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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.

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An