ee70 review. electrical current circuit elements an electrical circuit consists of circuit elements...

EE70 Review

Electrical Current

t

t

tqdttitq

dt

tdqti

0

)()()(

)()(

0

Circuit Elements

An electrical circuit consists of circuit elements such as voltage sources, resistances, inductances and capacitances that are connected in closed paths by conductors

Reference Directions

The voltage vab has a reference polarity that is positive at point a and negative at point b.

“downhill: resistor”“uphill: battery”


Energy is transferred when charge flows through an element having a voltage across it.

2

1

)(

)()()(t

t

dttpw

titvtp Watts

Joules

Power and Energy

• If current flows in the passive configuration the power is given by p = vi

• If the current flows opposite to the passive configuration, the power is given by p = -vi

Current is flowing in the passive

configuration

Reference Direction

Dependent Sources

Resistors and Ohm’s Law

Riv

iRv

abab

a

b

The units of resistance are Volts/Amp which are called “ohms”. The symbol for ohms is omega:

A

LR

is the resistivity of the material used to fabricate the resistor. The units of resitivity are ohm-

meters (-m)

Resistance Related to Physical Parameters

Power dissipation in a resistor

R

vRivip

22

Kircohoff’s Current Law (KCL)

• The net current entering a node is zero

• Alternatively, the sum of the currents entering a node equals the sum of the currents leaving a node

Kircohoff’s Current Law (KCL)

Series Connection

cba iii

Kircohoff’s Voltage Law (KVL)

The algebraic sum of the voltages equals zero for any closed path (loop) in an electrical circuit.

Kircohoff’s Voltage Law (KVL)

KVL through A and B: -va+vb = 0 va = vb

KVL through A and C: -va - vc = 0 va = -vc

Parallel Connection

Equivalent Series Resistance

eqiRRRRiiRiRiRvvvv )( 321321321

eqR

v

RRRv

R

v

R

v

R

viiii

321321321

111

Equivalent Parallel Resistance

Circuit Analysis using Series/Parallel Equivalents

1. Begin by locating a combination of resistances that are in series or parallel. Often the place to start is farthest from the source.

2. Redraw the circuit with the equivalent resistance for the combination found in step 1.

total321

111 v

RRR

RiRv

total321

222 v

RRR

RiRv

Of the total voltage, the fraction that appears across a given resistance in a series circuit is the ratio of the given resistance to the total series resistance.

Voltage Division

total21

2

11 i

RR

R

R

vi

total21

1

22 i

RR

R

R

vi

For two resistances in parallel, the fraction of the total current flowing in a resistance is the ratio of the other resistance to the sum of the two resistances.

Current Division

Node Voltage Analysis

svv 1

03

32

4

2

2

12

R

vv

R

v

R

vv

03

23

5

3

1

13

R

vv

R

v

R

vv

Node Voltage Analysis

Mesh Current Analysis

Thévenin Equivalent Circuits

oct vV


t

tsc R

Vi


sc

oc

i

vRt


Norton Equivalent Circuits

scn iI

Norton Equivalent Circuits

Source Transformations

tLL

LL

Lt

tLLL

Lt

tL RR

dR

dPR

RR

VRiP

RR

VI

02

2

Maximum Power Transfer

The superposition principle states that the total response is the sum of the responses to each of the independent sources acting individually. In equation form, this is

nT rrrr 21

Superposition Principle


VVvRR

Rv s 515

105

5

21

21

Current source open circuit


Voltage source short circuit

VVVvvv

VAARR

RRiRiv

T

seqs

66.1166.65

66.6)33.3)(2(510

)5)(10()2(

21

21

212

Req

The input resistance Ri is the equivalent resistance see when looking into the input terminals of the amplifier. Ro is the output resistance. Avoc is the open circuit voltage gain.

