Integrator circuit using operational amplifier

[vvkannan]

I read in a book that perfect integration using inverting op-amp is possible only when the time constant of the circuit(R1Cf) is lesser than the time period of the input signal.
I have trouble in understanding this concept.

Lets assume i give a square wave as input at the inverting terminal.Now the output would ramp down when the input is positive and since the time constant is less than time period would not the capacitor charge soon (before the cycle completes)and behave as an open circuit making the op-amp work as though in open-loop configuration?
Now since the open loop gain is very high would it not saturate?
How then can it integrate properly?
On the other hand if time constant is greater than time period of the input signal can't it integrate properly?

 

[Valkarie]

The key to understanding this concept is to understand that the output of an inverting op-amp is actually a low-pass filtered version of the input signal. This means that the output will not respond to changes in the input signal that occur faster than the time constant of the circuit (R1Cf). When the time constant is less than the time period of the input signal, the output will closely follow the input signal and the circuit will integrate properly.If the time constant is greater than the time period of the input signal, then the output will not be able to keep up with the changing input signal, and the circuit will not be able to integrate properly. The output will lag behind the input and may even saturate if the time constant is too large.

 

[piercas]

I can understand your confusion about the concept of perfect integration using an inverting op-amp. Let me explain it in simpler terms.

An integrator circuit using an operational amplifier is a circuit that performs mathematical integration on the input signal. This means that the output of the circuit is the integral of the input signal over time. In order for this integration to be accurate, the time constant of the circuit (R1Cf) needs to be smaller than the time period of the input signal.

Now, in the scenario you described where a square wave is given as input, the output of the circuit will indeed ramp down when the input is positive. However, the capacitor will not charge quickly enough to behave as an open circuit before the cycle completes. This is because the time constant is smaller than the time period, allowing the capacitor to charge and discharge multiple times during one cycle.

If the time constant is larger than the time period, the capacitor will not have enough time to charge and discharge properly, leading to inaccurate integration. This is because the capacitor will not have enough time to fully charge or discharge before the input signal changes.

In conclusion, in order for an integrator circuit using an inverting op-amp to work properly, the time constant needs to be smaller than the time period of the input signal. This allows the capacitor to charge and discharge accurately, resulting in a proper integration of the input signal. I hope this explanation helps you understand the concept better.