chapter 5 - ele232_feb11 [compatibility mode]

Chapter 5 : AC Analysis of BJT

Norsabrina SihabFaculty of Electrical Engineering,Universiti Teknologi MARAPulau PinangTel : 04-3823355Email : [email protected]

1

Chapter 5 – AC Analysis of BJT

Learning Outcome

2

Learning Outcome

At the end of this chapter, students able to:Analyze the small signal of single stage BJT amplifier circuitsCalculates the transistor parameters such as voltage gain Av, Ca cu ates t e t a s sto pa a ete s suc as o tage ga v,current gain Ai, input impedance Zi and output impedance Zo.

Electronics 1 Updated May 2011Norsabrina Sihab

Chapter 5 – AC Analysis of BJT3

IntroductionIntroduction

Increasing the power of an AC signal is called amplification.I th l l k it k i l bIncrease the power level - make it a weak signal become more stronger.Amplification circuit called amplifier eg. Bipolar transistorAmplifie p ope tiesAmplifier properties:

1. Voltage gain, AV

2. Current gain, Ai

3 I t i d Z3. Input impedance, Zi

4. Output impedance, Zo

Good Amplifier:h l1. High Voltage gain, AV

2. High Input impedance, Zi

3. Low Output impedance, Zo


4. High Bandwidth, BW


IntroductionIntroduction

In this chapter we learn how to analyze and work with a single lifistage amplifier.

To examine ac response of BJT Amplifiers by reviewing models frequently used to represent the transistor in ac domain.A lit d f th i t i l ill d t i h th tAmplitude of the input signal will determine whether to use small signal or large signal analysis techniques.



Transistor Equivalent Model

• A model is an equivalent circuit that represents the AC characteristics of the transistor.A model uses circuit elements that approximate the behavior of• A model uses circuit elements that approximate the behavior of the transistor.

• There are two models commonly used in small signal AC analysis of a transistor:of a transistor:

1. Hybrid -π equivalent model (will be more emphasized)2. re-model



1. Hybrid-π Model only this model being use in ELE232

The hybrid π model is most useful for analysis of high f t i t li tifrequency transistor applications.

At lower frequencies the hybrid π model closely approximate the re parameters, and can be replaced by ethem.

bivg = β

e

bm

mVrr

ivg

261+= )(ββ

π

π

Ee I

r =



2 d l2. re model

BJT b i ll t t ll d d i th f th d lBJTs are basically current-controlled devices, therefore the re-model uses a diode and a current source to duplicate the behavior of the transistor.

One disadvantage to this model is its sensitivity to the DC level ThisOne disadvantage to this model is its sensitivity to the DC level. This model is designed for specific circuit conditions.

The use of re model then becomes more desirable because an important parameter of the equivalent circuit is determined by theimportant parameter of the equivalent circuit is determined by the actual operating conditions but one must still turn to the data sheets for some of the other parameters of the equivalent circuit.

The re model also failed to include the feedback term, which in some e ,cases can be important.

The re model is really a reduced version of the hybrid-π model used extensively for high-frequency analysis.



2. re Transistor Model2. re Transistor Model

b cib

βre

e

βib rce=ro

+

Vbe

_e

Transistor ac equivalent circuit for common emitter

re

e

ie

c

rce=ro

+

Vbe

ie

b_

Transistor ac equivalent circuit for common base



1a) Common Emitter - by passed RE1a) Common Emitter by passed RE

1. Replace all capacitors with short circuit and set dc source to zero.

1R cR

ccVcircuit and set dc source to zero.

R2R1 Rc

iCoC

Simplified AC equivalent circuit

2R EReC

inv ov

rbb’

b cib

2. Substitute BJT with hybrid-π model.

rπ

e

gmvπ rce

+

Vπ

_

By passed RE


e

Hybrid-π transistor model


Common Emitter - by passed RE

3. Determine the BJT model parameters : Zi Z A and Ai

rbb’b c

+

ib

parameters : Zi, Zo, Av, and Airπ

e

gmvπ rce

+

Vπ

_

R1//R2 Rc

Simplified hybrid-π equivalent circuit of a CE amplifier.

Zi ZoZb

Input impedance

Amplifier with the hybrid-πequivalent circuit

bbbbbb

b

b

rrrriV )1()(Zwhere

Z||R||RZ

''

21i

++=+

==

=

βπ


ebbbb

b rrii

)1( Zwhere ' ++ β


Common Emitter - by passed RE

Output impedance:Voltage gain:

S t Vi t

b

bb

rrandiwhererr

)1(Vg)(i

)r||(RVgVVA

'b

ceCm

i

ov

+==+

−==

ββπ

πSet Vi to zero. Therefore Vπ=0. gmVπ=0 also.

