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EE 332

DEVICES AND CIRCUITS II

Lecture 7

Small-Signal Modeling and LinearAmplification (2)

Small-Signal Current Gain and

Amplification Factor of BJT

==

int

11

poQCiF

FC

I

Frmgo

o > FforiCIM, however, o and o are

assumed to be equal.

TV

CEV

AV

CI

CEV

AV

TV

CI

ormgF

+=

+==

Amplification factor is given by:

ForVCE

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Small Signal Operation of BJT

++++=

=+=

...

3

61

2

211

expexp

TVbe

v

TVbe

v

TVbe

v

CI

TVbe

v

TVBE

V

SIciC

ICi

=

TVBE

v

SI

Ci exp

+++== ...

3

61

2

21

TVbe

v

TVbe

v

TVbe

v

CI

CI

Cici

For linearity, ic

should be proportional to vbe

V005.02 =

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C-E Amplifier Voltage Gain: Example Problem: Calculate voltage gain

Given data: F=100, VA =75 V, Q-point is(1.45 mA, 3.41 V),R1 = 10

k,R2 = 30 k, R3 = 100 k,RC= 4.3 k,RI= 1k.

Assumptions: Transistor is in active region, O =F. Signals are low

enough to be considered small signals.

Analysis:

( )dB3.42130==

+=

rB

RI

R

rB

R

LRmgvA

mS0.58)mA45.1(4040 ===C

Img

k72.1mA45.1

)V025.0(100===

CI

TVo

r

k1.54mA45.1

V14.3V75=

+=

+=

CI

CEV

AV

or

k5.721

== RRB

R

k83.3

3

== R

C

R

o

r

L

R

( )mV57.8

)(V)005.0( =

+

rR

rB

RI

R

iv

C-E Amplifier Input Resistance

Input resistance, the total resistance

looking into the amplifier at

coupling capacitorC1 represents

total resistance presented to source.

B 21xi

in

rRRrRR

rB

R

xv

)(xixv

===

=

C-E Amplifier Output Resistance

Output resistance is the total equivalent

resistance looking into the output of the

amplifier at coupling capacitorC3. Input

source is set to 0 and test source is

applied at output.

CCRorRR

mg

orCR

==

++=

xv

out

bevx

vxvxi

xi

But vbe=0.

As ro>> RC.

Small-Signal Model Simplification

If we assume

GenerallyR3 >>RCand load resistor

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Sample Analysis of C-E Amplifier

Problem: Find voltage gain, input

and output resistances.

Given data: F= 65, VA =50 V

Assumptions: Active-region

operation,VBE=0.7 V, small signal

o eratin conditions.

Analysis: To find the Q-point,

dc equivalent circuit is

constructed.

A24566

A24165

A71.3

==

==

=

IE

IB

IC

IB

I

5)4106.1()1(510 =+++B

IFBE

VB

I

V67.3

0)5()4106.1(4105

=

=

CEV

EI

CEV

CI

Small Signal Model of MOSFET

Using 2-port y-parameter network,

The port variables can represent either

time-varying part of total voltages and

currents or small changes in them away

from - oint values.

dsv

22gsv

21di

dsv

12gsv

11gi

yy

yy

+=

+=

0

0ds

vgsv

gi

11pointQ

=

=

=

=

GSv

Gi

y

0

0gsvds

v

gi

12pointQ

=

=

=

=

DSv

Giy

TNV

GSV

DI

GSvDi

y

=

=

=

=

2

0ds

vgsv

di

21pointQ

DSVDS += 0gsv

ds pointQ

DI

vDi

y =

==

1vd

i

22

End of Lecture 7

Sample Analysis of C-E Amplifier

(contd.)Next we construct the ac

equivalent and simplify it.

0.84

in

in)3out

( =+

==

RI

R

RRRmg

ivov

vA

S31064.940 ==C

Img

k64.6==

CI

TVo

r

k223=+

=

CI

CEV

AV

or

k23.6in

== rBRR

k57.9out

==o

rC

RR

36