It can be mathematically proven that when we have a resistor in the negative feedback of an op-amp, the input voltages in the inverting and non-inverting terminals get equal.

I'm wondering if we can say the same when we have a capacitor rather than a resistor in the feedback (like the integrator op-amp circuit).

enter image description here

If yes, can you tell me how we could prove that mathematically?

  • \$\begingroup\$ Depends on the frequency. At infinite frequency, yes. At DC, no. Everything else is somewhere in between. (In theory, and assuming an ideal opamp, etc...) \$\endgroup\$
    – Kyle B
    Sep 15, 2022 at 21:03
  • 2
    \$\begingroup\$ It's not 'mathematically proven', it's assumed to be approximately true, when the opamp is operating within its limits (frequency of operation, output voltage and current). Changing the type of feedback component doesn't change that. \$\endgroup\$
    – Neil_UK
    Sep 15, 2022 at 21:04
  • \$\begingroup\$ Tomas, I think @Neil_UK writes it out very well. Proofs start with axioms/premises and are achieved by applying valid logic. In this case, one starts with an axiom that the circuit is arranged in such a way that the opamp can control its output such that the two inputs are equal to each other. It also includes other axioms related to the ideality of an opamp. But given these, one can write out a provable dynamical time-domain statement. Of course if any of the axioms are wrong, then the proof obviously fails. \$\endgroup\$
    – jonk
    Sep 15, 2022 at 21:33

2 Answers 2


First of all, in your diagram, the current through and voltage across the feedback capacitor are in correct. See my diagram.


simulate this circuit – Schematic created using CircuitLab

The voltage \$v_{P}-v_{N}\$ determined by the quality of the differential amplifier input stage, and the open loop gain of the amplifier. So even if the input stage is perfect, then mathematically$$v_{P}-v_{N}=\frac{v_{out}}{Avol}$$ whether there is a resistor, capacitor, inductor, or other network in the feedback path. The only condition is that the amplifier must be operating in its linear region.

This is not a proof, It is the way the amplifier is designed.


I'm wondering how the feedback can decrease the Vp-Vn to zero.

Sometimes I think the word "feedback" gets us off track. The input current divides into \$I_{N}\$ and \$I_{f}\$. The currents must flow somewhere. If the output cannot absorb the current \$I_{f}\$, then \$I_{N}\$ will increase causing \$V_{N}\$ to increase. This in turn drives the output voltage more negative. This allows the output to absorb more current thus reducing \$I_{N}\$. Notice that this mechanism is the same for any feedback impedance, not just capacitors.

  • \$\begingroup\$ Of course, in order to keep the output of the amplifier within the linear region, we need the keep Vp-Vn as small as possible (almost zero). But the amplifier does not do it itself. We should do it by adding the feedback. Now the question is how can feedback make it happen? In case of resistor feedback, it is almost clear. However, in case of capacitor, I'm wondering how the feedback can decrease the Vp-Vn to zero. \$\endgroup\$
    – Tomas
    Sep 16, 2022 at 6:26
  • \$\begingroup\$ @Tomas: See my edit \$\endgroup\$
    – RussellH
    Sep 16, 2022 at 14:28

With an ordinary 2 resistor feedback amplifier configuration, as frequency is increased, the difference voltage between the op amp's inputs increases. This happens because, as frequency increases, the op amp's open loop gain falls off at -20 dB/decade and so the difference voltage amplitude between the inputs must increase in order to try and keep the output at a constant amplitude.

With the integrator shown. As frequency increases, not only does the open loop gain fall off at -20 dB/decade but also the output falls off at -20 dB/decade. This means that the the difference voltage between the op amp's inputs stays at a fairly constant amplitude.

In the case of a conventionally configured 2 resistor feedback amplifier the peak to peak difference voltage between the inputs can become quite large.

For example, assuming the open loop gain is on its -20 dB/decade roll-off stage of its frequency response (-90 degrees).

Then, for a non-inverting configuration, at the closed loop -3 dB frequency, The peak to peak difference voltage between the inputs will be 0.707 times the peak to peak input signal.

For an inverting amplifier with the same feedback beta, at it closed loop -3 dB frequency, the open loop gain will have the same value as for the non-inverting configuration. For the same magnitude of input signal, its output will be 1-beta times smaller and because its open loop gain is the same at this frequency, the peak to peak difference voltage between its inputs will also be 1-beta times smaller or 0.707*(1-beta)*the magnitude of the input signal.


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