C+

Io

ebbb

b

eb

rrii

rrandiwhere

)r||(R))1((

)r||(RA

)1(Vg

'

ceCv

m

++−

=∴

+

ββ

βββ ππ

rce Rc

+

Vo

-

ebb rr )1()r||(R

'

ceC

++−

=β

β

cer||RZ Co == o

IV

Zo

oI

ivi Z

ZAA =

Current gain without RL:

//RZiZ

vAiA =

Current gain with RL:


oZ L//RoZ


Exercise 1

For the circuit in figure below obtain the + 20Vthe i. Emitter current, IE and reii. Input impedance, Zi, and the

output impedance, Zoiii V lt i A d th t F10

F10μ82k 2k7

iii. Voltage gain, Av and the current gain, Ai

Given that the parameters of the ov

F10μ

1rT

120R33k

transistor is as follows. β=100, rbb’=10Ω, rce=∞Ω and VBE=0.7V. State any assumptions made.

inv

ov

F47μ

18k

120R

1k5

Answer : IE=1.79mA, re=14.53Ω, Zi=1.343kΩ, Zo=2.7kΩ, Av= -168.92, Ai= -90.75

Zi Zo


Chapter 5 – AC Analysis of BJT

Exercise

13

A basic common emitter BJT amplifier in below has a quiescent point of the collector current, ICQ = 2 mA. Answer

Exercise

the following :i.Draw the ac equivalent circuitii.Input impendance, Zi

Answer : 478.9Ωiii.Output impedance, Zo

Answer : 2.107kΩiv.Voltage gain, AV

Answer : -151.1v Current Gain A Answer : 34 35v.Current Gain, Ai

Answer : -34.35Use VT = 26mV, Ω=Ω== 100,50,120 bbrkcerβ

+20V

Vo

2.2 k671 IC

C CVi

1.54 k CE

Ci Co


1.54 k

Zi

Zo


1b) Common Emitter – un-bypassed RE1b) Common Emitter un bypassed RE

ccV

1R cR

C C

ovinv

1rT

R

iC oC

R

Ac equivalent circuit

invov2R

eC

LR


Hybrid-π equivalent circuit of the Common Emitter Amplifier.


Common Emitter – un-bypassed RECommon Emitter un bypassed RE

( )//// LRRrvgVVoltage gainInput impedance

( )

( ))1(Vmg

)1)1('(////

errandbiwhereERrbbrbiLRcRcervmg

iVoV

vA

+==+++

−==

βπβπ

βππ

E1' ))((////

iRrri

iVZ

ZRRZ

b

bbb

b

bb

bin

121

+++==

=

βπ

( )

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

////

LRcRcerERrbbrbi

LRcRcerbivA

−+++

−=

ββπ

βE1' )( Rrrbb 1+++= βπ

Output impedance ( )1)1(' ERrbbr

Lcce+++

=βπ

Output impedanceSet Vi to zero. Therefore Vπ=0. gmVπ=0 also.

cceo

oo Rr

ivz //==

//RZiZ

vAiA =

Current gain:


oL//RoZ


Exercise 2

For the circuit below, obtain the +20V

i. Emitter current, IE and re

ii. Input impedance, Zi, and the output impedance, Zo

iii Voltage gain A and the current

F10μ82k2k7


Given that the parameters of the

F10μ

1rT

120R 33kGiven that the parameters of the transistor is as follows. β=100, rbb’=10Ω, rce=∞Ω and VBE=0.7V. State any assumptions made

invov

F47μ

18k120R

1k5

33k

State any assumptions made.

Answer : IE=1.79mA, re=14.53Ω, Zi= 7.08 kΩ, Zo=2.7kΩ, Zi Zo


Av= -18.35 , Ai= -52.04


Exercise 3

For the circuit below obtain the voltage gain, Av, the current gain, Ai, the input impedance Z and the output impedance Z given that theinput impedance, Zi, and the output impedance, Zo, given that the parameters of the transistor is as follows. β=100, rbb’=75Ω, rce=30kΩ and VBE=0.7V. State any assumptions made.

Answer : re=10.28Ω, Av= -14.136, Zin=8.362kΩ, Zout=2.724kΩ, Ai= -43.346V20V20

3k3k82F10μ

R180k33

F10μ

1rT

invov

k18

R180

R820F47


F47μ


2) Common Base Amplifier2) Common Base Amplifier

rce

cegmvπ+

Vπ

_RE

Rcrπ

e cieib

b

Ac equivalent circuit of a common

rbb’

b’E

Ac equivalent circuit of a common base amplifier.

Hybrid-π equivalent circuit of the Common Base Amplifier

bZi ZoZe

the Common Base Amplifier.



