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The EF ampli er of Fig. 3-26(a) utilizes a Si transistor with negligible leakage current and 59. Also, VCC 15 V; VL 3 V (VL is the dc component of vL ), and RE 1:5 k. Calculate (a) RB ; b the output impedance Zo , and (c) the input impedance Zin . Ans: a 339 k; b 1:185 k; c 50:98 k
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The ampli er of Fig. 6-31 has an adjustable emitter resistor RE , as indicated, with 0  1. Assume that hre hoe % 0 and Cc ! 1, and nd expressions for (a) the current-gain ratio Ai iL =is , (b) the voltagegain ratio Av vL =vs , and (c) the input impedance Zin . Ans: hfe RB RL ; RC RL RB hie hfe 1 RE hfe RC RB RL ; b Av RC RL fRS RB RS RB hie hfe 1 RE g RB hie hfe 1 RE c Ri RB hie hfe 1 RE a Ai
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RL RB
_ Zin
Fig. 6-31
For the CB ampli er of Problem 6.19, use SPICE methods to nd (a) the input impedance Zin and (b) the output impedance Zo . (Netlist code available at author website.) Ans: a 24:81 ; b 2:195 k The exact small-signal equivalent circuit for the CC ampli er of Fig. 6-2(a) is given by Fig. 6-21. Find the Thevenin equivalent for the circuit to the right of terminals b; b, assume that hre hoe % 0, and show that the circuit of Fig. 6-2(c) results. (Hint: The conversion from CE to CC h parameters was worked out in Problem 6.14.) Apply the CC h-parameter model of Fig. 6-16 to the ampli er of Fig. 6-2(a) to nd an expression for the voltage-gain ratio Av vE =vi . Evaluate Av if hic 100 ; hrc 1; hfc 100; hoc 10 5 S, and RE 1 k. Ans: Av hfc RE = hic hoc RE 1 hrc hfc RE % 0:999 Find an expression for Ro in the CC ampli er of Fig. 6-21; use the common approximations hrc % 1 and hoc % 0 to simplify the expression; and then evaluate it if R1 1 k; R2 10 k; hfc 100, and hic 100 . Ans: Ro hic = hoc hic hfc hrc % hic =hfc 1  The cascaded ampli er circuit of Fig. 6-24(a) matches a high-input-impedance CC rst stage with a highoutput-impedance CE second stage to produce an ampli er with high input and output impedances. To 0 illustrate this claim, refer to Fig. 6-24(b) and determine values for (a) Zin Rin ; b Zin ; c Zo , and (d) Zo0 if RS 5 k and all other circuit values are as given in Problem 6.24. Ans: a 29:18 k; b 818:2 ; c 5 k; d 9:99  To illustrate the e ect of signal-source internal impedance, calculate the voltage-gain ratio Av vL =vs for the cascaded ampli er of Fig. 6-24(a) if RS 20 k and all other values are as given in Problem 6.24; then compare your result with the value of Av found in Problem 6.24. Ans: Av 58:61, which represents a reduction of approximately 40 percent
CHAP. 6]
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For the ampli er of Fig. 6-32, nd expressions for (a) the voltage-gain ratio Av vL =vs and current-gain ratio Ai iL =is . Assume that hre hoe % 0. Ans: a Av hfe RC RL =f RC RL hfe 1 RE hie g; b Ai hfe RC = RC RL
+ VCC RC CC RL +
(b) the
CC iL + LL _
_ Rin
RE _V
Fig. 6-32
Find expressions for (a) Rin and (b) Ro for the ampli er of Fig. 6-32 if hre hoe % 0. Ans: a Rin hie hfe 1 RE ; b Ro RC Suppose v2 is replaced with a short circuit in the di erential ampli er of Fig. 3-22. Find the input impedance Rin1 looking into the terminal across which v1 appears if RB 20 k; RE 1 k; hie 25 ; hfe 100; and hre hoe % 0. Ans: 9:11 k For the Darlington-pair emitter-follower of Fig. 6-33, hre1 hre2 hoe1 hoe2 0. In terms of the (non0 zero) h parameters, nd expressions for (a) Zin ; (b) the voltage gain Av  vE =vs ; c the current gain Ai  ie2 =iin ; d Zin ; and (e) Zo (if the signal source has internal resistance RS ). Ans: a c e
0 Zin hie1 hfe1 1 hie2 hfe2 1 RE ;
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