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I've been trying to model a circuit where I'm getting the gain of a non-inverting op amp connected to two probes. I've isolated the following feedback circuit and am trying to work out the transfer function for it.

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    \$\begingroup\$ Jim Brown - Hi, Please don't remove the question's content after it has been answered. Part of the point of Stack Exchange is to collect questions and their answers for the future. I have reversed that edit. Attempting to remove the question is classed as vandalism and is not allowed. Please don't do that again. Thanks. \$\endgroup\$
    – SamGibson
    Oct 26, 2023 at 14:50

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The determination of the transfer function is quite complicated if you consider the open-loop gain of this op-amp - otherwise there is no fun : )

I have used the fast analytical circuits techniques or FACTs to obtain the complete transfer function, ignoring, of course, the internal op-amp ac response, only its open-loop gain \$A_{OL}\$:

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For a very large open-loop gain, the dc gain simplifies to that of a non-inverting configuration \$\frac{R_1}{R_2}+1\$ and the pole is pushed to infinity. If the capacitor is now increased to 1 µF with an open-loop gain of 100k, despite the high-value of the capacitor, the pole will be located at 1.24 MHz. This circuit is a classic to improve robustness of op-amp-based circuits to capacitive loading. It is discussed here.

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  • \$\begingroup\$ With pleasure but in your sketch - now it is alone - we don't know where is the stimulus (the input) and where is the response (the output)? In lack of op-amp, it is a simple \$RC\$ low-pass filter. \$\endgroup\$ Oct 26, 2023 at 11:11
  • \$\begingroup\$ Perhaps, but the op-amp adds its active part and changes the ac characteristics: you cannot dissociate the two. \$\endgroup\$ Oct 26, 2023 at 14:13

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