Voltage-Amplifier Model

Ideally, an amplifier produces an output signal with identical waveshape as the input signal, but with a larger amplitude.

tvAtv ivo

Voltage Gain

i

oi i

iA

L

iv

ii

Lo

i

oi R

RA

Rv

Rv

i

iA

Current Gain

i

o

P

PG

L

iviv

ii

oo

i

o

R

RAAA

IV

IV

P

PG 2

Power Gain

Operational Amplifier

Operational amplifiers are almost always used with negative feedback, in which part of the output signal is returned to the input in opposition to the source signal.

Summing Point Constraint

In a negative feedback system, the ideal op-amp output voltage attains the value needed to force the differential input voltage and input current to zero. We call this fact the summing-point constraint.


1. Verify that negative feedback is present.

2. Assume that the differential input voltage and the input current of the op amp are forced to zero. (This is the summing-point constraint.)

3. Apply standard circuit-analysis principles, such as Kirchhoff’s laws and Ohm’s law, to solve for the quantities of interest.


The Basic Inverter

Applying the Summing Point Constraint

1

2

in R

R

v

vA o

v

Inverting Amplifier

Summing Amplifier

1

2

in

1R

R

v

vA o

v

Non-inverting Amplifiers

10

111

2

R

R

v

vA o

vin

Voltage Follower

0

0

tqdttitqt

t

0

0

1tvdtti

Ctv

t

t

C

qv

v

qC

Capacitance

Capacitances in Parallel

Capacitances in Series

321

1111

CCCCeq

Inductance

The polarity of the voltage is such as to oppose the change in current (Lenz’s law).

Inductance

Series Inductances

321 LLLLeq

Parallel Inductances

321

11

LLLLeq

Mutual InductanceFields are aiding Fields are opposing

Magnetic flux produced by one coil links the other coil

Discharge of a Capacitance through a Resistance

0

R

tv

dt

tdvC CC

0 tvdt

tdvRC C

C

KCL at the top node of the circuit:

dt

dvC

dt

dqiCvq

v

qC C

iC iR

R

tvi CR

)(


0 tvdt

tdvRC C

C

stC Ketv

0 stst KeRCKse

tvRCdt

tdvC

C 1

We need a function vC(t) that has the same form as it’s derivative.

Substituting this in for vc(t)

RCs

1

RCtC Ketv

RCtiC eVtv iC Vv 0


Solving for s:

Substituting into vc(t):

Initial Condition: Full Solution:


RCtiC eVtv iC Vv 0

RCtC Ketv

To find the unknown constant K, we need to use the boundary conditions at t=0. At t=0 the capacitor is initially charged to a voltage Vi and then discharges through the resistor.

Charging a Capacitance from a DC Source through a Resistance


KCL at the node that joins the resistor and the capacitor

Current into the capacitor:

dt

dvC c

Current through the resistor:

R

Vtv SC )(

0)()(

R

Vtv

dt

tdvC SCC


0)()(

R

Vtv

dt

tdvC SCC

Rearranging:

SCC Vtvdt

tdvRC )(

)(

This is a linear first-order differential equation with contant coefficients.


The boundary conditions are given by the fact that the voltage across the capacitance cannot change instantaneously:

0)0()0( CC vv


stC eKKtv 21)(

SCC Vtvdt

tdvRC )(

)(

Try the solution:

Substituting into the differential equation:

Gives:

Sst VKeKRCs 12)1(


Sst VKeKRCs 12)1(

For equality, the coefficient of est must be zero:

RCsRCs

101

Which gives K1=VS


RCtS

stC eKVeKKtv /

221)(

sSSC VKKVeKVv 220

2 0)0(

Substituting in for K1 and s:

Evaluating at t=0 and remembering that vC(0+)=0

RCtSS

stC eVVeKKtv /

21)(

Substituting in for K2 gives:

tssC eVVtv


DC Steady State

dt

tdvCti C

C)(

)(

In steady state, the voltage is constant, so the current through the capacitor is zero, so it behaves as an open circuit.

DC Steady State

dt

tdiLtv L

L)(

)(

In steady state, the current is constant, so the voltage across and inductor is zero, so it behaves as a short circuit.

