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For an inverting voltage follower, C1 would be part of an LP filter. But in the depicted non-inverting voltage follower circuit, what is the purpose of C1 (if any)?

schematic

simulate this circuit – Schematic created using CircuitLab

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The purpose of the resistor R2 is to eliminate the DC offset caused by the op-amp input bias currents. If the bias currents are exactly matched, then the voltage drop across each 1K resistor (R1 and R2) is the same and the output voltage is equal to the input voltage.

Fine, but if the op-amp has significant input capacitance or there is a lot of stray capacitance on the traces then the feedback signal has a lag and the AC output may be higher than the input. In an extreme case (very high resistance and an op-amp with a lot of input capacitance), it could oscillate. Putting the capacitor across the resistor deals with that.

1K is such a low value that this would typically not be a problem in practice.

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  • \$\begingroup\$ With "very high resistance" you mean what resistance exactly? (R2?) How does the capacitor help with the oscillation? \$\endgroup\$ – apriori Aug 15 '15 at 13:23
  • \$\begingroup\$ A stray capacitance at the inverting input will add additional phase shift to the loop gain. In any case, this will degrade stability of the closed-loop circuit - in some critical cases (unity gain configuration) this can lead to oscillations. A feedback capacitor will introduce some phase enhancement to the loop gain (high pass effect). This is a kind of compensation and helps to stabilize the circuit. \$\endgroup\$ – LvW Aug 15 '15 at 13:38
  • \$\begingroup\$ With zero ohms for R1 and R2, the circuit (as drawn) would behave so why have R1 and R2? If the sine wave source is in fact a non-zero ohm impedance then R1 should not equal R2. In an inverting configuration nobody seems to worry about not putting a cap across the feedback resistor yet, the input parasitic capacitance to ground is still present and would still potentially cause oscillation on a small number of op-amps. Maybe I can't see the wood for the trees. This comment is also sent to @DaveTweed. Once Spehro has explained my errors I'd be quite happy to delete my own post. \$\endgroup\$ – Andy aka Aug 15 '15 at 16:48
  • \$\begingroup\$ @Andyaka I did not see an error in your post. I don't know why they put R1 there- one reason might be for ESD or overvoltage protection if V1 spikes beyond the supply rails. Or so a cap could be added from the non-inverting input to ground for a LPF. Anyway, the question was why C1 and I tried to answer that. Tough audience today I guess. \$\endgroup\$ – Spehro Pefhany Aug 15 '15 at 19:31
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This Answer has been edited (see below the original answer) because it has received down votes and was closed off. I requested that it be re-opened because I think the down-voters could not understand what I was trying to say.


If OA1 is assumed to be an ideal op-amp then R1, R2 and C1 serve no purpose at all. You might just as well short the resistors and leave the cap off. Then again, V1 is shown as a voltage source so the op-amp is totally redundant too - you might as well connect the sine wave voltage source to whatever connects to the op-amp's output: -

enter image description here


EDIT SECTION


I am still saying the addition of the capacitor has no practical purpose and this may call into question Spehro's answer because the reasons I give contradict it. Consider a simple inverting op-amp configuration: -

enter image description here

Then ask yourself the question, does this work (i.e. not oscillate) without adding a feedback capacitor thus: -

enter image description here

The answer is "YES".

In other words it doesn't oscillate because of parasitic input capacitance to ground unless of course it is a totally badly designed PCB. If the question asked was related to the inverting configuration (why a capacitor was placed across R2) the great and the good would say that it was acting as a low pass filter.

I asked Spehro what he thought and he said "I did not see an error in your post" so, given he is a top contributor to this site I can only conclude that some amount of bad-thinking has been going on.

So, down-voters, please do consider what I'm saying and try looking a little bit further past the end of your noses.

And, if you think I'm being rude or insulting, that's the price for me taking the time to ressurect this answer from the grave in order to give the readers of stack-exchange a little more to think about.

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  • \$\begingroup\$ Andy - sorry, but I am confused about your update. At first, are you speaking of inverting or non-inv. circuits (note you have shown positive feedback!) ? And, secondly, I think the question is not "oscillate yes/no" but "stability improved yes/no" ? Am I wrong? \$\endgroup\$ – LvW Aug 16 '15 at 8:44
  • \$\begingroup\$ @LvW - oops I have picked up a wrong diagram from the web - thank you for pointing this out - the inputs should be swapped - I will update immediately. \$\endgroup\$ – Andy aka Aug 16 '15 at 9:04
  • \$\begingroup\$ @LvW hopefully done and I'd appreciate your feedback on why people would add a capacitor across the feedback resistor in a non-inverting configuration but choose not to do it when the config is inverting - maybe I'm in danger of learning something here!! \$\endgroup\$ – Andy aka Aug 16 '15 at 9:07
  • \$\begingroup\$ Andy-Stability properties depend on loop gain only. And the loop gain for inverting and non-inv. configurations is, of course, identical for equal resistors in both cases. Hence, as far as stability and phase compensation with a feedback cap is concerned, I see no difference between both cases. Why do you think, people don`t do that for inverting configurations? \$\endgroup\$ – LvW Aug 16 '15 at 9:28
  • \$\begingroup\$ Andy-a "drastic" example for an inverting configuration with a feedback capacitor (for stability reasons) is the classical inverting differentiating circuit. \$\endgroup\$ – LvW Aug 16 '15 at 9:46

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