Impossible to make a log pot from a pot for series resistanc

[fishfude]

Hi folks,

I'm looking to make a log pot to replace the course cut-off pot [R3] in this schematic:



The pot is acting as a series variable resistor. I get from GEOFEX's 'Secret Life of a Pot' article that I can make a log pot from a linear pot IF I'm using it as a voltage divider and not a series variable resistor. I believe him.

What I'm wondering is what would happen if I used the voltage divider arrangement, would the pot behave as I want or do some other electronic principles kick in to mess up my signal?

Thanks,

ff

[induction]

Replacing a variable resistor with a voltage divider is very unlikely to get you what you want. Log pots are usually not too hard to find, but if you must substitute you're probably best off using a linear pot. The same resistance values are available to you, they will just be distributed over the pot differently. (The sonic variation will be more concentrated at lower settings.) Since the pot is used as a coarse adjustment, you will probably find it easy enough to dial in what you want.

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.

[Lucifer]

I echo what Induction said, but there is something you can try.

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%.

To save you doing the maths, at 0%, 10%, 20, 30, 40, 50, 60, 70, 80, 90 and 100%, the resistance is 91k, 190k, 289k, 388k, 486k, 583k, 680k, 775k, 866k, 950k, 1Meg. These are slightly different to the original 0, 100k, 200k, 300k, etc.

You could try other values of resistor to see what resistance values you get. You probably won't get a logarithmic range, but at least it won't be linear.

Have fun. Good luck.

[commathe]

What is that circuit? Some sort of filter?

[fishfude]

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.

yeah that wouldn't be ideal.

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)

What is that circuit? Some sort of filter?

Yes! It's an Active Low-Pass Resonant Filter. It's a snippet from Ray Wilson's Weird Sound Generator synth.


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.

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)?

[induction]

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)?

Again, a voltage divider and a variable resistor are not the same thing, and they won't behave the same. You can certainly try it, but I wouldn't get my hopes up. You will be changing the topology of the circuit, and the result will almost certainly be different to that of a log pot wired as a variable resistor.

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.

First a clarification. If you wired it the way it's drawn in the schematic, then lower settings on the pot have higher resistance. As you increase the pot setting, you decrease the resistance. 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.

But if you're going to replace the pot, you could just as easily replace it with a log pot.

Why don't you want to use a log pot? Are they hard to find where you are?

[Lucifer]

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)

Hi FF,
Not quite, as the fixed resistor (eg, the suggested 100k) would be in parallel with the pot resistance between pins 1 and 2. So you get the first part of the pot resistance (between pins 2 and 3) in series with the above parallel bit.

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.

I believe that the techniques described by RG to create the log response was to put a fixed resistor across pins 1 and 2 of a linear pot, but in this configuration you are getting an 'altered linear' response instead of a fully logarithmic response. But, like I said earlier, if you do some simple maths like I did above, with different shunt resistors, you might be able to achieve a response that is acceptable to your ears.

[dwmorrin]

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.

[fishfude]

Hey folks,

Thanks for all the extremely helpful replies.

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.

Ah, yes, obviously, thanks Lucifer.

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.

Yes, I had actually reversed the wiring because it seemed more natural to me at the time. That's a nice suggestion actually, I'm gonna measure it to help me find the correct log pot to use.

Why don't you want to use a log pot?

I can actually get Log pots easy enough. I'm treating every build I'm doing as a learning experience and I got sucked down the modifying potentiometers rabbit hole a bit, and I was really interested to see if I could make the series log pot from a linear.

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.

Thanks for the suggestion, I think ill do that.

cheers,

ff