Installing selectable resistance switch for pickupss

Ken B.

New member
I saw video on YouTube about installing a 250k/500k dual pot for selectable resistance on a guitar pickup for more tonal options. Then I saw this article on Seymour's website about making your own "dual pot".

https://www.seymourduncan.com/blog/tips-and-tricks/create-a-two-stage-potentiometer

OK, so...one question: why not use a 4 or 5 position slide switch, and install multiple different resistors for 250k, 333k, 500k, 750k, and 1000k settings?

Another question about this article, why does the author use a 500K pot, but then install a 510k resistor with the switch and say that this now gives him a 250k/500k switchable pot? Shouldn't he have used a 250k resistor?
 
I saw video on YouTube about installing a 250k/500k dual pot for selectable resistance on a guitar pickup for more tonal options. Then I saw this article on Seymour's website about making your own "dual pot".

https://www.seymourduncan.com/blog/tips-and-tricks/create-a-two-stage-potentiometer

OK, so...one question: why not use a 4 or 5 position slide switch, and install multiple different resistors for 250k, 333k, 500k, 750k, and 1000k settings?

Another question about this article, why does the author use a 500K pot, but then install a 510k resistor with the switch and say that this now gives him a 250k/500k switchable pot? Shouldn't he have used a 250k resistor?

Hi,

1)Why not? Probably because lowering a typical tone pot already does what you evoke - the tone capacitor being really involved from 3.5 to 0/10 approximatively: from 10/10 to 3.5, a tone pot is mostly varying the resistive load ("mostly" being not synonymous of "only", of course: I'm not saying that a tone cap can't be heard at all but just that its presence or absence is not much noticeable during 2/3 of the curve on a tone control).

Illustration of what I mean below (it's about a bass guitar and it's a sim... but my archives are full of lab results like that: I've just not enough time to search and share 'em right now).

https://www.talkbass.com/attachments/jazzpu-jpg-jpg.2971673/

2) 500k in parallel with a 500k pot = a total load of 252,47k. And 250k in parallel with 500k = 166,66k. You can check it by using a calculator like the one shared below. :-)

https://www.allaboutcircuits.com/tools/parallel-resistance-calculator/

FWIW.HTH.
 
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Can't edit my answer so I'll just add a footnote.

For a load of 1M, one can use a 1M volume pot and a no-load tone control.

Obtaining a load of 250k just requires to set the tone control aforementioned @ 333k : in parallel with the 1M volume, it will give an overall value of 249k. And @ 333k, the cap of the tone pot doesn't yet reshape the sound drastically...

Etc.
 
freefrog,

thanks for that info. I clicked the link. Was there a calculator?

Also, are the tone and volume pots supposed to match in resistance
 
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freefrog,

Also, are the tone and volume pots supposed to match in resistance

No. The volume control is wired as a voltage divider, whilst the tone control is wired as a rheostat i.e. a variable series resistance between the capacitor and ground. A 250k vs 1Meg volume control of the same taper will still respond the same way (tonal effects aside). A 250k vs 1Meg tone control of the same taper will feel different in use; the range will be all bunched up with the 1Meg (which works for some).
 
freefrog,

[...]

Also, are the tone and volume pots supposed to match in resistance

Spaghetti Bolo has already replied but politeness wants me to answer too. :-)

No, volume and tone pots are not supposed to match: one can use whatever value from 100k to 1M for both (I've even mounted a 50k tone once by accident instead of a 500k. The tone was dull but still usable).

To add my two cents to what our fellow member shared in post 9: something to keep in mind is also that a 1M pot makes the sound brigther when full up but somehow darker when set half way (because the resistance in series with the pickup is higher). So a 1M volume pot might require a treble bleed cap.

FWIW
 
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