The steps in determining the forced response for RLC circuits with dc sources are:

1. Replace capacitances with open circuits.

2. Replace inductances with short circuits.

3. Solve the remaining circuit.

DC Steady State

RL Transient Analysis

S

S

Vdt

tdiLRti

tvRtiV

)(

)(

0)()(

SVdt

tdiLRti

)()( steKKtiTry 21)(

Sst VesLKRKRK )( 221

AV

R

VKVRK S

S 250

10011

R

LssLKRK

022

steKKtiTry 21)( SVdt

tdiLRti

)()(

RL Transient Analysis

)()()(

)()(

)(

)()()(

tftxdt

tdxR

tvti

dt

tdi

R

L

tvtRidt

tdiL

t

t

RC and RL Circuits with General Sources

First order differential equation with constant coefficients

Forcing function

RC and RL Circuits with General Sources

The general solution consists of two parts.

The particular solution (also called the forced response) is any expression that satisfies the equation.

In order to have a solution that satisfies the initial conditions, we must add the complementary solution to the particular solution.

)()()(

tftxdt

tdx

The homogeneous equation is obtained by setting the forcing function to zero.

The complementary solution (also called the natural response) is obtained by solving the homogeneous equation.

0)()(

txdt

tdx

Integrators produce output voltages that are proportional to the running time integral of the input voltages. In a running time integral, the upper limit of integration is t .

Integrators and Differentiators

dttvRC

tvt

o in

0

1

dt

dvRCtvo

in

Differentiator Circuit

dt

tdvti

Cdt

tdiR

dt

tidL s )(

)(1)()(

2

2

Differentiating with respect to time:

tvvdttiC

tRidt

tdiL sC

t

01

0

Second–Order Circuits

dt

tdv

Lti

LCdt

tdi

L

R

dt

tid s )(1)(

1)()(2

2

L

R

2

LC

10

Dampening coefficient

Undamped resonant frequency

dt

tdv

Ltf s )(1)(

)()()(

2)( 2

02

2

tftidt

tdi

dt

tid

Define:

Forcing function

Second–Order Circuits

02

2

202

2

202

2

tidt

tdi

dt

tid

solutionaryComplement

tftidt

tdi

dt

tid

solutionParticular

Solution of the Second-Order Equation

02

:

0)2(

:

02

:)(

20

2

20

2

20

2

ss

equationsticCharacteri

Kess

Factoring

KesKeKes

KetiTry

st

ststst

stC

02 202

2

tidt

tdi

dt

tid

Solution of the Complementary Equation

0

20

21 s

20

22 s

Dampening ratio

Roots of the characteristic equation:

Solution of the Complementary Equation

1. Overdamped case (ζ > 1). If ζ > 1 (or equivalently, if α > ω0), the roots of the characteristic equation are real and distinct. Then the complementary solution is:

tstsc eKeKti 21

21

In this case, we say that the circuit is overdamped.

2. Critically damped case (ζ = 1). If ζ = 1 (or equivalently, if α = ω0 ), the roots are real and equal. Then the complementary solution

is

tstsc teKeKti 11

21

In this case, we say that the circuit is critically damped.

3. Underdamped case (ζ < 1). Finally, if ζ < 1 (or equivalently, if α < ω0), the roots are complex. (By the term complex, we mean that the roots involve the square root of –1.) In other words, the roots are of the form:

nn jsjs 21 and

in which j is the square root of -1 and the natural frequency is given by:

220 n

For complex roots, the complementary solution is of the form:

teKteKti nt

nt

c sincos 21

In this case, we say that the circuit is underdamped.

