Change pot value as variable resistor

[calciferspit]

I know how to lower a pot's value when using it as a potentiometer, but I'm thinking it would be different using the pot as a variable resistor. I'd like to use a 390r resistor to 'remove' ~ that value from a 1k pot. Would I put the resistor across 1&3, 1&2, or inline to 1 or 3? I don't really care what happens to the taper.

[JakeFuzz]

I am not quite sure what you mean by removing a value but I think I know what you are asking. Just to be sure you are asking if you just want to use a pot as a variable resistor (like connecting lugs 1&2 or 2&3 to the circuit) how do you change the maximum resistance value to lower than what is specified by the pot.

You would use the same technique that most use to lower the value of their pots by placing a resistor across the outer (1&3) of the pot. This "adds" resistance in parallel to the resistance value of the pot. It works the same for the variable resistor because we are still sweeping between the outer ends of the trace (and therefore the overall effective resistance will be the same) you are just not using one of the outer lugs.

Putting resistors across 1&2 (or 2&3) is actually very interesting. It does very strange things to the values of the pot and the tapers. I used to have an Excel file that plotted the values as a function of pot rotation, it was very cool but probably not too useful.   

[calciferspit]

yes that what i'm asking. Thank you for your reply. I'd like a pot with about 600-700r used as a variable resistor. Its actually for a gain control at r4 of the weener wah. since i'm removing the the 390r, i figure I'd reuse it on the 1k pot I have to make it's range more usable (o-~600r vs 0-1k). so resistor across 1&3, 1&3 to board, and 1&2 tied together should do it.

[JakeFuzz]

yes that what i'm asking. Thank you for your reply. I'd like a pot with about 600-700r used as a variable resistor. Its actually for a gain control at r4 of the weener wah. since i'm removing the the 390r, i figure I'd reuse it on the 1k pot I have to make it's range more usable (o-~600r vs 0-1k). so resistor across 1&3, 1&3 to board, and 1&2 tied together should do it.

So if you massage the equation a bit you can get the exact resistor value you need to put across lugs 1&3 to get 600 ohms. it comes out to:

R2 = (R1*Rf)/(R1-Rf)

Where R2 is the resistor on the outer lugs; R1 is the resistance of the pot; Rf is the resistance value you want (600r). Solving this tells you to get a 600 ohm value with a 1K pot you need to put a 1.5K ohm resistor across the out lugs.


EDIT: Also you don't need to tie the pot wiper (lug 2) to one of the out lugs. You can but it will change the taper in a weird way. When you do that now you are putting the 1.5K resistor in parallel with whatever shows up across it and lug 3 in your example. I would connect 1 and 2 to the board and just leave 3 hanging (connected to the other end of the 1.5K of course)

[calciferspit]

So it does. That makes total sense. I was thinking about this the wrong way. Gotta love the maths. thanks again.

[mgwhit]

Using a pot as a variable resistor (rheostat) involves connecting either lugs 1 & 2 or 3 & 2 into your circuit.  The remaining lug is frequently connected back to lug 2 just to provide a path of last resort in case of a component failure (e.g. the wiper stops conducting).

If you're wiring input to lug 1, output from lug 2 and lug 3 back to lug 2 for a failsafe (as above), I don't think it matters if you put the parallel resistor across 1 & 2 or 1 & 3.  Should be electrically identical.

In a scenario where you were wiring input to lug 1, output from lug 2 and leaving lug 3 unconnected you need to put the resistor across lugs 1 & 2.  Putting it across lugs 1 & 3 messes with the taper so much that the high value of the rheostat may occur somewhere in the sweep other than the end.

[mgwhit]

I just went back to read Paul's posts.  (I spent a long time thinking about this.)  His math and explanation are totally correct, but I'm somewhat concerned with using lugs 1 & 3 as a rheostat.  If you are designing your own circuit and you know that you will be using a pot with a parallel resistor, you can design with lugs 1 & 3, but most commercial boards use lugs 1 & 2 as a rheostat (or 3 & 2).  If you design for lugs 1 & 3 and then don't use a parallel resistor, you end up with a non-working rheostat.

If you are, however, trying to modify the Drive control of a Tube Screamer-type overdrive that uses a lug 1 & 2 rheostat I would recommend putting the parallel resistor across lugs 1 & 2 or you will drive yourself crazy.

(Paul, let me know what you think.)

Edit: Although now that I think about it, most decent boards (like the Green Bean, Snarkdoodle, Ego Driver, Boneyard, Slow Loris, etc.) do the failsafe trick, so it probably doesn't matter a hill of beans.

[calciferspit]

Time to experiment. I can't fry anything here so I might as well try the options and find what functions best. Honestly, a 1k rheostat could be completely usable as-is. I haven't tried it yet. My choice was just based on the fact that at 600r, the weener is as clean as it will get, but doesn't get dirty until it gets down to about 300r. I want a usable range. Has anybody done an R4 gain pot on their weener?

[JakeFuzz]

Matt, Now that I think about it I think your way is simpler. Yeah the 1&3 resistor does make the pot taper weird. I have a meeting here in a few minutes but I will do some plots when I get back. The equations should be fairly easy. Thinking out loud:

For the 1&2 resistor we will call the output an "effective" resistance:

Reff = (R(A)*R2)/(R2+R(A))

Where R(A) is the resistance of the pot as a function of angular position of the shaft and R2 is the value of the resistor you use from lug 1 to 2.


Now for the 1&3 resistor things get a little wild:

Reff = (R(A)*(Rt-R(A)+R2))/(Rt+R2)

This condition holds as long as one of the outer lugs is left open. Now Rt is the total value of the pot.

I will plot these when I get back and we should see some pretty wild looking tapers. Math is FUN!  


EDIT: Actually looking at these now the 1&3 is just going to be a quadratic taper. The 1&2 is going to be the strange one. Cool, I am excited to see what these will look like.

[midwayfair]

Is there a reason it has to be exactly 390R for your gain pot? You could use a 500Ohm pot in series with a 510R resistor. These are commonly available parts and waaaay less thinking.

[calciferspit]

Is there a reason it has to be exactly 390R for your gain pot? You could use a 500Ohm pot in series with a 510R resistor. These are commonly available parts and waaaay less thinking.

none at all. I was just trying to hack with what I had on hand. a 1k pot and few resistors. I'm looking to get a pot with a range around 600r-ish. If I had a 500r pot, i'd just use that. I just subtracted 390r (because I have that value resistor) from the 1k and thought i could get a 610r pot. not that easy it turns out. I'm going to just try the 1k plain and see if I like the range, otherwise I'll use a 1.5k resistor to fudge the pot into a 600r.

[JakeFuzz]

Well when you look at them they both look very similar. I thought the tapers would be more extreme than that but it looks like either one will work fine! These plots use 1KB as the pot value and a 1.5K resistor as the parallel resistance in both cases.

EDIT: I see now. The more extreme values give very different tapers between the two. The endpoints are the same but some of them look a little crazy. And I've invented the most useless taper in the world by shorting the outer lugs of the pot you can get a bump in the middle of the sweep... weird.

[calciferspit]

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