Page 101 - Schaum's Outline of Theory and Problems of Electric Circuits

AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
90

Fig. 5-38 [CHAP. 5

v C 21 v C v C v C v 2
þ þ þ ¼ 0 ð40Þ
3 6 3 8
Substituting v C ¼ ð3=5Þv 2 from (39) into (40) we get v 2 ¼ 10 V. Then

v C ¼ 6V
i 1 ¼ð21 v C Þ=3000 ¼ 0:005 A ¼ 5mA
R in ¼ 21=i 1 ¼ 21=0:005 ¼ 4200
¼ 4:2k

5.14 In the circuit of Fig. 5-38 change the 21-V source by a factor of k. Show that v C , i 1 , v 2 in
Problem 5.13 are changed by the same factor but R in remains unchanged.

Let v s ¼ 21k (V) represent the new voltage source. From the inverting amplifier we have [see (39)]
v 2 ¼ ð5=3Þv C
Apply KCL at node C to obtain [see (40)]
v C v s v C v C v C v 2
þ þ þ ¼ 0
3 6 3 8
Solving for v C and v 2 , we have
v C ¼ð6=21Þv s ¼ 6k ðVÞ and v 2 ¼ ð10=21Þv s ¼ 10k ðVÞ
i 1 ¼ðv s v C Þ=3000 ¼ð21 6Þk=3000 ¼ 0:005k A
R in ¼ v s =i 1 ¼ 21k=0:005k ¼ 4200
These results are expected since the circuit is linear.

5.15 Find v 2 and v C 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:
ð6Þð3Þ 6
R Th ¼ ¼ 2k
and v Th ¼ ð21Þ¼ 14 V
6 þ 3 3 þ 6
Replace the circuit to the left of node C by the above v Th and R Th and then apply KCL at C:
v C 14 v C v C v 2
þ þ ¼ 0 ð41Þ
2 3 8
For the inverting amplifier we have v 2 ¼ ð5=3Þv C or v C ¼ 0:6 v 2 , which results, after substitution in (41),
in v 2 ¼ 10 V and v C ¼ 6V.