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vL 1:995vS 1:995 vS vS
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vL 1:995vS 413:2 vbe 4:828 10 3 vS
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In the CB ampli er of Fig. 6-19(a), let R1 R2 50 k, RC 2:2 k, RE 3:3 k, RL 1:1 k, CC CB ! 1; hrb % 0; hib 25 ; hob 10 6 S; and hfb 0:99. Find and evaluate expressions for (a) the voltage-gain ratio Av vL =vs and (b) the current-gain ratio Ai iL =is .
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(a) With hrb 0, the CB h-parameter model of Fig. 6-1(b) can be used to draw the small-signal circuit of Fig. 6-19(b). By Ohm s law at the input mesh, ie vs hib 1
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Ohm s law at the output mesh requires that   RC RL hfb ie 1 vL kRC kRL hfb ie hob RC RL hob RC RL Substitution of (1) into (2) allows the formation of Av : Av RC RL hfb vL vs hib RC RL hob RC RL 2:2 103 1:1 103 0:99 29:02 25 2:2 103 1:1 103 10 6 2:2 103 1:1 103
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(b) By current division at node E, ie Current division at node C gives iL RC hfb ie 1=hob kRC hfb ie 1=hob kRC RL RC RL hob RL RC 4 RE i RE hib s 3
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CHAP. 6]
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SMALL-SIGNAL MIDFREQUENCY BJT AMPLIFIERS
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+ VCC RC R1 iL CC
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CC i s R2 RE _ (a) +
+ LL _
is +
RE _ Zin
Vsen_ + E 2 ie 0V hib
ic iL + LL _
hfb ie
0 (b)
Fig. 6-19
Now substitution of (3) into (4) allows direct calculation of Ai : Ai RE RC hfb iL is RE hib RC RL hob RL RC 3:3 103 2:2 103 0:99 0:655 3:3 103 25 2:2 103 1:1 103 10 6 1:1 103 2:2 103
Let vS sin 2000t V and apply SPICE methods to the small-signal equivalent circuit of Fig. 6-19(b) to solve Problem 6.19.
(a) The netlist code below describes the circuit:
Prb6_20.CIR vs 1 0 SIN( 0V 1V 1kHz ) RE 1 0 3.3kohms Rhib 1 2 25ohms Vsen 2 0 DC 0V Fhfb 3 0 Vsen -0.99 Rhob 3 0 {1/1e-6S} RC 3 0 2.2kohms RL 3 0 1.1kohms .TRAN 5 us 1ms .PROBE .END
After executing hPrb6_20.CIRi, the traces of the input voltage vS V 1 and the output voltage vL V 3 of Fig. 6-20(a) are generated using the Probe feature of PSpice. Since the input voltage
SMALL-SIGNAL MIDFREQUENCY BJT AMPLIFIERS
[CHAP. 6
has been conveniently selected at 1 V peak, the voltage gain is simply equal to the peak value of vL , or Av 29:02 as marked on Fig. 6-20(a). (b) The resulting instantaneous waveforms for iS I vs) and iL I(RL) are shown by the upper plot of Fig. 6-20(b). The current gain is determined by the ratio of maximum or peak values of output current iL to input current iS as displayed by the lower plot of Fig. 6-20(b) where Ai 654:6 10 3 .
Fig. 6-20
Use the CC h-parameter model of Fig. 6-16 to nd expressions for the current-gain ratios (a) Ai0 ie =ib and (b) Ai ie =ii for the ampli er of Fig. 6-2(a).
(a) The equivalent circuit is given in Fig. 6-21. ie RE vec hfc ib
ii +
At the output port,  1 kR hoc E  hfc RE i hoc RE 1 b
E + ie + hfe ib hoc
Lec RE LE _
b B hic R1R2 RB = R1 + R2
ib + hre Lec _
_ b Rin
_ C Ro
Fig. 6-21 and Ai0 is obtained directly from (1) as Ai0 hfc ie ib hoc RE 1
CHAP. 6]
SMALL-SIGNAL MIDFREQUENCY BJT AMPLIFIERS
(b) With RTh Rin hic hrc hfc RE = hoc RE 1 , current division at node B gives ib 1 RB ii RB 1 ie Ai0 RB Rin ie RB Rin Ai hfc RB RB Ai A0 RB Rin i RB hic hrc hfc RE = hoc RE 1 hoc RE 1 hfc RB RB hic hoc RE 1 hrc hfc RE
In the two-stage ampli er of Fig. 6-22, the transistors are identical, having hie 1500 , hfe 40, hre % 0, and hoe 30 S. Also, Ri 1 k; RC2 20 k; RC1 10 k; RB1  R11 R12 5 k R11 R12 and RB2
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