Replacing a tone pot with a resistor?

[spilla]

Id like to replace a tone pot with a resistor/cap. I'd like it to be 50's style wiring. Is this the way i'd wire it up (pic below)?

[spilla]

btw, i editted that pic above. I'm just hoping to get confirmation it's correct. Cheers

[Mike]

This would give you the sound of the tone pot rolled all the way back, I believe. Is that what you're after?

[Mike]

A simpler solution - use a 250K pot instead of a 500K. The load is what makes the difference in tone, not the cap.

[spilla]

This would give you the sound of the tone pot rolled all the way back, I believe. Is that what you're after?

No, thats not what im after. i'd like to remove my tone pot but not change the tone of the guitar. Removing the tone pot makes it a bit brighter. Also it currently has 50's style wiring, which i believe effects how dark it gets when you roll the volume down. I'd also like to keep it the way 50's wiring effects the vol pot (doesnt go dark as quickly as standard wiring).

I think i'll just order some 500k trimmers and wire it as it currently is. I jsut didnt want to wait for shipping. Cheers!

[spilla]

Ive been looking into this more. Found this helpful comment:

Solder a 22n cap in series with a 470K resistor between the hot of the pickup and ground. That's your simulated tone pot on '10.'

Technically you don't really need the cap because its not doing much when the pot is at max but lots of people think it cant be right without the cap so its better to just put it in and not worry about the 'what ifs' of your imaginary tone pot.

So looks like that drawing will work but i dont really need the use cap. From what ive read the cap only effects the signal when the tone pot is lowered. So if i'm just setting and forgetting i wont need the cap. I believe the Fralin website suggests just using the resistor by itself too.

Going to try it this weekend when i have some free time.

[Mike]

Interesting - logic tells me that resistor = pot turned all the way "down", but apparently not.

[Markus]

Interesting - logic tells me that resistor = pot turned all the way "down", but apparently not.

I'm no electrical engineer, but if I understood correctly when the pot is on 10, the resistance between the hot and the ground is as high as possible, i.e. less signal gets sent to ground (more volume - or brighter tone in the case of a tone pot). Thus if you put a constant high resistor there, it will be like having tone on 10. When the pot is on 0, the resistance is low and the whole signal gets sent to ground

Edit: Fralin pickups have a pretty good guide to these things

https://www.fralinpickups.com/2017/0...tone-pots-101/

[Mike]

Ah, so my thinking is wrong - it's unlike a volume pot, where you want resistance to be zero. Thanks!

[beneharris]

Ah, so my thinking is wrong - it's unlike a volume pot, where you want resistance to be zero. Thanks!

I had the same thoughts you did. I never run tone pots, so I haven't had to think about that in quite some time

[Markus]

Ah, so my thinking is wrong - it's unlike a volume pot, where you want resistance to be zero. Thanks!

I think it works about the same with volume, i.e. when the volume 10 the resistance on the ground is at its maximum (and resistance to the input minimum, basically 0) - and conversely when volume is 0 the resistance on the ground is low and input high.

Don’t take my word for it though - I’ve always sucked at electronics

[spilla]

I think it works about the same with volume, i.e. when the volume 10 the resistance on the ground is at its maximum (and resistance to the input minimum, basically 0) - and conversely when volume is 0 the resistance on the ground is low and input high.

Don’t take my word for it though - I’ve always sucked at electronics

Yep, this. The lower you turn the pot (the less resistance) more of the signal is sent to ground.

[bigdave]

Removing the tone and use a treble bleed mod on the vol pot. Having no tone does make things a tad brighter but nothing a nominal knob or slider on a EQ wont cure.