ch 14 - alternating currentf

8/18/2019 Ch 14 - Alternating Currentf

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24 AlternatingCurrent


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24. Alternating

Currents

24.3 Transmission

of electrical energy24.4 Rectification

24.2

Transformer

24.1 Characteristicsof alternating current

period,

frequency,peak value

and root-mean

square value

For resistive load

= P₀/2

I=I₀ sin t

Advantage of ac

and high voltages

Distinguish

between peak andrms value …

Half-wave and

full-wave

http://www.magnet.fsu.edu/education/tutorials/java/ac/index.html

Principle of

operation

2

o

rms

I I

s

P

P

s

P

s

I

I

V

V

N

N

effect of single

capacitance on

smoothing

Single diode and

bridge rectifier

http://www.magnet.fsu.edu/education/tutorials/java/ac/index.htmlhttp://www.magnet.fsu.edu/education/tutorials/java/ac/index.html


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Terms you are likely to encounter

http://upload.wikimedia.org/wikipedia/commons/4/49/Types_of_current_by_Zureks.svghttp://upload.wikimedia.org/wikipedia/commons/4/49/Types_of_current_by_Zureks.svg


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Alternating Current AC) Electricity

by Ron Kurtus (revised 2 June 2009)

Alternating current (AC) electricity is the type ofelectricity commonly used in homes and businessesthroughout the world. While direct current (DC)

electricity flows in one direction through a wire, ACelectricity alternates its direction in a back-and-forthmotion. The direction alternates between 50 and 60times per second, depending on the electrical system of

the country.

http://www.school-for-champions.com/science/ac.htm

http://www.school-for-champions.com/science/ac.htmhttp://www.school-for-champions.com/science/ac.htm


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Alternating Current AC) Electricity

by Ron Kurtus (revised 2 June 2009)

AC electricity is created by an AC electric generator,which determines the frequency. What is special aboutAC electricity is that the voltage can be readily changed,

thus making it more suitable for long-distancetransmission than DC electricity. But also, AC canemploy capacitors and inductors in electronic circuitry,allowing for a wide range of applications.

http://www.school-for-champions.com/science/ac.htm

http://www.school-for-champions.com/science/ac.htmhttp://www.school-for-champions.com/science/ac.htm


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24.1 Characteristics of a.c.

Current flow inone direction.

An alternating current (a.c.) is anelectric current that periodicallyreverses its direction in thecircuit, with a frequency findependent of the constants ofthe circuit.

http://www.toptenz.net/top-10-format-wars.php

timetime

currentcurrent

http://www.toptenz.net/top-10-format-wars.phphttp://www.toptenz.net/top-10-format-wars.php


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Terms a) Peak or maximumcurrent (Io, or Im) is themaximum current or

amplitude of current.b) One cycle is one

alternation.

c) Frequency (f) is the

number of cyclesoccurring per second.

d) Period (T) is the time forone complete cycle.

T = 1/f

Equations of graph?

amplitude

one cycle


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Equation of sinusoidal alternating current

a) The instantaneous current is the current at any time t is given by,

i = I0 sin ( t + )

where I0 = peak or maximum current

t = time

= 2f = angular frequency

= phase

f = frequency of a.c.


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A.C. flowing in a resistor,

The voltage from the sourceis

v = Vo sin t

Current though the load,

Where I = Vo /R

The current and voltage arein phase.

t I R

vi

o sin


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Power

The instantaneous power developed

across the resistor is

P = i2R

= P0 sin 2t

where P0 = Io²R

=peak or maximum power

http://www.phys.unsw.edu.au/PHYS1111/acc/acc.html

http://www.ngsir.netfirms.com/englishhtm/Rms.htm

http://www.phys.unsw.edu.au/PHYS1111/acc/acc.htmlhttp://www.ngsir.netfirms.com/englishhtm/Rms.htmhttp://www.ngsir.netfirms.com/englishhtm/Rms.htmhttp://www.phys.unsw.edu.au/PHYS1111/acc/acc.html


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Mean power

The mean power is equal

to half the maximum power.

2

2

.sin

2

2

0

2

o

o

T

o

P

RI

dt t I T R P

Io²

Io²/2

The equation i2 = Io2 sin2 t is

also a sinusoidal curve with a

mean value Io2/2.


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Review: Mean value


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Exercise 24.2

Determine the mean of the curvesshown.

Soln

a) = 0

b)

= Po/2

0

Po

Power

Po/2

a)

b)


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Exercise 24.3

Sketch a graph of an alternatingcurrent of amplitude 20 A, 50 Hz anda phase of /2 radian. What is theequation of the graph?

time/ms

current

5 10 15 20 25


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Pulse generator[oscillator]

Provides alternatingcurrent of different

frequency andamplitude.


