Chapter 1 BJT Amplifier

8/14/2019 Chapter 1 BJT Amplifier

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Bipolar Junction Transistor (BJT)

and Linear Amplifier

Chapter 1BJT Amplifiers

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Outline

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IntroductionBJT Linear AmplifierGraphical Analysis & AC Equivalent CircuitSmall-signal hybrid- equivalent circuit

Hybrid- Equivalent Circuit and Early EffectExpanded Hybrid- Equivalent CircuitAC Load Line AnalysisMaximum Symmetrical Swing Common-emitter AmplifierCommon-collector AmplifierCommon-base Amplifier


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Introduction

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Analog Electronic Circuits Analog Electronic Circuits produce ananalog signals

Linear amplifier circuitMagnifies input signal & produce output

signal that is larger & directlyproportional to input signalElectronic

CircuitSignalInput

SignalOutput


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BJT Linear AmplifierBJT needs to be biased with DC voltage atquiescent point (Q-point) where BJT is biased inforward active region

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Linear Amplifier

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Time-varying output voltage is directlyproportional to & larger than time-varying inputvoltage


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Graphical Analysis & ACEquivalent Circuit

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A common -emitter circuit with time-varyingsignal source in series with the base DC source


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Graph plots collector current (i c) vs collector-emitter potential (v CE ).For different base currents (i B) different curves are obtained.

AC base current superimposed on I BQ - time-varying base current inducesac collector current superimposed on I CQ

AC collector-emitter voltage = output voltage - is larger than the input =amplificationLinear amplifier = ac added / superimposed on dc ==> superposition ==>only if ac is small ==>small signal analysis .

Line between VCC /R C and V CC = dc load line

Q-point is chosen wheredistance between i B curves aresimilar / even so thatamplification properties arelinear.


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Contd time-varying signals linearly related & superimposedon dc values)

If signal source, v s = 0:

(4)

(3) (2)

(1)

be BEQ BE

ceCEQCE

cCQC

b BQ B

vV v

vV vi I i

i I i

(6) loop)E-(C

(5) loop)E-(B

CEQC CQCC

BEQ B BQ BB

V R I V

V R I V

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Contd For B-E loop, considering time varying signals:

Rearrange:

Base on (5), left side of (7) is 0. So:

For C-E loop, considering time varying signals:

9Base on (6), left side of (11) is 0. So:

(7) )()( be BEQ Bb BQ BE B B s BB vV Ri I v RivV

(8) sbe Bb BEQ B BQ BB vv RiV R I V

(9) be Bb s v Riv

(11) (10)

ceccCEQC CQCC

ceCEQC cCQCE C C CC

v RiV R I V vV Ri I v RiV )()(

(12) 0 cecc v Ri9 2013/2014 EKT204


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I B versus V BE Characteristic

b BT

be BQ B i I V

v I i )1(

Time-varying signal source, v s appliedto base - time-varying base currentcomponent

==> there is a time-varying base-emitter component

Figure shows exponential relationship

between i B vs v BE

IF MAGNITUDES of time-varyingsignals superimposed on dc Q-pt aresmall => develop linear relationshipbetween ac v BE and ac i B

This relationship corresponds to slopeof curve at the Q-pt.

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Rules for AC Analysis

Replacing all capacitors by short circuits

Replacing all inductors by open circuits

Replacing dc voltage sources by ground connections

Replacing dc current sources by open circuits

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AC Equivalent Circuitfor Common Emitter

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EquationsInput loop

Output loop

beT

BQb

be Bb s

vV

I i

v Riv

bc

ceC c

ii

v Ri

00.026 V

The DC voltage sources have been set equal to zero or ground(V CC =0).Only ACc condition are to be considered.


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Small Signal AC Equivalent Circuit

ac input signal voltages and currents are in the orderof 10 percent of Q-point voltages and currents.

e.g. If dc current is 10 mA, the ac current (peak-to- peak) < 0.1 mA.

