EE203 Notes-Slides Hasan-Raghib Chapter-5F PDF

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Common Emitter Amplifier The common emitter amplifier circuit is shown below, where the input is applied on the base and the output is VCC taken from the collector. DC Analysis Capacitors are considered open circuit for DC. The DC circuit is simplified Rsig as shown below RBB  R1 // R2 R2 VBB  VCC R1  R2 VBB  VBE IB  RBB  (1  ) RE VCE  VCC  I C RC  I E RE g m  40I C , r    / g m ro | VA | / I C

vsig

C1

+

C2

vo

R2

RE

CE

RL -

VCC IC

RC

+

IB

AC Analysis For AC all capacitors and DC sources are short circuit. The resulting circuit is as shown bellow. Replacing the transistor by its small signal low frequency model, we obtain the circuit shown below. Rsig The input resistance is: Rin  RBB // r

RC

R1

v sig

VEC II

+ R BB

VBB

V BE

-

I RE

-

IE

+

vo RBB

RC

RL -

111

ib

b

+

+ R BB

v sig

v in

r

-

+

g m vbe

R sig

c v be -

ro

R L vo

RC

-

e

R in

R out

To obtain the voltage gain vo   gm vbe ( ro // RC // RL )

vin  vbe

Thus av   gm ( ro // RC // RL ) The output resistance can be obtain as

VCC

Rout  RC // ro

RC

R1

If the capacitor across C1 R E is removed, the circuit shown is obtained. The DC Rsig analysis is the same as before R2 but for ac analysis the vsig equivalent circuit is obtained. Rin  RBB // Rin

Neglecting the current in ro vin ibr  (1  )ib RE  ib ib Rin  r  (1  )RE

Rin 

The voltage gain is

vo -

c +

+

+

r vbe

Rsig RBB

v sig

vo   gmvbe ( RC // RL )

RL

RE

ib

b

+

C2

v in

-

i b

ro

RC

RL

e RE

-

Rin

Rin

vo -

Rout  Rout 111

vbe  gm vbe ) RE r  ( RC // RL ) v g r (R // RL ) av  o   m  C  r  (1  )R E v in r  (1  g m r ) RE

v in  v be  (

The output resistance can be obtain as  Rout  RC // Rout  can be calculated and Rout

from the circuit shown: vx  (ix  ib )ro  (ib  ix ) RE ( Rsig // RBB  r ) ib  ( ib  ix ) RE  0

b

ib

c +

r vbe

Rsig RBB

 RE i x ib  (Rsig // R BB  r )  R E

-

ib

ix

ro

vx

e RE

 Rout

( RE  ro ) RE ix v x  ix (ro  RE )  ( Rsig // RBB  r )  RE v ( ro  RE ) RE R out  x  ( ro  RE )  ix ( Rsig // RBB  r )  RE

111

Common Base Amplifier The common base amplifier circuit is shown below, where the input is applied on the emitter and the output is taken from the collector. V CC

RC

R1

+

C2

C1

vo

CE

R2

R sig

RE

v sig

RL

-

The DC analysis is exactly the same as common emitter amplifier. For AC analysis all capacitors and DC sources are short circuit. The resulting circuit is as shown Replacing the transistor by its small signal low frequency T-model the following equivalent circuit is obtained

+

Rsig

RE

RC

vsig

RL -

Input resistance Rin  RE // Rin if we neglect the current in ro , then Rin 

vo

 v be  re  ie

111

ro -

e

vbe +  i e

c

+

ie re

R sig

RC

RE

v sig

vo

RL

b

Rin

 R out

R in

Rout

To obtain the voltage gain vo  ie ( RC // RL ) (neglecting the current in ro ) vin  iere

Thus

av 

 (RC // R L )  g m (RC // R L ) re

The output resistance can be obtain as  Rout  RC // Rout  can be calculated from the circuit shown: and Rout

ro

e R sig

RE

re

ie

c

-

vbe

 ie

b

ix

vx

+  Rout

vx  (ix  ie )ro  ie re

(Rsig // RBB // re )(ix  ie )  ie re ie 

 ix ( Rsig // RBB // re ) ( Rsig // RBB // re)  re

R out 

(ro  re )(R sig // R BB // re ) vx  ro   ( Rsig // RBB // re )  re ix 121...