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SIGNAL GENERATOR

Other names: Oscillator,

low voltage a.c., pulse generator

Provides different waveforms in theform of a.c.

Allow different frequencies andamplitudes of sinusoidal wave forms tobe selected.

Variation in voltage.

3/28/2016 B. H. KHOO 16

symbol


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Root mean square current.

= Irms²R

where

Irms = root mean square current

For d.c., P = I²R

The root mean square [rms] currentis the a.c. that has the same heatingeffect in a given resistor as a direct

current (d.c.)

R I

P

Power Mean

o

2

2

2

orms

I

I


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Mathematics corner [for info]

cos 2 = cos2 - sin2

= 1 - 2 s i n2

T as

I

t t

T

I

dt t T

I

dt t I T

I

o

T o

T o

T

o

2

2

]2

2sin[

2

)2cos1(2

.sin1

2

0

2

0

2

2

0

22


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Exercise 24.4

For the alternating voltageshown, determine,

a) peak voltage,

b) frequency,c) root mean square

voltage,

d) state the equation ofthe sinusoidal voltage.


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Exercise 24.5

An alternating current of

i = [5.0/A] sin [100 t/s]

passes through a load of resistance 20 . The current is measured in

ampere and the time in second. Calculate,a) the peak current,

b) the root mean square current,

c) the frequency of the current,

d) the period,

5.0 A

t I io

sin

2

o

rms

I I

A I rms

54.32

5

= 100, =2f f = 15.9 Hz

ms

T

8.62

100

22


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Exercise 24.5

An alternating current of

i = 5.0 sin [100 t/s]

Calculate,

e) the instantaneous current at 70ms,

f) the maximum power dissipatedg) the mean power dissipated,

(Ans. a) 5.0 A, b) 3.54 A , c) 15.9 Hz,d) 62.8 ms e) 3.2 A, f) 500 Wg) 250 W)

e) I = 5 sin 100(0.07)

= 3.3 A

f) Po = Io²R = 5²20

=500W

g)

= 500/2 = 250 W


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Exercise 24.5a) Sketch the graph of i versus time.

b) Sketch the graph of theinstantaneous power versus time.

c) Sketch the graph of double the

peak current but half thefrequency.

time/ms

5

-5

06331


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How to calculate root mean square current.Start from the back of the phrase.

a) Find the sum of

square of thequantity,

b) Find the mean of a),and

c) Take the square rootof b),

22

2

2

1

1

2 .....) N

N

i

i I I I I a

2

2

2

)

)

I I c

N

I I b

rms


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Exercise 24.6Find a) the mean voltage, andb) the mean squarevoltage and c) root meansquare voltage.

(Ans. a) 3.33 V b) 33.3 V2, c) 5.77 V)

Soln.

a) 3 = 10[1]

= 3.33 V

b) 3= 100[1]

= 33.3 V2

c) Vrms = 33.3 = 5.77 V

2

10

6 84

voltage/V

t/s


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Self-test 24.1

1) What is analternating current?

2) What is the peakvoltage?

1) An alternating current (a.c.)is an electric current that

periodically reverses itsdirection in the circuit, witha frequency f independentof the constants of thecircuit.

2) Peak or maximum voltage isthe maximum voltage oramplitude of voltage.


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Self-test 24.1

3) Distinguish between thepeak and root meansquare voltage.

3) Peak voltage is the maximum voltage whileroot mean square voltage is the alternatingvoltage that has the same heating effect in a

given resistor as a direct voltage. The peak voltage is larger than rms voltage.


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Self-test 24.14) What is meant by

a) frequency,

b) the root mean squarecurrent of an a.c.?

4a) Frequency (f) is thenumber of cyclesoccurring per second.

b) The root mean square

[rms] current is the a.c.that has the sameheating effect in a givenresistor as a direct

current (d.c.)


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Self-test 24.15) For an alternating voltage,v=[20/V]sin [200t/ms]

Determine

a) the peak voltage,

b) the rms voltage,c) frequency of the alternating

voltage.

5a) 20 V

b) 20/ 2 =14.1 V

c) 2 f = 200 /10¯³

f = 100 kHz


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PYP 24.1

The magnetic flux density B of the field

due to a long straight wire is given by

An overhead power cable carries analternating current of 2000 A r.m.s. Atwhat distance would the peakmagnetic flux density due to thecurrent in the cable be 100 T?

