How to make a log pot's taper more log?

[asmith]

Recently I finished my humbucker/humbucker Les-Paul-copy guitar's rewiring, according to this scheme:



Note: the "B Tone" (Bridge Tone) and "N Tone" (Neck Tone) potentiometers are wired as simple 'variable resistors.'

In the scheme, these variable resistors act as controls to do one of two things, depending on how the scheme's switches are set. They either act as 'Spin-a-Split' controls, or as 'normal' tone controls. (A 'Spin-a-Split' control sets the volume of one coil in a humbucker while the other coil's volume is unaffected; this lets me 'dial in' whether a pickup is either a humbucker, or a single coil, or some point in between.)

The potentiometers are both rated at 500k ohms, and have 'log' tapers.

I really like my wiring scheme. However, I would like to change the 'sweep' of these controls. In both 'Spin-a-Split' mode and 'normal tone control' mode, the controls all have the most noticeable change in tone between 1 and 5 on the 'dial.' Don't get me wrong: I like the tonal changes from 5 to 10! (And, those changes from 5 to 10 are more prominent on a tube amp than they are on a solid state amp, interestingly.) But, still: I would rather have the range of tones I get from 1 to 5 on the control right now happen over, say, 1 to 8 (and the tonal range I get from 5 to 10 right now therefore happen from, say, 8 to 10).

To change this 'sweep,' I think I need to change the taper of my pots, to make them even more log than they are already.

But can I do this by somehow soldering resistors across the potentiometers' lugs and rewiring the potentiometers?

(I don't mind losing 100k resistance or so over the whole potentiometer in doing this.)

Thanks in advance, Nutz.

[newey]

Obvious typo, but your diagram lists "N tone" as "V tone". Not sure of an answer to your question, but the first thought that came into my mind was an antilog taper and wire the pot backwards?.

[asmith]

Sept 23, 2022 16:06:30 GMT -5 newey said:

Obvious typo, but your diagram lists "N tone" as "V tone".

🤦

[Yogi B]

Sept 23, 2022 15:05:49 GMT -5 asmith said:

To change this 'sweep,' I think I need to change the taper of my pots, to make them even more log than they are already.

But can I do this by somehow soldering resistors across the potentiometers' lugs and rewiring the potentiometers?

(I don't mind losing 100k resistance or so over the whole potentiometer in doing this.)

Unfortunately not, no — a parallel resistor can only make the taper less log (or equivalently, more anti-log).

The average resistance of parallel resistances is the harmonic mean of the resistances. With the harmonic mean being less than or equal to the other two Pythagorean means (arithmetic & geometric) for any set of positive numbers, you could say it is 'weighted' towards the lower numbers within the dataset.

For example: you say you wouldn't mind losing 100k, so the total resistance of the pot goes from 500k to 400k. In order to do this you'd need a 2 Meg resistor (1/(1/400k − 1/500k) = 2Meg).

Assuming you currently have a 10% taper pot such that at "5" the resistance is 50k — with 50k being much smaller than the full value of 500k the 2Meg resistor has much less of an effect upon it (again, the harmonic mean is 'weighted' such that smaller numbers have greater influence over the result). The 50k will drop, but only by around 1.2k, down to 48.8k.

So, looking at the combined 500k pot and 2Meg resistor, we have a max value of 400k and value at halfway of 48.8k, or a taper of 12.2% — 'worse' (closer to linear, 50% taper) than the 10% we started with.

---

See also: the last section/graph of R. G. Keen's The Secret Life of Pots.

[JohnH]

I've explored those issues too with pots and humbuckers. As youve noted, with both spin-a-split and normal tone control, and also with series blending, all the action is in the first half if the turn 1-5, (or 5.5) with just a little bit extra to add in the second part 5-10. And you can't spread this out better with parallel resistors as Yogi notes.

But there can be ways using pot surgery, starting with a 250k log pot instead of 500k. Assuming your pots are 10% taper, the 1-5 range that you are using is up to around 50k. A 250k log will give you much nicer control 1-5, getting to 25k, then to 50k at about 6 or so. Its all better except it stops at 250k and we want that last bit up to 500k.

Starting with a full-size 250k log, you can open it up and take out the track. Then scrape part of the track. Keep a meter across the ends of the track measuring resistance, and you have very precise control as you watch resistance increase with each scrape, k by k.

You need to visualise which end of the track is the 5-10 part. I've done lots of these and never lost one yet.

There are a few options.

If you work on about the upper third, which is the high end, you can stretch the max ohms from 250k up to around 400k, and I did exactly that on my LP which came with 250k pots. Work on the outer edges and corners of the track with gentle scrapes.

And/or, you can scrape a complete cut at just before 10 and make it no-load. I really like a no-load pot with my humbuckers. So you get to 250k at 9.5 (or a bit more if you stretched it as above), then it jumps to infinite and you get a tad more treble than you had before, or no residusl split action. I like that extra bit. But if you want to end with a pure 500k instead of infinite, then put a 470k or 560k fixed resistor across your pot, which will remain in circuit past the no-load cut. The last increment 250 to 500, or 250 to infinite is really a small step. Obviously, you cant then subdivide it, but its small step.

You can of course just buy a no-losd pot, which will also give you a nice detent at 10.

[asmith]

Sept 24, 2022 16:58:43 GMT -5 JohnH said:

You can of course just buy a no-losd pot, which will also give you a nice detent at 10.

I suppose buying a 250k no-load pot is the way to go, then. Or maybe even less than 250k, if all the action happens from 0–100k or so?

Yogi B, thanks very much for your clarification. 🙇

[JohnH]

Id suggest the 250k, because the tonal step from 250k to 500k, or infinite, is quite small and if you don't have any finer steps in between, then there's not much lost. But 100k still has a significant change above that value that it can't control.

[asmith]

Sept 28, 2022 7:32:12 GMT -5 JohnH said:

Id suggest the 250k, because [good reasons].

Great!

...

...Uh... know anywhere that sells push-push DPDT switchpots rated at 250k with a no-load end at '10' and a corresponding detent?

[newey]

Oct 23, 2022 10:07:02 GMT -5 asmith said:

know anywhere that sells push-push DPDT switchpots rated at 250k with a no-load end at '10' and a corresponding detent?

Sorry, nope. You'd have to modify a regular push/pull pot (which can probably be done. And, no detent.

[Deleted]

Push pull pots are easy to split
Bend the table that joins the main ring and pull up.
Avoid pushing the switch down as it makes it harder to clip the arm back in to place.

Normally it's a blank between the Pot and the Switch side. So will need to bend the tabs on the top to remove the pot PCB . Then just need to cut the right side and put it all back together

watch out when putting the arm back in to the switch, not to click it down in error.