barcode fonts for ssrs SMALL-SIGNAL MIDFREQUENCY BJT AMPLIFIERS in Software

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Fig. 6-25
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form has been implemented using the FFT feature of PSpice to determine the value of the fundamental frequency component of vL as shown in the lower plot of Fig. 6-25(a). Then Av 1:003 100:3 0:01
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The negative sign indicates that vL has a 1808 phase shift with respect to vS . (b) Use the Probe and FFT features of PSpice to plot the Fourier spectra of the input current I(CC1) and the output current I(RL) as shown by Fig. 6-25(b). The current gain is found as the ratio of the marked spectra fundamental component values of Fig. 6-25(b : Ai 200:6 10 3 734:3 273:2 10 6
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The negative sign indicates a 1808 phase shift between iS and iL .
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The cascaded ampli er of Fig. 6-26(a) is built up with identical transistors for which hre hoe % 0, hfe 100; and hie 1 k. Let RE1 1 k, RC1 10 k, RE2 100 , RC2 RL 3 k, and Cc CE ! 1. Determine (a) the overall voltage-gain ratio Av vL =vs , and (b) the overall current-gain ratio Ai iL =is .
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(a) The small-signal equivalent circuit is given in Fig. 6-26(b). From the results of Problem 6.7 with hoe 0 and RC replaced with RC2 , Av2 hfe RL RC2 100 3 103 3 103 150 hie RL RC2 1 103 3 103 3 103
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From the results of Problems 6.48, in which RC , RL , and RE are replaced with RC1 , hie , and RE1 , respectively, Av1 hfe RC1 hie 100 10 103 1 103 0:891 RC1 hie hfe 1 RE1 hie 11 103 100 1 1 103 1 103
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SMALL-SIGNAL MIDFREQUENCY BJT AMPLIFIERS
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+ VCC RC1 is Cc Q1 +
Cc iL
Q2 RL RC2 _V (a)
+ LL _
is +
B1 hie
E1 +
hfeib1 +
B2 ib2 RC1
C2 iL hfeib2 RC2 RL + LL _
RE1 _
_ Rin
_ E2 (b) Ro
Fig. 6-26
Thus;
Av Av1 Av2 0:891 150 133:6
(b) From the results of Problem 6.48 with RC and RL replaced with RC1 and hie , respectively, Ai1 hfe RC1 100 10 103 90:91 RC1 hie 10 103 1 103
Now, by current division at the output network, RC2 RC2 RL hfe RC2 i 100 3 103 Ai2 L 50 ib2 RC2 RL 3 103 3 103 Ai Ai1 Ai2 90:91 50 4545:4 iL hfe ib2
Hence, and
In the cascaded CB-CC ampli er of Fig. 6-27(a), transistor Q1 is characterized by hrb1 hob1 % 0, hib1 50 , and hfb1 0:99. The h parameters of transistor Q2 are hoc2 % 0, hrc2 1, hic2 500 , and hfc2 100. Let RL RE2 2 k; RB1 30 k; RB2 60 k; R1 50 k; R2 100 k; RE1 5 k; and CB Cc ! 1. Find (a) the overall voltage-gain ratio Av vL =vS and (b) the overall current-gain ratio Ai iL =iS .
(a) The small-signal equivalent circuit is shown in Fig. 6-27(b). RB R1 kR2 , Av1 From the results of Problem 6.19, with
hfb1 RB hic2 0:99 33:3 103 500 9:75 hib1 RB hic2 50 33:3 103 500
By the results of Problem 6.44,
CHAP. 6]
SMALL-SIGNAL MIDFREQUENCY BJT AMPLIFIERS
+ VCC RE2 Cc RL iL + LL _
RE1 is +
RB2 Q1
Q2 Cc R2
Cc RB1 CB
(a) is +
E1 ie1
E2 iL + LL _
RE1 _ Rin
hib1
hfb1ie1
R1||R2
hic1
hfc2ib2 C2
B1 (b)
Fig. 6-27
Av2 Thus;
hfc2 RE2 kRL 100 1 103 0:995 hic2 hrc2 hfc2 RE2 kRL 500 1 100 1 103 Av Av1 Av2 9:75 0:995 9:70
(b) Based on the results of Problem 6.19, Ai1 hfb2 RE1 RB 0:99 5 103 33:3 103 0:966 RE1 hib1 RB hic2 5 103 50 33:3 103 500
By current division at node E2 , hfc2 RE2 iL 100 2 103 Ai2 50 ib2 RE2 RL 2 103 2 103 Then; Ai Ai1 Ai2 0:966 50 48:3
Use the CE h-parameter model to calculate the output voltage vo for the ampli er of Fig. 3-22, thus demonstrating that it is a di erence ampli er. Assume identical transistors with hre hoe % 0.
The small-signal circuit is given in Fig. 6-28. KVL, Let a hie hfe 1 RE and b hfe 1 RE ; then, by 1 2 3
v1 aib1 bib2 v2 b1 ib1 aib2 vo hfe RC ib1 ib2 Solving (1) and (2) simultaneously using Cramer s rule gives a2 b2
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