Circuits with Parallel L and C

t

nL tiidttvL

tvRdt

tdvC

0

)()0()(1

)(1)(

We can replace the circuit with it’s Norton equivalent and then analyze the circuit by writing KCL at the top node:

dt

dvC

dt

dqiivdt

Li

R

Vi

t

cLLR 0

)0(1


dt

tdi

Ctv

LCdt

tdv

RCdt

tvd

dt

tditv

Ldt

tdv

Rdt

tvdC

atingdifferenti

tiidttvL

tvRdt

tdvC

n

n

t

nL

)(1)(

1)(1)(

)()(

1)(1)(

:

)()0()(1

)(1)(

2

2

0


)()()(

2)(

)(1)(

12

1

)(1)(

1)(1)(

20

2

0

2

tftvdt

tdv

dt

tvd

dt

tdi

CtffunctionForcing

LCfrequencyresonantUndamped

RCtcoefficienDampening

dt

tdi

Ctv

LCdt

tdv

RCdt

tvd

n

n


L

RcircuitSeries

RCcircuitParallel

tftvdt

tdv

dt

tvd

2

2

1

)()()(

2)( 2

0

2

This is the same equation as we found for the series LC circuit with the following changes for :

LL Lj IV

90 LLjZ L

LLL Z IV

Complex Impedances-Inductor

Complex Impedances-Inductor

CCC Z IV

90111

CCjC

jZC

RR RIV

Complex Impedances-Capacitor

Complex Impedances-Capacitor

RR RIV

Impedances-Resistor

Impedances-Resistor

0321 VVV

We can apply KVL directly to phasors. The sum of the phasor voltages equals zero for any closed path.

The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.

outin II

Kirchhoff’s Laws in Phasor Form

Power in AC Circuits

Z

VIwhereI

Z

V

Zm

mmm

0VI

For >0 the load is called “inductive” since Z=jL for an inductor

For <0 the load is “capacitive” since Z=-j/C for a capacitor

Load Impedance in the Complex Plane

Z

X

Z

R

jXRZZ

)sin(

)cos(

Power for a General Load

)cos(

)cos(

PF

IVP rmsrms

currentvoltage

Power angle:

If the phase angle for the voltage is not zero, we define the power angle :

AC Power Calculations

WIVP cosrmsrms

VARIVQ sinrmsrms

Average Power:

Reactive Power:

Apparent Power: VAIV RMSRMS

2rmsrms22 IVQP

Power Triangles

Average power

Reactive power

Average power

Apparent power

The Thevenin equivalent for an ac circuit consists of a phasor voltage source Vt in series with a complex impedance Zt

Thevenin Equivalent Circuits

The Thévenin voltage is equal to the open-circuit phasor voltage of the original circuit.

ocVV t

We can find the Thévenin impedance by zeroing the independent sources and determining the impedance looking into the circuit terminals.


The Thévenin impedance equals the open-circuit voltage divided by the short-circuit current.

scsc

oc

I

V

I

V ttZ


Norton Equivalent Circuit

The Norton equivalent for an ac circuit consists of a phasor current source In in parallel with a complex impedance Zt

scII n

The Thevenin equivalent of a two-terminal circuit delivering power to a load impedance.

Maximum Average Power Transfer

If the load can take on any complex value, maximum power transfer is attained for a load impedance equal to the complex conjugate of the Thévenin impedance.

If the load is required to be a pure resistance, maximum power transfer is attained for a load resistance equal to the magnitude of the Thévenin impedance.

Maximum Average Power Transfer

Transfer Functions

The transfer function H(f ) of the two-port filter is defined to be the ratio of the phasor output voltage to the phasor input voltage as a function of frequency:

in

out

V

VfH

First-Order Low Pass Filter

RCf

ffjfRCjfH

fCjRfCjfCj

fCjR

BBin

out

inout

in

2

1

)/(1

1

21

1)(

2

12

1

2

1

2

1

V

V

VIV

VI

Half power frequency

BB

B

f

fff

ffjfH

arctan1

01

1

1

2

21

1

BfffH

Bf

ffH arctan


For low frequency signals the magnitude of the transfer function is unity and the phase is 0. Low frequency signals are passed while high frequency signals are attenuated and phase shifted.


