7
\$\begingroup\$

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.

\$\endgroup\$
  • \$\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
  • 2
    \$\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
5
\$\begingroup\$

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!!!

|improve this answer|||||
\$\endgroup\$
  • \$\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
  • 1
    \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.