Common Base AmplifierCommon Base Amplifier

Input Impedance Output impedance

)()(//

'' +=

+==

=

ππ bbbbbi

eEi

rrrriVZ

ZRZSet Vi to zero. Therefore Vπ=0. gmVπ=0 also.

Voltage gain

)()( 11 +=

+==

ββbee ii

Z

Voltage gain

cceo

o Rrivz //==

( ))(

//

' π

π

rriRrvg

VVA

bbb

ccem

i

ov +

==

1≈= iZAA

Current gain, Ai

oi

( ) ( ))(

//)(

//)1(

''

ππ

ββββ

rrRr

rriRcriA

rrandiwhere

bb

cce

bbb

cebv

eb

+=

+=

+==Vgm


1≈=Lo

vi //RZAA)()( '' ππ rrrri bbbbb ++


Exercise 4Exercise 4

For the circuit below, obtain the Voltage gain, Av, the input impedance, Ziand the output impedance Z given that the parameter of the transistor isand the output impedance, Zo, given that the parameter of the transistor is as follows. β=100,rb’b=10Ω,rce=∞ and VBE=0.7V. State any assumptions made.

Answer : re=10.28Ω, Av=257.14, Zin=10.27Ω, Zout=2.7kΩ, Ai=0.98

. F10μ

F100μ

F47μ

A common base Amplifier


A common base Amplifier.


3) Common Collector Amplifier3) Common Collector Amplifier

RER1 R2

ac equivalent circuit of a CC amplifier.


Hybrid-π equivalent circuit of a CC amplifier.


Common Collector AmplifierCommon Collector Amplifier

Voltage gainInput Impedance

ERbiVgbbrrbiiVerrandbiwhere

bZRRiZ

)()'()1(Vmg

//2//1

++++==

=βπβπ EReirbbrbi

ERei

iVoV

vA)'( ++

==π

iERbibibbrerbi

biERbiVmgbbrrbi

biiV

bZ

)()')1((

)()'(

++++=

+++==

ββ

ππ

ERERbirbbrbi

ERbi

)1()1()'(

)1(

+=

++++

=

ββπ

β

biERbbrerbi

bi))1(')1(( ++++

=ββ

ERrbbr )1()'( +++ βπ

ERbbrer )1(')1( ++++= ββ



Common Collector Amplifier

Vi=0

rbb’ rπb e

gmvπ RE

+ Vπ -

ib ie

R1//R2

+

Vo

Output impedanceSet Vi to zero.

rbb’ rπ eib i

c--

π−+−

==

=)'('

'//

eibbrrbi

xixV

oZ

oZERoZ

rbb rπ e

gmvπ RE

+ Vπ -

ib ie

+

Vo

βπ

ππ

++

=

++

=

)'(

)'(

bibibbrrbiVmgbibbrrbi

Zo=RE//Zo’c

--

rbb’ rπ eib ieZo’β

π

β

++

=

+

1'bbrrbibi

e

gmvπ

+ Vπ -

b e

VX

ix

zCurrent gain


Zo’c

LRzzAA

o

ivi //

=


Exercise 5Exercise 5

For the circuit in figure below obtain the V20the i. Emitter current, IE and re

ii. Input impedance, Zi, and the output impedance, Zo

7k2k82p p , o


iC

Given that the parameters of the transistor is as follows. β=100, rbb’=10Ω, rce=∞Ω and VBE=0.7V.

invovk18 0k1

bb ce BE

State any assumptions made.

Answer : IE=2.53mA, re=10.28Ω, Zi=12.89kΩ, Zo=10.27Ω, Av= 0.99, Ai= 1.24k

Common collector Amplifier circuit.


i o v i


TutorialTutorial

1. Determine Zin, Zo, AV and AVS β=150 r =10Ω r =50kΩβ=150,rb’b=10Ω,rce=50kΩand VBE=0.7V. State any assumptions made.



TutorialTutorial

2. Determine Zin, Zo, AV, Ai, and AVS. Vo when V 25 V β 120 10Ω Ω d V 0 7VVs=25mV. β=120,rb’b=10Ω,rce=∞Ω and VBE=0.7V. State any assumptions made.

A I 7 81 A 3 33Ω Z 412 2Ω Z 1 8kΩ A

Vcc = +24V

Answer: IE=7.81mA, re= 3.33Ω, Zin = 412.2Ω , Zo = 1.8kΩ, AV= -338.27, Ai = -119.79, and AVS =-137.75. Vo = 3.45V

Ci

Co

R1240k

RC1k8

Vo

VRS

600R

RE1k0

Vi

CE

VS

600Rβ=120

RL3k3


Zi Zo

chapter 5 - ele232_feb11 [compatibility mode]

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