Magnitude Bode Plot for First-Order Low Pass Filter

1. A horizontal line at zero for f < fB /10.

2. A sloping line from zero phase at fB /10 to –90° at 10fB.

3. A horizontal line at –90° for f > 10fB.

First-Order High-Pass Filter

fRCfwhere

ffj

ffj

fRCj

fRCj

fRCj

fCjR

R

BB

B

in

out

inininout

2

1

)/(1

)/(

12

2

2

11

1

2

1

V

V

VVVV


BB

B

B

BB

B

B

B

ffff

fH

ff

fffH

ffff

ff

ffj

ffjfH

arctan90arctan

90

1

arctan1

90

1

2

2in

out

V

V

21 B

B

ff

fffH

BfffH arctan90


Bode Plots for the First-Order High-Pass Filter

0log10)log(20

)log(20

2

BBB

BB

fffffHffFor

fffHffFor

0

90

arctan90

0log10)log(20

)log(20

1log10)log(20

2

2

fHffFor

fHffFor

f

ffH

fffffHffFor

fffHffFor

fffffH

B

B

B

BBB

BB

BB

Series Resonance

LCf

LCf

CfLf

resonanceFor

fCjRfLjfZ s

2

1

)2(

1

2

12

:

2

12)(

0

22

00

0

For resonance the reactance of the inductor and the capacitor cancel:

Series Resonance

CRfQ

LCffrom

CfLSubstitute

R

LfResistance

resonance at inductance of ReactanceQ

s

s

0

020

2

0

2

1

2

1

)()2(

1

2

Quality factor QS

Series Resonant Band-Pass Filter

f

f

f

fjQ

f

f

f

fjQ

R

f

f

f

fjQ

R

fZ

ss

R

s

sR

s

s

s

s

0

0

0

0

0

0

1

1

1

1

/

)(

V

VVIV

VVI

ffffjQss

R

001

1

V

V

Series Resonant Band-Pass Filter

Parallel Resonance

fLjfCjRZ p 2121

1

At resonance ZP is purely resistive:

LC

fLfjCfj

2

1212 000

Parallel Resonance

CRfQ

LCffrom

CfLSubstitute

Lf

R

resonance at inductance of Reactance

ResistanceQ

P

P

0

020

2

0

2

2

1

)()2(

1

2

Quality factor QP

Parallel Resonance

f

f

f

fjQ

RZ

P

Pout0

0

1

IIV Vout for constant current,

varying the frequency

Second Order Low-Pass Filter

f

f

f

fjQ

ffjQ

LCfff

f

R

Lfj

fRC

j

fH

LCfff

f

R

Lfj

fRC

j

fC

jfLjR

fC

j

ZZZ

Z

S

S

in

out

inininCLR

Cout

0

0

0

00

0

00

0

1

)/(

2

121

2)(

2

121

2

22

2

V

V

VVVV

200

2

0

0012

002

0

00

0

1

1

90

1

ffffQ

ffQfH

ffffQTanffffQ

ffQ

ffffjQ

ffjQfH

S

s

sS

s

s

s

in

out

V

V


Second Order High-Pass Filter

At low frequency the capacitor is an open circuit

At high frequency the capacitor is a short and the inductor is open

Second Order Band-Pass Filter


At high frequency the inductor is an open circuit

Second Order Band-Reject Filter


At high frequency the inductor is an open circuit

First-Order Low-Pass Filter

ffB

Bi

f

ffi

f

i

f

ff

ff

f

ff

f

f

ff

i

f

i

o

CRf

ffjR

R

CfRjR

R

Z

ZfH

CfRj

RZ

R

CfRj

RfCj

RZ

Z

Z

V

VfH

2

1

)/(1

1

21

1)(

21

21

2

1111

)(

A low-pass filter with a dc gain of -Rf/Ri


iiB

B

B

i

f

ii

if

i

i

ii

if

ii

f

i

f

ffi

ii

i

f

i

o

CRf

ffj

ffj

R

R

CRfj

CRfj

R

R

CRfj

CRfj

CfjR

R

Z

ZfH

RZCfj

RZ

Z

Z

v

vfH

2

1

)/(1

)/(

21

2

21

2

2

1)(

2

1

)(

A high-pass filter with a high frequency gain of -Rf/Ri

N

BA

dA

AB

Magnetic flux passing through a surface area A:

