I'm designing a 2nd order low-pass Butterworth filter using a Sallen-Key topology (for reminder) :


simulate this circuit – Schematic created using CircuitLab

Now, I am trying to find the optimum values for the components, given that :

  • \$ \omega_0 \$ is known
  • \$ Q \$ is known (\$ \frac{1}{\sqrt{2}} \$ if I am remembering well)
  • The input impedance should be maximum (with reasonable component values, of course)

Is there a tool available that is able to give the optimum values with these constraints, or am I better off coding a little program that can give me the answer ?


In principle, I know what you mean. However, the answer is not easy because it depends - more or less - also on the corresponding operating frequency. Here are some basic rules:

  • The values of the resistors and impedances (at the operating frequencies) should be large (small) if compared with the output (input) impedance of the opamp.

  • Preferred component ranges are: R=(1E2...1E5)ohms and C=(0.1...1000)nF.

  • More than that, preferred maximum "component spreading" values are 0.1, 1, or 10.

  • If the input impedance of the filter circuit is an important design parameter depends on specific application requirements. If necessary, an additional buffer can be used or the (known) signal source resistance is to be considered as part of the design process.

  • In your specific example, there is another criterion which matters: For frequencies far above the pole frequency (stop band of the lowpass) a part of the input signal is coupled DIRECTLY to the output of the opamp (through C1). Because for rising frequencies the opamps open-loop gain continuously decreases, we have an increase in the opamps output impedance - and there will be a remarkable unwanted signal voltage drop across this output impedance. Hence, the attenuation in the stop band becomes worse for high frequencies (stop band). For this reason, the feedback capacitor should be as low as possible. This is a well-known (but not always documented) disadvantage of the Sallen-Key lowpass topology.

  • As you can see, there are several (partly conflicting) optimization strategies. Therefore, there is no general answer to your question.

  • \$\begingroup\$ I see what your point, thank you ! Maybe in the future I'll try to develop a tool for such optimizations. It's only mathematics after all. \$\endgroup\$
    – Albits
    Jun 2 '16 at 10:50

I do not know what you mean by 'optimum' values, but a Butterworth filter has a flat response, so it's construction is easy. In this case C1=C2 and R1=R2, so the roll-off frequency is 1/(2*pi)RC. You can add a third pole by adding a R3 and C3 on the outputs equal to R1 and C1.

The upper limit or R depends on the type of op-amp used. With jfet or mosfet op-amps R can be as high as 10M ohm. The math for variations of this circuit would fill up a long page, so suggested reading for elaborate details when Q is peaked or R1!=R2 and C1!=C2: wikipedia.org/wiki/Sallen–Key_topology

  • \$\begingroup\$ What I'm saying is that, from my point of view, we have four unknowns (R1, R2, C1 and C2) and our parameters are \$ \omega_0 \$ and Q. To that, I add one more constraint of having the highest possible impedance. I think there should be a set of values that gives the most "desirable" characteristics, right ? Without having R1=R2 and C1=C2 as it happens often, I think we can reach better results. \$\endgroup\$
    – Albits
    Jun 1 '16 at 21:40
  • \$\begingroup\$ For C1=c2 and R1=R2 the gain of the opamp must be A=3-(1/Qp). (Butterwort: Qp=0.7071) \$\endgroup\$
    – LvW
    Jun 2 '16 at 9:28

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.