0195136047.pdf

412 TRANSISTOR AMPLIFIERS

RL 2 =RD+RL (8.5.10)

RL 3 =RS+RG (8.5.11)

The frequenciesωZ,ωL 1 ,ωL 2 , andωL 3 are known as break frequencies, at which the behavior of

|Av|changes. Forωabove all break frequenciesAv∼=Av 0 , which is the midband value of the

gain. The largest break frequency is taken to beωL, the low 3-dB frequency.

The amplifier gain at higher frequencies nearωHis found from the analysis of Figure 8.5.2(b).

Neglecting the second-order term inω^2 in the denominator, the gain can be found to be

Av=

vL

vs

∼=Av^0 ωH(ωZ−jω)

ωZ(ωH+jω)

(8.5.12)

where

Av 0 =

−gmRGRDL

RS+RG

(8.5.13)

RDL=ro‖RD‖RL=

roRDRL

roRD+roRL+RDRL

(8.5.14)

RG=R 1 ‖R 2 =

R 1 R 2

R 1 +R 2

(8.5.15)

ωZ=

gm

Cgd

(8.5.16)

ωH=

1

RA[Cgs+Cgd( 1 +gmRDL+RDL/RA)]

(8.5.17)

RA=RS‖RG=

RSRG

RS+RG

(8.5.18)

An illustrative example of both low- and high-frequency designs is worked out next.

EXAMPLE 8.5.1

Consider a CS JFET amplifier with the following parameters:R 1 =350 k;R 2 =100 k,RSS=

1200 , RD= 900 , RL= 1000 , RS= 2000 , ro=15 k, gm= 6 × 10 −^3 S,Cgs= 3

pF,Cgd=1 pF, andωL= 2 π×100 rad/s. Discuss the low- and high-frequency designs.

Solution

For the low-frequency design,

RG=R 1 ‖R 2 =

R 1 R 2

R 1 +R 2

=

350 ( 100 )

450

= 77 .78 k

RL 1 =

RSS

1 +gmRSS

=

1200

1 +( 6 × 10 −^3 × 1200 )

=

1200

8. 2

= 146. 34 

RL 2 =RD+RL= 900 + 1000 = 1900 

RL 3 =RS+RG= 2000 +77,780=79,780

Noting thatRL 1 is the smallest, from Equation (8.5.6),

CS=

1

RL 1 ωL

=

1

146. 34 ( 2 π× 100 )

= 10. 87 μF