I am a bit confused regarding the gain effect over the cutoff of the active low pass filter. While I was checking the AC sweep of an active low pass filter with Gain = 5 and Gain = 10, I noticed that the cutoff is shifted:

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I expected the gain to only change but can not figure out why the cutoff is changing as well. Also, should I go to another stage just for gain?

  • 1
    \$\begingroup\$ I am afraid that the ciruits do not work in reality (unstable). Gain values are too large. Be aware that the magnitude display of an ac analysis does NOT reveal possible instabilities. Look at the phase diagram - and you will observe anomalies...or perform a TRAN analysis. \$\endgroup\$
    – LvW
    Commented Mar 4, 2021 at 11:54
  • \$\begingroup\$ This looks like a Sallen-Key lowpass filter structure but the component values might be incorrect. Have you studied: en.wikipedia.org/wiki/Sallen%E2%80%93Key_topology note how all lowpass filters use the opamp in the unity gain configuration. \$\endgroup\$ Commented Mar 4, 2021 at 12:29
  • \$\begingroup\$ It's often a bad idea to ask for gain in a filter (Sallen keys can do that but are very sensitive to tolerances). MFB filters can't do gain, as a rule. Also you are using a somewhat ideal opamp which only has a single pole rolloff, a real opamp will probably be GBWP starved in such a situation \$\endgroup\$ Commented Mar 4, 2021 at 14:00
  • \$\begingroup\$ In case @LvW's hints are vague, the phase needs to have a negative slope. A positive slope means a right-half plane root, which is unstable. Also, the Sallen-Key topology uses (what you have noted as) R3 and R4 for the quality factor, too. That's what you're actually seeing there: a softer corner frequency. \$\endgroup\$ Commented Mar 4, 2021 at 14:13
  • \$\begingroup\$ Yes - a rising phase in the pole frequency region indicates instability. More than that, the value of the phase slope at the pole frequency is a good (rough) indication for the available phase margin. \$\endgroup\$
    – LvW
    Commented Mar 4, 2021 at 14:23

1 Answer 1


Assume that everything is ideal LTI ZIC.
Try to derive the transfer function first to understand it.
Hint: Remember the golden rule of op amp and use nodal analysis at the input terminal of it.

Second-order LPF Sallen-Key

The result will be like this:

The Transfer function

Ideal second-order LPF LTI ZIC (such as series RLC) will have prototype transfer function like this:

Ideal second-order LPF LTI ZIC

  1. The G factor doesn't change the pole frequency (natural frequency). Even though that's true, it still change most of the filter characteristics, which we usually design using known topology (such as Butterworth, Chebyshev, …). It changes asymptotic gain (of course), 3dB cut-off frequency, bandwidth, the damping ratio/quality factor, thus the phase margin or stability and so on.
    Both of the circuit natural frequency is 53.636 rad/s, thus the pole frequency at 8.53 Hz.

  2. Now lets try to take a look at the characteristics equation of the pole. By Descartes' rule of signs, at least one sign changes of the term will results in positive roots, thus instabilities. We need to make sure that all of the terms have the same sign. The only possible term is on the middle one because there is a subtraction of \$1 - G\$. And we don't want the second term to be zero, it would be an oscillator, so it must be strictly positive.

    \$ \frac{1}{C1R2} + \frac{1}{C1R1} + \frac{1}{C2R2}(1-G) \gt 0 \$

    \$ \frac{1}{C1R2} + \frac{1}{C1R1} + \frac{1}{C2R2} \gt \frac{1}{C2R2} G \$

    \$ \frac{C2}{C1} + \frac{C2R2}{C1R1} + 1 \gt G \$

    \$ \frac{C2}{C1} \left( 1 + \frac{R2}{R1} \right) + 1 \gt G \$

    In both of your circuit you have

    \$ \frac{1u}{10u} \cdot \left(1 + \frac{31.6k}{1.1k} \right) + 1 = 3.9\overline{72} \$

    While you have G factor of 6 on the first circuit and 11 on the second one. This indicates that both of the circuits is unstable. No wonder in the bode plot, at infinite frequency you get phase lead of +180 degrees instead of the correct phase lag of -180 degrees.

    G factor needs to be less than \$3.9\overline{72}\$ in order for the circuit to be stable. R3 in your circuit (in mine is R4) must be no more than \$59454.5\ \Omega\$.

My favorite site to simulate ideal Sallen-Key filter:


  • \$\begingroup\$ Short comment to 1.: The cut-off frequency does depend on the G-factor because the pole-Q depends on G (and with it the filter characteriszics: Butterworth, Chebyshev,...). However, it is the pole frequency which does not depend on G. \$\endgroup\$
    – LvW
    Commented Mar 4, 2021 at 14:10
  • \$\begingroup\$ @LvW Thanks so much for pointing it out, I've mixed it up. \$\endgroup\$
    – Unknown123
    Commented Mar 4, 2021 at 18:58
  • \$\begingroup\$ Please note that I use Descartes' rule of signs to detect positive real root. In general case, Routh–Hurwitz stability criterion should be used to also account for the positive imaginary conjugate pairs. Just because all the terms are positive it still doesn't imply that the system is stable. \$\endgroup\$
    – Unknown123
    Commented Mar 10, 2021 at 13:34

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