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Small-signal hybrid- equivalent circuit

BQ F CQ

CQ

T F

BQ

T

b

be

I I

I V

I V

r iv

,

g m =I CQ /V T

r = VT/I CQ

v be = i b r

r = diffusion

resistance /base-emitter inputresistance

1/r = slope of i B

V BE curve

14

Phasor signals are shown in parentheses.

r g mCommon-Emitter Current Gain, as constant;2013/2014 EKT204


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)( b

b

I

i

Using common-emitter current gain ( ) parameter

Small-signal hybrid- equivalent circuit

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How to construct Small-signal hybrid-

Place a terminal for the transistor

Common Terminal as ground

B

E

C

We know that

i across B ib

i across C i b

i across E (+1)i b

r between B -E

R C

R B

v s

v O

V BB

V CC

B

E

C

ibr

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C bemceo RV g V V s B

be V Rr r

V

BC m

s

ov Rr

r R g

V V

A

gain,voltagesignalSmall

Output signal voltage

Input signal voltage17

Small-signal Voltage Gain

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Problem-Solving Technique:BJT AC Analysis

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1. Analyze circuit with only dc sources to find Qpoint.

2. Replace each element in circuit with small-signal model, including the hybrid model forthe transistor.

3. Analyze the small-signal equivalent circuitafter setting dc source components to zero.


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Transformation of Elements

Element DC Model AC Model

Resistor R R

Capacitor Open C

Inductor Short L

Diode +V g, r f r d = V T /I D

Independent Constant

Voltage Source

+ V S - Short

Independent ConstantCurrent Source

IS Open

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Example 1

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Determine the small signalvoltage gain, including theeffect of the transistoroutput resistance, r o.

Assume the transistor &circuit parameter are =100,VCC=12V, V BE=0.7V,RC=6k, RB=50k andVBB=1.2V.


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21

Solution 1Do the dc analysis to find Q-point values

Determine the small signal output resistance ,

Do the ac analysis to find the small signal base-emitter input resistance,r , transconductance, g m and then voltage gain, AV

V k m R I V V

mA I I

A RV V

I

C CQCC CEQ

BQCQ

B

BE BB BQ

6)6)(1(12 then,

1)10(100thatso

10507.02.1

k m I

V r

CQ

Ao 50

1

50

4.11506.2

6.2)6(5.38)(

/5.38026.0

1

6.21

)026.0(100

m

Rr r

R g V V

A

V mAm

V

I g

k m I

V r

BC m

S

oV

T

CQm

CQ

T

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Hybrid- Equivalent Circuitand Early Effect

transconductanceparameter

current gainparameter

r o =V A/I CQ

r o = small-signaltransistor output

resistance V A = early voltage22 2013/2014 EKT204


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Early Voltage(V A )

Early Effect

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Example 2 If the Early voltage, V A is 50 V, reconsider thecircuit in example 1 and determine the small-signal voltage gain including the effect of thetransistor output resistance, r 0 .

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Solution 3

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C omo Rr V g V

r R R Ri 21 C oo Rr R

sS

V Rr R R

r R RV

21

21

C oS

m s

ov Rr Rr R R

r R R g

V V

A

21

21


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Expanded Hybrid- EquivalentCircuit

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h-Parameter Model for npn

fe

bie

h

r r r h

ooe

re

r r h

r

r h

11

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T-Model of an npn BJT

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4 Equivalent 2-port Networks

Voltage Amplifier

Current Amplifier

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Cont

Transconductance Amplifier

Transresistance Amplifier

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Small Signal Equivalent Circuit

If

The coupling capacitor is assumed to be a shortcircuit. (If signal source >>> 2kHz)

TH cC R

fC Z

2

1


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ExampleDetermine the small voltage gain for the circuit below if=100, VCC=5V, V BE=0.7Vand V A=100V


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SolutionDo the dc analysis to find Q-point values

V k m R I V V mA I I

Ak R

V V I

V V R R R

V

k k

k k R R

R R R

C CQCC CEQ

BQCQ

TH

BE TH BQ

CC TH

TH

31.6)6)(95.0(12 then,95.0)5.9(100thatso

5.99.5

7.0756.0

756.0)12(1003.6

9.5100

)3.6)(7.93(

21

2

21

21


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SolutionDetermine the small signal output resistance ,

Do the ac analysis to find the small signal base-emitter input resistance,r , transconductance, g m and then voltage gain, AV

k m I V

r CQ

Ao 3.10595.0100

1635.087.1

87.1)3.105//6(5.36

////

)//(

/5.36026.095.0

74.295.0

)026.0(100

k m

R Rr Rr

r R g V V

A

V mAm

V

I g

k m I

V r

S TH

TH oC m

S

oV

T

CQm

CQ

T


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The basic common-emitter circuit used inprevious analysis causes a serious defect :

If BJT with V BE =0.7 V is used, I B=9.5 A & I C=0.95 mA

But, if new BJT with V BE =0.6 V is used, I B=26 A & BJTgoes into saturation; which is not acceptable Previouscircuit is not practicalSo, the emitter resistor is included: Q-point is stabilizedagainst variations in , as will the voltage gain, A V

AssumptionsCC acts as a short circuitEarly voltage = ==> r o neglected due to open circuit

Basic Common-Emitter Amplifier

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Circuit with Emitter ResistorTo improve dc biasing designAV with R E less dependent on.VA infinite, r o open circuit.

