Make a 250KA pot more like 100K?

[chuggydown]

I want to use a concentric pot for controls that would best be served by 100KA and 250KA pots but the closest I can find is a concentric dual 250KA.   Can something be done to make a 250K pot act more like a 100K?

Thanks

[HotBluePlates]

Maybe.


What makes you think 100kΩ is a better value than 250kΩ? What is the circuit?

[sluckey]

I want to use a concentric pot for controls that would best be served by 100KA and 250KA pots but the closest I can find is a concentric dual 250KA.   Can something be done to make a 250K pot act more like a 100K?

Dual concentric means two stacked pots that are controlled individually by an outer and inner shaft. Is this what you really mean?

Or are you talking about a more common dual ganged pot where two stacked pots are both operated simultaneously by a single shaft?

If the latter, then AES has dual 100Ks in log or linear taper.

[chuggydown]

The values are stock for a Clinesmith lap steel guitar and the controls work well. One control is close to where I pick and I want to move it.  I do mean duel concentric pot.
Thanks
Jim

[HotBluePlates]

If they are the volume & tone control, then the only change you'll get by switching to 250kΩ for both is maybe a hair more brightness or output if the volume control was 100kΩ to start with. In which case, you turn down the volume and/or tone slightly to compensate.



If the Tone pot is the 100kΩ, replacing it with a 250kΩ ought to have little impact other than a possible slight brightening. Again, you just turn down the Tone control a little.


If you think having a 100kΩ pot is the only workable alternative, you can place a 180kΩ resistor across that section, from outer-lug to outer-lug. When turned all the way up, the total resistance is 104kΩ which is probably within tolerance of the existing pot. I don't think the altered pot will exactly follow a log taper anymore, as it will only match a 100kΩ pot's resistance at zero and full-up.


Your choice. You could try both.

[chuggydown]

Thank you! Very helpful.

[jojokeo]

I don't think the altered pot will exactly follow a log taper anymore

It should keep some or most of the taper since it's all just ratios through the dial-

[HotBluePlates]

I don't think the altered pot will exactly follow a log taper anymore

It should keep some or most of the taper since it's all just ratios through the dial-

I know what you mean, but others have pointed out how a resistor can be added to a linear taper pot to fake other tapers, but also that the the composite pot doesn't behave 100% like a true log taper pot.


Of course, here we want to change the value but not the taper, and I'm suggesting that the taper will be impacted in some way.


My usual knee-jerk is to go find the math formula to help calculate what the effect will be (and we can still do that by way of establishing an equivalent circuit and the equations that goven its behavior). But it's late, and it seems faster to me to just cut to the chase with a practical experiment.


I took a Bourns 500kΩ audio taper pot, and rigged a resistance across its outer lugs to mimic the change from a 250kΩ to 100kΩ total resistance (that's 40% of the initial resistance of the pot). My pot is 10% taper, which I found by measuring the resistance at half-rotation. Interestingly, though marked as "500kΩ" when I measured actual total resistance of the pot, I found it to be a hair under 400kΩ. Mid-rotation measured ~40kΩ.


I calculated the total resistance for my "new pot" at 40% of 400kΩ, or 160kΩ. I could have calculated the resistance to parallel with the outer lugs, but just hooked up a decade resistance box and turned knobs until I measured 160kΩ (that happened at 270kΩ paralleled with the total pot resistance; if you do the math, you'll find it matches my experimental result).


I twisted the pot knob until it was at half-rotation, and got 50.4kΩ; this is ~31% of my total pot resistance.


This result tells me the pot taper has changed from at least a 10% taper to something closer to a 30% taper (meaning closer to the action of a linear taper pot).


I would have tested more points, but the pot was not mounted on a panel or with a dial indicating percent rotation, so any other measurements would be ballpark at best.


I gather from this that the faked "100kΩ pot" created by paralleling across a 250kΩ pot will have a faster taper which will seem less smooth in its sweep. This may not be a problem for the original poster because he very well might leave the control in a set-n-forget position (like a lot of us do with guitar controls).

[jojokeo]

Thanks for taking the trouble to do that HBP. Could you say that the taper changed by the appx amount that the overall pot's resistance changed? (Again thinking ratio's here)

What lost me was you went from 500k (actually measured at 400k) then went down to 250k then 100k. What if you tried your experiment starting from the beginning w/ 500k unchanged (10% taper @ 1/2 rotation), But you need the pot's actual resistance measurement.

So, maybe you have a pot that's closer to 470k measured? Then you can take a 470k (measured) resistor across it and measure the resistance it at it's 1/2 rotation point. Could this be closer to the 20% taper since we changed it's ratio by half? Therefore the taper changes accordingly? Or does it actually try to stay closer to the original taper? My guess is that it does indeed change but closer to only 15% or maybe 20%...?

[HotBluePlates]

What lost me was you went from 500k (actually measured at 400k) then went down to 250k then 100k. What if you tried your experiment starting from the beginning w/ 500k unchanged (10% taper @ 1/2 rotation), But you need the pot's actual resistance measurement.

I was trying not to bog the post down with math. 


Original Poster wanted to make a 250kΩ pot into a 100kΩ. 100k/250k = 0.4, or 40%.


I have a 500kΩ pot, but upon measurement find it's 400kΩ. Cool, let's work with what it is, not what it should be. Same-change for this pot as the original poster's 250kΩ->100kΩ means my 400kΩ must be 400kΩ * 0.4 = 160kΩ.


Everything from there is straightforward.

Could you say that the taper changed by the appx amount that the overall pot's resistance changed? (Again thinking ratio's here)

While I think the whole line of questioning is based on a misunderstanding, we'll hold off on this.


If you really wanted an answer to this problem, deriving an equivalent circuit, applying the math for various cases, and plotting the results would be the way to go. That is, if you want to really know, "How bad do I make things when I try to fake a certain pot?"


Me, I'd just buy the right pot, or accept the flaws of the available pot... 

[jojokeo]

Me, I'd just buy the right pot, or accept the flaws of the available pot... 

Agreed! My speculation was just that, and at that point I didn't care any longer what the original posted values were - only the result of the experiment or hypothesis on "how the taper is affected" using "normal and simple" values to get a better/simpler to understand result. No big deal  ;)

[HotBluePlates]

I sat down with Excel and tabulated the resultant resistance to ground from the wiper when a 250kΩ pot had a 180kΩ resistor between lugs 1 & 3.


Long story short, the audio taper gets flattened to be closer to a linear taper, with the exception of a slight bend at the top of the rotation.


In essence, you look at the resistance from the wiper to ground as the actual resistance from wiper to lug 3 (the grounded lug) in parallel with the resistance from wiper to lug 1 (the "top" of the pot), which then has the 180kΩ resistor in series. That is, going from the wiper towards the top of the pot, electrons then see the 180kΩ is series as another path to ground.


When the pot is set to a low number (1, 2, 3 on the control), the small resistance from wiper to ground is only slightly lowered by the high-resistance parallel path through the external resistor. But as you turn the pot higher, that external resistance path becomes more significant, both lowering the total resistance to ground and also making the pot taper less curved.


Bottom-line: the pot will seem to "turn up too fast" but that may not matter if you set it for a desired sound and leave it.

[jojokeo]

Great to know and excellent description, thank you. The only time I can think of doing it was in a difficult situation that I didn't want to take apart a bunch of stuff to simply replace a pot once during a tweaking phase.