CHAP. 5]

AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS

v2 ¼ ð9=5Þv1 ¼ ð9=5Þ

77

  5 vs ¼ 1:5vs 6

vo ¼ ð6=1:2Þv2 ¼ 5ð1:5vs Þ ¼ 7:5vs v is ¼ i1 ¼ s ðAÞ ¼ 0:166vs ðmAÞ 6000 if ¼ 0 (b) Rf ¼ 40 k. From the inverting op amps we get vo ¼ 5v2 and v2 ¼ ð9=5Þv1 so that vo ¼ 9v1 . Apply KCL to the currents leaving node B. v1  vs v1 v1  vo þ þ ¼0 ð19Þ 1 5 40 Substitute vo ¼ 9v1 in (19) and solve for v1 to get v1 ¼ vs v2 ¼ ð9=5Þv1 ¼ 1:8vs vo ¼ ð6=1:2Þv2 ¼ 5ð1:8vs Þ ¼ 9vs v  v1 ¼0 is ¼ s 1000 Apply KCL at node B. if ¼ i1 ¼

v1 v ðAÞ ¼ s ðAÞ ¼ 0:2vs ðmAÞ 5000 5000

The current i1 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 is drawn from vs 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: v1 dv þC 2 ¼0 R dt and

dv2 1 v ¼ RC 1 dt

from which

v2 ¼ 

1 RC

ðt v1 dt

(20)

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 1 k, C ¼ 1 mF, and v1 ¼ sin 2000t. Assuming v2 ð0Þ ¼ 0, ﬁnd v2 for t > 0.

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An