[Ans.: 5.7 m]

Soln.

peak current Io

= 20002

= 2828 A

d = 5.7 m

d

I B

o

2

d

x x

2

)2828(10410100

76

d

I

B o

2


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PYP 24.2

Ans A

P = ½ Po =

Is independent of frequency R

V o

2

2


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24.2 Transformer

A transformer changes i.e. transforms an alternating p.d from one valueto another of greater (step-up) or smaller value (step-down) usingthe mutual induction principle.

http://www.tucsontransformer.com/img/general-electric-unit-substation-transformer-44-1012-1.htmhttp://www.tucsontransformer.com/img/general-electric-unit-substation-transformer-44-1012-1.htm


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Power Transformer

2.5 MVA General Electric Unit Substation Transformer

http://www.tucsontransformer.com/img/general-electric-unit-substation-transformer-44-1012-1.htmhttp://www.tucsontransformer.com/img/general-electric-unit-substation-transformer-44-1012-1.htmhttp://www.tucsontransformer.com/img/general-electric-unit-substation-transformer-44-1012-1.htmhttp://www.tucsontransformer.com/img/general-electric-unit-substation-transformer-44-1012-1.htm


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500 MVA Single-phaseautotransformers


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Substation Equipment:Power Transformers

http://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation_equipment/power_transformers.html

http://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation.html#Equipmenthttp://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation_equipment/power_transformers.htmlhttp://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation_equipment/power_transformers.htmlhttp://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation_equipment/power_transformers.htmlhttp://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation_equipment/power_transformers.htmlhttp://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation.html#Equipment


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ElectricityFlow on the

Farm

http://www.wisconsinpublicservice.com/business/farm_voltage_electricity.aspx

http://www.wisconsinpublicservice.com/business/farm_voltage_electricity.aspxhttp://www.wisconsinpublicservice.com/business/farm_voltage_electricity.aspx


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Transformer (electrical appliants)

Used in laboratory power supply.


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Description

a) A simple transformer

consists of two coils, theprimary and the secondarycoils wound over a coremade of magnetically softmaterial.

b) There is no electricalconnection between the

primary and secondarycoils, but the soft iron core

provides a magnetic link between them.

http://www.electricityforum.com/products/trans-s.htm

http://www.electricityforum.com/products/trans-s.htmhttp://www.electricityforum.com/products/trans-s.htm


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Description

a) An alternating voltage applied to the primary coil produces an a.c.through it, which produces an alternating magnetic flux in the corethreading the secondary coil.

b) An alternating voltage is induced in the secondary coil.

c) Frequency of secondary voltage is the same as the primary voltage.


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Transformer

Transformers are designed so that all the magnetic fluxproduced by the primary coil passes through thesecondary. The primary coil is connected to an a.c.source.

http://www.daviddarling.info/encyclopedia/E/electromagnetic_induction.html

CROa.c.source

primary

coil secondarycoil


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http://www.tpub.com/content/doe/h1011v4/css/h1011v4_48.htm

http://www.tpub.com/content/doe/h1011v4/css/h1011v4_48.htmhttp://www.tpub.com/content/doe/h1011v4/css/h1011v4_48.htm


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Principle [how it works]

When current rises in theprimary coil, the magneticfield through the secondarycoil due to this currentincreases.

The changing flux through thesecondary causes e.m.f to beinduced in the secondary coil

As the current reversesdirection, the emf in thesecondary reversesdirection.

the frequency of thesecondary is the same asthe primary.

http://www.daviddarling.info/encyclopedia/E/electromagnetic_induction.html


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Main causes of

energy loses

a) R esistance of coil. Power dissipated in theresistor is i2R where R is the resistance of theresistor and i the current passing through the

coil. This is reduced by using low resistancethick copper wire.

b) Eddy current. The changing flux in the core willcause and induced current called eddy currentto flow. Laminating the core reduces theenergy losses due to eddy current.


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Main causes of energy loses [ ]

c) Flux leakage occurs when the changing flux fromthe primary threads the secondary. Efficient coredesign to ensure that all the primary flux is linkedwith the secondary.

d) Hysteresis loss. Magnetization of the core isrepeatedly changed from one direction another and back again. This requires energy and causes thecore to get hot. This is reduced by using softmagnetic material for the core.

Commercial transformer has an efficiency of 95% to99%. An ideal transformer has an efficiency of 100%.


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eddy current

changing field

induced changing

field


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Laminated iron core

magnetic field

produced by thecurrent.

induced

current

[eddycurrent]

induced

changing

field

coilSolid iron

core

I


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Function of soft iron core :a) concentrates the magnetic flux,

b) laminated to reduce eddy current losses.

