Dual gang pots, different values per gang?

Pickups, wiring schemes, switch techniques and onboard active electronics for guitars and basses
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Prostheta
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Post by Prostheta »

Grrr. Need to source or construct a dual gang pot with a 500k track and a 50k track for a dual passive/active circuit in a bass.
I suspect I will need to open up a dual 500k pot and rag a single 50k pot for the wafer unless anybody has any brighter ideas or good sources.
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iq01221
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Post by iq01221 »

And resistors in parallel to get 50K? It's a single calc, and no opened pots :idea:

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Post by rocklander »

get a dual gang 500K and just put a 56KR across one pair of terminals?
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Post by Prostheta »

That would alter the taper I'm afraid. I might try it if all else fails.
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rocklander
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Post by rocklander »

not sure.. but to be fair, I don't see where you mentioned the taper initially...
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Post by Prostheta »

True. The configuration is a volume and a tone control. Values out by one decade each so passive runs through the 500k and active through the 50k gangs. Capacitor adjusted by one decade also of course.
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cpm
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Post by cpm »

you can try to change values in one part of the circuit so both pots are matched
schematic?

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Post by Prostheta »

I don't believe so. :hmmm:
Perhaps sucking the parallel resistor and seeing how it feels in use is a good idea. It might end up being a linear 500k for the "passive" gang and a more logged one for the paralleled "active" gang. I'll do some groundwork.
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Post by roseblood11 »

it's very easy to replace one of the wafer, I've done that a few times without problems

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Post by Prostheta »

Any recommendations on which brands are most friendly for disassembly, ie. CTS, etc? FWIHR on Geofex some shaft mountings are less than perfect to reassemble. :hmmm:
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Post by roseblood11 »

if it's for a pedal, use the 16mm Alphas. Many people have proven that it's quite easy to disassemble them. And if you break them, they're cheap at least...

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Post by Prostheta »

They're actually for an instrument which will be have both passive and active modes but one common control set, hence why I first went straight to the CTS question.
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