C bo R I V )(tage,output volac


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Common-Emitter Amplifierwith Emitter Resistor

ac output voltage

Input voltage loop

Input resistance, R ib

Input resistance to amplifier, Ri

Voltage divider equation of V in to V s

Remember : Assume VA is infinite , r o is neglected

C bo R I V

E bbbin R I I r I V

E b

inib Rr I

V R 1

ibi R R R R 21

s

S i

iin V

R R

RV

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E

C

E

C V

E si

si

i

E

C

ib

in

s

C

s

C b

s

oV

s si

iin

ibi

E

b

inib

E bbbin

R R

R R

A

r R R R

R R R

Rr R

RV

V R

V R I

V V

A

V R R

RV

R R R R

Rr I

V R

R I I r I V

)1(

)1( and if

)1()(

////,resistanceinput

effectrulereflectionresistance)1(

)(

21


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Example

Determine the small voltage gain for the circuitwith an emitter resistor below if =100,VBE=0.7Vand V A=.


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Solution

0.5

)1( and if

53.4

06.8////

6.41)1(

/1.83

2.1

81.4;16.2

21

V

E si

s

oV

ibi

E b

inib

CQ

A

o

T

CQm

CQ

T

CEQCQ

A

r R R R

V V A

k R R R R

k Rr I V

R

I V

r

V mAV

I g

k I V

r

V V mA I


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Circuit with Emitter BypassCapacitor

Use to effectively to short out a portion @ all of emitterresistance by the ac signal .


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R S

R 1

R 2 R E

R C

vs

vO

C C

V CC

C E

B C

E

V o

V s R C

RS

r r oR 1||R 2

g m V

Emitter bypass capacitor, C E provides a s h o r t c i r c u i t toground for the ac signals

Common-Emitter Amplifierwith Emitter Bypass Capacitor

Small-signal hybrid-

equivalent circuit

Emitter bypass capacitor isused to short out a portion orall of emitter resistance bythe ac signal. Hence no REappear in the hybrid- equivalent circuit

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Common-emitter Amplifier withEmitter Bypass Capacitor

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AC Load Line - KVL on C-E loop

1

11

1

1- Slope

)(

Assuming

0

E C

E C c E cC cce

ec

E eceC c

R R

R Ri Ri Riv

ii

Riv Ri

49

Visualized the relationshipbetween small-signalresponse & transistorcharacteristics

Occurs when capacitorsadded in transistor circuit


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Example (DC & AC Load Line)

Determine the dc and ac load line. V BE=0.7V, =150, VA=

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Solution

To determine dc Q-point, KVL around B-E loop

k R R-

R I R I V V V

mA I I mA I I

A R R

V V I

R I V R I R I V R I V

E C

E EQC CQCEQ

BQ EQ BQCQ

E B

EB BQ

E BQ EB B BQ E E EB B BQ

1511 Slope

53.6)( point,-QFor

9.0)1( & 894.0Then

96.5)1(

)1(

52


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)//()//)((

/4.34

36.4

53.6;894.0

LC c LC meo

CQ

Ao

T

CQm

CQ

T

ECQCQ

R Ri R Rv g vv

I V

r

V mAV

I g

k I

V r

V V mA I

53


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Maximum Symmetrical Swing

When symmetrical sinusoidal signal applied to theinput of an amplifier, the output generated is alsoa symmetrical sinusoidal signal

AC load line is used to determine maximumoutput symmetrical swing

If output is out of limit, portion of the output signal willbe clipped & signal distortion will occur

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Maximum Symmetrical Swing

Steps to design a BJT amplifierWrite DC load line equation (relates of I CQ & V CEQ)Write AC load line equation (relates ic , vce ; vce = - icReq ,Req = effective ac resistance in C-E circuit)Generally, ic = I CQ I C (min), where I C (min) = 0 or someother specified min collector currentGenerally, vce = V CEQ V CE(min), where V CE(min) issome specified min C-E voltageCombination of the above equations produce optimumI CQ & V CEQ values to obtain maximum symmetricalswing in output signal

56

E l (M i S i l S i )


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Example (Maximum Symmetrical Swing )

Determine the maximum output symmetrical swing for the ac loadline in previous figure.

