# Practical Opamp integrator problem The opamp has an unity gain of 1 MHz.So when the resistor in parallel acts a feedback to bias the opamp stable.So if not for the capacitor,the circuit would have been a inverting amplifier with gain 100 => bandwidth would be 10Khz (from GBP).

However, when i use a capacitor to implement a low pass filter ,the range of frequency over which it can integrate is set between 1/ 2pi(100 * 160p) and 1/2pi(10k * 160p) ie; between 100Khz and 10Mhz respectively.

My question is will the Gain bandwidth product have any significant role in limiting the bandwidth down to 10Khz (if it does so).

Also,if thats the case then if i input the signal within (100KHz -10MHz) range in the hope of getting a integrated output ,will i get one? Shouldn't the input be attenuated because of the opamp frequency limitation?

what will be the bandwidth ? is it 10K or 100kHz

ps: i have assumed a one pole model for the opamp (obvious simple case)

## 1 Answer

The answer is simple: For an ideal opamp your circuit provides a lowpass function with max gain of 40 dB and a cut-off (-3dB) at app. 100kHz. However, the real opamp (GBW=1MHz) allows a gain of 40 dB only up to app. 10kHz. Hence, it is the open loop gain of the opam that dominates and determines the frequency response of the whole circuit (3dB cut-off at app. 10kHz)