (α) reaches to unity when the base current reduces to zero.

The base current becomes zero only when it is thin and lightly doped. The

practical value of the amplification factor varies from 0.95 to 0.99 in the

commercial transistor.

Collector Current :

The Base Current is because of the recombination of the electrons and holes in

the base region, So that, the whole Emitter Current will not flow through the Base

to Collector Current. The collector current increase slightly because of the leakage

current flows due to the minority charge carrier. The total collector current

consists;

i. The large percentage of Emitter Current that reaches the Collector terminal,

i.e., α. IE.

ii. The leakage current Ileakage. The minority charge carrier is because of the

flow of minority charge carrier across the collector-base junction as the

junction is heavily reversed. Its value is much smaller than αIE.

Total collector current,

The above expression shows that if IE = 0 (when the emitter circuit is open) then

still a small current flow in the collector circuit called leakage current. This leakage

current is represented by as ICBO, i.e., collector-base current with emitter circuit is

open.

The leakage current is also abbreviated as ICO i.e., the collector current with

emitter circuit open.

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Characteristics of Common Base (CB) Configuration

The characteristic diagram of determining the common base characteristic is

shown in the figure below.

The Emitter to Base voltage VEB can be varied by adjusting the potentiometer R1.

A series resistor RS is inserted in the Emitter Circuit to limit the Emitter current IE.

The value of the Emitter change to a large value even the value of a potentiometer

slightly change. The value of Collector voltage changes slightly by changing the

value of the potentiometer R2.

Input Characteristic : In common base configuration, It is the graph between

the Emitter-Base voltage VEB ( Input Voltage ) and Emitter Current IE ( Input

Current ) at constant Collector Base Voltage VCB ( Output Voltage ) is called Input

Characteristic of Common Base Configuration.

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The following points are taken into consideration from the characteristic curve.

i. For a specific value of VCB, the curve is a diode characteristic in the forward

region. The PN emitter junction is forward biased.

ii. The emitter current IE increases with the small increase in emitter-base

voltage VEB. It shows that input resistance is small.

Input Resistance: In common base configuration, It is the ratio of change in

Emitter-Base Voltage ( Input Voltage ) to the resulting change in Emitter Current

( Input Current ) at constant Collector Base Voltage VCB ( Output Voltage ) is

known as Input Resistance.

The value of collector base voltage VCB increases with the increases in the

collector-base current. The value of input resistance is very low, and their value

may vary from a few ohms to 10 ohms.

Output Characteristic : In Common Base configuration, It is the graph

between Collector Base Voltage VCB ( Out Voltage ) and Collector Current ( Out

Current ) and at constant Emitter Current IE ( Input current ) is called Output

Characteristic.

The CB configuration of PNP transistor is shown in the figure below.

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The following points from the characteristic curve are taken into consideration.

i. The active region of the collector-base junction is reverse biased, the

collector current IC is almost equal to the emitter current IE. The transistor is

always operated in this region.

ii. The curve of the active regions is almost flat. The large charges in

VCB produce only a tiny change in IC The circuit has very high output

resistance ro.

iii. When the emitter current is zero, the collector current is not zero. The

current which flows through the circuit is the reverse leakage current, i.e.,

ICBO. The current is temperature depends and its value range from 0.1 to

1.0 μA for silicon transistor and 2 to 5 μA for germanium transistor.

Output Resistance : In Common Base configuration, It is the ratio of change in

Collector - Base Voltage ( Output Voltage ) to the change in Collector Current

( Output Current ) at constant Emitter Current IE ( Input Current ) is known as

output resistance.

The output characteristic of the change in collector current is very little with

the change in VCB with the change in collector-base voltage. The output resistance

is very high of the order of several Kilo-Ohms ( KΩ ).

2. Common Emitter Configuration

The configuration in which the Emitter is connected between the Collector and

Base is known as a Common Emitter Configuration. The Input is connected

between Emitter and Base, and the Output circuit is taken from the Collector and

Emitter. Thus, the Emitter is common to both the Input and the Output Circuit, and

hence the name is the common emitter configuration. The common emitter

arrangement for NPN and PNP transistor is shown in the figure below.

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Current Amplification Factor (β) : In Common Emitter Configuration, The

Current Amplification Factor is defined as the ratio of the Output Current and Input

Current. In common emitter amplification, the output current is the Collector

Current IC, and the input current is the Base Current IB.

In other words, the ratio of change in collector current with respect to base

current is known as the base amplification factor. It is represented by β (beta).

Relation Between Current Amplification Factor (α) & Current

Amplification Factor (β)

We Know that

Now

Substituting the value of ΔIB in equation (1),

The above equation shows that the when the α reaches to unity, then the β

reaches to infinity. In other words, the current gain in a common emitter

configuration is very high, and because of this reason, the common emitter

arrangement circuit is used in all the transistor applications.

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Collector Current: In CE configuration, the input current IB and the output

current IC are related by the equation shown below.

If the base current is open (i.e., IB = 0). The collector current is current to the

emitter, and this current is abbreviated as ICEO that means collector- emitter

current with the base open.

Putting the ICEO in Equation No. - 3

Characteristics of Common Emitter (CE) Configuration

The characteristic of the Common Emitter Transistor Circuit is shown in the figure

below.

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The Base to Emitter voltage varies by adjusting the potentiometer R1 and the

Collector to Emitter voltage varied by adjusting the potentiometer R2. For the

various setting, the current and voltage are taken from the milli ammeters and

voltmeter. On the basis of these readings, the input and output curve plotted on

the curve.

Input Characteristic : In Common Emitter Configuration, It is the graph between

Base-Emitter voltage VBE ( Input Voltage ) and Base Current IB ( Input Current )

at constant Collector Emitter Voltage VCE ( Output Voltage ) is called Input

Characteristics.

The curve for common emitter configuration is similar to a forward diode

characteristic. The base current IB increases with the increases in the Base

Emitter Voltage VBE. Thus the input resistance of the CE configuration is

comparatively higher that of CB configuration.

Input Resistance: In Emitter Configuration, the ratio of change in Base-Emitter

Voltage VBE ( Input Voltage ) to the change in Base Current ∆IB ( Input Current ) at

constant Collector-Emitter Voltage VCE ( Output Voltage ) is known as Input

Resistance, i.e.,

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Output Characteristic : In Common Emitter Configuration, It is the graph

between Collector-Emitter Voltage VCE ( Output Voltage ) and Collector Current

IC ( Output Current ) at a constant Base Current IB ( Input Current ) is called

Output Characteristic.

The following points from the characteristic curve are taken into consideration.

i. In the Active Region, the collector current increases slightly as collector-

emitter VCE current increases. The slope of the curve is quite more than

the output characteristic of CB configuration. The output resistance of the

common base connection is more than that of CE connection.

ii. The value of the collector current IC increases with the increase in VCE at

constant voltage IB, the value β of also increases.

iii. In the active region IC = βIB, a small current IC is not zero, and it is equal to

reverse leakage current ICEO.

Output Resistance: In Common Emitter Configuration, It is the ratio of the

variation in Collector Emitter Voltage ( Output Voltage ) to the Collector Current

( Output Current ) at Constant Base Current IB ( Input Current ) is called Output

Resistance ro.

The output resistance of the common base connection is more than that of CE

connection.

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3. Common Collector Configuration

The configuration in which the Collector is Common between Emitter and Base is

known as Common Collector Configuration. In Common Collector

Configuration, the Input Circuit is connected between Emitter and Base and the

Output is taken from the Collector and Emitter. The collector is common to both

the input and output circuit and hence It is called as Common Collector

Connection or Common Collector Configuration.

Current Amplifier Factor (Y) : In Common Collector Configuration, the Current

Amplification Factor is defined as the ratio of the Output Current to the Input

Current. In Common Collector Configuration, the Output Current is Emitter Current

IE, whereas the Input Current is Base Current IB.

Thus, the ratio of change in Emitter Current to the change in Base Current

is known as the Current Amplification Factor for Common Collector Configuration.

It is expressed by the Y ( Gama ).

Relation Between ( Gama ) Υ and ( Alpha ) α

The Y is the current amplification factor of common collector configuration and

the α is current amplification factor of common base connection.

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We Know that

Substituting the value of ΔIB in above first equation;

The above relation shows that the value of Y is nearly equal to β. This circuit is

mainly used for impedance matching because of this arrangement input

resistance is High, and output resistance is very Low.

Collector Current We Know that

Input Characteristic : In Common Collector Configuration, It is the graph between

Collector Base Voltage VCB ( Input Voltage ) and Base Current IB ( Input Current )

at constant Emitter Collector Voltage VCE (Output Voltage).

The value of the Output Voltage VCE changes with respect to the Input

Voltage VCB and IB With the help of these values, Input Characteristic Curve is

drawn. The Input Characteristic Curve is shown below.

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Input Resistance: In Collector Configuration, the ratio of change in Collector-

Base Voltage VCB ( Input Voltage ) to the change in Base Current ∆IB ( Input

Current ) at constant Collector-Emitter Voltage VCE ( Output Voltage ) is known as

Input Resistance, i.e.

Output Characteristic : In Common Collector Configuration, It is the graph between

Emitter-Collector Voltage VCE ( Output Voltage ) and Emitter Current IE (Output

Current) at constant Input Current IB.

If the input current IB is zero, then the collector current also becomes zero,

and no current flows through the Transistor.

The transistor operates in active region when the base current increases and

reaches to saturation region. The graph is plotted by keeping the base current

IB constant and varying the Emitter-Collector Voltage VCE, the values of output

current IE are noticed with respect to VCE. By using the VCE and IE at constant

IB the output characteristic curve is drawn.

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Output Resistance: In Common Collector Configuration, It is the ratio of the

variation in Collector Emitter Voltage ( Output Voltage ) to the Emitter Current

(Output Current) at Constant Base Current IB ( Input Current ) is called Output

Resistance ro.

Difference Between Common Base Configuration, Common Emitter

Configuration and Common Collector Configuration:

Sr.

No.

Parameter Common Base

Configuration

Common Emitter

Configuration

Common Collector

Configuration

1. Voltage Gain High, Same as CE High Less than Unity

2. Current Gain Less than Unity High High

3. Power Gain Moderate High Moderate

4. Phase

inversion No Yes No

5. Input

Impedance Low (50 Ohm) Moderate (1 K Ohm) High (300 K Ohm)

6. Output

Impedance High (1 M Ohm) Moderate (50 K) Low (300 Ohm)

DC LOAD LINE

The DC load represents the desirable combinations of the Collector Current and

the Collector-Emitter Voltage. It is drawn when No Signal is given to the Input, and

the transistor becomes bias. It is used to determine the correct DC operating point,

often called the Q point.

Load line is locus of operating point of the transistor. It is a graph plotted

between VCE and IC. It helps to decide respective values of collector current and

collector to emitter voltage to operate transistor in any particular mode or region.

Consider a NPN Transistor Circuit is used as Common Emitter

configuration and no Input Signal is applied to the Circuit as shown in figure. For

this circuit, DC condition will obtain, and the output characteristic of such a circuit

is shown in the figure below.

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The DC load line curve of the above circuit is shown in the figure below.

By applying Kirchhoff’s voltage law to the collector circuit,

The above equation shows that the VCC and RC are the constant value, and it is

the first-degree equation which is represented by the straight line on the output

characteristic. This load line is known as a DC load line. The Output Characteristic

is used to determine the locus of VCE and IC point for the given value of RC. The

end point of the line are located as Point A and Point B.

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To obtain A

When the Collector Current IC = 0, then Collector Emitter Voltage is maximum and

will be equal to the VCC. This gives the maximum value of VCE. This is shown as

This gives the point A, which means (OA = VCC) on the Collector Emitter Voltage

axis ( X- Axis ) shown in the above figure.

To obtain B

When Collector Emitter Voltage VCE = 0, the Collector Current is maximum and is

equal to VCC/RC. This gives the maximum value of VCE. This is shown as

This gives the point B (OB = VCC/RC) on Collector Current axis ( Y- Axis ), shown

in the above figure.

By adding the points A and B, the DC load line is drawn. With the help of

load line, any value of collector current can be determined.

TRANSISTOR AS AN AMPLIFIER IN CE CONFIGURATION

The Common Emitter NPN Amplifier Circuit is shown in the figure below.

The source VBB is applied to the input circuit in addition to the signal. The

VBB battery provides the forward bias voltage to the emitter-base junction of the

transistor. The magnitude of the forward bias voltage should be such that it should

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keep the emitter-base junction always in the forward biased regarding the polarity

of the signal source.

The source VCC is applied to the Output Circuit. The VCC battery provides

the reverse bias voltage to the emitter-collector junction of the transistor. The

magnitude of the reverse bias voltage should be such that it should keep the

emitter-collector junction always in the reverse biased regarding the polarity of the

signal.

Current Gain and Voltage Gain of Common Emitter Amplifier

The Current Gain of the common emitter amplifier is defined as the ratio of change

in collector current to the change in base current.

