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In an answer to a question, I was told that the feedback capacitor appears as a load to the op-amp in an inverting integrator topology. If that is the case, we should see a pole with the output resistance, which should erode the phase margin as the capacitance becomes larger, pushing the op-amp towards instability. I was not so sure about this, so I tried to draw the block diagram of the inverting integrator(Ri and C are the integration resistor and capacitors while ro is the output resistance of the op-amp.)

BD

There is a pole in the forward path, but a zero that cancels it.

From a different point of view, as the capacitance becomes larger, the integrator tends to become a voltage follower at a certain frequency, which is a stable topology as mentioned here. Which is the correct explanation in this case? If I use a larger capacitor am I pushing my integrator towards instability?

Inverting integrator

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    \$\begingroup\$ I don't see how it becomes a voltage follower. If you short C the voltage doesn't follow anything. \$\endgroup\$ Apr 5, 2022 at 8:15
  • \$\begingroup\$ Indeed, if we imagine C shorted through for higher frequencies, then op amp must output 0 to make inputs equal. Op amp output will sink all Iin coming through Rin. \$\endgroup\$
    – Ilya
    Apr 5, 2022 at 8:20
  • \$\begingroup\$ For "higher frequencies" (when C acts as a short) the open-loop gain is drastically reduced which means: The input voltages are NOT equal anymore. \$\endgroup\$
    – LvW
    Apr 5, 2022 at 8:38
  • \$\begingroup\$ @Sambeet P., your first box (input function) is not correct. For r_out=0 this box must contain a simple 1st-order lowpass; 1/(1+sRinC). \$\endgroup\$
    – LvW
    Apr 5, 2022 at 8:42
  • \$\begingroup\$ I changed the block diagram. Thank you for the comment. There appears to be a pole-zero cancellation in the forward path.But as the effect of pole is cancelled , should I be worried about stability here? \$\endgroup\$ Apr 5, 2022 at 10:08

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... the feedback capacitor appears as a load to the op-amp in an inverting integrator topology.

Yes, the capacitor appears as a load ...

... the integrator tends to become a voltage follower ...

As @ilya pointed ... It is not a "voltage follower", it should be a "current follower" ...

If the "opamp" can, it pushes only the current "asked" from the input circuit, i.e. \$ Iin = Vin / Rin \$ ...

Here is a Maple sheet with a picture, EE&O ..., which shows the behavior.
Into account : opamp gain versus f, opamp output impedance (rs=0..100).

enter image description here

enter image description here

I noted just a "problem" ... when rs is greater than ~ 200 Ohm vs Rin=1000 Ohm ...

enter image description here

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  • \$\begingroup\$ Should I be worried about stability as I increase the capacitance though, as the block diagram above shows a pole-zero cancellation? \$\endgroup\$ Apr 6, 2022 at 4:26
  • \$\begingroup\$ EE&O ... I add a picture in the answer ... thus, no worries, I think :-) \$\endgroup\$
    – Antonio51
    Apr 6, 2022 at 10:37

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