CHAP. 5]

87

AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS

Because of high gain, saturation occurs quickly at jv2 j ¼ 105 jvd j ¼ 10 V

jvd j ¼ 104 V

or

We may ignore the linear interval and write  v2 ¼ where vd ¼ vþ  v ¼ sin t (V).

þ10 V 10 V

vd > 0 vd < 0

One cycle of the output is given by  þ10 V 0<t< v2 ¼ 10 V  < t < 2

For a more exact v2 , we use the transfer characteristic of the op amp in Fig. 5-7. 8 < 10 vd < 104 V 5 4 v2 ¼ 10 vd 10 < vd < 104 V : þ10 vd > 104 V Saturation begins at jvd j ¼ j sin tj ¼ 104 V. output v2 is then given by v2 v2 v2 v2

Since this is a very small range, we may replace sin t by t.

¼ 105 t ¼ 10 ¼ 105 ðt  Þ ¼ 10

 104 104   104  þ 104

The

< t < 104 s < t <   104 s < t <  þ 104 s < t < 2  104 s

To appreciate the insigniﬁcance of error in ignoring the linear range, note that during one period of 2 s the interval of linear operation is only 4  104 s, which gives a ratio of 64  106 .

5.7

Repeat Problem 5.6 for vþ ¼ sin 2t (V) and v ¼ 0:5 V. The output voltage is 10 V

when vþ > v

v2 ¼ 10 V

when vþ < v

v2 ¼

Switching occurs when sin 2t ¼ 1=2. cycle of v2 is given by

This happens at t ¼ 1=12, 5/12, 13/12, and so on.

v2 ¼ 10 V v2 ¼ 10 V

Therefore, one

1=12 < t < 5=12 s 5=12 < t < 13=12 s

Figure 5-34 shows the graphs of vþ , v , and v2 .

5.8

In the circuit of Fig. 5-35 vs ¼ sin 100t.

Find v1 and v2 .

At nodes B and A, vB ¼ vA ¼ 0. Then, 30 v ¼ 0:6 sin 100t ðVÞ 20 þ 30 s 100 100 v2 ¼  v ¼ ð0:6 sin 100tÞ ¼ 2 sin 100t ðVÞ 30 1 30 100 v ¼ 2 sin 100t ðVÞ v2 ¼  20 þ 30 s v1 ¼

Alternatively,

5.9

Saturation levels for the op amps in Fig. 5-31 are þVcc ¼ 5 V and Vcc ¼ 5 V. The reference voltage is vo ¼ 1 V. Find the sequence of outputs corresponding to values of vi from 0 to 1 V in steps of 0.25 V.

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An