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By (1.16), hie vbe ib vce 0
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If vce 0 (short-circuited) in the network of Fig. 6-1(b), then vcb vbe , so that, by KVL around the E; B loop, vbe hib ie hrb vcb hib ie hrb vbe h 1 ie rb vbe hib   1 hfb 1 hrb hob vbe hib
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which gives KCL at node B then gives
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ib 1 hfb ie hob vcb Now, (1) and the given approximations, hie By (1.17), hre If ib 0, then ic ie in Fig. 6-1(b). By KVL,
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hib hib % hib hob 1 hfb 1 hrb 1 hfb vbe vce ib 0
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vce vcb hrb vcb hib ie 1 hrb vcb hib ie
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SMALL-SIGNAL MIDFREQUENCY BJT AMPLIFIERS
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KCL at node C then gives ic ie hfb ie hob vcb h ie ob 1 hfb hib hob v vbe 1 hfb ce
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Substituting (5) into (4) with vcb vce vbe gives vce 1 hrb vce vbe
After rearranging, (3) and the given approximations lead to hre By (1.18), hfe ic ib vce 0 6 hrb 1 hfb hib hob h h % ib ob hrb hib hob hrb 1 1 hfb 1 hfb
By KCL at node B of Fig. 6-1(b), with vce 0 (and thus vcb veb vbe , ib 1 hfb ie hob vcb 1 hfb ie hob vbe Solving (2) for vbe with ie ib ic and substituting now give ib 1 hfb ib ic hib hob i ic 1 hrb b
After rearranging, (6) and the given approximations lead to hfe By (1.19), hoe hfb 1 hrb hib hob hfb % 1 hfb 1 hrb hib hob 1 hfb ic vce ib 0
If ib 0, then ic ie . Replacing ie with ic in (4) and (5), solving (4) for vcb , and substituting into (5) give   hob vce hib ic ic 1 hfb 1 hrb 1 hrb After rearranging, (7) and the given approximations lead to hoe hob hob % 1 hfb 1 hrb hib hob 1 hfb
Apply the de nitions of the z parameters given by (1.10) through (1.13) to the CB h-parameter circuit of Fig. 6-1(b) to nd values for the z parameters in terms of the CB h parameters.
The circuit of Fig. 6-1(b) is described by the linear system of equations      hib hrb ie v eb hfb hob vcb ic By (1.10) and Fig. 1-8, z11 veb ie ic 0
CHAP. 6]
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Setting ic 0 in (1) yields vcb hfb i hob e 3
Substituting (3) into the rst equation of (1) and applying (2) yield z11 hib By (1.12) and (3), z21 By (1.11), z12 hrb hfb hob
hfb vcb ie ic 0 hob veb ic ie 0
Setting ie 0 in (1), solving the two equations for vcb , and equating the results give veb i c hrb hob Finally, by (1.13), z22 from which z12 hrb hob
vcb ic ie 0
Letting ie 0 in the second equation of (1) and applying (5) yield z22 1=hob directly.
For the CE ampli er of Fig. 3-17, assume that hre hoe % 0; hie 1:1 k; hfe 50; Cc ! 1; RF 100 k; RS 5 k; and RC RL 20 k. Using CE h parameters, nd and evaluate 0 expressions for (a) Ai iL =iS ; b Ai0 iL =ib ; c Av vL =vS , and (d) Av vL =vbe .
(a) The small-signal equivalent circuit for the ampli er is given in Fig. 6-18. voltages, vs vbe vce vbe vbe 0 RS RF hie vbe vce R RL hfe ib C v 0 RF RC RL ce
RS +
By the method of node 1 2
ic +
C iL + LL _
hie _ ib
hfe ib
Fig. 6-18 Rearranging (1) and (2) and substituting ib vbe =hie lead to 2 1 1 1 6 RS RF hie 6 4 hfe 1 hie RF 3 1 2v 3   S 7 vbe RF 7 4 RS 5 1 R RL 5 vce 0 C RC RL RF
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[CHAP. 6
The determinant of coe cients is then      1 1 1 1 R RL 1 hfe 1 C R RF hie RF RF hie RF RC RL  S     1 1 1 1 10 10 1 50 1 6 10 10 6 4:557 10 6 5 100 1:1 100 10 10 100 1:1 100 By Cramer s rule,    1 R RL 1 1 10 3 vS C vS RF 100 10 RC RL vbe 1 4:828 10 3 vS RS 5 103 4:557 10 6     hfe 1 1 50 vS 10 3 vS 2 RF hie 100 1:1 1:995vS vL vce RS 5 103 4:557 10 6  Ai iL vL =RL R S vL 5 103 1:995vS 0:501 iS vS vbe =RS RL vS vbe 20 103 vS 4:828 10 3 vS Ai0 iL vL =RL h v 1:1 103 1:995vs ie L 22:73 ib vbe =hie RL vbe 20 103 4:828 10 3 vS Av
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