For a constant magnetic flux density perpendicular to the surface:

The flux linking a coil with N turns:

Flux Linkages and Faraday’s Law

Faraday’s law of magnetic induction:

dt

de

The voltage induced in a coil whenever its flux linkages are changing. Changes occur from:

• Magnetic field changing in time

• Coil moving relative to magnetic field

Faraday’s Law

Lenz’s law states that the polarity of the induced voltage is such that the voltagewould produce a current (through an external resistance) that opposes the original change in flux linkages.

Lenz’s Law

Lenz’s Law

intensityfieldMagneticH HB

ytpermeabiliRelativer0

70 104

AmWb

Ampère’s Law: idlH

Magnetic Field Intensity and Ampère’s Law

The line integral of the magnetic field intensity around a closed path is equal to the sum of the currents flowing through the surface bounded by the path.

Ampère’s Law

Magnetic Field Intensity and Ampère’s Law

iHldlength lincrementa the as

direction samethe in points and magnitude

constant a has H field magnetic the If

productdotHdld

l

lH

)cos(

In many engineering applications, we need to compute the magnetic fields for structures that lack sufficient symmetry for straight-forward application of Ampère’s law. Then, we use an approximate method known as magnetic-circuitanalysis.

Magnetic Circuits

magnetomotive force (mmf) of an N-turn current-carrying coil

IN F

reluctance of a path for magnetic flux

A

R

RF

Analog: Voltage (emf)

Analog: Resistance

Analog: Ohm’s Law

R

INr

NI

r

R

r

R

A

l

rARl

2

21211

2

2

22

2

RF

F

R

Magnetic Circuit for Toroidal Coil

A Magnetic Circuit with Reluctances in Series and Parallel

a

aa

a

aa

cba

ab

cba

ba

totalc

ba

ctotal

AB

AB

RR

R

dividercurrentRR

R

R

Ni

RR

RR

)(

111

A Magnetic Circuit with Reluctances in Series and Parallel

1

11

1

111

i

iL

2

22

2

222

i

iL

2

12

2

21

1

21

1

12

iiiiM

Self inductance for coil 1

Self inductance for coil 2

Mutual inductance between coils 1 and 2:

Mutual Inductance

Mutual Inductance

21222

12111

Total fluxes linking the coils:

Currents entering the dotted terminals produce aiding fluxes

Mutual Inductance

Circuit Equations for Mutual Inductance

dt

diL

dt

diM

dt

de

dt

diM

dt

diL

dt

de

iLMi

MiiL

22

122

211

11

2212

2111

Ideal Transformers

tvN

Ntv 1

1

22

)()( 12

12 ti

N

Nti

tptp 12

Ideal Transformers

Mechanical Analog

12

121

2

12

11

221

1

22

)()( Fl

lFti

N

Nti

dl

ldtv

N

Ntv

d1

d2

LL ZN

NZ

2

2

1

1

1

I

V

Impedance Transformations

Semiconductor Diode

Shockley Equation

breakdown reverse reaching until-for TDsD VvIi

1exp

T

DsD nV

vIi mV

q

kTVT 26

TVfor exp

D

T

DsD v

nV

vIi

DDSS vRiV

Assume VSS and R are known. Find iD and vD

Load-Line Analysis of Diode Circuits

DDSS vRiV DDSS vRiV

Load-Line Analysis of Diode Circuits

The ideal diode acts as a shortcircuit for forward currentsand as an open circuit withreverse voltage applied.

iD > 0 vD < 0 diode is in the “on” state

vD < 0 ID = 0 diode is in the “off” state

Ideal Diode Model

Assumed States for Analysis of Ideal-Diode Circuits

1. Assume a state for each diode, either on (i.e., a short circuit) or off (i.e., an open circuit). For n diodes there are 2n possible combinations of diode states.

