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I was trying to understand the 2nd answer from this post here, with the circuit of the post shown below. The 2nd answer has equations including open loop gain I wasn't able to understand (1st question).

It has the following equations:

Loop Gain = \$ A_{ol} \frac{R_1}{R_1 + R_F} \frac{1}{s/w_p +1}\$ and \$A_{ol} = \frac{A_{DC}}{(S/W_1+1)(S/W_2+1)}\$

The original post tries to explain why one may want to have a capacitor in parallel with the feedback resistor.

I know that it is to kill the high frequency noise generated from the op-amp.

But if you look at the output of the op amp and see how it would go to the inverting input through the feedback resistor, doesn't the feedback resistor and the parasitic capaitance, Cp, form a low pass filter? (2nd question)

enter image description here

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  • \$\begingroup\$ All op-amps have a upper roll-off point due to parasitic capacitance. The parasitic capacitance of Rf is > zero. \$\endgroup\$ – Sparky256 Oct 4 '18 at 21:53
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    \$\begingroup\$ To your question: Yes - the feedback function is a lowpass filter.....and that is the problem because it adds additional phase shift to the loop gain and, thus, reduces the stability margin. \$\endgroup\$ – LvW Oct 5 '18 at 13:45
  • \$\begingroup\$ Continue: What makes a capacitor across RF ? It reduces the loop gain for higher frequencies and can bring the magnitude of the loop gain below unity before the phase reaches the critical value of -180 deg. . \$\endgroup\$ – LvW Oct 5 '18 at 16:08
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You don't want to put the capacitor there between the negative input and ground, because that node is a virtual ground. In a perfect world, that capacitor would have no effect, because the negative input stays at the same level as ground. In reality, there are likely to be stability issues. The capacitor in that schematic represents the parasitic capacitance of the input.

By putting the capacitor in parallel with the feedback resistor, you're not trying to fight the operation of the op amp any more, but reducing its gain at higher frequencies using the feedback path. Even at frequencies where the op amp doesn't respond significantly, it's still a high frequency bypass to a low impedance. This gives greater control overall, including compensating for the parasitic capacitance.

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  • \$\begingroup\$ In your answer, it's not apparent what "there" is. Can you clarify? \$\endgroup\$ – Scott Seidman Oct 5 '18 at 13:25

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