Page 97 - Schaum's Outline of Theory and Problems of Electric Circuits

AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
86
ck [CHAP. 5
v 2 R i
¼ 5 where c ¼ ð36Þ
v s 1 þ ck 5 þ 6R i
(b) For R i ¼ 1k
, c ¼ 1=11 which, substituted into (36), gives
5k
v 2
¼ ð37Þ
v s 11 þ k
For R i ¼1 we get c ¼ 1=6 and so
5k
v 2
¼ ð38Þ
v s 6 þ k
Table 5-6 gives values of v 2 =v s in (37) and (38) versus k. Note that (38) is identical with (32).

Table 5-6

v 2 =v s
k R i ¼ 1k
R i ¼1
1 0:31 0:71
10 2:38 3:12
100 4:51 4:72
1000 4:95 4:97
1 5:00 5:00

(c) Comparing the two columns in Table 5-6 we see that the smaller R i reduces the overall gain G .
However, as the open-loop gain k increases, the effect of R i is diminished. As k becomes very
large, v 2 =v 1 approaches 5 unless R i ¼ 0.

5.5 Let again R 1 ¼ 1 k
and R 2 ¼ 5 k
in the circuit of Fig. 5-33. Replace the circuit to the left of
node A including v s , R 1 , and R i by its The ´ venin equivalent. Then use (5) to derive (36).

The The ´ venin equivalent is given by
R i v s R i v s
v Th ¼ ¼
R 1 þ R i 1 þ R i
R 1 R i R i
R Th ¼ ¼
R 1 þ R i 1 þ R i
where the resistors are in k
:
From (5),
k
v 2 ¼ð1 bÞ v Th
1 þ bk
R Th R i 5ð1 þ R i Þ
where b ¼ ¼ and 1 b ¼
R Th þ R 2 6R i þ 5 6R i þ 5
Therefore,
5ð1 þ R i Þ k R i 5R i k
v 2 ¼ v s ¼ v s
6R i þ 5 1 þ R i k=ð6R i þ 5Þ 1 þ R i 6R i þ 5 þ R i k
which is identical with (36).

5 þ
5.6 Find the output voltage of an op amp with A ¼ 10 and V ¼ 10 V for v ¼ 0 and v ¼ sin t (V).
cc
Refer to Figs. 5-7 and 5-8.