The Voltage Gain is defined as the product of the current gain and the ratio of the

output resistance of the collector to the input resistance of the base circuits.

Numerical-1: In Common Base Configuration, When ΔIE = 0.5 mA, ΔVCB =

0.07 V and ΔIC = 0.95 mA.

Calculate:

i. Input Resistance (ri )

ii. Current Amplification factor (α)

Solution :

Given : ΔIE = 0.5 mA ΔVCB = 0.07 V ΔIC = 0.95 mA

i. Input Resistance ( ri ) = ΔVCB / ΔIE

= 0.07 V / 0.5 mA = 0.14 x 103 Ω

Input Resistance ( ri ) = 140 Ω

ii. Current Amplification factor (α) = ΔIC / ΔIE

= 0.95 / 0.5

α = 1.9

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Numerical-2: In Common Emitter Configuration, When Voltage drop

across 10 KΩ resistor connected in collector is 20 V.

Calculate:

i. Collector Current IC

ii. Current Amplification Factor (β) when Base Current IB = 0.05 mA

Solution : Given Voltage Drop across Collector Resistor = 20 V

Collector Resistor = 10 KΩ

i. Collector Current (IC) = Voltage Drop across Collector Resistor / Collector Resistor

= 20 V / 10 KΩ

IC = 2 mA

ii. Current Amplification Factor (β) = IC / IB = 2 mA / 0.05 mA

β = 40

Numerical- 3: When α of Transistor is 0.9, Calculate β of that transistor.

Solution : Given α = 0.9

We know that β = α / ( 1- α ) = 0.9 / ( 1 - 0.9 ) = 0.9 / 0.1

β = 9

Numerical- 4: Calculate Emitter Current in a Transistor. If β = 50 and IB =

0.05 mA.

Solution : Given : β = 50 IB = 0.05 mA

IC = β IB = 50 x 0.05 mA = 2.5 mA

IC = 2.5 mA

Emitter Current IE = IC + IB = 0.05 mA + 2.5 mA = 2.55 mA

IE = 2.55 mA

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Fill IN THE BLANKS:

1. A transistor has …………… PN Junctions.

2. The base of a transistor is …………. doped.

3. The ……………. has the biggest size in a transistor.

4. The collector of a transistor is …………………… doped.

5. A transistor is a ………………… operated device.

6. The emitter of a transistor is…………….. doped.

7. The input impedance of a transistor is …………….

8. Most of the majority carriers from emitter pass through base region to ………...

9. In a transistor, ............. = IC + IB .

10. The value of α of a transistor is ………….. than 1.

11. The output impedance of a transistor is …………….

12. The relation between β and α is β = ……………..

13. The most commonly used transistor arrangement is Common ……….

arrangement.

14. The output impedance of a transistor connected in Common Base arrangement

is the ……………..

15. The phase difference between the input and output voltages in a common base

arrangement is ………. .

16. The phase difference between the input and output voltages of a transistor

connected in common emitter arrangement is …………….

17. The voltage gain in a transistor connected in Common Emitter arrangement is

the …………...

18. Power gain in a transistor connected in Common Emitter arrangement is ……….

19. The phase difference between the input and output voltages of a transistor

connected in common collector arrangement is …………..

20. If the value of α is 0.9, then value of β is ……… .

ANSWERS:

1) Two 2) Lightly 3) Collector 4) Moderately

5) Current 6) Heavily 7) Low 8) Collector

9) IE 10) Less 11) High 12) α / (1 – α )

13) Emitter 14) Highest 15) 0O 16) 180O

17) Highest. 18) Highest 19) 0O 20) 90

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Fill IN THE BLANKS

21. In a transistor, signal is transferred from a ………. Resistance to ……..

Resistance circuit.

22. In Transistor, ………….. is very Thin.

23. The leakage current in CE arrangement is …………. than that in CB

arrangement.

24. When transistors are used in switching Mode, they usually operate in the ……….

and ………. region.

25. A current ratio of IC / IE is usually less than one and is called ………….. .

26. Beta’s ( β ) current amplification factor is ratio of ………… .

27. In a transistor, collector current is controlled by ………….. Current.

28. Under saturation conditions, IC has …………….. value.

29. When transistor is operating in active region, the emitter-base junction is

…………...

30. The arrow in a Transistor Symbol indicates the direction of current in ………… .

31. The most heavily doped region in a transistor is …………….. .

32. In Common Emitter configuration of the transistor, the circuit has ……….. gain.

ANSWERS:

21) Low, High 22) Base 23) More 24) Saturation, cutoff

25) Alpha (α ) 26) IC / IB 27) Base 28) Maximum

29) Forward Biased 30) Emitter 31) Emitter 32) High

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TRANSISTOT BIASING CIRCUITS

TRANSISTOR BIASING CIRCUITS

TRANSISTOR BIASING

Transistor Biasing is the process of setting a transistors DC operating voltage or

current conditions to the correct level so that any AC input signal can be amplified

correctly by the transistor.

Or

The proper flow of zero signal collector current and the maintenance of proper

collector-emitter voltage during the passage of signal is known as Transistor

Biasing. The circuit which provides transistor biasing is called as Transistor

Biasing Circuit.

Biasing is the process of providing DC voltage which helps in the

functioning of the circuit. A transistor is based in order to make the emitter base

junction forward biased and collector base junction reverse biased, so that it

maintains in active region, to work as an amplifier.

Need for DC biasing

If a signal of very small voltage is given to the input of Transistor, it cannot be

amplified. For proper functioning of Transistor as Amplifier, the below two

conditions have to be satisfied:

1. The input voltage should exceed cut-in voltage for the transistor to be ON.

2. The Transistor should be in the active region, to be operated as

an amplifier.

If appropriate DC voltages and currents are given through Transistor by external

Sources, so that Transistor operates in active region and superimpose the AC

signals to be amplified, then this problem can be avoided.

The given DC voltage and currents are so chosen that the transistor

remains in active region for entire input AC cycle. Hence DC biasing is needed.

OPERATING POINT

When a line is drawn joining the saturation and cut off points, such a line can be

called as Load line. This line, when drawn over the output characteristic curve,

makes contact at a point called as Operating point.

This operating point is also called as quiescent point or Q-point. There

can be many such intersecting points, but the Q-point is selected in such a way

that irrespective of AC signal swing, the transistor remains in the active region.

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The following graph shows how to represent the operating point.

The operating point should not get disturbed as it should remain stable to achieve

faithful amplification. Hence the quiescent point or Q-point is the value where

the Faithful Amplification is achieved.

Faithful Amplification

The process of increasing the signal strength is called as Amplification. This

amplification when done without any loss in the components of the signal, is

called as Faithful amplification.

Faithful amplification is the process of obtaining complete portions of

input signal by increasing the signal strength. This is done when AC signal is

applied at its input.

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In the above graph, the input signal applied is completely amplified and

reproduced without any losses. This Amplification is called as Faithful

Amplification.

The operating point is so chosen such that it lies in the active region and it

helps in the reproduction of complete signal without any loss.

If the operating point is considered near saturation point, then the

amplification will be as under.

If the operation point is considered near cut off point, then the amplification will be

as under.

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Hence the placement of operating point is an important factor to achieve faithful

amplification. But for the transistor to function properly as an amplifier, its input

circuit (i.e., the base-emitter junction) remains forward biased and its output circuit

(i.e., collector-base junction) remains reverse biased.

The amplified signal thus contains the same information as in the input

signal and the strength of the signal is also increased.

Basic factors for Faithful Amplification

To get faithful amplification, the following basic conditions must be satisfied.

1. Proper zero signal collector current

2. Minimum proper base-emitter voltage (VBE) at any instant.

3. Minimum proper collector-emitter voltage (VCE) at any instant.

The fulfillment of these conditions ensures that the transistor works over the

active region having input forward biased and output reverse biased.

1. Proper Zero Signal Collector Current

Let us consider a NPN transistor circuit as shown in the figure below. The base-

emitter junction is forward biased and the collector-emitter junction is reverse

biased. When a signal is applied at the input, the base-emitter junction of the NPN

transistor gets forward biased for positive half cycle of the input and hence it

appears at the output.

For negative half cycle, the same junction gets reverse biased and hence

the circuit doesn’t conduct. This leads to unfaithful amplification as shown in

the figure below.

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Let us now introduce a battery VBB in the base circuit. The magnitude of this

voltage should be such that the base-emitter junction of the transistor should

remain in forward biased, even for negative half cycle of input signal. When no

input signal is applied, a DC current flows in the circuit, due to VBB. This is known

as zero signal collector current IC.

During the positive half cycle of the input, the base-emitter junction is more

forward biased and hence the collector current increases. During the negative half

cycle of the input, the input junction is less forward biased and hence the collector

current decreases. Hence both the cycles of the input appear in the output and

hence faithful amplification results, as shown in the below figure.

Hence for faithful amplification, proper zero signal collector current must flow. The

value of zero signal collector current should be at least equal to the maximum

collector current due to the signal alone.

2. Proper Minimum VBE at any instant

The minimum base to emitter voltage VBE should be greater than the cut-in

voltage for the junction to be forward biased. The minimum voltage needed for a

silicon transistor to conduct is 0.7 V and for a germanium transistor to conduct is

0.3 V. If the base-emitter voltage VBE is greater than this voltage, the potential

barrier is overcome and hence the base current and collector currents increase

sharply.

Hence if VBE falls low for any part of the input signal, that part will be

amplified to a lesser extent due to the resultant small collector current, which

results in unfaithful amplification.

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3. Proper Minimum VCE at any instant

To achieve a faithful amplification, the collector emitter voltage VCE should not fall

below the cut-in voltage, which is called as Knee Voltage. If VCE is lesser than

the knee voltage, the collector base junction will not be properly reverse biased.

Then the collector cannot attract the electrons which are emitted by the emitter

and they will flow towards base which increases the base current. Thus the value

of β falls.

Therefore, if VCE falls low for any part of the input signal, that part will be

multiplied to a lesser extent, resulting in unfaithful amplification. So if VCE is

greater than VKNEE the collector-base junction is properly reverse biased and the

value of β remains constant, resulting in faithful amplification.

Factors affecting the operating point

The main factor that affect the operating point is the temperature. The operating

point shifts due to change in temperature.

As temperature increases, the values of ICE, β, VBE gets affected. So the

main problem which affects the operating point is temperature. Hence operating

point should be made independent of the temperature so as to achieve stability.

To achieve this, biasing circuits are introduced.

Stabilization

The process of making the operating point independent of temperature

changes or variations in transistor parameters is known as Stabilization.

Once the stabilization is achieved, the values of IC and VCE become

independent of temperature variations or replacement of transistor. A good

biasing circuit helps in the stabilization of operating point.

Need for Stabilization

Stabilization of the operating point has to be achieved due to the following

reasons.

1. Temperature dependence of IC

2. Individual variations

3. Thermal runaway

Let us understand these concepts in detail.

1. Temperature Dependence of IC

As the expression for collector current IC is

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IC = βIB + ICEO

= βIB + (β+1) ICBO

The collector leakage current ICBO is greatly influenced by temperature variations.

To come out of this, the biasing conditions are set so that zero signal collector

current IC = 1 mA. Therefore, the operating point needs to be stabilized i.e. it is

necessary to keep IC constant.

Individual Variations

As the value of β and the value of VBE are not same for every transistor, whenever

a transistor is replaced, the operating point tends to change. Hence it is

necessary to stabilize the operating point.

Thermal Runaway

As the expression for collector current IC is

IC = β IB + ICEO

= β IB + (β+1) ICBO

The flow of collector current and also the collector leakage current causes heat

dissipation. If the operating point is not stabilized, there occurs a cumulative effect

which increases this heat dissipation.

The self-destruction of such an un-stabilized transistor is known

as Thermal run away.

In order to avoid thermal runaway and the destruction of transistor, it is

necessary to stabilize the operating point, i.e., to keep IC constant.

Stability Factor

It is understood that IC should be kept constant in spite of variations of ICBO or ICO.

The extent to which a biasing circuit is successful in maintaining this is measured

by Stability factor. It denoted by S.

By definition, the rate of change of collector current IC with respect to the

collector leakage current ICO at constant β and IB is called Stability factor.

S = dIC / dICO at constant IB and β

Hence we can understand that any change in collector leakage current

changes the collector current to a great extent. The stability factor should be as

low as possible so that the collector current doesn’t get affected. S=1 is the ideal

value.

The general expression of stability factor for a CE configuration can be

obtained as under.

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IC = β IB + (β+1) ICO

Differentiating above expression with respect to IC, we get

1 = β dIB / dIC + (β+1) dICO / dIC

Or

1 = β dIB / dIC + (β+1) / S

Since dICO / dIC = 1 / S

Or

S = ( β + 1 ) / 1−β(dIB / dIC)

Hence the stability factor S depends on β, IB and IC.

TRANSISTOR BIASING

Transistors are one of the largely used semiconductor devices which are used for

wide variety of applications including amplification and switching. However to

achieve these functions satisfactorily, transistor has to be supplied with certain

amount of current and/or voltage. The process of setting these conditions for a

transistor circuit is referred to as Transistor Biasing.