Lamination . The core of a transformer is formed of a piles of thin iron or steel stampings (thin sheets) called lamination. These are oxidizedon the surface or lightly varnished to increase the electricalresistance from one to another.

F f


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For a transformer,

where

v s = secondary voltage

v p= primary voltage

Ns = number of turns in secondary

Np = number of turns in primary.

#vs and v p must both be either peak voltage or both rms.

*For Ns> N p, it’s a step uptransformer and

if N p> Ns it’s a step downtransformer.

p

s

p

s

N N

vv


48/75

For an ideal transformer Power output = power input

where i p = primary current

i s

= secondary current

For ideal transformer there is notenergy lost, the input energy iscompletely transformed into theoutput energy.

s

p

p

s

i

i

v

v

p P s s

vivi


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24.3 Transmission of electrical energy

Transformer play anessential part in thetransmission of electricity.

Power plants are usually placedsome distance from towns.

Electricity needs to be

transmitted over long distance.There is always power loss intransmission lines due to theirresistance (I²R).

pylon

T i i f l i l

http://www.t2.unh.edu/spring99/Image13.gif


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Transmission of electrical energy

http://www.t2.unh.edu/spring99/pg4.html

pylon

http://www.t2.unh.edu/spring99/Image13.gifhttp://www.t2.unh.edu/spring99/pg4.htmlhttp://www.t2.unh.edu/spring99/Image13.gifhttp://www.t2.unh.edu/spring99/pg4.htmlhttp://www.t2.unh.edu/spring99/Image13.gifhttp://www.t2.unh.edu/spring99/Image13.gif


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TransmissionIn Britain a network of cables, called the national grid, links all the powerstations. It allows the demand for electricity to be shared out betweenthe power stations. Most of the cables in the grid system are carriedoverhead on pylons. Underground cables are more expensive anddifficult to maintain. They are used in cities and where the scenery must

not be spoilt.


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Advantages of using electrical energy

1. Electrical energy is the easiest form of energy to transmit, and distributed by cables.

2. For many modern appliances, electrical energy is the only form of energythat can be used.

3. Electrical energy can be converted efficiently into any one of the other forms of energy.


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How electrical energy is transmitted?

Electricity is sent over long distance using cables. Transmission is doneusing alternating current at high voltages to reduces energy losses incables.

1. The voltage is step up to high voltage before transmitted frompower station.

2. This ensures that the current flowing in the cables is small andthe rate of power dissipated in cables are minimum (I²R).


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How electrical energy is transmitted?

3. Through the national grid, the voltage is lowered in stages atreceiving substations depending on the need of the customer.

4. The national grid is made up of close network of cables that joinreceiving substations

E l 24 7


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Example 24.7

The output power P and output

voltage V from a power station isconnected to a factory by cablesof total resistance R. Calculate

a) the current flowing in the

circuit,b) the power dissipated incables,

c) the power input to thefactory.

factory

R

power station

~

P, V


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Advantages of a.c. in transmission

1. Direct current are less easy to generate thanalternating currents.

2. Alternating e.m.fs are more convenient to step upand to step down.

3. Alternating current is just as suitable for heatingas direct current. The heating effect does not depend ondirection of current. e.g. (a) lighting: filament lampsdepend on the heating effect, gas discharge lamps runas well on alternating current. (b) small motors invacuum cleaners can use a.c.


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Advantages of a.c. in transmission [2]

4. Transmission using alternating current is more efficient than

d.c. transmission. This is because high voltage transmission ismore efficient than low voltage transmission.

5. In high voltage and low current transmission of electricalpower, low currents require thinner and therefore cheaper

cables.

Disadvantage.

For use of high voltage the high cost of the substation insulationneeded. Cost of transmitting a.c. is lower than direct current.


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Practical transmission system

The energy loss in the cables can be reduced in twobasic ways:

(a) By reducing the resistance of the cables.

(b) By reducing the current flowing.

Large reductions in the resistance of the cables canonly be brought about by making the cables verythick. This is not practical for several reasons.


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Practical transmission system [2]

1. Thicker cables are more expensive as more materials are required,expensive to manufacture and installed.

2. Thicker cables may not be slung from pylons.

3. It is more difficult and costly to insulate high voltage cables than to be laid underground.


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PYP 24.4

Ans: C


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24.4 Rectification1. Rectification is the conversion ofalternating current (a.c.) to directcurrent (d.c.)

