In principle, you could use RG's trick to create a reverse log pot out of a linear one and then wire it backwards. Then your taper would be correct, but the pot would work backwards. It would probably be confusing to have the two frequency pots work in opposite directions.
If you put a 100k resistor (as an example) between terminals 1 and 2 on the course pot, you can work out what the resistance of the pot will be for 10% rotation, 20%, etc, all the way up to 100%.
What is that circuit? Some sort of filter?
fishfude wrote:What I guess my question boils down to is; if I use a voltage divider arrangement I can get a log response curve, but what i'm wondering is will the signal be affected in some way differently, than if I just used a log pot (in series arrangement)?
What I noticed (and why I'm trying to do this) was that the interesting frequency sweeps all seem to be bunched up towards the lower part of the pot (0 ~ 250k) so I was thinking if I could do something like what Ray has done with the rez pot (half the overall value and create an anti-log response curve) but get a log curve.
fishfude wrote:If you put a 100k resistor (as an example) between terminals 1 and 2 on the course pot, you can work out what the resistance of the pot will be for 10% rotation, 20%, etc, all the way up to 100%.
Would that not just be a series resistor after the section of the pot between terminals 3 and 2? (apologies if this is a really ignorant misunderstanding)
So, if we call the pot resistance between pins 2 and 3 on the pot R1, and let's call the remaining resistance (from pin 2 to pin 1) of the pot R2, then you have the equation for the overall resistance as R = R1 + 1/(1/R2 +1/100k). For 0% rotation, the resistance R = 0 + 1/(1/1M +1/100k), and so on for 10% rotation where R = 100k +1/(1/900k +1/100k), then similarly for 20% rotation, etc all the way up to 100% rotation, where R = 1M (the 100k resistor is completely shorted out by the pot wiper at this point). Try different resistors instead of the 100k that I used, and you may be delighted.
So unless you reversed the wiring, you are interested in the higher resistance range. A log pot will spread this section out over more of the sweep like you want. You could also dial the pot to the maximum setting that you would ever use and measure the resistance between pins 2 and 3. Then replace the pot with a series resistor of that value and a pot of 1M minus that value.
Why don't you want to use a log pot?
I've built the MFOS Weird Sound Generator before, and yes, that filter is cramped, but it is also a very simple active filter... that is the topology of the circuit severely limits what you can expect from it.
I've had several questions for MFOS mods, and I've also just emailed Ray directly through the website.
He always responded to me, including schematics with modifications drawn in. I think he would be very happy to hear out your question.
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