The biasing in transistor circuits is done by using two DC sources VBB and

VCC. It is economical to minimize the DC source to one supply instead of two

which also makes the circuit simple.

The commonly used methods of transistor biasing are

1. Base Resistor method

2. Collector to Base bias

3. Biasing with Collector feedback resistor

4. Voltage-divider bias

All of these methods have the same basic principle of obtaining the required value

of IB and IC from VCC in the zero signal conditions.

1. Base Resistor Method ( Fix Bias Method )

In this method, a resistor RB of high resistance is connected in base, as the name

implies. The required zero signal base current is provided by VCC which flows

through RB. The base emitter junction is forward biased, as base is positive with

respect to emitter.

The required value of zero signal base current and hence the collector

current (as IC = βIB) can be made to flow by selecting the proper value of base

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resistor RB. Hence the value of RB is to be known. The figure below shows a base

resistor method of biasing circuit.

Let IC be the required zero signal collector current. Therefore,

IB = IC / β

Considering the closed circuit from VCC, base, emitter and ground, while applying

the Kirchhoff’s voltage law, we get,

VCC = IB RB+ VBE

Or

IB RB = VCC − VBE

Therefore

RB = (VCC − VBE) / IB

Since VBE is generally quite small as compared to VCC, the former can be

neglected with little error. Then,

RB = VCC / IB

We know that VCC is a fixed known quantity and IB is chosen at some suitable

value. As RB can be found directly, this method is called as fixed bias method.

Stability factor

S = ( β + 1 ) / 1−β(dIB / dIC)

In fixed-bias method of biasing, IB is independent of IC so that,

dIB / dIC = 0

Substituting the above value in the previous equation,

Stability factor, S = β + 1

Thus the stability factor in a fixed bias is (β+1) which means that IC changes (β+1)

times as much as any change in ICO.

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Effect of Temperature on Operating Point

T↑ ICO↑ IC↑ T↑

There is no check to increase in IC due to increase in temperature. Hence

operating point is not stable against rise in temperature.

Effect of change in β on Operating point

IC = β IB

IB is fixed by selecting a value of RB, So, IC is completely depends upon β. Any

rise in value of β ( Due to replacement of Transistor by another Transistor ) will

bring corresponding increase in IC and hence the operating point will shift. The

Operating Point is Not stable against β.

Advantages

i. The circuit is simple.

ii. Only one resistor RB is required.

iii. Biasing conditions are set easily.

iv. No loading effect as no resistor is present at base-emitter junction.

Disadvantages

i. The stabilization is poor as heat development can’t be stopped.

ii. The stability factor is very high. So, there are strong chances of thermal

run away.

Hence, this method is rarely employed.

2. Collector to Base Bias ( Collector Feedback Bias Circuit )

The collector to base bias circuit is same as base bias circuit except that

the base resistor RB is returned to collector, rather than to VCC supply as shown in

the figure below.

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This circuit helps in improving the stability considerably. If the value of

IC increases, the voltage across RL increases and hence the VCE also increases.

This in turn reduces the base current IB. This action somewhat compensates the

original increase.

The required value of RB needed to give the zero signal collector current

IC can be calculated as follows.

Voltage drop across RC will be

VC = ( IC + IB ) RC ≅ IC RC

From the figure,

IC RC + IB RB + VBE = VCC

Or

IB RB = VCC − VBE − IC RC

Therefore

RB = ( VCC − VBE − IC RC ) / IB

Or

RB = (VCC − VBE−ICRC) x β / IC

Applying KVL we have

( IB + IC ) RC + IBRB + VBE = VCC

Or

IB ( RC + RB ) + IC RC + VBE = VCC

Therefore

IB = ( VCC − VBE − ICRC ) / ( RC+RB )

Since VBE is almost independent of collector current, we get

dIB / dIC = − RC / ( RC + RB )

We know that

S = ( β + 1 ) / 1− β ( dIB / dIC)

Therefore

S = ( β + 1 ) / 1 + β ( RC / ( RC + RB ) )

This value is smaller than (1+β) which is obtained for fixed bias circuit. Thus there

is an improvement in the stability.

This circuit provides a negative feedback which reduces the gain of the

amplifier. So the increased stability of the collector to base bias circuit is obtained

at the cost of AC voltage gain.

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T↑ ICO↑ IC↑ ICRC↑ VCE↓ IB↓ IC↓

This biasing checks the tendency of IC to increase with rise in temperature.

Effect of Change in β

If β of Transistor increases due to replacement of Transistor, IC will increase. So,

the denominator part of the stability factor equation will decrease, but this is small

compared to change that would take place in fixed biasing circuit. Hence this

circuit is better than the fixed biasing circuit.

Advantages

i. The circuit is simple as it needs only one resistor.

ii. This circuit provides some stabilization, for lesser changes.

Disadvantages

i. The circuit doesn’t provide good stabilization.

ii. The circuit provides negative feedback.

3. Fixed bias with emitter resistor:

The fixed bias circuit is modified by attaching an external resistor to the emitter.

This resistor introduces Negative Feedback that stabilizes the Q-point.

From Kirchhoff's voltage law, the voltage across the base resistor is

VRB = VCC – IE RE – VBE

From Ohm’s Law the base current is

IB = VRB / RB

The way feedback controls the bias point is as follows. If VBE is held

constant and temperature increases, emitter current increases. However, a larger

IE increases the emitter voltage VE = IERE, which in turn reduces the voltage

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VRB across the base resistor. A lower base-resistor voltage drop reduces the base

current, which results in less collector current because IC = β IB. Collector current

and emitter current are related by IC = α IE with α ≈ 1, so the increase in emitter

current with temperature is opposed, and the operating point is kept stable.

Similarly, if the transistor is replaced by another, there may be a change in

IC (corresponding to change in β-value, for example). By similar process as

above, the change is negated and operating point kept stable.


T↑ ICO↑ IC↑ IE↑ IE. RE↑ VBE↓ IB↓ IC↓

This circuit has tendency to check any rise in IC with rise in temperature. This is

due to emitter resistance connected in the circuit.

Effect of Change in β

If the Transistor is replaced by another Transistor having different value of β, any

rise in β will increase IC and hence IE. It will increase IE. RE, reducing base emitter

voltage to reduce IB and hence IC.

β ↑ ICO↑ IC↑ IE↑ IE. RE↑ VBE↓ IB↓ IC↓

The Operating Point is stable against change in temperature as well β of the

Transistor.

Advantages:

i. The circuit has the tendency to stabilize operating point against changes

in temperature and β-value.

Disadvantages:

i. As β-value is fixed for a given transistor, this relation can be satisfied

either by keeping RE very large, or making RB very low.

a) If RE is of large value, high VCC is necessary. This

increases cost as well as precautions necessary while

handling.

b) If RB is low, a separate low voltage supply should be

used in the base circuit. Using two supplies of different

voltages is impractical.

ii. In addition to the above, RE causes AC feedback which reduces the

voltage gain of the amplifier.

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4. Voltage Divider Bias Method

Among all the methods of providing biasing and stabilization, the voltage divider

bias method is the most prominent one. Here, two resistors R1 and R2 are

employed, which are connected to VCC and provide biasing. The resistor

RE employed in the emitter provides stabilization.

The Name Voltage Divider comes from the voltage divider formed by

R1 and R2. The voltage drop across R2 forward biases the base-emitter junction.

This causes the base current and hence collector current flow in the zero signal

conditions. The figure below shows the circuit of voltage divider bias method.

Suppose that the current flowing through resistance R1 is I1. As base current IB is

very small, therefore, it can be assumed with reasonable accuracy that current

flowing through R2 is also I1.

Now let us try to derive the expressions for collector current and collector

voltage.

Collector Current, IC

From the circuit, it is evident that,

I1 = VCC / ( R1 + R2 )

Therefore, The voltage across resistance R2 is

V2 = R2 xVCC / ( R1 + R2)

Applying Kirchhoff’s voltage law to the base circuit,

V2 = VBE + VE

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V2 = VBE + IERE

IE = ( V2 − VBE ) / RE

Since IE ≈ IC,

IC = ( V2 − VBE ) / RE

From the above expression, it is evident that IC doesn’t depend upon β. VBE is

very small that IC doesn’t get affected by VBE at all. Thus IC in this circuit is almost

independent of transistor parameters and hence good stabilization is achieved.

Collector-Emitter Voltage, VCE

Applying Kirchhoff’s voltage law to the collector side,

VCC = ICRC + VCE + IERE

Since IE ≅ IC

=ICRC + VCE + ICRE

=IC (RC + RE) + VCE

Therefore,

VCE = VCC − IC ( RC + RE )

RE provides excellent stabilization in this circuit.

V2 = VBE + IC RE

Suppose there is a rise in temperature, then the collector current

IC decreases, which causes the voltage drop across RE to increase. As the

voltage drop across R2 is V2, which is independent of IC, the value of

VBE decreases. The reduced value of IB tends to restore IC to the original value.


Rise in temperature increase IC and hence IE. It will increase IE. RE, reducing base

emitter voltage to reduce IB and hence IC. Thus the circuit has tendency to check of

IC rise with temperature.

T ↑ ICO↑ IC↑ IE↑ IE. RE↑ VBE↓ IB↓ IC↓

Effect of β on Operating Point

If the Transistor is replaced by another Transistor having different value of β, any

rise in β will increase IC and hence IE. It will increase IE. RE, reducing base emitter

voltage to reduce IB and hence IC.

β ↑ ICO↑ IC↑ IE↑ IE. RE↑ VBE↓ IB↓ IC↓

In this circuit, the Operating Point is stable against change in temperature as well

β of the Transistor. This circuit is commonly used in AC Amplifier Circuit.

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Advantages

i. It is very simple method of transistor biasing.

ii. The biasing conditions can be very easily set.

iii. It provides better bias stabilization.

iv. The resistor RE introduces a negative feedback. So all the advantages

of negative feedback are obtained.

Fill IN THE BLANKS:

1. Transistor biasing represents ………….. conditions.

2. Operating point represents …………………. signal values of IC and VCE.

3. For faithful amplification by a transistor circuit, the value of VBE should

……………. for a silicon transistor.

4. The circuit that provides the best stabilization of operating point is

……………….. bias.

5. The point of intersection of DC and AC load lines represents ……………….

6. The operating point is also called …………….

7. For proper amplification by a transistor circuit, the operating point should be

located at the …………… of the DC load line.

8. The purpose of resistance in the emitter circuit of a transistor amplifier is to Limit

the change in …………….. .

9. The base resistor method is generally used in …………… Circuit.

10. If the value of collector current IC increases, then the value of VCE ……………… .

11. If the temperature increases, the value of VCE …………………..

12. Ideally, for linear operation, a transistor should be biased so that the Q-point is

at the ……….. of DC Load Line.

13. Voltage Divider Bias is the most ……………… circuit.

14. Improper Biasing of a Transistor Circuit leads to ………….. in Output Signal.

15. The thermal runway can be avoided by checking the increase of ……………….

with respect to temperature.

ANSWERS:

1) DC 2) Zero 3) 0.7 V 4) Potential divider

5) Operating Point 6) Quiescent point 7) Middle 8) Emitter Current

9) Switching 10) Decrease 11) Decrease 12) Centre

13) Common Bias 14) Distortion 15) Collector Current

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FIELD EFFECT TRANSISTORS


FIELD EFFECT TRANSISTOR

A Field Effect Transistor (FET) is a three-terminal semiconductor device with

source, drain, and gate. The charge carries are electrons or holes, which flow from

the source to drain through an active channel. This flow of electrons from source

to drain is controlled by an electric field (by applying the voltage across the gate

and source terminals). FET is also called as Uni-polar Transistor as it involves

single carrier type operation.

In ordinary transistors both holes and electrons take part, due to which

these are called the bipolar transistors. Such transistors have two main drawbacks

namely low input impedance because of forward biased emitter junction and

considerable noise level. Both of these drawbacks can be overcome to a great

extent by using field effect transistor (FET), which is an electric field (or voltage)

controlled device. FET’s have all the advantages of Vacuum tubes and ordinary

transistors (BJTs), so that FETs are replacing both the vacuum tubes and BJTs in

applications.

There are two types of FETs are available.

1. Junction Field Effect Transistor (JFET)

2. Metal Oxide Semiconductor FET (MOSFET)

1. Junction Field Effect Transistor

The Junction FET transistor is a type of field-effect transistor that can be used as

an electrically controlled switch. The Electric Energy flows through an active

channel between sources to drain terminals. By applying a reverse Bias Voltage to

the terminal, the channel is strained so the electric current is switched off

completely.

The functioning of Junction Field Effect Transistor depends upon the flow of

majority carriers (electrons or holes) only. Basically, JFETs consist of an N type

or P type silicon bar containing PN junctions at the sides.

Following are some important points to remember about FET −

i. Gate − By using diffusion or alloying technique, both sides of N type bar

are heavily doped to create PN junction. These doped regions are called

Gate (G).

