Page 88 - Schaum's Outline of Theory and Problems of Electric Circuits

AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
CHAP. 5]
77
5
v 2 ¼ ð9=5Þv 1 ¼ ð9=5Þ v s ¼ 1:5v s
6
v o ¼ ð6=1:2Þv 2 ¼ 5ð 1:5v s Þ¼ 7:5v s
v s
i s ¼ i 1 ¼ ðAÞ¼ 0:166v s ðmAÞ
6000
i f ¼ 0
(b) R f ¼ 40 k
. From the inverting op amps we get v o ¼ 5v 2 and v 2 ¼ ð9=5Þv 1 so that v o ¼ 9v 1 . Apply KCL
to the currents leaving node B.
v 1 v s v 1 v 1 v o
þ þ ¼ 0 ð19Þ
1 5 40
Substitute v o ¼ 9v 1 in (19) and solve for v 1 to get

v 1 ¼ v s
v 2 ¼ ð9=5Þv 1 ¼ 1:8v s
v o ¼ ð6=1:2Þv 2 ¼ 5ð 1:8v s Þ¼ 9v s
v s v 1
i s ¼ ¼ 0
1000
Apply KCL at node B.
v 1 v s
i f ¼ i 1 ¼ ðAÞ¼ ðAÞ¼ 0:2v s ðmAÞ
5000 5000
The current i 1 in the 5-k
input resistor of the first op amp is provided by the output of the second op amp
through the 40-k
feedback resistor. The current i s drawn from v s is, therefore, zero. The input resistance of
the circuit is infinite.

5.11 INTEGRATOR AND DIFFERENTIATOR CIRCUITS
Integrator
By replacing the feedback resistor in the inverting amplifier of Fig. 5-13 with a capacitor, the basic
integrator circuit shown in Fig. 5-23 will result.

Fig. 5-23

To obtain the input-output relationship apply KCL at the inverting node:
1
v 1 dv 2 dv 2
þ C ¼ 0 from which ¼ v 1
R dt dt RC
ð
1 t
and v ¼ v dt (20)
2
1
RC 1
In other words, the output is equal to the integral of the input multiplied by a gain factor of 1=RC.

EXAMPLE 5.17 In Fig. 5-23 let R ¼ 1k
, C ¼ 1 mF, and v 1 ¼ sin 2000t. Assuming v 2 ð0Þ¼ 0, find v 2 for t > 0.