Pot values when closed

[Crazy_Fool]

Hi all,

This is a question that may very well have zero real world application but it makes me wonder. What is a reasonable value in ohms for a volume pot that is at 10/open? I have a drawer full of desoldered pots and there is clear variation amongst these between lug two and three when set to ten. Amongst them there are values that range from .2 ohms to 5ohms. My curiosity is that- if we consider the incredibly small amount of power passing through a pot - what is the influence of a few ohms on such a small signal. E.g. lifting 500 lbs is nothing for an olympic weightlifter(or a few ohms across the plates of a power tube), but how would a toddler handle that weight (or the signal from a pickup deal with a few ohms of resistance before it even hits the output jack)?

It might be that the affect of resistance is not liniar with values that small or I just have no idea of what I'm talking about. Regardless, I'd like to know it resistance at this point is worth considering.

[The Dude]

It depends on pot value. If it's a 10 ohm pot and you get 5 ohms from wiper to end when turned fully to that end, it's a problem. On guitar pots (usually 250k or 500k), that much variation is normal. I believe most pot tolerances are 20-30%.

[Crazy_Fool]

It depends on pot value. If it's a 10 ohm pot and you get 5 ohms from wiper to end when turned fully to that end, it's a problem. On guitar pots (usually 250k or 500k), that much variation is normal. I believe most pot tolerances are 20-30%.

Thanks. It is indeed a 500K pot and it reads 500K on the button. I've changed mountains of pots over the years without ever checking the values other than hot to ground, so there's a good chance that I have a lot of pots in guitars with variation on that side of the sweep. At the same time I am rather curious about how much signal would be held back when say there is a ~250mv current from a passive guitar pickup pass through a 500K pot set to 10 but where there is still ~5ohms of resistance. It may be within normal parameters to have such resistance. Doing a little google-fu I cam cross a voltage calculator. Now it may be that I'm completely barking up the wrong tree on this and massively wrong but it appears as though if we assume a humbucker to have an output of 250mv and in condition 1 we have a pot with a minimum resistance of 5 ohms we would get: I = V / R = 250×10^-3V / 5Ω = 0.05A
P = V×I = 250×10^-3V×0.05A = 0.0125W

Now for 2, if we change out the pot for one where the minimum resistance is .5 ohms (but leave everything else as it was from 1 we would get
I = V / R = 250×10^-3V / 0.5Ω = 0.5A
P = V×I = 250×10^-3V×0.5A = 0.125W

I can accept that I'm just completely wrong and using this formula so far out of context that it's absurd. At the same time, such a small amount of resistance appears to have a pretty large effect on the amount of power that can pass through the pot. Feel free to school me on where my logic is failing if I'm missing something.

Thanks

[The Dude]

Firstly, I'm not sure what you're really after solving (guitar output, power dissipation across the pot, other?), I do see problems with your theory. One is that the pickup output in standard guitar wiring is across the entire potentiometer (500K) regardless of where the wiper is set. So, signal level would vary from wiper to ground and would not be constant (your 250mv). The pot is a variable voltage divider.

[35L6]

I often turn the volume down just a hair to soften the tone . I've never measured the ohms but it could have a subtle effect . On the other hand , if you have a pot that measures exactly 500k , keep it you may never see another !

[Chuck H]

It depends on pot value. If it's a 10 ohm pot and you get 5 ohms from wiper to end when turned fully to that end, it's a problem. On guitar pots (usually 250k or 500k), that much variation is normal. I believe most pot tolerances are 20-30%.

And I'll add... It has a lot to do with the impedance of the circuit the pot is used in. For a guitar circuit with an impedance of something like 6k to 15k a series resistance of 5 ohms is basically meaningless for tone. That said, guitar amps have very high gain. And some modern, high gain amps are high gain versions of an already high gain device. So almost any signal bleed in the system can make for audible amplification of signal. This applies to when a pot is full off rather than any affect such low resistance has on the sound when the pot is full on. This can be annoying. You turn your guitar all the way down and there's still a tiny signal getting through. I've seen this happen so I try to choose pots on the lower end of your measured results for the "off" lugs. Without as much concern about the "on" lugs.