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ii. Source − It is the entry point for majority carriers through which they enter

into the semiconductor bar.

iii. Drain − It is the exit point for majority carriers through which they leave the

semiconductor bar.

iv. Channel − It is the area of ( N Type or P Type ) material through which

majority carriers pass from the source to drain.

There are two types of JFETs commonly used in the field semiconductor

devices: N-Channel JFET and P-Channel JFET.

1) N-Channel JFET

N channel JFET consists of an N-Type bar at the sides of which two P-Type

layers are doped. The channel of electrons constitutes the N channel for the

device. Two ohmic contacts are made at both ends of the N-channel device,

which are connected together to form the gate terminal.

Following figure shows the crystal structure and schematic symbol of an N-

channel JFET.

The Source and Drain terminals are taken from the other two sides of the bar.

The potential difference between Source and Drain terminals is termed as VDS

and the potential difference between Source and Gate terminal is termed as VGS.

The charge flow is due to the flow of electrons from source to drain.

Whenever a positive voltage is applied across Drain and Source terminals,

electrons flow from the Source ‘S’ to Drain ‘D’ terminal, whereas conventional

Drain current ID flows through the Drain to Source. As current flows through the

device, it is in one state.

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When a negative polarity voltage is applied to the Gate terminal, a

depletion region is created in the channel. The channel width is reduced, hence

increasing the channel resistance between the Source and Drain. Since the Gate-

Source junction is reverse biased and no current flows in the device, it is in off

condition.

So basically if the voltage applied at the Gate terminal is increased, less

amount of current will flow from the Source to Drain.

The N channel JFET has a greater conductivity than the P channel JFET.

So the N channel JFET is a more efficient conductor compared to P channel

JFET.

2) P-Channel JFET

P channel JFET consists of a P-Type bar, at two sides of which N-Type layers are

doped. The Gate terminal is formed by joining the ohmic contacts at both sides.

Like in an N channel JFET, the source and drain terminals are taken from the

other two sides of the bar. A P-type channel, consisting of holes as charge

carriers, is formed between the source and drain terminal.

A Negative voltage applied to the Drain and Source terminals ensures the flow of

current from Source to Drain terminal and the device operates in ohmic region. A

Positive voltage applied to the Gate terminal ensures the reduction of channel

width, thus increasing the channel resistance. More Positive is the gate voltage;

less is the current flowing through the device.

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Characteristics of N and P Channel Junction FET Transistor

Given below is the characteristic curve of the N and P Channel Junction Field

Effect transistor and different modes of operation of the transistor.

Cutoff Region: When the voltage applied to the Gate terminal is enough

positive for the channel width to be minimum, no current flows. This causes the

device to be in cut off region.

Ohmic Region: The current flowing through the device is linearly proportional

to the applied voltage until a breakdown voltage is reached. In this region, the

transistor shows some resistance to the flow of current.

Active Region: When the drain-source voltage reaches a value such that the

current flowing through the device is constant with the drain-source voltage and

varies only with the gate-source voltage, the device is said to be in the Active

region.

Break Down Region: When the drain-source voltage reaches a value that

causes the depletion region to break down, causing an abrupt increase in the

drain current, the device is said to be in the breakdown region. This breakdown

region is reached earlier for a lower value of drain-source voltage when gate-

source voltage is more positive.

Applications of Junction Field Effect Transistor

1. The junction field effect transistor (JFET) is used as a constant current

source.

2. The JFET is used as a buffer amplifier.

3. The JFET is used as an electronic switch.

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4. The JFET is used as a phase shift oscillator.

5. The JFET is used as high impedance wide band amplifier.

6. The JFET is used as a voltage variable resistor (VVR) or voltage

development resistor (VDR).

7. The JFET is used as a chopper.

2. MOSFET ( Metal-Oxide Semiconductor Field-Effect Transistors )

MOSFET stands for Metal Oxide Silicon Field Effect Transistor or Metal Oxide

Semiconductor Field Effect Transistor. This is also called as IGFET meaning

Insulated Gate Field Effect Transistor. The FET is operated in both Depletion and

Enhancement modes of operation.

This type of FET transistor has three terminals, which are Source, Drain,

and Gate. The voltage applied to the Gate terminal controls the flow of current

from Source to Drain. A unique feature of this FET is its gate construction. Here,

the gate is completely insulated from the channel. When voltage is applied to the

gate, it will develop an electrostatic charge. At this point of time, no current is

allowed to flow in the gate region of the device. Also, the gate is an area of the

device, which is coated with metal. Generally, silicon dioxide is used as an

insulating material between the gate and the channel. Due to this reason, it is also

known as Insulated Gate FET ( IGFET ).

Construction of a MOSFET

The construction of a MOSFET is a bit similar to the FET. An oxide layer is

deposited on the substrate to which the gate terminal is connected. This oxide

layer acts as an insulator (sio2 insulates from the substrate), and hence the

MOSFET has another name as IGFET. In the construction of MOSFET, a lightly

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doped substrate, is diffused with a heavily doped region. Depending upon the

substrate used, they are called as P-type and N-type MOSFETs.

The voltage at gate controls the operation of the MOSFET. In this case,

both positive and negative voltages can be applied on the gate as it is insulated

from the channel. With Negative gate bias voltage, it acts as Depletion

MOSFET while with Positive gate bias voltage it acts as an Enhancement

MOSFET.

Classification of MOSFETs

Depending upon the type of materials used in the construction, and the type of

operation, the MOSFETs are classified as in the following figure.

The N-channel MOSFETs are simply called as NMOS. The symbols for N-

channel MOSFET are as given below.

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The P-channel MOSFETs are simply called as PMOS. The symbols for P-

channel MOSFET are as given below.

Construction of N- Channel MOSFET

Let us consider an N-channel MOSFET to understand its working. A lightly doped

P-type substrate is taken into which two heavily doped N-type regions are

diffused, which act as source and drain. Between these two N+ regions, there

occurs diffusion to form an N-Channel, connecting drain and source.

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A thin layer of Silicon dioxide (SiO2) is grown over the entire surface and holes

are made to draw ohmic contacts for drain and source terminals. A conducting

layer of aluminum is laid over the entire channel, upon this SiO2 layer from

source to drain which constitutes the gate. The SiO2 substrate is connected to

the common or ground terminals.

Because of its construction, the MOSFET has a very less chip area than

BJT, which is 5% of the occupancy when compared to bipolar junction transistor.

This device can be operated in modes. They are depletion and enhancement

modes.

Working of N - Channel Depletion Mode MOSFET

If the NMOS has to be worked in depletion mode, the gate terminal should be at

Negative Potential while drain is at positive potential, as shown in the following

figure.

When No voltage is applied between Gate and Source, some current flows due to

the voltage between Drain and Source. Let some Negative voltage is applied

at VGG, then the minority carriers i.e. holes, get attracted and settle

near SiO2 layer. But the majority carriers, i.e., electrons get repelled.

With some amount of Negative Potential at VGG a certain amount of drain

current ID flows through Source to Drain. When this Negative Potential is further

increased, the electrons get depleted and the current ID decreases. Hence the

more Negative the applied VGG, the lesser the value of drain current ID will be.

The channel nearer to drain gets more depleted than at source (like in FET)

and the current flow decreases due to this effect. Hence it is called as Depletion

mode MOSFET.

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Working of N-Channel MOSFET (Enhancement Mode)

The same MOSFET can be worked in enhancement mode, if we can change the

polarities of the voltage VGG. So, let us consider the MOSFET with Gate Source

voltage VGG being Positive as shown in the following figure.

When No voltage is applied between Gate and Source, some current flows due to

the voltage between Drain and Source. Let some Positive Voltage is applied

at VGG, then the minority carriers i.e. holes, get repelled and the majority carriers

i.e. electrons gets attracted towards the SiO2 layer.

With some amount of Positive Potential at VGG a certain amount of Drain

current ID flows through Source to Drain. When this Positive Potential is further

increased, the current ID increases due to the flow of electrons from Source and

these are pushed further due to the voltage applied at VGG. Hence the more

Positive the applied VGG, the more the value of Drain current ID will be. The

current flow gets enhanced due to the increase in electron flow better than in

depletion mode. Hence this mode is termed as Enhanced Mode MOSFET.

P - Channel MOSFET

The construction and working of a PMOS is same as NMOS. A lightly doped N-

substrate is taken into which two heavily doped P+ regions are diffused. These

two P+ regions act as source and drain. A thin layer of SiO2 is grown over the

surface. Holes are cut through this layer to make contacts with P+ regions, as

shown in the following figure.

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Working of PMOS

When the Gate terminal is given a Negative Potential at VGG than the Drain

Source voltage VDD, then due to the P+ regions present, the hole current is

increased through the diffused P channel and the PMOS works in Enhancement

Mode.

When the Gate terminal is given a Positive Potential at VGG than the Drain

Source voltage VDD, then due to the repulsion, the depletion occurs due to which

the flow of current reduces. Thus PMOS works in Depletion Mode. Though the

construction differs, the working is similar in both the type of MOSFETs. Hence

with change in voltage polarity both of the types can be used in both the modes.

Drain Characteristics

The drain characteristics of a MOSFET are drawn between the drain

current ID and the drain source voltage VDS. The characteristic curve is as shown

below for different values of inputs.

Actually when VDS is increased, the drain current ID should increase, but due to

the applied VGS, the drain current is controlled at certain level. Hence the gate

current controls the output drain current.

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Advantages of MOSFET

1. MOSFETs provide greater efficiency while operating at lower voltages.

2. Absence of gate current results in high input impedance producing high

switching speed.

3. They operate at lower power and draws no current.

Disadvantages of MOSFET

1. The thin oxide layer make the MOSFETs vulnerable to permanent damage

when evoked by electrostatic charges.

2. Overload voltages makes it unstable.

Applications of MOSFET

1. MOSFET amplifiers are extensively used in radio frequency applications.

2. It acts as a passive element like resistor, capacitor and inductor.

3. DC motors can be regulated by power MOSFETs.

4. High switching speed of MOSFETs make it an ideal choice in designing

chopper circuits.

Comparison between BJT, FET and MOSFET

TERMS BJT FET MOSFET

Device type Current controlled Voltage controlled Voltage Controlled

Current flow Bipolar Uni-Polar Uni-Polar

Operational

modes No modes Depletion mode only

Both Enhancement and

Depletion modes

Input

impedance Low High Very high

Output

resistance Moderate Moderate Low

Operational

speed Low Moderate High

Noise High Low Low

Thermal

stability Low Better High

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C-MOS-FET ( Complementary Metal Oxide Semiconductor Field

Effect Transistor )

In CMOS technology, both N-Type MOSFETs and P-Type MOSFETs are used to

design logic functions. The same signal which turns ON a transistor of one type is

used to turn OFF a transistor of the other type. This characteristic allows the

design of logic devices using only simple switches, without the need for a pull-up

resistor.

In this circuit, Two MOSFETs (P-channel MOSFET and N-channel-

MOSFET) are connected in series so that source of P-channel device is

connected to a Positive voltage supply + VDD and the source of N-channel device

is connected to the ground. Gates of both the devices are connected as a

common input and drain terminals of both the devices are connected together as a

common output.

CMOS (complementary metal oxide semiconductor) logic has a

few desirable advantages:

1. CMOSFET has High input impedance.

2. The outputs of CMOSFET actively drive both ways.

3. CMOS logic takes very little power.

4. The Speed of Operation of CMOS Technology is very high.

5. It has good speed to power ratio compared to other logic types.

6. CMOS gates are very simple.

7. CMOS has very high Noise Margin.

Applications

1. CMOS technology are used as image sensors (CMOS sensors),

2. These are used as data converters.

3. These are used as RF circuits (RF CMOS).

4. These are used as highly integrated transceivers for many types of

communication.

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Fill IN THE BLANKS:

1. A FET is a ………………. Controlled Device.

2. A bipolar transistor is a ………… Controlled Device.

3. FET acts as constant current source in …………… region.

4. In the …………… region, the FET can be used as Voltage variable resistor.

5. The drain of FET is analogous to ………….. of BJT.

6. FETs are …………….. Noisy to BJTs at high frequencies.

7. In P-channel FET, the current is due to ……………….

8. In a FET, the channel is ……………. doped.

9. In a FET, gate is ……………… doped.

10. A FET Depends on the variation of a ………………. for its operation.

11. N-channel FETs are superior to P-channel FETs because Mobility of electrons is

……………… than that of holes.

