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RG v id R S R G ri i
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rds _ D id iL
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mLgs _ S + RS RD RL
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Fig. 7-14 But by Ohm s law, id vgs rds RS RD kRL 2
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Substituting (2) into (1) and solving for vgs yield vgs Now voltage division gives vL RD kRL vgs rds RS RD kRL 4 RG rds RS RD kRL vi RG ri rds  1 RS RD kRL 3
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and substitution of (3) into (4) and rearrangement give Av vL RG RD RL vi RG ri f RD RL rds  1 RS RD RL g 5
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SMALL-SIGNAL MIDFREQUENCY FET AND TRIODE AMPLIFIERS
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[CHAP. 7
With  gm rds and the given values, (5) becomes Av 2 10 3 30 103 160 2 2 0:272 160 5 f 2 2 30 60 1 3 2 2 g
(b) The current gain is found as Ai (c) iL vL =RL A R ri 0:272 160 5 v G 22:4 ii vi = RG ri RL 2
RL is disconnected, and a driving-point source is added such that vdp vL . With vi deactivated (shortcircuited), vgs 0 and Ro RD k rds RS RD rds RS 2 103 30 103 3 103 1:89 k RD rds RS 2 103 30 103 3 103
Note that when RS is not bypassed, the voltage- and current-gain ratios are signi cantly reduced.
Use SPICE methods to determine the voltage gain for the CG ampli er of Example 7.3. RS 2 k and vi 0:25 sin 2 103 t V for computational purposes.
The netlist code that follows describes the circuit:
Prb7_8.CIR vi 1 0 SIN(0V 0.25V 1kHz) RS 1 0 2kohm RD 2 0 1kohm rds 1 2 30kohm G 1 2 (1,0) 2e-3 .TRAN 1us 1ms .PROBE .END
Execute hPrb7_8.CIRi and use the Probe feature of PSpice to give the resulting waveforms for vi and vo shown by Fig. 7-15. The voltage gain is found as the ratio of the marked peak values. Av vi 0:492 1:97 vo 0:250
Fig. 7-15
CHAP. 7]
SMALL-SIGNAL MIDFREQUENCY FET AND TRIODE AMPLIFIERS
Find a small-signal equivalent circuit for the two parallel-connected JFETs of Fig. 7-16 if the devices are not identical.
VDD RD iD iD1 Q1 ri ii +
iD2 Q2 RL
CC S RG CS
+ LL _
Fig. 7-16 By KCL, iD iD1 iD2 1
Since the parallel connection assures that the gate-source and drain-source voltages are the same for both devices, (1) can be written as iD f1 vGS ; vDS f2 vGS ; vDS Application of the chain rule to (2) yields id iD % diD gm1 gm2 vgs @iD1 @vGS Q @iD2 @vGS Q   vds @vDS @iD2 Q 2
rds1
1 rds2
where
gm1
gm2
rds1
@vDS @iD1 Q
rds2
Equation (3) is satis ed by the current-source circuit of Fig. 7-1(a) if gm gm1 gm2 and rds rds1 krds2 .
In the circuit of Fig. 7-16, RS 3 k; RD RL 2 k; ri 5 k; and RG 100 k. Assume that the two JFETs are identical with rds 25 k and gm 0:0025 S. Find (a) the voltage-gain ratio Av vL =vi , (b) the current-gain ratio Ai iL =ii , and (c) the output impedance Ro .
(a) The small-signal equivalent circuit is given in Fig. 7-17, which includes the model for two parallel JFETs as determined in Problem 7.9. By voltage division,
ii +
G + RG
iL + LL _
2gmLgs
1 2 rds
_ S Ro
Fig. 7-17
SMALL-SIGNAL MIDFREQUENCY FET AND TRIODE AMPLIFIERS
[CHAP. 7
vgs Now let Req 1 rds kRD kRL 2
RG 100 v v 0:952vi RG ri i 100 5 i
rds RD RL 25 2 2 103 962  2RL RD rds RL RD 2 2 2 25 2 2
Then, by Ohm s law, vL 2gm vgs Req ; with (1) and (2), this gives Av vL RG 2gm R 2 0:0025 0:952 962 4:58 vi RG ri eq
(b) The current-gain ratio is Ai (c) iL vL =RL A R ri 4:58 100 5 240:4 v G RL 2 ii vi = RG ri With vi deactivated (short-
We replace RL with a driving-point source oriented such that vdp vL . circuited), vgs 0; thus, Ro RD k 1 rds 2
RD rds 2 25 103 1:72 k 2RD rds 2 2 25
Move capacitor CS from its parallel connection across RS2 to a position across RS1 in Fig. 4-33. Let RG 1 M; RS1 800 ; RS2 1:2 k; and RL 1 k. The JFET is characterized by gm 0:002 S and rds 30 k. Find (a) the voltage-gain ratio Av vL =vi ; b the currentgain ratio Ai iL =ii ; c the input impedance Rin , and (d) the output impedance Ro .
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