0195136047.pdf

406 TRANSISTOR AMPLIFIERS

R 1 R 2 RD RL

(b)

RS

+

−

vS

+

−

vGS

+

−

vL

gmvGS

Output

Input

S

GD

iS ro

iL

Rin = R 1 ||R 2 Ri −∞ Ground

v 1

Figure 8.4.1Continued

Further analysis yields

Av 1 =

−gmroRF

ro+RF

(8.4.3)

Ai=

−gmroRFRin

RL(ro+RF)

=

Rin

RL

Av 1 (8.4.4)

where

RF=RD‖RL=

RDRL

RD+RL

(8.4.5)

The CS JFET amplifier is capable of large voltage and current gains.

EXAMPLE 8.4.1

A JFET for whichVA=80 V,VP=4 V, andIDSS=10 mA has a quiescent drain current of 3

mA when used as a common-source amplifier for whichRD=RSS=1kandRL=3k. For

the case of fully bypassedRSS, find the amplifier’s voltage gainAv 1. Also determine the current

gainAiifR 1 =300 kandR 2 =100 k.

Solution

From Equations (7.4.5) and (7.4.4),

ro=

VA

IDQ

=

80

3 × 10 −^3

=26,666.7

gm=

2

VP

√

IDSSIDQ=

2

4

√

10 × 10 −^3 × 3 × 10 −^3 = 2. 7386 × 10 −^3 S

RF=RD‖RL=

1

( 1 / 1000 )+( 1 / 3000 )

= 750 

Av 1 =

−gmroRF

ro+RF

=

− 2. 7386 ( 10 −^3 )(26,666.7) 750

26,666.7+ 750

∼=− 2