12. A FET differs from a bipolar transistor as it has ………… input impedance.

13. In a JFET, gates are always ……………. biased.

14. The input resistance of JFET ideally should be ………………. .

15. For a JFET, above the pinch-off voltage, the Drain current remains …………..

16. A JFET can operate in Only ………….. mode.

17. The name field effect is related to the Depletion layers of a …………… .

18. A FET has ………… terminals.

19. A FET is also called …………… Transistor.

20. A MOSFET is sometimes called ……………… FET.

21. In a JFET, Drain Current will be maximum when VGS is …………….

22. A JFET can be cut off with the help of ……………… .

ANSWERS:

1) Voltage 2) Current 3) Ohmic 4) Ohmic

5) Collector 6) Less 7) Holes 8) Lightly

9) Heavily 10) Magnetic Field 11) Greater 12) High

13) Reverse 14) Infinity 15) Constant 16) Depletion

17) JFET 18) Three 19) Uni-Polar 20) Insulated gate

21) Zero 22) VGS

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UNIT – XIII

INTRODUCTION TO ELECTRICAL MACHINES


TRANSFORMER

A Transformer is a Static Electrical Device which transfers AC Electrical Power

from one circuit to the other circuit with the voltage level can be altered, that

means voltage can be increased ( Step-Up ) or decreased ( Step-Down )according

to the requirement at the constant frequency through the process of

Electromagnetic Induction.

Principal of Operation

Transformer works on the principle of Faraday’s Law of Electromagnetic

Induction which states that “ the magnitude of voltage is directly proportional to

the rate of change of flux.”

When one winding (also known as a coil) which is supplied by an alternating

electrical source. The Alternating Current through the winding produces a

continually changing and alternating flux that surrounds the winding. If another

winding is brought close to this winding, some portion of this alternating flux will

link with the second winding. As this flux is continually changing in its amplitude

and direction, there must be a changing flux linkage in the second winding or coil.

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According to Faraday’s Law of Electromagnetic Induction, there will be

an EMF induced in the second winding. If the circuit of this secondary winding is

closed, then a current will flow through it. This is the basic working principle of a

transformer.

The winding which receives Electrical Power from the source is known as

the ‘Primary Winding’. In the diagram above this is the ‘First Coil’.

The winding which gives the desired output voltage due to mutual induction

is commonly known as the ‘secondary winding’. This is the ‘Second Coil’ in the

diagram above.

A transformer that increases voltage between the primary to secondary

windings is defined as a Step-Up Transformer. Conversely, a transformer that

decreases voltage between the primary to secondary windings is defined as a

Step-Down Transformer.

Whether the transformer increases or decreases the voltage level depends

on the relative number of turns between the primary and secondary side of the

transformer. If there are more turns on the primary coil than the secondary coil

than the voltage will decrease (step down). If there are less turns on the primary

coil than the secondary coil than the voltage will increase (step up).

Construction of Single Phase Transformer

Where:

VP – is the Primary Voltage

VS – is the Secondary Voltage

NP – is the Number of Primary Windings

NS – is the Number of Secondary Windings

Φ (phi) – is the Flux Linkage

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A single phase voltage transformer basically consists of two electrical coils of wire,

one called the “Primary Winding” and another called the “Secondary Winding”.

The “Primary” side of the transformer as the side that usually takes power, and the

“Secondary” as the side that usually delivers power.

These two coil windings are electrically isolated from each other but are

magnetically linked through the common core allowing electrical power to be

transferred from one coil to the other. When an electric current passed through the

primary winding, a magnetic field is developed which induces a voltage into the

secondary winding.

In other words, for a transformer there is no direct electrical connection

between the two coil windings, so it is also called an Isolation Transformer.

Generally, the primary winding of a transformer is connected to the input voltage

supply and converts or transforms the electrical power into a magnetic field. While

the job of the secondary winding is to convert this alternating magnetic field into

electrical power producing the required output voltage as shown in above figure.

The difference in voltage between the Primary and the Secondary Windings

is achieved by changing the number of coil turns in the Primary Winding ( NP )

compared to the number of coil turns on the Secondary Winding ( NS ).

Turn Ratio :

It is defined as the ratio of Number of Turns in Primary Coil to Number of Turns of

Secondary Coils.

If NS > NP, the transformer is called Step-Up Transformer

If NS < NP, the transformer is called Step-Down Transformer

As the transformer is basically a linear device, a ratio now exists between

the number of turns of the Primary Coil divided by the number of turns of the

Secondary Coil. This ratio is also called as ratio of Transformation, more

commonly known as a transformers “Turns Ratio”, ( TR ).

This turns ratio value dictates the operation of the transformer and the

corresponding voltage available on the secondary winding. The turns ratio, has no

units, such as 3 : 1 (3-to-1). This means that if there are 3 volts on the Primary

Winding, there will be 1 volt on the Secondary Winding, 3 volts-to-1 volt.

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TYPE OF LOSSES IN A TRANSFORMER

There are various types of losses in the transformer such as copper loss, iron

loss, hysteresis loss, eddy current loss, stray loss, and dielectric loss. The

hysteresis losses occur because of the variation of the magnetization in the core

of the transformer and the copper loss occurs because of the transformer winding

resistance.

Copper Loss Or Ohmic Loss

These losses occur due to ohmic resistance of the transformer windings. If IP and

IS are the Primary and the Secondary current. RP and RS are the resistance of

Primary and Secondary Winding then the copper losses occurring in the primary

and secondary winding will be IP2RP and IS2RS respectively.

Therefore, the Total Copper Losses will be

These losses varied according to the load and known hence it is also known as

variable losses. Copper losses vary as the square of the load current.

Iron Losses

Iron losses are caused by the alternating flux in the core of the transformer as this

loss occurs in the core it is also known as Core loss.

Iron loss is further divided into Two Losses

1. Hysteresis Loss

2. Eddy Current Loss.

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1. Hysteresis Loss

The core of the transformer is subjected to an alternating magnetizing force, and

for each cycle of emf, a hysteresis loop is traced out. Power is dissipated in the

form of heat known as hysteresis loss and given by the equation shown below:

Where

KȠ is a proportionality constant which depends upon the volume

and quality of the material of the core used in the transformer,

f is the supply frequency,

Bmax is the maximum or peak value of the flux density.

The iron or core losses can be minimized by using silicon steel material for the

construction of the core of the transformer.

2. Eddy Current Loss

When the flux links with a closed circuit, an emf is induced in the circuit and the

current flows, the value of the current depends upon the amount of emf around the

circuit and the resistance of the circuit.

Since the core is made of conducting material, these EMFs circulate

currents within the body of the material. These circulating currents are

called Eddy Currents. They will occur when the conductor experiences a

changing magnetic field. As these currents are not responsible for doing any

useful work, and it produces a loss (I2R loss) in the magnetic material known as

an Eddy Current Loss.

The eddy current loss is minimized by making the core with thin laminations.

Stray Loss

The occurrence of these stray losses is due to the presence of leakage field. The

percentage of these losses are very small as compared to the iron and copper

losses so they can be neglected.

Dielectric Loss

Dielectric loss occurs in the insulating material of the transformer that is in the oil

of the transformer, or in the solid insulations. When the oil gets deteriorated or the

solid insulation gets damaged, or its quality decreases, and because of this, the

efficiency of the transformer gets affected.

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TRANSFORMER EFFICIENCY

The Efficiency of the transformer is defined as the ratio of useful Output Power to

the Input Power. The Input and Output Power are measured in the same unit. Its

unit is either in Watts (W) or KW. Transformer efficiency is denoted by Ƞ.

Where,

VS – Secondary terminal voltage

IS – Full load secondary current

CosϕS – Power Factor of the load

Pi – Iron losses = hysteresis losses + eddy current losses

PC – Full load copper losses = IC2Res

The transformer will give the maximum efficiency when their copper loss is equal

to the iron loss.

DC MACHINE

A DC Machine is an electromechanical energy alteration device. The working

principle of a DC machine is when electric current flows through a coil within a

magnetic field, and then the magnetic force generates a torque which rotates the

dc motor.

The DC machines are classified into two types:

1. DC Generator : DC Generator is a Device which convert Mechanical

Energy in to DC Electrical Energy.

2. DC Motor : DC Motor is a Device which converts DC Electrical Energy into

Mechanical Energy.

Working Principle of DC Generator

The DC Generator working principle is based on Faraday’s laws

of electromagnetic induction. When a conductor is located in an unstable

magnetic field, an electromotive force gets induced within the conductor. The

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induced e.m.f magnitude can be measured from the equation of the

electromotive force of a generator.

If the conductor is present with a closed lane, the current which is induced

will flow in the lane. In this generator, field coils will generate an electromagnetic

field as well as the armature conductors are turned into the field. Therefore, an

electromagnetically induced electromotive force (e.m.f) will be generated within

the armature conductors. The path of induced current will be provided by

Fleming’s right-hand rule.

Working Principle of a DC Motor

The DC motor is the device which converts the direct current into the mechanical

work. It works on the principle of Lorentz Law, which states that “the current

carrying conductor placed in a magnetic and electric field experience a

force”. And that force is called the Lorentz force. The Flemming left-hand rule

gives the direction of the force.

Fleming Left Hand Rule

If the thumb, middle finger and the index finger of the left hand are displaced from

each other by an angle of 90°, the middle finger represents the direction of the

magnetic field. The index finger represents the direction of the current, and the

thumb shows the direction of forces acting on the conductor.

Construction of DC Machine

A DC Generator can be used as a DC Motor without any constructional changes

and vice versa is also possible. Thus, a DC Generator or a DC Motor can be

broadly termed as a DC Machine. These basic constructional details are also

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valid for the construction of a DC Motor. Hence, It is commonly called as DC

machine.

Basic Structure of Electrical Machines

The DC machine is constructed by some of the essential parts like Yoke, Pole

core & pole shoes, Pole coil & field coil, Armature core, Armature winding,

commutator, brushes & bearings.

1. Yoke ( Frame ) : The main function of the yoke in the machine is to offer

mechanical support intended for poles and protects the entire machine from the

moisture, dust, etc. The materials used in the yoke are designed with cast iron,

cast steel otherwise rolled steel.

2. Poles and Pole Shoes: Poles are joined to the yoke with the help of bolts or

welding. They carry field winding and pole shoes are fastened to them. Pole shoes

serve two purposes; (i) they support field coils and (ii) spread out the flux in air

gap uniformly.

3. Field winding: These are usually made of copper. Field coils are former wound

and placed on each pole and are connected in series. These are wound in such a

way that, when energized, these form alternate North and South poles.

4. Armature Core: Armature core is the rotor of a DC Machine. It is cylindrical in

shape with slots to carry armature winding. The armature is built up of thin

laminated circular steel disks for reducing eddy current losses. It may be provided

with air ducts for the axial air flow for cooling purposes. Armature is keyed to the

shaft.

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5. Armature Winding: It is usually a former wound copper coil which rests in

armature slots. The armature conductors are insulated from each other and also

from the armature core. Armature winding can be wound by one of the two

methods; lap winding or wave winding. Double layer lap or wave windings are

generally used. A double layer winding means that each armature slot will carry

two different coils.

6. Commutator: Physical connection to the armature winding is made through a

commutator-brush arrangement. The function of a commutator, in a DC

Generator, is to collect the current generated in armature conductors. Whereas, in

case of a DC Motor, commutator helps in providing current to the armature

conductors. A commutator consists of a set of copper segments which are

insulated from each other. The number of segments is equal to the number of

armature coils. Each segment is connected to an armature coil and the

commutator is keyed to the shaft.

7. Brushes: Brushes are usually made from carbon or graphite. They rest on

commutator segments and slide on the segments when the commutator rotates

keeping the physical contact to collect or supply the current.

CLASSIFICATION DC MACHINE

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1. Separately excited DC machines: In separately excited DC Machines, the

field winding is supplied from a separate power source. That means the field

winding is electrically separated from the armature circuit. Separately excited DC

generators are not commonly used because they are relatively expensive due to

the requirement of an additional power source or circuitry. They are used in

laboratories for research work, for accurate Speed Control of DC Motors.

2. Self-excited DC machines: In this type of DC Machine, Field Winding and

Armature Winding are interconnected in various ways to achieve a wide range of

performance characteristics (for example, field winding in series or parallel with

the armature winding).

In a self-excited type of DC Generator, the field winding is energized by

the current produced by themselves. A small amount of flux is always present in

the poles due to the residual magnetism. So, initially, current induces in the

armature conductors of a DC Generator only due to the residual magnetism. The

field flux gradually increases as the induced current starts flowing through the field

winding.

Self-excited machines can be further classified as –

i. Series wound DC Machines – In this type of DC Machine, field

winding is connected in series with the armature winding. Therefore, the

field winding carries whole of the load current (armature current). That is

why series winding is designed with few turns of thick wire and the

resistance is kept very low (about 0.5 Ohm).

ii. Shunt wound dc machines – In this type of DC Machine, field

winding is connected in parallel with the armature winding. Hence, the full

voltage is applied across the field winding. Shunt winding is made with a

large number of turns and the resistance is kept very high (about 100

Ohm). It takes only small current which is less than 5% of the rated

armature current.

iii. Compound wound dc machines – In this type of DC Machine, there

are two sets of field winding. One is connected in series and the other is

connected in parallel with the armature winding. Compound wound

machines are further divided as -

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a) Short shunt – field winding is connected in parallel with only the

armature winding.

b) Long shunt – field winding is connected in parallel with the

combination of series field winding and armature winding.