2. A rectifier is a conductor which islargely unidirectional.

3. Ideal rectifier or diode.

a) Must have a zero resistance whenthe current flow in one directionand

b) Must have an infinite resistancewhen the current flows inopposite direction.

current

voltage


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Forward biased

• Direction of conventional current isthe same as direction of arrow of diode.

• Rectifier conducts and has a zeroresistance.

•A real diode has low resistance

http://www.gadgetjq.com/tach_install.htm

d b d

http://www.gadgetjq.com/tach_install.htmhttp://www.gadgetjq.com/tach_install.htm


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Reversed biased

Direction of conventionalcurrent is opposite to that of thearrow of the diode.

The diode is non-conducting,and has an infinite resistance.

A real diode has a highresistance and negligible current

flows.


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Why do we need to rectify a.c.?

D.C is required for

a) battery charging

b) operating of CRO

c) operation of GM tube.

d) operation of X-ray tube

e) operation of radio receivers and transmitters.

H lf W R tifi ti


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Half Wave Rectification

During the first half cycle when P ispositive, the diode is.......................biased and is ................................

During the second half cycle when Q isnegative, the diode is ........................biased, and is ..................................

http://www.antonine-education.co.uk/physics_a2/options/module_9/Topic_3/topic_3.htm

forward

conducting

reversed

non-conducting

P

Q

http://www.antonine-education.co.uk/physics_a2/options/module_9/Topic_3/topic_3.htmhttp://www.antonine-education.co.uk/physics_a2/options/module_9/Topic_3/topic_3.htm


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Half Wave Rectification [2]

The output is a pulsatingunidirectional (direct) current.The rectifier conducts only during.................. the cycle.

The disadvantage is that only.................. cycle contributes tothe rectification.

This is adequate for a crude circuit, forexample the low voltage fan motorfor a hair dryer.

half

half


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Full wave Rectification

In the case above we see that both forward and reverse half cyclesare rectified.

· Two half-wave rectifiers are placed back to back.

· The load is connected to a centre tapping of the transformer.

· This is called a centre-tap full-wave rectifier.

· It always needs a transformer with a centre tap.


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BridgeRectifier The arrows show the forward andreverse half cycles:

www.antonine-education.co.uk/.../TOPIC_3.HTM

http://www.eleinmec.com/article.asp?18

X

Y

Ho it orks?

http://www.antonine-education.co.uk/.../TOPIC_3.HTMhttp://www.antonine-education.co.uk/.../TOPIC_3.HTM


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How it works?

During the first half cycle, when terminal X of the supplyis positive, diodes ........ and ........ are conducting, anddiodes .......... and ............. are reversed biased.

During the second half cycle, when terminal Y is positive,

diodes .......... and ......... are conducting, and diodes.......... and .......... are reversed biased.

In both half cycle. the current through the load are in the................. direction.

B D

A C

B A C

D

same

How it works?


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How it works?

Power utilised is ...................... that achieved with halfwave rectification.

The output is .............................. with an average voltageof

= 2/3 Vo

where Vo is the peak voltage.

doubled

doubled


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Alternative diagram

http://ocw.weber.edu/automotive-technology/ausv-1320-automotive-electronics/12-diodes/rectification

Smoothing


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SmoothingThe pulsating output produced by

both half-wave and full-wave

rectifiers can be made moresteady (smooth) by puttingsuitable capacitor in parallelwith the load.

http://en.wikipedia.org/wiki/Bridge_rectifier

http://en.wikipedia.org/wiki/File:Diode_bridge_smoothing.svghttp://en.wikipedia.org/wiki/File:Diode_bridge_smoothing.svghttp://en.wikipedia.org/wiki/Bridge_rectifierhttp://en.wikipedia.org/wiki/Bridge_rectifierhttp://en.wikipedia.org/wiki/File:Diode_bridge_smoothing.svghttp://en.wikipedia.org/wiki/File:Diode_bridge_smoothing.svg


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Smoothing [2]When the current flows through the load in bothhalf cycles the capacitor charges, and when thevoltage across the load decreases the capacitor

discharges.If the time constant CR is large the capacitorrecharges before it has completely discharges i.e.use a capacitor with large capacity.

The output is ripple voltage at twice the inputfrequency.


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Summary1. Alternating currents can be rectified using

diodes;

2. A single diode will carry out half wave

rectification;3. Two diodes connected to a centre tapped

transformer well carry out full wave rectification;

4. Four diodes in a bridge circuit form a bridge

rectifier.5. Capacitors are used to smooth rectified AC.

ch 14 - alternating currentf

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