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I performed the AC analysis of the following circuit: enter image description here

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

enter image description here

The pictures shows that the response decreased by almost -40db per decade but it then started to increase uptill a certain point which according to me indicates the presence of a zero. But there is no zero in the T.F of the sallen-key topology. So please can someone explain why this happens?

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  • \$\begingroup\$ 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 . \$\endgroup\$ – Autistic Apr 15 '17 at 3:14
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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.

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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.

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