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By Cramer s rule applied to the system consisting of (1) and (2), ie 2 = , where     RE RL RE RL rc re rb re 1 rc RE RL RE RL 2 rc v B
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[CHAP. 6
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Now, by Ohm s law, vL RE kRL ie Then Av R E R L 2 RE RL
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vL RE RL rc = RE RL vB rb re 1 rc RE RL = RE RL rc re RE RL = RE RL
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(b) The input impedance can be found as Zin RB k vB =ib . Now, in the system consisting of (1) and (2), by Cramer s rule, ib 1 = , where   RE RL v 1 re 1 rc RE RL B     RE RL RE RL   RB rb re 1 rc R B rc re RE RL RE RL     Hence, Zin RB k v RE RL RE RL 1 B RB rb re 1 rc rc re RE RL RE RL
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Answer the following questions relating to a CE-connected transistor: (a) How are the input characteristics (iB versus vBE ) a ected if there is negligible feedback of vCE (b) What might be the e ect of a too-small emitter-base junction bias (c) Suppose the transistor has an in nite output impedance; how would that a ect the output characteristics (d) With reference to Fig. 3-9(b), does the current gain of the transistor increase or decrease as the mode of operation approaches saturation from the active region
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(a) The family of input characteristics degenerates to a single curve one that is frequently used to approximate the family. (b) If IBQ were so small that operation occurred near the knee of an input characteristic curve, distortion would result. (c) The slope of the output characteristic curves would be zero in the active region. (d) iC decreases for constant iB ; hence, the current gain decreases.
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Use a small-signal h-parameter equivalent circuit to analyze the ampli er of Fig. 3-10(a), given RC RL 800 , Ri 0; R1 1:2 k, R2 2:7 k; hre % 0; hoe 100 S; hfe 90, and hie 200 . Calculate (a) the voltage gain Av and (b) the current gain Ai .
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(a) The small-signal circuit is shown in Fig. 6-13, where RB R1 R2 = R1 R2 831 . division in the collector circuit, iL RC 1=hoe h i RC 1=hoe RL 1=hoe RL RC fe b By current
ii +
hie E
hfe ib
LL = Lce _
_ Zin Z in
Fig. 6-13
CHAP. 6]
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The voltage gain is then Av  hfe RL RC vL RL iL 90 800 2 173:08 vi hie ib hie RC RL hoe RL RC 200 1600 100 10 6 800 2 RB i RB hie i i RB iL RB hie 831 200 173:08 34:87 A Ai  L 800 1031 ii RB hie ib RL RB hie v ib 1
(b) By current division,
For the ampli er of Example 6.5, use SPICE methods to determine the voltage gain Av vL =vi .
Execute the le hEx6_5.CIRi of Example 6.5, then use the Probe feature of PSpice to generate the instantaneous waveforms of input voltage vi and output voltage vL shown by Fig. 6-14. The peak values of vi and vL are marked. Hence, Av vL V 1:1528 4:61 Lm 0:250 vi Vim
Fig. 6-14
Suppose the emitter-base junction of a Ge transistor is modeled as a forward-biased diode. Express hie in terms of the emitter current.
The use of transistor notation in (2.1) gives iB ICBO evBE =vT 1 Then, by (6.5), 1 @i 1 B I evBEQ =vT hie @vBE Q VT CBO 2 1
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[CHAP. 6
But, by (1) and Problem 2.1, IBQ ICBO evBEQ =vT 1 % ICBO evBEQ =vT IEQ IBQ 1 VT 1 IEQ 3 (4)
and Equations (2), (3), and (4) imply
hie
For the CB ampli er of Problem 3.12, determine graphically
(a) hfb and (b) hob .
By (6.13),
(a) The Q point was established in Problem 3.12 and is indicated in Fig. 3-16. i 3:97 2:0 10 3 0:985 hfb % C iE vCBQ 6:1 V 4 2 10 3 (b) By (6.14), hob % iC 3:05 2:95 10 3 12:5 S vCB IEQ 3 mA 10 2
Find the input impedance Zin of the circuit of Fig. 3-10(a) in terms of the h parameters, all of which are nonzero.
The small-signal circuit of Fig. 6-13, with RB R1 R2 = R1 R2 , is applicable if a dependent source hre vce is added in series with hie , as in Fig. 6-1(a). The admittance of the collector circuit is given by G hoe and, by Ohm s law, vce By KVL applied to the input circuit, ib vi hre vce hie 2 hfe ib G 1 1 1 RL RC
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