I often turn the volume down just a hair to soften the tone . I've never measured the ohms but it could have a subtle effect . On the other hand , if you have a pot that measures exactly 500k , keep it you may never see another !

Any pot turned down by hand even "just a hair" will have a series resistance in the tens of thousands of ohms. Especially audio pots typically used in guitars. All pots exhibit something I'll call "pad lag" for lack of any term I know for the issue. That the wiper isn't actually off the zero or ten pads the moment you begin adjustment. So say you have an actual "electrical travel" of 1.5 to 9.5. Even with the pot set to 9.3 (just one fifth of a /10 increment off the pad) you'll already have something like 25k resistance when using a linear taper pot and nearer 30k when using an audio pot. If you want to see what I'm talking about clip a 500k pot to your meter and try to adjust it in 5 ohm increments

[Helmholtz]

You wrongly applied formulas, as the 250mV don't appear across the "end resistance".

E.g. if you have 250mV across the entire pot, the current delivered by the PU is 0.25/500k = 0.5µA - independent of wiper position.
That current produces a voltage drop across a 5R end resistance of 2.5µV.
So power consumed by the end resistance is 1.25 pW (1 pico watt = 1E-12 W)

[J M Fahey]

Besides lots of measuring,I have actually made pots so I know the innards and practical problems.

A "perfect" pot should measure zero ohm between wiper and matching end terminal when fully turned clockwise or ccw, but in practice that is not the case, wiper can´t actually touch terminal which is held by a small metal rivet because it would be bent backwards quickly and lose contact on track, so it´s designed to be very close but never touch it.

Meaning you always have a mm or so of resistive track, so it´s never zero ohm but , say, 1% of total resistance.

In those particular cases where you NEED zero ohm, then before riveting a drop of silver conducting paint is applied at each end, so wiper rests on it at each end.

Problem is that said drop is never an exact size, typically "too large", end result is that a volume pot is still "zero" even if wiper moves from 0 to 1.

[Crazy_Fool]

You wrongly applied formulas, as the 250mV don't appear across the "end resistance".

E.g. if you have 250mV across the entire pot, the current delivered by the PU is 0.25/500k = 0.5µA - independent of wiper position.
That current produces a voltage drop across an end resistance of 5R of 2.5µV.
So power consumed by the end resistance is 1.25 pW (1 pico watt = 1E-12 W)

Thanks for the insight all. The physics of electricity is clearly out of my wheelhouse. From the in context summary above I can see that if we take 250mv as an arbitrary value coming from a humbucker and pass it through 5R at the pot we lose .0000025 volts or .00000625% if the signal. With that consideration, I think I can live with a few ohms at the end of the sweep.

Thanks for the help all.

[Helmholtz]

... or .00000625% if the signal.

No.
Absolute signal loss in the example is 2.5µV = 2.5 E-6 V.
Relative signal loss is 2.5µV/250mV = 10E-6 or 1E-3 %.
Same result is obtained from 5R/500k.

Anyway the loss is insignificant.

[mozz]

I have seen many rheostats that actually go to zero. Not in a guitar though.

[Crazy_Fool]

No.
Absolute signal loss in the example is 2.5µV = 2.5 E-6 V.
Relative signal loss is 2.5µV/250mV = 10E-6 or 1E-3 %.
Same result is obtained from 5R/500k.

Anyway the loss is insignificant.

Ah, I shouldn’t have replied while mowing the lawn at the same time. Regardless, I appreciate you clearing that up and demonstrating why the values I was seeing weren’t an issue at all. My thought is that - it’s better to ask questions even if silly and improbable rather than accept the status quo as no one has addressed the topic.

With that I chose a Dunlop Super Pot reading 501K closed (lug 2 and 3) and 2.5R open. It had the nicest travel.