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200 Hz, hie hfe 1 RE 300 101 1000 268:7 F !L RE hie 2 200 1000 300
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Let C1 ; CE ! 1 in the capacitor-coupled ampli er of Fig. 6-22. hre2 0. Find an expression for the voltage-gain ratio Av s .
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The rst-stage ampli er can be replaced with a Thevenin equivalent, and the second stage represented 0 by its input impedance, as shown in Fig. 8-17. Av follows from voltage division and (6.46) if RL , hfe , and hie are replaced with RC1 ; hfe1 ; and hie1 , respectively:
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0 Av
Req hfe1 RC1 hfe1 RC1 Req Req Ri hie1 hie1 Req Ri hie1 R11 R12 hie1 R11 R12 R11 R12
where
Req hie1 kRB1 hie1 kR11 kR12
Zo1 is given by (6.50) with hoe replaced with RC1 (and with hre1 hoe1 0): Zo1 RC1
+ AL Li _ Zin2
Lo1 _
Fig. 8-17 The second-stage input impedance is given by (6.47) if hie is replaced with hie2 kRB2 hie2 kR21 kR22 : Zin2 Now, from (2) of Problem 8.3, Vo1 sZin2 C2 0 Av Vi sC2 Zin2 Zo1 1 and rearranging yields the rst-stage gain as Av1 Vo1 sZin2 C2 0 Av Vi sC2 Zin2 Zo1 1 6 5 hie2 RB2 hie2 R21 R22 hie2 RB2 hie2 R21 R22 R21 R22 4
The second-stage gain follows directly from (6.46) if RL is replaced with RC2 : Av2 Consequently, the overall gain is
0 Av Av1 Av2 Av
hfe2 RC2 hie2
hfe2 RC2 sZin C2 sC2 Zin2 Zo1 1 hie2
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[CHAP. 8
Substituting (1) into (7) and simplifying yield the desired gain: Av s hfe1 hfe2 RC1 RC2 Req sZin2 C2 hie1 hie2 Req Ri sC2 Zin2 Zo1 1 8
In the cascaded ampli er of Problem 8.7 (Fig. 6-22 with C1 ; CE ! 1), let hie1 hie2 1500 , hfe1 hfe2 40; C2 1 F; Ri 1k; RC1 10 k; RC2 20 k, and RB1 RB2 5 k. Determine (a) the low-frequency gain, (b) the midfrequency gain, and (c) the low-frequency cuto point.
(a) Letting s ! 0 in (8) of Problem 8.7 makes apparent the fact that the low-frequency gain Av 0 0. (b) The midfrequency gain is determined by letting s ! 1 in (8) of Problem 8.7: From (2), (3), and (4) of Problem 8.7, Av 1 lim Av s
hfe1 hfe2 RC1 RC2 Req Zin2 hie1 hie2 Req Ri Zin2 Zo1 h R 1500 5000 1153:8  Req ie1 B1 6500 hie1 RB1 Zo1 RC1 10 k h R 1500 5000 1153:8  Zin2 ie2 B2 6500 hie2 RB2
Then (c)
Av 1
40 40 10 103 20 103 1153:8 1153:8 7881:3 1500 1500 2153:8 1153:8 10 103
The low-frequency cuto point is computed from the lag term in (8) of Problem 8.7: fL !L 1 1 14:3 Hz 2 2C2 Zin2 Zo1 2 1 10 6 1153:8 10 103
The two coupling capacitors in the CB ampli er of Fig. 6-15 are identical and cannot be neglected. Assume hrb hob 0. (a) Find an expression for the voltage-gain ratio VL =VS . (b) Find an expression for the midfrequency voltage-gain ratio.
(a) The small-signal low-frequency equivalent circuit is given in Fig. 8-18. Applying Ohm s law in the Laplace domain, we obtain IS VS sCC VS 1=sCC RE hib = RE hib sCC RE hib = RE hib 1
CC iL
+ RE hib hfb ie RC RL
LL _
Fig. 8-18
CHAP. 8]
FREQUENCY EFFECTS IN AMPLIFIERS
Voltage division then gives Ie RE RE sCC I V hib RE S hib RE sCC RE hib = RE hib 1 S 1
By current division at the output, VL RL IL RL shfb RL RC CC Ie RC h I RC 1=sCC RL fb e sCC RL RC 1 2
Substituting (1) into (2) and rearranging lead to the desired voltage-gain ratio: Av s
2 RE RL RC hfb CC s2 VL VS hib RE sCC RE hib = RE hib 1 sCC RL RC 1
(b) Letting s ! 1 in (3) leads to the midfrequency gain: Av 1 RL RC hfb hib RL RC 4
The two coupling capacitors in the CB ampli er of Fig. 6-15 are identical. Also, hrb hob 0. (a) Find an expression for the current-gain ratio Ai s that is valid at any frequency. (b) Find an expression for the midfrequency current-gain ratio.
(a) The low-frequency equivalent circuit is displayed in Fig. 8-18. RE I Ie hib RE S and IL shfb RL RC CC Ie RC h I RC 1=sCC RL fb e sCC RL RC 1 By current division, 1
Substituting (1) into (2) and dividing by IS give the desired current-gain ratio: Ai s shfb RL RC RE CC IL Ii hib RE sCC RL RC 1 3
(b) The midfrequency current-gain ratio is found by letting s ! 1 in (3): Ai 1 hfb RL RC RE hib RE RL RC 4
On a common set of axes, sketch the asymptotic Bode plots (Mdb only) for the voltage- and current-gain ratios of the CB ampli er of Fig. 6-15, and then correct them to exact plots. Assume that the coupling capacitors are identical and that, for typical values, 1 ( CC RE khib ( CC RL RE .
The Laplace-domain transfer functions that serve as bases for Bode plots of the voltage- and currentgain ratios are, respectively, (3) of Problem 8.9 and (3) of Problem 8.10. Under the given assumptions, inspection shows that the two transfer functions share a break frequency at ! 1= CC RL RC and the voltage-gain transfer function has another at a higher frequency. Moreover, the voltage plot rises at 40 db per decade to its rst break point, and the current plot at 20 db per decade. With !1v !1i !Li 1 CC RL RC and !2v !Lv RE hib CC RE hib
the low-frequency asymptotic Bode plots of voltage and current gain are sketched in Fig. 8-19. The given assumption assures a separation of at least a decade between !1v and !2v and between ! 1 and !1v . Since the parameter values are not known, the sketches were made under the assumption that Kb 1 in both plots. When values become known, the Bode plots must be shifted upward by 20 log Kbv 20 log
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