Adjusting The Taper of Audio Pots

Posted: April 22, 2014 in Effects, Ideas To Be Developed

For my projects, I usually only use audio (log) potentiometers.  The only problem with the taper of a log pot is that the last 1/4 of clockwise rotation changes values very quickly.  I’m working on a project where a 500k pot has 82k resistors on both outside lugs in series.  This reduced the total value to about 200k and the last part of the taper stayed within a few k-ohms of 200.  There just wasn’t enough of a difference for the pot to even be used beyond the 3 ‘o clock position.  I searched a bit and found some articles on how to change a linear pot to a logarithmic using resistors.  There was no information on adjusting a log taper, so I pulled out the DMM, some resistors, and went to work.

http://www.geofex.com/article_folders/potsecrets/potscret.htm

Using the idea of putting a resistor of a value that is the same or less than 1/5th of the total value, I tried some different resistors across lugs 2 and 3 to pull it in an anti-log direction.  My estimation was that it would make the taper more gradual.

It worked.  There was a catch.  The adjusted resistance of the 500k pot dropped from 200k to 185k and the 1/2 position’s resistance shifted from 65k to 54k.  However, the travel from 1/2 way to fully counter clockwise was much more gradual.  I didn’t chart it, but I would assume that it either became a slight anti-log S-curve or a the curve became much closer to a linear.  For the 500k with resistors on either side, a value of 1k or lower was perfect (there wasn’t much of a difference at that point).  I have a lot of resistors under 1k that I rarely use.  Now, I’ll have a use for them from time to time.  The one I’m soldering in is a 56 Ohm.

Someone may wonder why I don’t just get a linear pot when I need one?  The answer is simple: I don’t use enough of them to stock up.  I like the way log pots work for most guitar and effects.  If I can just use a 10 cent resistor to avoid stocking up on a bunch of $1.35 pots, I think the answer is simple.

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