Page 98 - Schaum's Outline of Theory and Problems of Electric Circuits

AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
CHAP. 5]
Because of high gain, saturation occurs quickly at 87
5
jv 2 j¼ 10 jv d j¼ 10 V or jv d j¼ 10 4 V
We may ignore the linear interval and write

þ10 V v d > 0
v 2 ¼
10 V v d < 0

þ
where v d ¼ v v ¼ sin t (V). One cycle of the output is given by

þ10 V 0 < t <
v 2 ¼
10 V < t < 2
For a more exact v 2 , we use the transfer characteristic of the op amp in Fig. 5-7.
8
< 10 v d < 10 4 V
5
v 2 ¼ 10 v d 10 4 < v d < 10 4 V
: 4
þ10 v d > 10 V
Saturation begins at jv d j¼j sin tj¼ 10 4 V. Since this is a very small range, we may replace sin t by t. The
output v 2 is then given by
5
v 2 ¼ 10 t 10 4 < t < 10 4 s
v 2 ¼ 10 10 4 < t < 10 4 s
5 4 4
v 2 ¼ 10 ðt Þ 10 < t < þ 10 s
v 2 ¼ 10 þ 10 4 < t < 2 10 4 s
To appreciate the insignificance of error in ignoring the linear range, note that during one period of 2 s
6
the interval of linear operation is only 4 10 4 s, which gives a ratio of 64 10 .
þ

5.7 Repeat Problem 5.6 for v ¼ sin 2 t (V) and v ¼ 0:5V.
The output voltage is
þ
v 2 ¼ 10 V when v > v
þ
v 2 ¼ 10 V when v < v
Switching occurs when sin 2 t ¼ 1=2. This happens at t ¼ 1=12, 5/12, 13/12, and so on. Therefore, one
cycle of v 2 is given by
v 2 ¼ 10 V 1=12 < t < 5=12 s
v 2 ¼ 10 V 5=12 < t < 13=12 s

þ
Figure 5-34 shows the graphs of v , v , and v 2 .
5.8 In the circuit of Fig. 5-35 v s ¼ sin 100t. Find v 1 and v 2 .
At nodes B and A, v B ¼ v A ¼ 0. Then,
30
v 1 ¼ v s ¼ 0:6 sin 100t ðVÞ
20 þ 30
100 100
v 2 ¼ v 1 ¼ ð0:6 sin 100tÞ¼ 2 sin 100t ðVÞ
30 30
100
Alternatively, v 2 ¼ v s ¼ 2 sin 100t ðVÞ
20 þ 30

5.9 Saturation levels for the op amps in Fig. 5-31 are þV cc ¼ 5 V and V cc ¼ 5 V. The reference
voltage is v ¼ 1 V. Find the sequence of outputs corresponding to values of v from 0 to 1 V in
i
o
steps of 0.25 V.