How to make a 100k pot act like a 10k pot?

[danjel]

I have a bunch of pcbs I made for a simple audio mixing circuit. The audio is fed into 10k pot acting as an attenuator and then summed with an opamp summing circuit (where the summing resistors are 4.7k).
This circuit works perfectly.

However, I ran out of 10K pots and I had to use 100K instead. This creates an exaggerated response curve.

I tried the simple trick of putting a 10k resistor across terminals 1 and 3 of the 100k potentiometer. This resulted in a response much closer tot he 10k pot BUT there is a huge jump at the last few degrees of the rotation.
I guess it is making a kind of s-curve response.

Is there an easy way to fix this or a better way to make a 100k pot perform exactly like a 10K?

 

[dougy83]

If I understand correctly, the signal in goes through the pot to ground, with the wiper connected through a 4k7 to the summing junction (ideally the -ve opamp input).

If you increase the value of the summing resistors, the effect of the pot value has less effect.

 

[danjel]

yeah that would be the simplest solution for sure. The only issue is that the traces to this point are quite long and increasing the summing resistors resulted in a lot more channel bleed due to increased capacitance between those lines and ground. I think my only real solution is to find proper value pots and remove these ones unfortunatley

 

[scarygood536]

just because I'm curious... When you parallel resistors say 2 -100 ohms the resistance would be 50 formula would be R\N (resistance\number of resistors {if all same resistance}) or 1/[(1\r1)+(1\r2)+(1/r3)]... can this theory be applied to reduce the resistance of the pot?

 

[MrAl]

Hello there,


There is no way to get the dynamic resistance of a high value linear pot to a lower value and still act as a linear pot that i know of, except to add an active element of some type.
For example a voltage follower would do it.

The way the resistance changes as the pot arm rotates is such that parallel resistances dont allow a linear resistance because the impedance of the pot is too high near the center.

The only way to get a 10k pot out of a 100k pot is to parallel 10 pots. That would give the equivalent of a 10k pot at all arm angles. Obviously 10 pots would be nuts so the active approach would have to be used.

A voltage follow can be made with an op amp section and no other resistors are required. A voltage follower can also be made with a single transistor too with no additional resistors, but only if the application can stand a 0.7v voltage drop in line with the signal.

 

[Nigel Goodwin]

yeah that would be the simplest solution for sure. The only issue is that the traces to this point are quite long and increasing the summing resistors resulted in a lot more channel bleed due to increased capacitance between those lines and ground. I think my only real solution is to find proper value pots and remove these ones unfortunatley

It sounds like you've designed the PCB very poorly?, the long traces should be at the 'virtual earth' points, which prevents any such problems - check the designs of commercial mixers, where all channels run closely next to each other with no worries. If the board is already made?, then cut the track near the pot sliders and move the resistors to there.

The 4.7K summing resistors also sound rather low for 10K pots, I'd have used 10K pots with 100K summing resistors, 4.7K will affect the linearity of the pots.