Question

An op-amp-based inverting integrator is measured at 10 kHz to have a voltage gain of -104 V/V. At what frequency is its gain reduced to -1 V/V? What is the integrator time constant?

Answer 1

The frequency at which the gain of the integrator is reduced to -1 V/V can be determined by the inverse relationship between gain and frequency, where the gain is inversely proportional to the product of frequency and the integrator's time constant.

Step-by-step explanation:

To determine at what frequency the gain of an op-amp-based inverting integrator is reduced to -1 V/V from -104 V/V at 10 kHz, we use the concept that the magnitude of the gain in the integrator decreases inversely with frequency. If a gain of -104 V/V corresponds to 10 kHz, then by the inverse proportion, the frequency needed for the gain to become -1 V/V can be calculated. The gain of an integrator is given by -1/(2πfRC), where R is the resistance, C is the capacitance, and f is the frequency. Given that the gain has decreased by a factor of 104, the frequency must therefore increase by the same factor.