Adjustable low-pass filter with logarithmic cut-off control via potentiometer

Question

Let's consider this simple RC low-pass filter:

We can make the cut-off frequency adjustable by replacing \$R\$ with a potentiometer. But then the cut-off frequency \$f\$ is proportional to \$\frac{1}{R}\$.

Now I'd like to make it adjustable by using a (linear) potentimeter as a variable resistor for \$R\$; so far so good. If \$p \in [0,1]\$ encodes the potentiometer position (e.g. \$p=0\$ for all the way to the left, \$p=1\$ all the way to the right), then we have \$R \propto p\$.

Now I was wondering, is there a simple way to have a logarithmic relationship between the potentiometer \$p\$ state and the cut-off frequency \$f\$, that is \$p \propto \log(f)\$? Or one that is at least roughly proportional for a large range of \$p\$ (that is, no exact relationship is needed, and it also doesn't have to hold for the whole range of \$p\$)?

It is definitely possible by using voltage controlled filters and exponential converters etc., as it is frequently done in synthesizers, but I wonder whether there is also a much simpler solution (with fewer components) for just filtering audio signals, such that the pot roughly behaves like our perception.

Answer 1

If you can't fidn a logarithmic pot then you can use the poor man's log pot:

R1, R2 form the pot with x being the wiper, and it varies from min to max. The value is 1 Ω, for exemplification. R3 is added in parallel and you can see in the top plot how various values affect the output: lower values make the curve more accentuated. The downside is the current: with a regular pot it would have been 1 A, flat, since the resistance doesn't change. With R3, the stakes are raised.

If the initial curve isn't to your liking then you can use another resistor in parallel with the other segment, in which case the output and the current vary like this:

From the midpoint it's a 1/x variation, closer to an exponential, but you'll have the level 0.5, too.