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AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
[CHAP. 5
Fig. 5-38 vC 21 vC vC vC v2 þ þ þ ¼0 3 6 3 8
ð40Þ
Substituting vC ¼ ð3=5Þv2 from (39) into (40) we get v2 ¼ 10 V. Then vC ¼ 6 V i1 ¼ ð21 vC Þ=3000 ¼ 0:005 A ¼ 5 mA Rin ¼ 21=i1 ¼ 21=0:005 ¼ 4200 ¼ 4:2 k
5.14
In the circuit of Fig. 5-38 change the 21-V source by a factor of k. Show that vC , i1 , v2 in Problem 5.13 are changed by the same factor but Rin remains unchanged. Let vs ¼ 21k (V) represent the new voltage source. From the inverting amplifier we have [see (39)] v2 ¼ ð5=3ÞvC Apply KCL at node C to obtain [see (40)] vC vs vC vC vC v2 þ þ þ ¼0 3 6 3 8 Solving for vC and v2 , we have vC ¼ ð6=21Þvs ¼ 6k ðVÞ
and
v2 ¼ ð10=21Þvs ¼ 10k ðVÞ
i1 ¼ ðvs vC Þ=3000 ¼ ð21 6Þk=3000 ¼ 0:005k A Rin ¼ vs =i1 ¼ 21k=0:005k ¼ 4200 These results are expected since the circuit is linear.
5.15
Find v2 and vC in Problem 5.13 by replacing the circuit to the left of node C in Fig. 5-38 (including the 21-V battery and the 3-k and 6-k resistors) by its The´venin equivalent. We first compute the The´venin equivalent: RTh ¼
ð6Þð3Þ ¼ 2 k 6þ3
and
vTh ¼
6 ð21Þ ¼ 14 V 3þ6
Replace the circuit to the left of node C by the above vTh and RTh and then apply KCL at C: vC 14 vC vC v2 þ þ ¼0 2 3 8
ð41Þ
For the inverting amplifier we have v2 ¼ ð5=3ÞvC or vC ¼ 0:6 v2 , which results, after substitution in (41), in v2 ¼ 10 V and vC ¼ 6 V.