AC MACHINE

AC Machines are Motors that convert AC Electric Energy to Mechanical Energy

and Generators that convert Mechanical Energy to AC Electric Energy.

AC MOTOR: The Motor that converts the Alternating Current into Mechanical

power by using an Electromagnetic Induction phenomenon is called an AC Motor.

This Motor is driven by an Alternating Current. The stator and the rotor are

the two most important parts of the AC Motors. The stator is the stationary part of

the Motor, and the rotor is the rotating part of the Motor.

The AC Motor may be single phase or three phase. The three phase AC

Motors are mostly applied in the industry for bulk power conversion from electrical

to mechanical. For small power conversion, the single phase AC Motors are

mostly used. The Single Phase AC Motor is nearly small in size, and it provides a

variety of services in the home, office, business concerns, factories, etc. Almost all

the domestic appliances such as refrigerators, fans, washing machine, hair dryers,

mixers, etc., use single phase AC motor.

The AC motor is mainly classified into Two Types.

1. Synchronous Motor

2. Induction Motor ( Asynchronous Motor )

Synchronous Motor :

The Motor that converts the AC Electrical Power ( Energy ) into Mechanical

Power ( Energy ) and is operated only at the Synchronous Speed is known as a

Synchronous Motor.

Construction of Synchronous Motor

The stator and the rotor are the two main parts of the synchronous motor. The

stator becomes stationary, and it carries the armature winding of the motor. The

armature winding is the main winding because of which the EMF induces in

the motor. The rotator carry the field windings. The main field flux induces in the

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rotor. The rotor is designed in two ways, i.e., the salient pole rotor and the non-

salient pole rotor.

The synchronous motor uses the salient pole rotor. The word salient

means the poles of the rotor projected towards the armature windings. The

rotor of the synchronous motor is made with the laminations of the steel. The

laminations reduce the eddy current loss occurs on the winding of the transformer.

The salient pole rotor is mostly used for designing the medium and low-speed

motor. For obtaining the high-speed cylindrical rotor is used in the motor.

Working Principle of a Synchronous Motor

When Supply is given to Synchronous Motor, a revolving field is set up. This field

tries to drag the rotor with it, but could not do so because of rotor inertia. Hence,

No starting torque is produced. Thus, inherently synchronous motor is not a Self-

Starting the Motor.

The rotor is excited by the DC supply. The DC supply induces the north and south

poles on the rotor. As the DC supply remains constant, the flux induces on the

rotor remains same. Thus, the flux has fixed polarity. The north pole develops on

one end of the rotor, and the south pole develops on another end.

The AC is sinusoidal. The polarity of the wave changes in every half cycle,

i.e., the wave remains positive in the first half cycle and becomes negative in the

second half cycle. The positive and negative half cycle of the wave develops the

north and south pole on the stator respectively.

When the rotor and stator both have the same pole on the same side, they

repel each other. If they have opposite poles, they attract each other. This can

easily be understood with the help of the figure shown below.

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The synchronous motor is not self-starting. The prime mover is used for rotating

the motor. The prime mover rotates the rotor at their synchronous speed. The

synchronous speed is the constant speed of the machine whose value depends

on the frequency and the numbers of the pole of the machine.

When the rotor starts rotating at their synchronous speed, the prime mover

is disconnected to the motor. And the DC supply is provided to the rotor because

of which the north and south pole develops at their ends

The north and south poles of the rotor and the stator interlock each other.

Thus, the rotor starts rotating at the speed of the rotating magnetic field. And the

motor runs at the synchronous speed. The speed of the motor can only be

changed by changing the frequency of the supply.

Induction Motor or Asynchronous Motor

The Machine which converts the AC Electric Power ( Energy ) into Mechanical

Power ( Energy ) by using an Electromagnetic Induction phenomenon in called an

Induction Motor.

Working Principle of an Induction Motor

In an Induction Machine the armature winding serve as both the armature winding

and field winding. When the stator windings are connected to an AC supply, flux is

produced in the air gap. The flux rotates at a fixed speed called synchronous

speed. This rotating flux induces voltages in the stator and rotor winding.

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If the rotor circuit is closed, the current flows through the rotor winding and react

with the rotating flux and a torque is produced. In the steady state, the rotor

rotates at speed very close to synchronous speed.

Single Phase Induction Motor Construction

The main parts of a single -phase induction motor are the Stator, Rotor, Winding.

The stator is the fixed part of the motor to which A.C. is supplied. The stator

contains two types of windings. One is the main winding and the other is the

Auxiliary winding. These windings are placed perpendicular to each other. A

capacitor is attached to Auxiliary winding in parallel.

As AC Supply is used for working of single -phase induction motor, certain

losses should be looked out for such as- Eddy current loss, Hysteresis loss. To

remove the eddy current loss the stator is provided with laminated stamping. To

reduce the hysteresis losses, these stampings are usually built with silicon steel.

The rotor is the rotating part of the motor. Here the rotor is similar to the squirrel

cage rotor. To get smooth, quite working of the motor, by preventing magnetic

locking of the stator and rotor, slots are skewed rather than being parallel.

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Rotor conductors are the aluminium or coppers bars, are placed in the slots

of the rotor. End rings made up of either aluminium or copper electrically shorts

the rotor conductors. In this single-phase induction motor slip rings and

Commutator are not used, so their construction becomes very simple and easy.

Numerical-1: A Voltage Transformer has 1500 turns of wire on its Primary

coil and 500 turns of wire for its Secondary coil. What will be the turns ratio

(TR) of the transformer?

Solution : Given NP = 1500 Turns, NS = 500 Turns

Turns Ratio (TR) of Transformer = NP / NS

= 1500 / 500 = 3 / 1

Turns Ratio (TR) = 3 : 1

This Ratio of 3:1 (3-to-1) simply means that there are three primary windings for

every one secondary winding. As the ratio moves from a larger number on the left

to a smaller number on the right, the primary voltage is therefore stepped down in

value.

Numerical -2: If 240 volts rms is applied to the primary winding of the

transformer having Turn Ratio of 3 : 1 (3 - to -1), what will be the resulting

secondary No Load Voltage?

Solution : Given Turn Ratio = 3 : 1 , VP = 240 V, VS = ?

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Fill IN THE BLANKS:

1. Power transformers are designed to have maximum efficiency at ………….Load.

2. Transformer core are laminated in order to minimize ……………. loss.

3. The Transformer ratings are usually expressed in …………….. .

4. An Ideal Transformer have ……….. Resistance at Primary and Secondary

Winding.

5. An Ideal Transformer have ………………. Iron Loss.

6. A Step-up Transformer ……………….. Voltage.

7. ………………. of rotation of motor is determined by Fleming’s left-hand rule.

8. The current drawn by the armature of DC motor is directly proportional to

………….. .

9. The field of an induction motor rotor rotates relative to the stator at ……………..

speed.

10. In an induction motor, rotor runs at a speed ………. than the speed of stator

field.

11. If the resistance of the field winding of DC Generator is increased, then the

output voltage ……………...

12. When an induction motor runs at rated load and speed, the iron losses are

……………..

13. ……………….. convert Mechanical Energy in to DC Electrical Energy.

14. ………………. converts DC Electrical Energy into Mechanical Energy.

15. In Series Wound DC Machine, field winding is connected in …………. with the

armature winding.

16. In …………… DC Machine, field winding is connected in Parallel with the

armature winding.

17. Starter is used in DC Motor to reduce ……………….. .

18. Rotor of Single Phase Motor is always ……………. type.

ANSWERS:

1) Maximum 2) Eddy Current 3) KVA 4) Zero

5) Zero 6) Increases 7) Direction 8) Torque

9) Synchronous 10) Lower 11) Increases 12) Negligible

13) DC Generator 14) DC Motor 15) Series 16) Shunt Wound

17) Starting Current 18) Cage

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PRACTICALS

EXPERIMENT NO. – 1

AIM: Operation and Use of measuring Instrument Voltmeter, Ammeter, CRO,

Wattmeter, Multi-meter and Other accessories.

1. VOLTMETER: Voltmeter is employed to measure the potential difference across

any two points of a circuit. It is connected in the parallel across any element in the

circuit. The resistance of voltmeter is kept very high by connecting a high resistance

in series of the voltmeter with the current coil of the instrument. The actual voltage

drop across the current coil of the voltmeter is only a fraction of the total voltage

applied across the voltmeter which is to be measured.

2. AMMETER: Ammeter is employed for measuring of current in a circuit and

connected in series in the circuit. As Ammeter is connected in series, the voltage

drop across ammeter terminals is very low. This requires that the resistance of the

Ammeter should be as low as possible. The current coil of ammeter has low current

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PRACTICALS

carrying capacity whereas the current to be measured may be quite high. So for

protecting the equipment a low resistance is connected in parallel to the current coil

and it is known as shunt resistance.

3. WATTMETER: The measurement of Real Power in AC circuits is done by using an

instrument using Wattmeter. The Real Power in AC circuits is given by expression

V.I. Cos

Where, Cos is Power Factor.

A Wattmeter has two coils, namely, Current Coil and Pressure Coil. The Current

Coil (CC) is connected in series with the load and the Pressure Coil (PC) is

connected across the load. Watt meters are available in dual range for voltages as

well as for current.

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PRACTICALS

4. CRO: Cathode-Ray Oscilloscope (CRO) is a common laboratory instrument that

provides accurate time and amplitude measurements of Voltage Signals over a

wide range of frequencies. Its reliability, stability, and ease of operation make it

suitable as a general purpose laboratory instrument.

In Cathode Ray Oscilloscope, this is the part from where electrons are born

initially. It produces a sharply focused beam of electrons which are accelerated to

high velocity. This focused beam of electrons strikes the fluorescent screen with

sufficient energy to cause a luminous spot on the screen.

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PRACTICALS

5. MULTIMETER: Multi-meter is a measuring instrument used to measure the

Current, Voltage and Resistance. These can be used to troubleshoot many

electrical types of equipment such as Domestic Appliances, Power Supplies etc.

Multi-meters are capable of many different readings, depending on the model.

Basic testers measure voltage, amperage, and resistance and can be used to

check continuity, a simple test to verify a complete circuit. More advanced Multi-

meters may test for all of the following values:

AC (Alternating Current) Voltage and Amperage

DC (Direct Current) Voltage and Amperage

Resistance (Ohms)

Capacity (Farads)

Conductance (Siemens)

Decibels

Duty cycle

Frequency (Hz)

Inductance (Henrys)

Temperature Celsius or Fahrenheit

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PRACTICALS

Accessories or special sensors can be attached to some Multi-meters for additional

readings, such as:

Light level.

Acidity.

Alkalinity.

Wind speed.

Relative humidity.

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PRACTICALS

EXPERIMENT NO. – 02 ( I )

AIM: Measurement of Resistance of a Voltmeter.

APPARATUS / COMPONENTs REQUIRED:

1. D.C. Power Supply ( 0 – 150 V ) (ONE)

2. Voltmeters ( 0 - 30 V ) (ONE)

3. Switch (ONE)

4. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :

1. Make the Connection as per the Circuit Diagram.

2. Insert all the metal plugs of the Resistance Box tightly ( i.e. R = 0 Ω ).

3. Switch On the Power Supply and note the deflection in the Voltmeter. Let it be θ0.

4. Take out the metal plugs from the Resistance Box in small steps, so that the

deflection in the voltmeter is half of the value noted in step 2 (i.e. θ0 / 2).

5. Note down the Value of Resistance of the Resistance Box at this stage and this

Resistance value of Resistance Box (R) is equal to the value of Voltmeter

Resistance RV.

OBSERVATIONS:

Value of Voltmeter Resistance RV = _____________ Ω

RESULT: Measurement of Resistance of a Voltmeter is measured.

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PRACTICALS

EXPERIMENT NO. – 02 ( II )

AIM: Measurement of Resistance of an Ammeter.


1. D.C. Power Supply ( 0 – 150 V ) (ONE)

2. Ammeters ( 0 - 30 mA ) (ONE)

3. Switch (ONE)

4. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :

1. Make the Connection as per the Circuit Diagram.

2. Insert all the metal plugs of the Resistance Box tightly ( i.e. R = 0 Ω ).

3. Switch On the Power Supply and note the deflection in the Ammeter. Let it be θ0.

4. Take out the metal plugs from the Resistance Box in small steps, so that the

deflection in the Ammeter is half of the value noted in step 2 (i.e. θ0 / 2).

5. Note down the Value of Resistance of the Resistance Box at this stage and this

Resistance value of Resistance Box (R) is equal to the value of Ammeter

Resistance RA.

OBSERVATIONS:

Value of Voltmeter Resistance RA = _____________ Ω

RESULT: Measurement of Resistance of an Ammeter is measured.

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PRACTICALS


AIM: To verify Thevenin’s Theorem.


