0195136047.pdf

8.3 BJT AMPLIFIERS 401

where

Rin=

[

total input resistance

seen by source

]

=RB

∥

∥Ri= RBrπ

RB+rπ

(8.3.4)

Voltage and current gains are given by

Av 1 =

vL

v 1

=

−gmRL[ro‖RC]

RL+[ro‖RC]

(8.3.5)

Ai=

iL

iS

=

−gm(ro‖RC)(rπ‖RB)

RL+(ro‖RC)

(8.3.6)

The CE configuration yields signal inversion since the gains can be seen to be negative.

EXAMPLE 8.3.1

Consider the transistor biased in Example 8.1.1. Given thatRL= 500 andVA=75 V for the

transistor, determine the ac voltage and current gains.

Solution

TakingVT= 25. 861 × 10 −^3 V, as indicated in Section 7.3, and applying Equation (7.3.7),

gm=

ICQ

VT

=

15 × 10 −^3

25. 861 × 10 −^3

= 0 .58 S

By using Equation (7.3.8),

ro=

VA

ICQ

=

75

15 × 10 −^3

= 5000 

TakingRC= 300 from the solution of Example 8.1.1,

ro‖RC= 5000 ‖ 300 =

5000 ( 300 )

5000 + 300

= 283 

Applying Equation (8.3.5), one gets

Av 1 =

− 0. 58 ( 500 )( 283 )

( 500 + 283 )

=− 104. 8

From Equation (7.3.9),

rπ=

β

gm

=

70

0. 58

= 120. 7 

and

Rin=R 1 ‖R 2 ‖rπ=

1

( 1 /R 1 )+( 1 /R 2 )+( 1 /rπ)

=

1

( 1 / 5444 )+( 1 / 1467 )+( 1 / 120. 7 )

= 109. 3 

where values forR 1 andR 2 are taken from the solution of Example 8.1.1. From Equation (8.3.6),

Ai=

− 0. 58 ( 283 )( 109. 3 )

500 + 283

=− 22. 9