o BJT amplifier parameter rep p e

o CE amplifiers

o CE with RE amplifiers

o EF amplifiers

o CB amplifiers

1

BJT amplifier parameter reVCC

RCIC

R1

VB β

RE

R20.7 +

+-

B+

ICRE

β

E --C E

BJT amplifier

R

EB

CECBC

BB

21B21

1CCBB

)2(7.0RR

IRI7.0RI

V

)1(R//RR,RR

RVV

−−−+⎟⎟⎠

⎞⎜⎜⎝

⎛+

β=++

β=

−−−−−−−−−−−−=+

=

CeC

ECECBBB

ImV26rfindto)2(fromIFind =

⎟⎠

⎜⎝ ββ

2

Example: Find IC and re VCC=12V

IC

8.2k

β=100

47k

0.2K15k

β=100

15R

+⎟⎠⎞

⎜⎝⎛ +=+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

β==

===+

×=+

=

7.0k2.0100

k37.11)mA(I7.0RR

I9.2V

k37.1115//47R,V9.24715

1512RR

RVV

CEB

CBB

B21

1CCBB

Ω=====

⎠⎝⎟⎠

⎜⎝ β

7.3mA03.7

mV26ImV26rmA03.7

k313.0V2.2)mA(I

100

CeC

CECBB

3

Types of BJT amplifiers

VVCC

( ) E ( E) (b) Common Emitter (CE) with R VCC

RC

R CCR1 vo

RC

Ri CCCB

R1 vo

(a) Common Emitter (CE)High AV and low Ri

(Voltage amplifications)

(b) Common Emitter (CE) with RELow AV and high Ri

(Stability applications)

RL

Ri CCCB

vi R2RE

RL

CE

i CB

vi R2 RE E

VCC

(a) common emitter (CE)VCC

(b) common emitter (CE) with RE

CC

Ri CB

R1

CC

RC

CC

R1 vo

(c) Emitter Follower (EF)Very low Ro and high Ri( Buffer applications)

(d) Common Base (CB)Very low Ri and high AV

(High Frequency applications)

RL

iCC

B

vi R2RE

vo RLRi

C

CE viR2 RE

CB

4

(c) emitter follower (EF) (d) common base (CB)

1. CE amplifiersp

•• When the input to the amplifier is at Base and the output of the When the input to the amplifier is at Base and the output of the amplifier is at Collector, Emitter becomes common to both input and amplifier is at Collector, Emitter becomes common to both input and

t t d th lifi i C E itt (CE) lifit t d th lifi i C E itt (CE) lifi

VCC

output and the amplifier is Common Emitter (CE) amplifieroutput and the amplifier is Common Emitter (CE) amplifier

RC

R

CC

CR1 voi C

iovo

RL

C

Ri CB

vi R2

o

R

Rovin

iin iLB

E

CERERin

5

CE amplifier analysis VCC

Ri RC

R

Ri

CC

CB

R1 vo

R

iin iLBC

E

iovoFirst replace VFirst replace VCCCC with ac ground and all capacitors with ac ground and all capacitors

shortshort, , replace BJT with it’s equivalent circuit, replace BJT with it’s equivalent circuit, Calculate ACalculate Avv ,R,Rii, R, Roo, A, Aii, from the equivalent circuit, from the equivalent circuit RL

CE

vi R2RE

Rin

Rovin

Evv ,, ii,, oo,, ii, f q, f qCE CE amplifier has the following ac circuit amplifier has the following ac circuit

propertiesproperties

in //RvR β

Coo

o RivR ==Output resistanceOutput resistance

( ) LCLCbov r

R//Rri

R//RivvA −=

××−

==β

βVoltage gainVoltage gain

eBinin

in r//RivR β==Input resistanceInput resistance

eebin rriv × βg gg g

invinino

iL

oL

i RRA

R

Rvv

vR

v

iiA =

×⎟⎠⎞⎜

⎝⎛

===Current gainCurrent gain

6

LLin

inin RRR

vi

Example:Example: .amplifierCEfolowingtheofII

A,vvA,

ivR,RFind

in

Li

in

ov

in

inino ===

VCC=12V

8.2k47k VO

===+

×=+

= k37.1115//47R,V9.24715

1512RR

RVV B21

1CCBB

ininin

15k

β=100

OVin

10k

Iin IL

Ω=====

+⎟⎠⎞

⎜⎝⎛ +=+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

β==

7.3mV26mV26rmA03.7V2.2)mA(I

7.0k2.0100

k37.11)mA(I7.0RR

I9.2V

eC

CEB

CBB

21

0.2K RORin

Ω7.3mA03.7I

rmA03.7k313.0

)mA(IC

eC

k2.8RivR Co

oo ===Output resistanceOutput resistance

( ) k10//k28R//RR//Riv LCLCb ×β−

k36.07.3100//k37.11r//RivR eBin

inin =×=== βInput resistanceInput resistance

( )8.121

k037.0k10//k2.8

rR//R

riR//Ri

vvA

e

LC

eb

LCb

in

ov −=−=−=

β××β

==Voltage gainVoltage gain

384k36.08.121RARvv

Rv

iA invinin

oL

oL =

×−==

×⎟⎠⎞⎜

⎝⎛

===Current gainCurrent gain

7

38.4k10RR

Rvi

ALL

ininin

i −=−=====Current gainCurrent gain

2. CE with RE amplifiersE mp f

The amplifier is Common Emitter (CE) amplifier but the Emitter The amplifier is Common Emitter (CE) amplifier but the Emitter capacitor Ccapacitor CEE is removedis removedcapacitor Ccapacitor CEE is removed.is removed.