SolutionFrom the dc & ac load line, the maximum negative swing in the I c is from0.894 mA to zero (I CQ). So, the maximum possible peak-to-peak ac

collector current:

The max. symmetrical peak-to-peak output voltage:

Maximum instantaneous collector current:

mA79.1)894.0(2(min))(2 C CQc I I i

V56.2)2||5)(79.1()||(|||||| LC ceqcce R Ri Riv

mA79.1894.0894.0||21

cCQC i I i

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C-C Small Signal Voltage Gain

CommonCollector Circuit

Small SignalEquivalent

Circuit

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)//(

)1(

E ooo

bo

Rr I V

I I

si

i

E o

E o

s si

iin

E oib

ibb E obobin

R R R

Rr r Rr

V R R

RV

Rr r R R I Rr r I V r I V

)//)(1()//)(1(

VV

A

gainvoltagesignalSmall

)]//)(1([

)]//)(1([

s

ov

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C-C Input & Output Impedance

x

xo I

V R

o E s

o

oo E s x

x

m

s

x

o

x

E

x x

s

m x

s

x

o

x

E

xm x

x s

s

x

o

x

E

xm x

r R R R Rr

R

Rr R R R Rr V I

r g

R R Rr V

r V

RV

V R R Rr

r g I

R R Rr V

r V

RV

V g I

V R R Rr

r V

R R Rr V

r V

RV

V g I

////1

//// therefore,

111////

1

that Note

////////

//// therefore,

////

////

nodeoutputatcurrentsSumming

21

21

2121

21

21

21

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ExampleCalculate

the small signal voltage gain,the input and output resistance.

Assume the transistor & circuit parameter are; =100, VCC=5V,VBE=0.7V, V A=80V and r o=100k.

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C-C Small Signal Current GainCan be determine by using the input resistance & the concept ofcurrent dividers.

)1(

then,R r andR //R R thatassumeIf

////

)1(

Therefore,

////)1()1(

////

Eoib21

21

21

21

21

21

21

i

E o

o

ibi

ei

o E o

oe

iib

bo

iib

b

i

ei

A

Rr r

R R R R R

I I

A

I Rr

r I

I R R R

R R I I

I R R R

R R I

I I A

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Common Base Amplifier

63

C B S ll Sig l E i l t


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C-B Small Signal EquivalentCircuit

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C-B Small Signal Voltage Gain

zero]approachesR [as )//(

////11

)//(

////1

then,

since 111

E]nodeatEquation[KCL 0)(

)//)((

S LC mV

S E S

LC m s

o

V

S E S

s

mS

s

S E

S

s

E m

LC mo

R R g A

R Rr

R R R g V V

A

R Rr

RV

V

r g R

V

R Rr V

RV V

RV

r V

V g R RV g V

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C-B Small Signal Current Gain

infinity]approachesREandzeroapproachesR [as 1

1

//1

then,

)(

//1

E]nodeatEquation[KCL 0

L

r g A

Rr

R R R

g I I

A

R R RV g I

Rr

I V

RV

r V V g I

mi

E LC

C m

i

oi

LC

C mo

E i

E mi

66

d


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Input Impedance

ei

ie

mmbi

r r

I V

R

r V V g

r V

V g I I

1

resistanceInput

1

inputat theKCL

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Output Impedance

C o

S E m

R R

RV

RV

r V

V g

resistanceoutputThe

0Vgmeans0,VimpliesThis

0

emitter at theKCL

zero.toequalset beenhasv

m

s

68


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Summary & ComparisonConfiguration Common

EmitterCommonCollector

Common Base

Voltage Gain Av > 1 Av 1 Av > 1

Current Gain Ai > 1 Ai > 1 Ai 1

InputResistance

Moderate(k)

High(50-100k)

Low()

OutputResistance

Moderate toHigh

Low Moderate toHigh

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Chapter 1 BJT Amplifier