Un-Faking a log pot

pixpop · 31st May 2006, 02:20 PM

I see lots of threads and info on how to make a linear pot into a log(ish) one. I need to go the other way.. I have a logarithmic pot, actually, I think it's a reverse log pot, and I need to make it more or less linear. If I could just make it more equally balanced either side of center, that would be OK too.

The pot is dual gang, labelled 100K-C, but measures about 84K.

I don't need both gangs for this application. Is there any trick I can play by wiring both gangs up?

In my application, I'm going to apply a negative voltage to one end, and a positive to the other. The wiper voltage should be zero in the center, and swing a similar amount negative and positive. It will be used as an offset control in a measuring device. It used to be a volume control in an audio amplifier.

Changing to a new (linear)pot would create mechanical and other complications that I'd rather avoid. The pot also has a switch that I need.

richie00boy · 31st May 2006, 02:26 PM

Well if you simple put the resistor at the opposite of the pins as 'normal' with a lin pot you get an antilog pot, so maybe you can find some value of resistance that approximately flattens your log pot.

pixpop · 31st May 2006, 03:57 PM

Duh.. yeah. I wonder why I thought it would be more complicated.

I put the wiper in the center of its travel, and measured the resistance between the wiper and the CW end. Then I added parallel resistance between the wiper and the CCW end, until both sides were the same. That gave me zero volts in the center. On either side of the center, the curve is a little unbalanced, but not enough to be important in this case.

Thanks for the suggestion.

31st May 2006, 02:20 PM	#1
pixpop diyAudio Member Join Date: Jan 2005 Location: Los Angeles	Un-Faking a log pot I see lots of threads and info on how to make a linear pot into a log(ish) one. I need to go the other way.. I have a logarithmic pot, actually, I think it's a reverse log pot, and I need to make it more or less linear. If I could just make it more equally balanced either side of center, that would be OK too. The pot is dual gang, labelled 100K-C, but measures about 84K. I don't need both gangs for this application. Is there any trick I can play by wiring both gangs up? In my application, I'm going to apply a negative voltage to one end, and a positive to the other. The wiper voltage should be zero in the center, and swing a similar amount negative and positive. It will be used as an offset control in a measuring device. It used to be a volume control in an audio amplifier. Changing to a new (linear)pot would create mechanical and other complications that I'd rather avoid. The pot also has a switch that I need.

31st May 2006, 02:26 PM	#2
richie00boy Did it Himself diyAudio Member Join Date: Nov 2003 Location: Gloucestershire, England, UK	Well if you simple put the resistor at the opposite of the pins as 'normal' with a lin pot you get an antilog pot, so maybe you can find some value of resistance that approximately flattens your log pot. __________________ www.readresearch.co.uk my website for UK diy audio people - designs, PCBs, kits and more

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31st May 2006, 03:57 PM	#3
pixpop diyAudio Member Join Date: Jan 2005 Location: Los Angeles	Duh.. yeah. I wonder why I thought it would be more complicated. I put the wiper in the center of its travel, and measured the resistance between the wiper and the CW end. Then I added parallel resistance between the wiper and the CCW end, until both sides were the same. That gave me zero volts in the center. On either side of the center, the curve is a little unbalanced, but not enough to be important in this case. Thanks for the suggestion.