1. Regulated Power Supply ( 0 – 30 V ) (ONE)

2. Digital Ammeters ( 0 - 10 mA ) (ONE)

3. Digital Multi-meter (ONE)

4. Resisters 1 KΩ (ONE)

5. Resister 330 Ω (THREE)

6. Bread Board (ONE)

7. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :

1. Connections are given as per the Circuit Diagram.

2. Switch On the Power Supply.

3. Set a particular value of Voltage using Regulated Power Supply and Note

down the corresponding Ammeter readings.

To find VTH

4. Remove the Load Resistance and measure the open circuit voltage using

Multi-meter (VTH).

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PRACTICALS

To find RTH

5. To find the Thevenin’s Resistance, remove the Regulated Power Supply and

short circuit it and find the RTH using Multi-meter.

6. Give the connections for equivalent circuit and set VTH and RTH and note the

corresponding Ammeter reading.

7. Verify Thevenins Theorem.

Thevenin’s Equivalent circuit:

OBSERVATION TABLE:

Theoretical and Practical Values

E

( Volts )

VTH

( Volts )

RTH

( Ohm )

Load Current ( IL ) mA

Main Circuit Equivalent Circuit

Theoretical

Values 10 5 495 3.34 3.34

Practical

Values 10 4.99 485 3.3 3.36

RESULT: Thevenin’s Theorem verified. (Theoretically and Practically)

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PRACTICALS


AIM: To verify Norton’s Theorem.


1. Regulated DC Power Supply ( 0 – 30 V ) (ONE)

2. Digital Ammeters ( 0 - 30 mA ) (ONE)

3. Digital Multi-meter (ONE)


5. Resisters 330 Ω (THREE)


7. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :





To Find IN:

4. Remove the Load Resistance and measure the short circuit the terminals.

5. For the same Voltage note down the Ammeter readings.

346


PRACTICALS

To find RN

6. To find the Norton’s Resistance, remove the Regulated Power Supply and

short circuit it and find the RN using Multi-meter.

7. Give the connections for equivalent circuit and set IN and RN and note the

corresponding Ammeter reading.

8. Verify Norton’s Theorem.

Norton’s Equivalent circuit

OBSERVATION TABLE:

Theoretical and Practical Values

E

( Volts )

IN

( mA )

RN

( Ohm )

Load Current ( IL ) mA

Main Circuit Equivalent Circuit

Theoretical

Values 10 10.10 495 3.34 3.34

Practical

Values 10 10.4 485 3.3 3.36

RESULT: Norton’s Theorem verified. (Theoretically and Practically)

347


PRACTICALS


AIM: Observation of Change in Resistance of a Bulb in Hot and Cold

conditions, using Voltmeter and Ammeter.


1. AC Power Supply

2. Auto Transformer (ONE)

3. Volt-Meter ( 0 – 230 V AC ) (ONE)

4. Multi-Meter (ONE)

5. Ammeters ( 0 - 200 mA ) (ONE)

6. Switch (ONE)

7. Bulb (ONE)

8. Connecting wires

CIRCUIT DIAGRAM:

348


PRACTICALS

PROCEDURE :

1. Make the Connection as per the Circuit Diagram ( A ).

2. Set the Multi-Meter in Resistance Mode and Measure the Resistance of Bulb in

Cold Situation.

3. Note down the Reading of Resistance in the Observation Table.

4. Now, Make the Connection as per the Circuit Diagram ( B ).

5. Set the Auto-Transformer at Minimum Position and Switch On the Power Supply.

6. By Changing the Position of Auto-Transformer, Note down the Reading of

Voltmeter and Ammeter in the Observation Table.

7. Calculate the Resistance of Bulb in different Hot Positions.

OBSERVATIONS TABLE:

Resistance of Bulb ( Cold Position ) = ________________ Ω

Sr. No. Voltage

( V ) Current ( mA )

Resistance of Bulb R = V/ I (Ω)

1.

50 V

2.

100 V

3.

150 V

4.

200 V

RESULT: We Observed that Resistance of Bulb will increase with Voltage, as

Temperature of Bulb Filament ( Metal ) increases, resistance will also increase.

349


PRACTICALS


AIM: Verification of Krichhoff's Current Law in a DC Circuit.


1. Regulated DC Power Supply ( 0 – 30 V ) (ONE)

2. Digital Ammeters ( 0 - 30 mA ) (THREE)

3. Resisters 1 KΩ (TWO)

4. Resister 330 Ω (ONE)

5. Resistor 220 Ω (ONE)


7. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :





4. Repeat the same procedure for different voltages.

5. Verify Krichhoff's Current Law.

350


PRACTICALS

OBSERVATION TABLE:

Theoretical Values

Sr. No. E

( Volts )

Current I1 = I2 + I3

(mA) I1 (mA) I2 (mA) I3 (mA)

1.

5 5.68 3.12 2.56 5.68

2.

15 17.05 9.37 7.68 17.05

3.

25 28.42 15.62 12.68 28.42

Practical Values

Sr. No. E

( Volts )

Current I1 = I2 + I3

(mA) I1 (mA) I2 (mA) I3 (mA)

1.

5 5.6 3.1 2.2 5.3

2.

15 17.2 9.4 7.6 17

3.

25 28 15.6 12.7 28.3

RESULT: Krichhoff's Current verified. (Theoretically and Practically)

351


PRACTICALS


AIM: Verification of Krichhoff's Current and Voltage Law in a DC Circuit.


1. Regulated DC Power Supply ( 0 – 30 V ) (TWO)

2. Digital Voltmeters ( 0 - 30 V ) (THREE)


4. Resister 330 Ω (ONE)

5. Resistor 220 Ω (ONE)


7. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :



3. Set a particular value of Voltage using Regulated Power Supplies (1 & 2) and

Note down the corresponding Voltmeter readings.

4. Repeat the same procedure for different voltages.

5. Verify Krichhoff's Voltage Law.

352


PRACTICALS

OBSERVATION TABLE:

Theoretical Values

Sr. No.

E1

( Volts )

E2

( Volts )

Voltage KVL

E1 = V1 + V2

(Volts) V1 (Volts) V2 (Volts) V3 (Volts)

1.

5 5 0.58 4.41 0.583 4.99

2.

10 10 1.16 8.83 1.17 9.99

3.

15 15 1.75 13.2 1.75 14.95

Practical Values

Sr. No.

E1

( Volts )

E2

( Volts )

Voltage KVL

E1 = V1 + V2

(Volts) V1 (Volts) V2 (Volts) V3 (Volts)

1.

5 5 0.6 4.4 0.56 5

2.

10 10 1.13 8.83 1.19 9.96

3.

15 15 1.72 13.20 1.78 14.92

RESULT: Krichhoff's Voltage verified. (Theoretically and Practically)

353


PRACTICALS


AIM: To find the Ratio of Inductance of a Coil having Air-Core and Iron-

Core respectively and to observe the effect of Introduction of a

Magnetic Core on Coil Inductance.


1. Power Supply ( AC )

2. Auto Transformer (ONE)

3. Volt-Meter ( 0 – 300 V AC ) (ONE)

4. Ammeters ( 0 - 5 A ) (ONE)

5. Switch (ONE)

6. Inductor Coil ( 300 V AC, 1A ) (ONE)

7. Iron Core (ONE)

8. Connecting wires

CIRCUIT DIAGRAM:

354


PRACTICALS

PROCEDURE :

1. Make the Connection as per the Circuit Diagram ( A ).

2. Set the Auto-Transformer at Minimum Position and Switch On the Power Supply.

3. By Changing the Position of Auto-Transformer, Note down the Reading of

Voltmeter and Ammeter in the Observation Table.

4. Calculate the Impedance, Inductive Reactance and Inductance for Each Reading

for Air Core in the Table.

5. Now, Make the Connection as per the Circuit Diagram ( B ). Insert Soft Iron Core

in the Coil.

6. Switch On the Power Supply and Repeat for all Voltage Reading as done in

previous Circuit Diagram (A).

7. Note down the Reading of Voltmeter and Ammeter in the Observation Table.

8. Calculate the Impedance, Inductive Reactance and Inductance for Each Reading

for Soft Iron Core in the Table.

OBSERVATIONS TABLE:

Core Sr. No.

Voltage ( V )

Current ( mA / A )

Z = V/I XL = √Z2-R2 L = XL / 2πf

Air Core

1.

100V

2.

200 V

Soft Iron Core

1.

100 V

2.

200 V

Ratio of Inductance of a Coil having Air-Core and Iron-Core = L Air Core / L Iron Core = ______

RESULT: The Ratio of Inductance of a Coil having Air-Core and Iron-Core calculated

and the effect of Introduction of a Magnetic Core on Coil Inductance are observed. When

Soft Iron Core inserted in the Inductive Coil, Inductance of the Coil increased.

355


PRACTICALS


AIM: Charging and Testing of a Lead - Acid Storage Battery.

Lead Acid Battery

The Lead-Acid Battery consists of six cells mounted side by side in a single case.

The cells are coupled together, and each 2.0V cell adds up to the overall 12.0V

capacity of the battery.

Lead-Acid Batteries are still preferred over other lightweight options owing to their

ability to deliver large surges of electricity (which is required to start a cold engine in

an automobile).

A completely charged Lead-Acid Battery is made up of a stack of alternating lead

oxide electrodes, isolated from each other by layers of porous separators.

All these parts are placed in a concentrated solution of sulfuric acid. Intercell

connectors connect the Positive end of one cell to the Negative end of the next cell

hence the six cells are in series.

Chemical Reaction for Discharging

When the battery is discharged, it acts as a galvanic cell and the following chemical

reaction occurs.

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PRACTICALS

Negative:

Pb(s) + HSO4– + H2O(l) –> 2e– + PbSO4(s) + H3O+(aq)

Positive:

PbO2(s) + HSO4–(aq) + 3H3O+(aq) + 2e– –> PbSO4(s) + 5H2O(l)

Lead sulfate is formed at both the electrodes. Two electrons are also transferred in

the complete reaction. The lead acid battery is packed in a thick rubber or plastic

case to prevent leakage of the corrosive sulphuric acid.

Lead Acid Battery Charging

The sulphuric acid existing in the lead discharge battery decomposes and needs to

be replaced. Sometimes, the plates change their structure by themselves.

Eventually, the battery becomes less efficient and should be charged or changed.

357


PRACTICALS

When car batteries spend considerable durations of time in their discharged states,

the lead sulfate build-up may become extremely difficult to remove. This is the

reason why lead-acid batteries must be charged as soon as possible (to prevent

building up of lead sulfate). Charging of the lead batteries is usually done by

providing an external current source.

A plug is inserted which is linked to the lead-acid battery and chemical

reaction proceeds in the opposite direction. In cases where the sulphuric acid in the

battery (or some other component of the battery) has undergone decomposition,

the charging process may become inefficient. Therefore, it is advisable to check the

battery periodically.

Chemical Reaction for Recharging

The chemical reaction that takes place when the lead-acid battery is recharging can

be found below.

Negative:

2e– + PbSO4(s) + H3O+(aq) –> Pb(s) + HSO4– + H2O(l)

Positive:

PbSO4(s) + 5H2O(l) –> PbO2(s) + HSO4–(aq) + 3H3O+(aq) + 2e–

While recharging, the automobile battery functions like an electrolytic cell. The

energy required to drive the recharging comes from an external source, such as an

engine of a car. It is also important to note that overcharging of the battery could

result in the formation of by-products such as hydrogen gas and oxygen gas. These

gases tend to escape from the battery, resulting in the loss of reactants.

Testing of Lead Acid Battery

There are many ways to test a battery but the most common and accurate are

measurement of specific gravity and battery voltage.

A hydrometer is an instrument which measures the specific gravity of a liquid

against that of water. When we use a hydrometer to test a lead acid battery, we

are actually measuring the amount of sulfuric acid in the electrolyte. After using the

battery for a while, you might get a low reading on the hydrometer. This means we

are missing the chemical which produces the electrons.

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PRACTICALS

So where did the sulfur go?

It actually never left our battery, as the chemical is just resting on the battery plates

and once you recharge the battery, the sulfur goes back to the electrolyte to

produce electrons again!

To measure the battery voltage we can use a Voltmeter.

Result: Charging and Testing of a Lead - Acid Storage Battery studied.

359


PRACTICALS


AIM: Measurement of Power and Power Factor in a Single Phase R-.L-.C.

Circuit and Calculation of Active and Reactive Powers in the Circuit.


1. AC Power Supply ( 0 – 230 V )

2. Experimental Study Board for R-L-C Series Circuit. (ONE)

3. Digital Ammeters ( 0 – 3 A ) (ONE)

4. Digital Voltmeters ( 0 – 150 V) (FOUR)

5. Wattmeter ( 150V, 2.5/5A) (ONE)

6. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :

1. Connect the circuit as shown in the Diagram.

2. Adjust the Rheostat (Resistance) for maximum resistance and the Auto

Transformer to the position of Zero-output Voltage and Switch On Power Supply.

3. Adjust the voltage across the circuit to about 70 V and note I, VS, VL, VC, VR and

Power.

4. Adjust the Rheostat for several settings and repeat step 3.

5. Adjust the Rheostat to the maximum setting and change the capacitance to 140 µF

and repeat step 4.

