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I built this circuit and simulated it in CircuitLab and was trying to make an active low pass filter. However, it shows a frequency response like a resonant low pass filter and I can't figure out why.

Here's the schematic: Here's the schematic And the frequency response: Plot of the frequency response

The gain of the frequencies below the resonant frequency is set by the ratio of R2/R1 (x10 with these values), as expected for an inverting op-amp circuit. But I can't figure out why there is the resonant peak at 800 Hz for this first-order circuit, or how to calculate the resonant frequency.

And the response of a few different cap values: Sweep of cap values

And the response of different values of R1: Sweep of R1 values

Changing the value of R1 has no effect on the resonant frequency, only on the gain of the circuit. I don't understand why this circuit behaves this way. Can anyone explain it to me?

I came up with this equation for the value of \$V_{out}\$:

$$V_{out} = -( \frac{A \times R2}{R1 + R2 + R1R2Cs - A \times R1} ) V_{in} $$

Does this equation look right? It makes sense to me that when \$(R1 + R2 + R1R2Cs) << A \times R1 ,\$ the result is approximately equal to R2 / R1. But I have no idea where the peak at 800 Hz comes from.

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  • \$\begingroup\$ It is obviously not a first-order circuit, as first order circuits cannot exhibit resonance. Suggest you try to include the pole in the op-amp response. It's usually at something like 10Hz. \$\endgroup\$ – Spehro Pefhany Jan 30 '14 at 18:26
  • \$\begingroup\$ How do I do that? Are you saying that the second order behavior is due to a parasitic capacitance? \$\endgroup\$ – nonex Jan 30 '14 at 18:54
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    \$\begingroup\$ No, it's because the op-amp has a capacitor inside it for frequency compensation (to make it stable with unity gain). Above the pole frequency, the op-amp gain drops by -20dB/decade. Sometimes there's a second pole. \$\endgroup\$ – Spehro Pefhany Jan 30 '14 at 19:22
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I think, if you assume that the internals of the op-amp contain, in effect, a first order low pass filter, you will have created for yourself what is known as a multiple feedback low-pass filter. I used a great simulation tool from Mister Okawa here to produce this: -

enter image description here

If you look closely at the circuit in the picture above and imagine that R2 and C2 are inside the op-amp, your circuit becomes the same. There is some hand-waving here because I'm taking a stab in the dark about what R2 and C2 actually are and "massaging" them numerically to fit close to producing a peak near 800 Hz.

C2 being 5pF is in the right order for the "conventional" stabilizing stage inside an old-fashioned op-amp like the TL081 and I guess R2 would be in the vicinity of a few kohm.

Anyway I'm convinced this is what is happening!!!

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  • \$\begingroup\$ Very nice and innovative application of a existing tool. \$\endgroup\$ – Spehro Pefhany Jan 31 '14 at 3:07
  • \$\begingroup\$ Agree with above comment. But does this mean that any active filter with any op-amp, or a TL081 at the least, is technically a second (or third)-order Sallen-Key? And could it also be due to parasitic capacitance of the transistors in the op-amp? \$\endgroup\$ – tjbtech May 14 '17 at 18:08
  • \$\begingroup\$ Scratch out "is technically" since Sallen-Key is a topology - replace with "has the response characteristics of". I'm mostly wondering whether this phenomenon would apply to most or even all other op-amps to some degree... \$\endgroup\$ – tjbtech May 14 '17 at 18:25
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    \$\begingroup\$ Well, it's more like an MFB topology and yes this can happen. If you push the design of any topology filter using op-amps you will find that the lack of infinite bandwidth will give problems like these and they can even turn oscillatory. \$\endgroup\$ – Andy aka May 14 '17 at 18:40
  • \$\begingroup\$ OK, so, yes, but the effect will often be negligible to non-existent, depending on - well, everything else, including the specs/design of the op-amp used - if I'm following correctly. \$\endgroup\$ – tjbtech May 14 '17 at 19:12

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