Low pass Sallen-Key with variable cut-off frequency

Question

I have a project that I would like to get some help with and would like some things confirmed. I want to construct a 2nd order Low pass Sallen-Key filter with variable cut-off frequency using one or two potentiometers (variable resistor). The topology can be found here: Sallen-Key topology. Low pass filter - Application one

The filter should be able to take away all frequencies above 5kHz and all frequencies above 20kHz. I.e. it should be able to take away all frequencies above 5kHz when the potentiometer is set for a specific value (ohm) And take away all frequencies above 20kHz when the potentiometer is set for another specific value.

First of all, anyone have any tips or advice in how to begin the design? What components are the most critical and how do I determine them? I´m not really asking for any specific answers, more of guidelines in how to proceed. I have no restrictions in any component values (except the natural of course).

Moreover, I need the phase shift to begin as near the cut-off frequency as possible due to the fact that a phase shift more or less equals a time shift and I don´t want any of the passing frequencies (as few as possible at least) to be affected of this time shift. This mean a want a high Q-value, correct?

Also, one question about the cut-off frequency. The wiki-page states that the "natural frequency" is: $$ f_0 = \frac{1}{2*pi\sqrt{R_1R_2C_1C_2}} $$

So if I have understood this correctly, to get the cut-off frequency I just go down three dB from this (natural frequency)?

Answer 1

The -3dB point is your cutoff frequency. It's just standard practice to define it that way. In order to find what your values should be, I'd go with equal element implementation (it's simpler, and you can correct for gain with a simple gain stage later if you need to). Choose R1=R2, C1=C2, and pick a value for either R or C. This yields the following formula for the cutoff frequency: $$f_0=\frac{1}{2\pi RC}$$ I generally choose a value of C initially, as it's easier to find or make a resistor with a strange value, whereas it's more difficult with capacitors. So, set your cutoff frequency equal to f0, and solve for R.

Here's an example: let's say I want a LPF with f0=250 Hz. I'll choose C to be 0.1 micro and solve for R. $$250=\frac{1}{2\pi RC} \rightarrow 250=\frac{1}{2\pi R(0.1x10^{-6})}\rightarrow R\approx6400\Omega.$$

From there, all you need to do is implement your circuit. Once you know what your value for R is supposed to be, you can use a dual-channel potentiometer that has the correct resistance within it's range in place of the two resistors (for the above example, something like a 10k ohm potentiometer would do the trick). This will allow you to change your cutoff frequency, since it's based upon both R and C.

Edit: As Matt Young suggested in the comments, adding a resistor in series with the potentiometer will set the maximum cutoff, and prevent shorts. It's an excellent addition to the circuit, and will keep some sanity when adding the potentiometers.

Answer 2

The example in Wiki gives you a low pass filter at 15.9kHz - so start with that.

C1 = C2 = 1nF

R1 = R2 = 10k

Because R1 = R2 and C1 = C2 then Q = 0.5

The values of C1 and C2 (=C) are fine, so the question is what value should the resistors be for a 5 to 20kHz filter. (a much simpler problem)

Provided we keep R1 = R2 = R

Then w = 1/CR

for 5kHz

     2 *pi*5000 = 1/CR
              R = 32K (approx)

for 20KHz

      2 * pi*20000 = 1/CR 
                 R = 8k (approx)

By using a ganged variable resistor you can change both resistance values at the same time which will maintain the Q value.

Answer 3

Without working out the math for you, this is often harder than it seems it should be. It's probably not solvable just by changing one resistor for a pot, and you probably don't want to start using varicaps. Do you really need continuous change, or do you really need two different cutoffs. If you need two cutoffs, I suggest gang-switching however many components you need to, or maybe even building two separate filters and switching the outputs.

Also, keep in mind that a two pole filter is not a magic device that cuts off frequencies like a brick wall. The cutoff slope will be -40dB/decade, which may or may not be enough to meet your need.

Another approach would be to use a switched capacitor filter with more poles, and change the clock frequency to change your cutoff.