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PRACTICALS

6. Compare the values of Phase Angle as obtained from the meter readings and from

the Phasor Diagrams. (From the phasor diagrams compute cos θ and θ as given in

the last two columns of the table).

7. Draw Phasor Diagrams showing VR, VL, VC, VS, & I for different sets of readings and

Calculate Active and Reactive Powers in the Circuit.

OBSERVATION TABLE:

Sr. No.

VS

(Volts)

Power

( Watts)

Current

(A)

Voltage across RLC

Power Factor

Cos θ =

P / (VS. I)

θ from

Meter

Reading

θ from

Phasor

Diagram VR

( Volts )

VL

( Volts )

VC

( Volts )

1.

2.

3.

Phasor Diagram:

Calculations:

Active Power = V. x I. Cosϕ = ________ Watts

Reactive Power Pr = V x I Sinϕ = ________ Volt Ampere Reactive ( VAR )

RESULT: Power and Power Factor in a Single Phase R-.L-.C. Circuit is measured and

Active and Reactive Powers in the Circuit is also calculated.

361


PRACTICALS


AIM: Plotting of V-I Characteristics of a PN Junction Diode.


1. Experimental Study Board for PN Junction Diode Characteristics (ONE)

2. DC Regulated Power supply (0 - 150 V variable) (0 - 2 V variable) (TWO)

3. Digital Ammeters ( 0 - 200 mA, 0 - 200 µA) (ONE)

4. Digital Voltmeter (0 - 20V) (ONE)

5. Connecting wires

CIRCUIT DIAGRAM:

Forward Biasing Circuit:

Reverse Biasing Circuit:

PROCEDURE :

1. Connect the Circuit as Shown in Circuit Diagram (Forward Biasing).

2. Connect Digital Voltmeter and Ammeter at as shown in the Circuit.

3. Switch ON the Power Supply.

4. Observe Output Voltages and Current for different Input Voltages.

5. Write down Output Voltages and Current in the table for different Input

Voltages.

6. Repeat the same procedure for Reverse Biasing Circuit.

7. Plot Voltage-Current Characteristics for both Forward Biasing and Reverse

Biasing in the Graph Paper.

362


PRACTICALS

OBSERVATION TABLE:

Sr. No.

Forward Biasing Reverse Biasing

VF ( Volts ) IF ( mA) VR ( Volts ) IR ( µ A )

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

GRAPH:

RESULT: V-I Characteristics of a PN junction diode plotted on the graph paper and

studied.

363


PRACTICALS


AIM: Plotting of V-I Characteristics of a Zener Diode.


1. Experimental Study Board for Zener Diode Characteristics (ONE)

2. DC Regulated Power supply (0 - 12 V variable) (ONE)

3. Digital Ammeters ( 0 - 200 mA) (ONE)


5. Connecting wires

CIRCUIT DIAGRAM:

Forward Biasing Circuit:

Reverse Biasing Circuit:

PROCEDURE :

1. Connect the Circuit as Shown in Circuit Diagram (Forward Biasing).



4. Observe Output Voltages and Current for different Input Voltages.

5. Write down Output Voltages and Current in the table for different Input

Voltages.

6. Repeat the same procedure for Reverse Biasing Circuit.

7. Plot Voltage-Current Characteristics for both Forward Biasing and Reverse

Biasing in the Graph Paper.

364


PRACTICALS

OBSERVATION TABLE:

Sr. No.

Forward Biasing Reverse Biasing

VF ( Volts ) IF ( mA) VR ( Volts ) IR ( mA )

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

GRAPH:

RESULT: V-I Characteristics of a Zener diode plotted on the graph paper and studied.

365


PRACTICALS


AIM: Observe the Output Waveform of Half-Wave Rectifier Circuit using

one Diode.


1. Experimental Study Board for Half-Wave Rectifier Circuit (ONE)

2. Dual Chanel CRO (ONE)

3. AC Power Supply

4. CRO Leads

5. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :

1. Connect the Circuit as Shown in Circuit Diagram

2. Connect CRO Channel – I at the Secondary of the Transformer (at terminal

AB) through CRO Leads as shown in the Circuit.

3. Connect CRO Channel - II at Load Resistor (RL) through CRO Leads.


5. Observe Input and Output Waveform of the Circuit on CRO with amplitude and

time period of the signal.

6. Draw the Input and Output waveform on the Graph Paper.

366


PRACTICALS

WAVE FORMS:

RESULT: Input and Output waveforms of Half-wave Rectifier Circuit using one Diode

are observed.

367


PRACTICALS


AIM: Observe the Output Waveform of Full-wave Rectifier Circuit using

two Diodes.


1. Experimental Study Board for Full-Wave Rectifier Circuit (ONE)


3. AC Power Supply

4. CRO Leads

5. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :

1. Connect the Circuit as Shown in Circuit Diagram.

2. Connect CRO Channel – I at the Secondary of the Transformer through CRO

Leads as shown in the Circuit.






368


PRACTICALS

WAVE FORMS:

RESULT: Input and Output waveforms of Full-wave Rectifier Circuit using Two Diodes

are observed.

369


PRACTICALS

EXPERIMENT NO. – 10 ( III )

AIM: Observe the Output of Waveform of Bridge-Rectifier Circuit using

four Diodes.


1. Experimental Study Board for Full-Wave Bridge Rectifier Circuit (ONE)


3. AC Power Supply

4. CRO Leads

5. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :









370


PRACTICALS

WAVE FORMS:

RESULT: Input and Output Waveforms of Full-wave Bridge Rectifier Circuit using Four

Diodes are observed.

371


PRACTICALS


AIM: Plotting of the Wave Shape of Full Wave Rectifier with Shunt

Capacitor Filter.


1. Experimental Study Board for Full-Wave Rectifier Circuit with Filters. (ONE)


3. AC Power Supply

4. CRO Leads

5. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :









372


PRACTICALS

WAVE FORMS:

RESULT: Input & Output waveforms of Full-wave Rectifier with Shunt Capacitor Filter

are observed on CRO and Plotted on Graph Paper.

373


PRACTICALS


AIM: Plotting of the Wave Shape of Full Wave Rectifier with Series

Inductor Filter.


1. Experimental Study Board for Full-Wave Rectifier Circuit with Filters. (ONE)


3. AC Power Supply

4. CRO Leads

5. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :









374


PRACTICALS

WAVE FORMS:

RESULT: Input & Output waveforms of Full-wave Rectifier with Series Inductor Filter are

observed on CRO and Plotted on Graph paper.

375


PRACTICALS


AIM: Plotting of Input and Output Characteristics and Calculation of

Parameters of Transistors in CE configuration.


1. Experimental Study Board for Transistors in CE configuration (ONE)

2. DC Regulated Power supply (0 - 30 V variable) (TWO)

3. Digital Ammeters ( 0 - 200 mA, 0 - 200 µA) (TWO)

4. Digital Voltmeters (0 - 20V) (TWO)

5. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :




4. Set Output Voltage VCE at 0 Volt and Observe Input Current IB (µ A) for

different Input Voltages VBE ( Volts ).

5. Repeat the same procedure for Output Voltage VCE at 10 Volts and 20 Volts.

6. Write down Input Voltages (VBE) and Current (IB) in the table for different

Output Voltages (VCE).

7. Plot Input Voltage-Current Characteristics on the Graph Paper.

8. Set Input Current (IB) at 0 µ A and observe Output Current (IC) for different

Output Voltage (VCE)

376


PRACTICALS

9. Repeat the same procedure for Input Current IB at 60 µ A and 120 µ A.

10. Write down Output Voltages (VCE) and Current (IC) in the table for different

Input Current IB.

11. Plot Output Voltage-Current Characteristics on the Graph Paper.

OBSERVATION TABLE: Input Characteristics

Sr. No.

VCE = 0 V VCE = 10 V VCE = 20 V

VBE ( Volts ) IB (µ A) VBE ( Volts ) IB (µ A) VBE ( Volts ) IB (µ A)

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

GRAPH:

377


PRACTICALS

OBSERVATION TABLE: Output Characteristics

Sr. No.

IB = 0 µ A IB = 60 µ A IB = 120 µ A

VCE ( Volts ) IC (m A) VCE ( Volts ) IC (m A) VCE ( Volts ) IC (m A)

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

GRAPH:

RESULT: Input and Output Characteristics of Transistors in CE Configuration are Plotted

on Graph Paper.

378


PRACTICALS


AIM: Plotting of Input and Output Characteristics and Calculation of

Parameters of Transistors in CB configuration.


1. Experimental Study Board for Transistors in CB configuration (ONE)

2. DC Regulated Power supply (0 - 10 V Variable) (TWO)

3. Digital Ammeters ( 0 - 100 mA ) (TWO)

4. Digital Voltmeter (0 - 10V) (TWO)

5. Connecting wires (Single strand, Multi strand)

CIRCUIT DIAGRAM:

PROCEDURE :


2. Connect Digital Voltmeters and Ammeters at as shown in the Circuit.


4. Set Output Voltage VCB at 0 Volt and Observe Input Current IE (m A) for

different Input Voltages VEB ( Volts ).

5. Repeat the same procedure for Output Voltage VCB at 4 Volts and 8 Volts.

6. Write down Input Voltages (VEB) and Current (IE) in the table for different

Output Voltages (VCB).


379


PRACTICALS

8. Set Input Current (IE) at 0 m A and observe Output Current (IC) for different

Output Voltage (VCB)

9. Repeat the same procedure for Input Current IE at 30 m A and 50 m A.

10. Write down Output Voltages (VCB) and Current (IC) in the table for different

Input Current IE.

11. Plot Output Voltage-Current Characteristics on the Graph Paper.

OBSERVATION TABLE: Input Characteristics

Sr. No.

VCB = 0 V VCB = 4 V VCB = 8 V

VEB ( Volts ) IE (m A) VEB ( Volts ) IE (m A) VEB ( Volts ) IE (m A)

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

GRAPH:

380


PRACTICALS

OBSERVATION TABLE: Output Characteristics

Sr. No.

IE = 0 m A IE = 30 m A IE = 50 m A

VCB ( Volts ) IC (m A) VCB ( Volts ) IC (m A) VCB ( Volts ) IC (m A)

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

GRAPH:

RESULT: Input and Output Characteristics of Transistors in CB Configuration are Plotted

on Graph Paper.

381


PRACTICALS


AIM: Plotting of V-I Characteristics of a FET (Field Effect Transistor).


1. Experimental Study Board for Transistors in CB configuration (ONE)

2. DC Regulated Power supply (0 - 30 V Variable) (TWO)

3. Digital Ammeters ( 0 - 200 mA ) (TWO)

4. Digital Voltmeters (0 - 20V) (TWO)

5. Connecting wires (Single strand, Multi strand)

CIRCUIT DIAGRAM:

PROCEDURE :


2. Connect Digital Voltmeters and Ammeters at as shown in the Circuit.


4. Set Input Voltage VGS at 0 Volt and Observe Output Current ID (m A) for

different Output Voltages VDS ( Volts ).

5. Repeat the same procedure for Input Voltage VGS at 3 Volts and 6 Volts.

6. Write down Output Voltages (VDS) and Current (ID) in the table for different

Input Voltages (VGS).


382


PRACTICALS

OBSERVATION TABLE:

Sr. No.

VGS = 0 V VGS = 3 V VGS = 6 V

VDS ( Volts ) ID (m A) VDS ( Volts ) ID (m A) VDS ( Volts ) ID (m A)

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

GRAPH:

RESULT: V-I Characteristics of a FET (Field Effect Transistor) are Plotted on Graph

Paper.

383


PRACTICALS


AIM: To determine the efficiency of Single Phase Transformer.


1. Single Phase Transformer 1KVA ( Dry Type ) (ONE)

2. Single Phase Auto Transformer 230V / ( 0-270 V ) (ONE)

3. Single Phase Resistive Load ( Variable ) (ONE)

4. Watt Meter ( 5 A / 300 V ) (ONE)

5. Digital Ammeter ( 0 - 10A ) (ONE)


7. AC Power Supply

8. Connecting wires

CIRCUIT DIAGRAM:

PROCEDURE :


2. Connect Watt Meter, Voltmeter and Ammeter at as shown in the Circuit.

3. Switch on the Power Supply and apply the rated voltage of secondary winding

by varying the Autotransformer.

4. Note down the Ammeter, Voltmeter, and Wattmeter reading for the no-load

condition.

384


PRACTICALS

5. Switch on the load and apply the load in steps up to the rated secondary

current.

6. Note down the Ammeter, Voltmeter and Wattmeter readings for each step of

the Load.

7. Reduce the load and switch off the supply.

8. Write down the readings in the Table and Calculate the Efficiency of the

Transformer.

OBSERVATION TABLE:

Sr. No

Input Power W1

Output Voltage

V2

Output Current I2

% Efficiency = V2 x I2 X100 W1

1. 200 W

2. 400 W

3. 600 W

4. 800 W

5. 1000 W

RESULT: The Efficiency of Single Phase Transformer is determined.

385

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