# Sallen-Key topology for Butterworth low pass filter

I performed the AC analysis of the following circuit:

This is a simple second order low pass filter satisfying the Butterworth filter requirements. According to me, after the break point there will be a continuous decrease in magnitude by -40 dB/decade but this was not the case as seen below:

The pictures shows that the response decreased by almost -40 dB per decade but it then started to increase until a certain point, which according to me indicates the presence of a zero. But there is no zero in the transfer function of the Sallen-Key topology. Why does this happen?

• If you double C2 to 20nF and Remove R4 and short R3 then you should get a unity gain Butterworth 2nd order LPF .You should be a bit better off at the high end .What I have done and has been done by others in 1972 is use an emmitter follower instead of an opamp .It is much more cost effective than a gold plated opamp . Apr 15, 2017 at 3:14

## 2 Answers

Up to about 100kHz, it all goes as expected.

Unfortunately, after that, the op-amp output impedance rises so much that it's unable to control the feed-forward path that comes through R1 and C2.

Consider increasing the impedance of the filter components, so use 8.2k and 1nF, which should push trouble up another decade in frequency. A faster op-amp would also help. Also, try an 'ideal' op-amp in simulation to see the problem go away.

Another one to consider is the 3rd order Sallen Key, which has an RC input, which puts a true passive 6dB/octave on the response, which may disguise the bad thing sufficiently to be acceptable.

If you were using a perfect op-amp then the gain would continue to decrease after 100kHz as expected. However, you are not using an ideal op-amp, and the actual capacitances within the op-amp itself, however small, really come into play at the higher frequencies. When these capacitances start playing a role in the behavior of the filter it becomes very apparent in the Bode plot, as you see. This is expected.