Explanation: In an op-amp integrator the effective input capacitance becomes CF×(1-Av). Where Av is the gain of op-amp. The gain is infinite for ideal op-amp. So, effective time constant of the op-amp integrator becomes very large which results in perfect integration. 5. The op-amp operating in open loop result in output of the amplifier to saturate at a voltage
A. Close to op-amp positive power supply
B. Close to op-amp negative power supply
C. Close to op-amp positive or negative power supply
D. None of the mentioned
Answer:A
Explanation: In an op-amp integrator the effective input capacitance becomes CF×(1-Av). Where Av is the gain of op-amp. The gain is infinite for ideal op-amp. So, effective time constant of the op-amp integrator becomes very large which results in perfect integration. 5. The op-amp operating in open loop result in output of the amplifier to saturate at a voltage
A. Close to op-amp positive power supply B. Close to op-amp negative power supply C. Close to op-amp positive or negative power supply D. None of the mentioned
Explanation: In practice, the output of op-amp never becomes infinite rather the output of the op-amp saturate at a voltage close to op-amp positive or negative power supply depending on the polarity of the input dc signal. 6. The frequency at which gain is 0db for integrator is