Change a pot value

[Peter Wilcox]

Stupid question, but I just need confirmation (or rejection) of my thinking. I've been working on a friend's guitar amp and have it back to working condition. However, one of the gain pots has a broken shaft, and can't be adjusted. Unfortunately its physical size is small - about 1/2" diameter - and I can't fit a regular sized pot in the space. Its value is 100K, and I don't want to spend $10 (price and shipping) on a replacement, since he's financially destitute, and got the amp free at the dump.

I do have a 250K pot that will fit. I was thinking to parallel a 180K resistor across it, to keep the same resistance in the circuit. I know it will change the effective taper, but otherwise shouldn't have any effect on the circuit. Correct?

Thanks for any thoughts.

[David King]

If that trick works it would be easy to test out with a VOM.
A 250K is pretty close to a 100K anyway so I doubt you'd hear much difference between them.

[Peter Wilcox]

I finally got around to checking it out with a meter. It works, but with a reverse log taper. The resistance initially climbs rapidly, then more slowly, especially after the midpoint of rotation. I drew out the circuit and calculated and plotted resistances for several points. The pot is linear.

Pretty good for a simpleton like me.

[David King]

I'm impressed Peter.
You can order the little 1/2" sealed pots from China on ebay, they run about $3 ea including postage.