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A 50 kHz square wave is to be amplified by an op-amp to have an output voltage swing ± 10 V. Two op-amp are available A has slew rate of 0.5 V/µs and B has slew rate of 13 V/µs. Then correct answer is:
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- Both op-amp A & B are suitable for amplified a 50 kHz square wave
- Op-amp A is suitable But B is not
- Op-amp B is suitable But B is not
- Both the op-amp A & B are not suitable for amplified a 50 kHz
Correct Option: C
Given that output voltage swing = ± 10 V and input wave is square wave with 50 kHz
Slew rate for op-amp A = 0.5 V/µs.
Slew rate for op-amp B = 13 V/µs.
Maximum output voltage after passing a square wave can be calculated as first
Check for op-amp A
| Slew rate of op-amp A = | ||
| 106 |
0.5 × 106 = 2π × 50 × 103. Vm
| or Vm = | ||
| 314 |
Vm = 1.59 V
Since this voltage is less than the output voltage swing i.e. ± 10 V.
Therefore op-amp A is not suitable for amplified a 50 kHz square wave.
Now check for op-amp B
| Vm = | ||
| 2πf |
| = | = 41.4 V | |
| 2 × 3·14 × 50 × 103 |
Since this voltage is higher than the output voltage swing i.e. ± 10 V.
Therefore op-amp B is suitable for amplifying 50 kHz square wave.
Hence alternative (C) is the correct choice.