2. Analyze the circuit to determine the current through the diodes assumed to be on and the voltage across the diodes assumed to be off.

3. Check to see if the result is consistent with the assumed state for each diode. Current must flow in the forward direction for diodes assumed to be on. Furthermore, the voltage across the diodes assumed to be off must be positive at the cathode (i.e., reverse bias).

4. If the results are consistent with the assumed states, the analysis is finished. Otherwise, return to step 1 and choose a different combination of diode states.

Half-Wave Rectifier with Resistive Load

The diode is on during the positive half of the cycle. The diode is off during the negative half of the cycle.

CVVCTIQ rL The charge removed from the capacitor in one cycle:

Half-Wave Rectifier with Smoothing Capacitor

r

L

V

TIC

2

The capacitance required for a full-wave rectifier is given by:

Full-Wave Rectifier

Full-Wave Rectifier

Clipper Circuit

NPN Bipolar Junction Transistor

Bias Conditions for PN Junctions

The base emitter p-n junction of an npn transistor is normally forward biased

The base collector p-n junction of an npn transistor is normally reverse biased

Equations of Operation

1exp

T

BEESE V

vIi

BCE iii

From Kirchoff’s current law:


E

C

i

i

Define as the ratio of collector current to emitter current:

Values for range from 0.9 to 0.999 with 0.99 being typical. Since:

EBBEBCE iiiiiii 01.099.0

Most of the emitter current comes from the collector and very little (1%) from the base.


1B

C

i

i

Define as the ratio of collector current to base current:

Values for range from about 10 to 1,000 with a common value being 100.

BC ii

The collector current is an amplified version of the base current.

Only a small fraction of the emitter current flows into the base provided that the collector-base junction is reverse biased and the base-emitter junction is forward biased.

The base region is very thin

Common-Emitter Characteristics

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vvv

BCBECE

CEBEBC

0

vBC

vCE

Common-Emitter Input Characteristics

1exp)1(

T

BEESB V

vIi

Common-Emitter Output Characteristics

100 forii BC

Amplification by the BJT

A small change in vBE results in a large change in iB if the base emitter is forward biased. Provided vCE is more than a few tenth’s of a volt, this change in iB results in a larger change in iC since iC=iB.

Common-Emitter Amplifier

Load-Line Analysis of a Common Emitter Amplifier (Input Circuit)

tvtiRtvV BEBBBB in

CECCCC viRV

Load-Line Analysis of a Common Emitter Amplifier (Output Circuit)

As vin(t) goes positive, the load line moves upward and to the right, and the value of iB increases. This causes the operating point on the output to move upwards, decreasing vCE An increase in vin(t) results in a much larger decrease in vCE so that the common emitter amplifier is an inverting amplifier

Inverting Amplifier

Except for reversal of current directions and voltage polarities, the pnp BJT is almost identical to the npn BJT.

PNP Bipolar Junction Transistor

BCE

BC

EB

EC

iii

ii

ii

ii

)1(

PNP Bipolar Junction Transistor

NMOS Transistor

Operation in the Cutoff Region

toGSD Vvi for 0

By applying a positive bias between the Gate (G) and the body (B), electrons are attracted to the gate to form a conducting n-type channel between the source and drain. The positive charge on the gate and the negative charge in the channel form a capacitor where:

Operation Slightly Above Cut-Off

DStoGSox

DS vVvLt

Wi )(

For small values of vDS, iD is proportional to vDS. The device behaves as a resistance whose value depends on vGS.

Operation Slightly Above Cut-Off

22 DSDStoGSD vvvvCi

2

KP

L

WC

Operation in the Triode Region

Vtvtv inGS 4)()(

Load-Line Analysis of a Simple NMOS Circuit

tvtiRv DSDDDD


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Two-Input CMOS NAND Gate

Two-Input CMOS NOR Gate

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