VVCC

RC CCR vC

iovo

RL

Ri CB

vi R

R1 vo

Rovi

iin iLBC

E

o

vi R2RE

Rin

vin

8

CE with RE amplifier analysis

First replace VFirst replace VCCCC with ac ground and all capacitors with ac ground and all capacitors shortshort, , replace BJT with it’s equivalent circuit, replace BJT with it’s equivalent circuit, Calculate ACalculate Avv ,R,Rii, R, Roo, A, Aii, from the equivalent circuit, from the equivalent circuitvv ,, ii,, oo,, ii, f q, f q

CE with RCE with REE amplifier has the following ac circuit amplifier has the following ac circuit propertiesproperties

( ) ( )ERer//BRbi

ERbierbi//BRiiinv

inR +=+

== βββ

Input resistance

( ) R//RR//Riβ

biini

Coo

o RivR ==Output resistanceOutput resistance

( )( ) ERer

LR//CR

ERerbiLR//CRbi

invov

vA+

−=+×

×−==

ββ

βVoltage gainVoltage gain

invinino

Lo

L RARvv

Rv

i ×⎟⎠⎞⎜

⎝⎛

9

Linv

L

in

inin

LinL

i RRA

Rv

Rv

RiiA =⎠⎝===Current gainCurrent gain

3 EF amplifiers3. EF amplifiers•• When the input to the amplifier is at Base and the output of the When the input to the amplifier is at Base and the output of the amplifier is at Emitter, then Collector becomes common to both amplifier is at Emitter, then Collector becomes common to both ppinput and output and the amplifier is Common Collector (CC) input and output and the amplifier is Common Collector (CC) amplifier or Emitter Follower (EF) amplifieramplifier or Emitter Follower (EF) amplifier

VCC

Ri

CC

CBR1

vov

iin

RLvi R2 RE

RoRin

vin

iL

10

oin

EF amplifier analysisFirst replace VFirst replace V with ac ground and all capacitors with ac ground and all capacitors First replace VFirst replace VCCCC with ac ground and all capacitors with ac ground and all capacitors shortshort, , replace BJT with it’s equivalent circuit, replace BJT with it’s equivalent circuit, Calculate ACalculate Avv ,R,Rii, R, Roo, A, Aii, from the equivalent circuit, from the equivalent circuitEFEF amplifier has the following ac circuit propertiesamplifier has the following ac circuit properties

( ) ( )LR//ERbierbi//BRinvR

+ ββInput resistanceInput resistance

( ) ( )1

R//RR//RivA LELEbo ==×

==β

[ ]( )LR//ERer//BRbi

//BRiniinR

+=

==

βVoltage gainVoltage gain

pp

( ) [ ] ⎤⎡

−==

β

β

)R//R()R//R(ii

v//Riv//RR

bo

Eoo

Eo

[ ] 1)R//R(r)R//R(riv

ALEeLEebin

v =+

=+×

==β

Output resistanceOutput resistance( ) [ ]

⎥⎦

⎤⎢⎣

⎡+=

−+×−

=ββ

β )R//R(r//Ri

)R//R(ri//R iBeE

biBeb

E

invinino

Lo

L RARvv

Rv

i ×⎟⎠⎞⎜

⎝⎛

11

Linv

L

in

inin

LinL

i RRA

Rv

Rv

RiiA =⎠⎝===Current gainCurrent gain

4. CB amplifiers

•• When the input to the amplifier is at Emitter and the output of When the input to the amplifier is at Emitter and the output of the amplifier is at Collector, then Base becomes common to both the amplifier is at Collector, then Base becomes common to both input and output and the amplifier is Common Base (CB) amplifierinput and output and the amplifier is Common Base (CB) amplifier

•• CB amplifier without CCB amplifier without CBB

12

CB amplifier analysisFirst replace VFirst replace VCCCC with ac ground and all with ac ground and all pp gg

capacitors shortcapacitors short, , replace BJT with it’s equivalent replace BJT with it’s equivalent circuit, Calculate Acircuit, Calculate Avv ,R,Rii, R, Roo, A, Aii, from the equivalent , from the equivalent circuitcircuit

CBCB amplifier has the following ac circuit amplifier has the following ac circuit

( )× R//RR//Riv βVoltage gainVoltage gain

mp f f gmp f f gpropertiesproperties

RCRB

vo

ie RL

ib ibre

B C

E

iL

iin

( )[ ]

⎥⎦

⎤⎢⎣

⎡+

=+×−

×−==

eB

LCeBbLCb

ino

vrRR//R

rRiR//Ri

vvA

ββ

β

I t i tI t i t RC

RE

Rin Ro

vin

e L

( )

⎤⎡

−+×−

==

B

beBb

Eein

Ein

Ri

rRi//Riv//RR

ββ

Input resistanceInput resistance

in Ro⎥⎦

⎤⎢⎣

⎡+= e

BE rR//R

β

Co RR = Output resistanceOutput resistanceov

13

L

inv

inin

Lo

in

Li R

RA

Rv

Rii

A ===Current gainCurrent gain

Effect of Base capacitor CB

•• CB amplifier with CCB amplifier with CBB

If CB is connected in the Base circuit RB

VCC

RC

If CB is connected in the Base circuit, RBis shorted by zero reactance of the capacitance. By substituting RB = 0 in the previous formula, we will have the f ll RC

RLRi

CC

R

R1 vo

C iiin

iLLCov r

R//RvvA ==Voltage gainVoltage gain

followings:

RLi

vi

R2

RE

CB

Rin Ro

vin

ieein rv

eEin r//RR =Input resistanceInput resistancein o

invLo

L RARv

iA

Co RR =Output resistanceOutput resistance

14

L

inv

inin

L

in

Li R

Rvi

A ===Current gainCurrent gain