SOLVED Modify Linear Pot to Log

J

jkjkjk

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Hi there,

I just finished up the Special Overdrive. It calls for a B100k volume potentiometer, but I'm finding that it uses up a whole bunch of volume in the first quarter turn. Seems like the solution is to switch to an A100k, but I don't have one on hand. I've read that I can solder a resistor between the middle and ground lug of the pot to make it more like a logarithmic pot. Can anyone advise on this - do I just add it to where the pot legs poke through the PCB? And which ones would be middle and ground? And what resistor would I add?

Thank you!
 
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Because adding a resistor in parallel to any two lugs of a pot results in a lower combined resistance, one needs to start with a higher value pot to achieve a taper change while retaining the same relative pot value. So if you want a about a 100K analog taper pot, you need to start with something about double that value. The rest is parallel resistance math.

This chart should help you get started. It assumes the pot having its taper adjusted has a linear taper.

Pot-Tapers.jpg
 
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Thanks for sharing this. If I use the existing 100k pot, and I put a 10k resistor, does that just mean I'm limiting the max sweep? Like if I'm not hitting "10" on my volume pot right now, would that be ok?
 
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jkjkjk said:
Thanks for sharing this. If I use the existing 100k pot, and I put a 10k resistor, does that just mean I'm limiting the max sweep? Like if I'm not hitting "10" on my volume pot right now, would that be ok?
Click to expand...
No, you're also changing the loading of that gain stage. That can have undesirable frequency response ramifications on top of whatever gain changes that loading change may bring. So as usual, it depends. ;)
 
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jkjkjk said:
Thanks for sharing this. If I use the existing 100k pot, and I put a 10k resistor, does that just mean I'm limiting the max sweep? Like if I'm not hitting "10" on my volume pot right now, would that be ok?
Click to expand...
Passinwind said:
No, you're also changing the loading of that gain stage. That can have undesirable frequency response ramifications on top of whatever gain changes that loading change may bring. So as usual, it depends. ;)
Click to expand...
Besides what @Passinwind states, when you do the math, a 10K resistor in parallel with a 100K pot will result in a <10K pot with the applied taper. If you add a 100K resistor, the result would be a ~50K pot with the applied taper. I imagine you really want a 100K pot with an analog taper, so I'd start with a 250K pot and apply a 160 reistor to get ~97K audio/analog taper pot.
 
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anevilspar7an said:
A good rule of thumb is if you have any volume control with a Linear taper, you'll want to swap it to a Logarithmic taper for a better sweep.
Click to expand...
Now I've learned, will definitely keep this in mind for future builds to have a spare logarithmic taper.

Nostradoomus said:
Try a C100K while you’re at it
Click to expand...

Unfortunately I don't have one of those available! Trying to figure out how to get a single potentiometer without it being dwarfed by shipping costs. Which may just mean ordering another batch of parts, haha.
 
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Ok, one more question here to pull on the wisdom of the group. Still toying around with the adding a parallel resistor idea: I have a 220k resistor available. If I add that to the pot in parallel, then the math suggests this would result in about a 67k audio taper pot. Is that ok to try to see if that’s enough headroom? I also have 100k and 150k resistors lying around.
 
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