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In the circuit of fig. shown the op-amp slew rate is SR = 0.5 V/µs. If the amplitude of input signal is 0.02 V, then the maximum frequency that may be used is:
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- 0.55 × 106 rad/s
- 0.55 rad/s
- 1.1 × 106 rad/s
- 1.1 rad/s
Correct Option: C
Max. output voltage
| Vo = Vi |  | – |  | = 24 × 0.02 | |
| 10 |
= 0.48 V
S.R. 2 π fmax. Vmax
or S.R.. Vo
| or | ||
| Vo |
| or = | = 1.1 × 106 rad/s. | |
| 0·48 |
