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hfb RL RC RE CC hib RE
for the current plot The exact plots are shown
Correction of the asymptotic plot requires only the application of Table 8-1. dashed.
0 For the CE ampli er of Fig. 3-10, determine (a) Zin , (b) Zin , and (c) Zo if CC ! 1 but CE cannot be neglected.
(a) The small-signal low-frequency equivalent circuit is given in Fig. 8-4. Using (8.11) and (8.13), we have
0 Zin
shie RE CE hie hfe 1 RE Vi hie hfe 1 ZE Ib sRE CE 1
0 Zin RB kZin 0 RB Zin 0 RB Zin
Substituting (1) into (2) and rearranging give Zin (c) RB shie RE CE hie hfe 1 RE sRE CE RB hie RB hie hfe 1 RE 3
With voltage source vi deactivated (shorted), KVL requires that Ib  so that 1 hfe Ib ZE khie hie  hfe ZE hie I 0 hie ZE hie b (4)
Since (4) can be satis ed in general only by Ib 0, the output impedance is simply Zo RC 5
In this particular case, (3) shows that the input impedance is frequency-dependent, while (5) shows that the output impedance is independent of frequency. In general, however, the output impedance does depend on frequency, through a nite-valued coupling capacitor CC . (See Problem 8.13.)
CHAP. 8]
FREQUENCY EFFECTS IN AMPLIFIERS
To examine the combined e ects of coupling and bypass capacitors, let the input coupling capacitor be in nitely large while the output coupling capacitor and the bypass capacitor have practical values in the CE ampli er of Fig. 3-10. For simplicity, assume hre hoe 0 and 0 RB ) Zin . (a) Find the voltage-gain ratio Av s vL =vi . (b) If CE 200 F, CC 10 F, Ri RE 100 ; RC RL 2 k; hie 1 k; and hfe 100, determine what parameters control the low-frequency cuto point and whether it is below 100 Hz. (c) Find an expression for the output impedance Zo .
ii Ri +
B hie
hfe ib E ie RC C
CC iL + LL _
Fig. 8-20 (a) The small-signal equivalent circuit is given in Fig. 8-20. ZE RE k Then, by KCL, Ie Ib hfe Ib hfe 1 Ib KVL around the input mesh requires that Vi Ri hie Ib ZE Ie Substituting (2) into (3) and solving for Ib yields Ib Vi Ri hie hfe 1 ZE 4 3 2 We rst de ne 1
1 RE sCE sRE CE 1
Current division at the collector node gives IL and Ohm s law and (5) yield V L R L IL RL RC h I RC RL 1=sCC fe b 6 RC h I RC RL 1=sCC fe b 5
Substituting (4) and (1) into (6) and rearranging now lead to the desired voltage-gain ratio: V Av s L Vi hfe RL RC CC s sRE CE 1 hfe 1 RE hie Ri   CE RE Ri hie sCC RC RL 1 s 1 hfe 1 RE Ri hie
(b) The Laplace-domain transfer function (7) is of the form T s Kb s 2 s 1 1 s 1 3 s 1
FREQUENCY EFFECTS IN AMPLIFIERS
[CHAP. 8
where
1 1 1 25 rad=s 1 CC RC RL 10 10 6 4000 1 1 1 50 rad=s !2 2 RE CE 100 200 10 6 1 hfe 1 RE Ri hie 101 100 100 1000 !3 509:1 rad=s CE RE Ri hie 3 200 10 6 100 1100 !1
Since there is at least a decade of frequency (in which the gain can attenuate from its midfrequency value) between !3 and the other (lower) break frequencies, !3 must be the low-frequency cuto !L . Then fL (c) !3 509:1 81:02 Hz < 100 Hz 2 2
As in Problem 8.12, Ib 0 if vi is deactivated; a driving-point source replacing RL would then see a frequency-dependent output impedance given by Zo Zdp RC 1 sCC 8
Assume that the coupling capacitors in the CS MOSFET ampli er of Fig. 4-25 are identical. Determine the voltage-gain ratio (a) for any frequency and (b) for midfrequency operation.
(a) The equivalent circuit is drawn in Fig. 8-21. By voltage division, where RG R1 kR2 R1 R2 R1 R2 1 RG sRG CC Vgs V V RG 1=sCC i sRG CC 1 1 Current division at the drain node yields IL from which Vo RL IL sgm RD RL rds CC = RD rds V sCC RD rds = RD rds RL 1 gs 3 sCC RD rds = RD rds gm Vgs RD krds g V RD krds 1=sCC RL m gs sCC RD rds = RD rds RL 1 2
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