I’d like to try changing out the potentiometers in my ms-20 filter - the resonance is fine, but the CV and cut off knobs have too little travel.
For the cut off, the filter is fully opened between 0-3, with 3-10 just being ‘open’.
Both are linear potentiometers. Am I right in thinking that swapping them to log would stretch out the range a little more?
What effect would increasing or decreasing the value have?
To be clear, presumably you’re talking about something like this:
There’s a pot here that sets the cutoff and there are attenuator pots on each CV.
Changing to a log pot would not change the range. For the cutoff the voltage range is between ±15 V (or ±12 V more likely in yours) regardless of the pot type, and for the CV it is from 0 to whatever the full CV is. But it would change the shape of the response. It would make it so changes are more gradual at the low end and faster at the high end, so what is “3” with a linear pot would be more like maybe “6” with a log pot.
An inverse log pot would make “3” become more like “2”.
Changing the value of the pot likewise has no effect on the range. The only effect changing the value would have is if you made it much larger than 100k (the value of the resistors after the pots) it would change the shape of the response, with faster changes at both ends and slower changes in the middle. Probably not what you want. If the value is less than or about equal to 100k then the response is close to linear (if the pot is linear).
“Opened” (for a low pass filter) means the cutoff frequency is higher than the input frequency and its main harmonics. What’s open for a low frequency input can be closed for a high frequency input. So it’s meaningless to say it’s fully opened between 0–3 without saying what frequency it’s opened for.
Usually the pot is wired so that turning up the pot turns up the cutoff frequency, so it becomes more open at the clockwise end (closer to 10). It sounds like yours is wired the other way if I understand what you’re saying correctly. And it sounds like you’re saying it never really cuts off, that you have substantial output even at 0. If that’s the case then I think something else is wrong. It’s designed so a 1 V change in the CV (including the voltage provided by the cutoff pot) produces a 1 octave change in the cutoff. If you’re using ±12 V that’s a 24 octave range! (30 for ±15 V!) That should be plenty large enough a range to fully cut off anything.
As for the CV inputs the pots attenuate them, making the filter less responsive to a CV than it would be if the CV went straight in (as on the third input above). You’d have to amplify the CV to make it more responsive. But again, a 5 V range should give you 5 octaves, and that in combination with the cutoff setting should be enough.
Here is a video of the behaviour, hopefully that makes it clearer. Regardless of frequency, the filter is mostly opened by only the first third of the rotation of the potentiometer, after which the filter is essentially ‘open’.
What I’d like is for the effect of turning the potentiometer to be more gradual. So that the ‘cut off’, filtered sounds are spread out across say 0-7 rather than 0-3 of the knob. It sounds like, from what you’ve said, swapping to a log pot would achieve this?
OK, I misunderstood. It’s closed at the CCW end, opens up by about “3”, and doesn’t change much above “3”.
So yes, changing to a log pot would make something like “6” be the old “3”. But, again, the overall cutoff range wouldn’t be affected. If the circuit is like the above Schmitz circuit, the range is very large, 24 octaves. It might be better to reduce that range. In the Barton 055 Sallen-Key filter, for example, the cutoff knob is wired like this:
With the fixed resistors added to the pot the voltage range is reduced from ±12 V to about -5 to +8 V, or about 13 octaves.
But what’s significant is not just the range but also the minimum cutoff value. If your pot is wired straight to ±12 V (of course I don’t know that, maybe it’s 0 to +12 V?) then your position “3” would be about -5 V. So using Barton’s version would move your “3” right down to “0”! Definitely not what you want. It would reduce the range but also raise the minimum cutoff frequency. Assuming a 100k pot it would be better to add say 100k on the CW end and no fixed resistor on the CCW end. Then you’d have a 12 octave range with -5 V at about “6” on the knob.
(But if your pot currently is wired to 0 to +12 V then “3” is about 3.6 V, which would be toward the upper end of the -5 to +8 range, so in that case Barton’s configuration might be exactly what you want.)
The minimum cutoff also is affected by the caps used in the filter. Both Barton and Schmitz use 1 nF caps. If one used for instance 10 nF then the minimum cutoff frequency would be 10x lower, or about 3.3 octaves.
So if the problem is that the range is too large, you can reduce it by adding fixed resistors to the cutoff pot, and if the problem is that the minimum cutoff value is too low you can raise it by reducing the filter capacitors or by using a smaller fixed resistor at the CW end and/or larger at the CCW end.
Another thing to check is whether the CV response really is 1 V/oct. If not, that would affect the range too. Easiest way to do that is if you can put the filter into self oscillation at high resonance, then you can see if the pitch it produces goes up 1 octave when you change the (un-attenuated) CV by 1 volt. But it should be close to 1 V/oct if the 1.8k resistor is correct.
One easy way to change the characteristic of a linear pot to resemble a log pot is to use some resistors in parallel.
Have a look here:
Although the graphs may seem daunting I suggest you read the full paper. You can try to adapt your circuit to your liking by experimenting with a few resistors.
Thanks! That’s really helpful.
I will try using a resistor to convert it to log, the article was great.
If that doesn’t work, I will try adding fixed resistors either side of the pot - I imagine I can work out what I need by taking voltage measurements each side of what I want the range to be, then play around until I get what I need.
I haven’t implemented 1v/oct yet, as it’s the polykit DCO so I don’t have pitch cv controlling the oscillator. I’ve been meaning to add in a midi to cv solution to sort that… once I’ve got my current tinkering out of the way I’ll sort that.