Dual to Single Gang Pot

[slashandburn]

I know that I know this. But my brain isnt working and my search skill seemingly also absent tonight.

I have a few dual gang pots lying around, and just noticed the values are almost perfect for the circuits Ive been playing where single gang pots are asked for.

So something very basic I need clarified if someone would be so kind.   I can just ignore one set of lugs and the jobs a good one, right?

I was about to look for that Secret Life of Pots article that probably answers my question, but figured Id be quicker asking. Excuse my laziness.


Iain

[italianguy63]

Yes.  They are just two pots on the same shaft.  Ignore one set.

MC

[slashandburn]

Thank you.      

I have a tendency to over think things.     Bridging them puts both wipers in parallel then?  Would bridging the lugs of a 100k dual gang pot be any different than putting a 100k resistor over a 100k pot?   Different taper?  

*edit, another mad thought that tells me its probably time for bed.   Is there a simple way to wire them in series?  Im short one 1M pot, and have might have a dual gang 500k, for example.

[aion]

Is there a simple way to wire them in series?  Im short one 1M pot, and have might have a dual gang 500k, for example.

It depends on the circuit (especially whether it's a voltage divider or a variable resistor) as to the exact wiring, but essentially just take the output lug of pot gang 1 (e.g. lug 2) and wire it to the equivalent input lug of gang 2. You'd end up with a log (audio) taper out of a dual linear, though, because turning both pots up at the same time will make the overall value go up exponentially rather than linearly.

[antonis]

just take the output lug of pot gang 1 (e.g. lug 2) and wire it to the equivalent input lug of gang 2. You'd end up with a log (audio) taper out of a dual linear, though, because turning both pots up at the same time will make the overall value go up exponentially rather than linearly.

I'm not sure about this...

If I've got your connection right, you connect 2 variable resistors in series with the same rotation so you have a double value pot wich increase or decrease with the double rate..

(if you want to make a kind of logarithmic pot (banana or hill) you have to connect in parallel to end lug ( upper or lower) and middle one a resistor with the 1/4 -1/5 value of the pot..)

[induction]

just take the output lug of pot gang 1 (e.g. lug 2) and wire it to the equivalent input lug of gang 2. You'd end up with a log (audio) taper out of a dual linear, though, because turning both pots up at the same time will make the overall value go up exponentially rather than linearly.

I'm not sure about this...

If I've got your connection right, you connect 2 variable resistors in series with the same rotation so you have a double value pot wich increase or decrease with the doube rate..

(if you want to make a kind of logarithmic pot (banana or hill) you have to connect in parallel to end lug ( upper or lower) and middle one a resistor with the 1/4 -1/5 value of the pot..)

You're correct for variable resistors, but you can connect voltage dividers in series for a quasi-log response. I've never tried it, but in principle, it's possible. There are a few ways to do it. Looking at the following series pot models:



Assuming linear pots, if r is the pot rotation percentage, the outputs are as follows:
A: Vout/Vin = r²
B: Vout/Vin = r/(2-r)
C: Vout/Vin = r/(1+r)

Here are some plots of the results:


As you can see, both A and B give a decent approximation to a log response, while C gives a decent approximation of an antilog response, but only gives a maximum of 50% output. The wrinkle is that, unlike with a normal single-gang pot, the input signal sees a resistance to ground that depends on the pot setting.

Is it useful or practical? No idea.

[antonis]

<Is it useful or practical? No idea.>

It doesn't have to be either...
(but it is very good idea.. )

[slashandburn]

Thanks.    Love this place.  Im glad I asked now.   Does the making a linear pot into an audio taper pot also then apply when wiring it as two parallel variable resistors? Im trying to get my head around whats happening both resistors (parallel) being controlled by the same shaft.

If thats the case then thats great. Makes life nice and easy and  saves me a headache.

<Is it useful or practical? No idea.>

It doesn't have to be either...
(but it is very good idea.. )

If it saves me a few quid in the short term and I learn a few things along the way, Ill be fairly content.   Some call it tight-fisted. I call it resourceful. Both are probably fair.

[induction]

Does the making a linear pot into an audio taper pot also then apply when wiring it as two parallel variable resistors? Im trying to get my head around whats happening both resistors (parallel) being controlled by the same shaft.

No, parallel variable resistors will give you a linear variable resistor with half the value of each pot.

The way to think about it is this. Parallel resistors give a total resistance of Rtot = R1*R2/(R1+R2).

E: If the two gangs are both wired to increase resistance clockwise (in goes to lug 1, out from lug 2, or vice-versa), then R1 = R2 = X*r, where X is the value of the pot, and r is the rotation percentage. So Rtot = R1²/(2*R1) = R1/2 = X*r/2

F: If the two gangs are both wired to increase resistance counterclockwise, then R1 = R2 = X(1-r), so Rtot = X(1-r)/2.

G: If you want to try something crazy and wire one to increase clockwise and the other to increase counterclockwise, you get Rtot = X(r-r²), which is a parabola. (The steps are left as an exercise for the reader.)



If you want to see the math behind the voltage divider plots above, let me know and I'll post it.

[induction]

Incidentally, if you reverse the order of the pots in example C, you get a decent full-range antilog approximation.

[antonis]

you get Rtot = X(r-r²), which is a parabola.

You're my man...!!!

(I'll replace my "deaf" son's Volume control pot with this... )

[induction]

That one works for voltage dividers, too. Maximum output voltage is 25% input voltage at 50% pot rotation.

[slashandburn]

Excellent stuff, thank you all very much!  Some of that went a little over my head but interesting stuff all the same.

Iain

[Hatredman]

Kinda thinking of using the J one on a filtre or wha and see what happens.

[MrStab]

good to see another Weegie on here, Iain!
i'm a Gorbals swamp inhabitant.

[slashandburn]

Mr Stab fae Glesga?  Aw naw man. Here, me slagging off the Old Firm, that was all just jokes. Honestly. Taken out of context.   

Truth be told, I'm an imposter. Just a bit up the road in Stirling. Moved back a few years ago. Couldnt hack the city life after the weans came along. Back home to Banjo Country.

[MrStab]

s'aw good, no Old Firm pish for me. hope that's an electric banjo plugged into a Tube Screamer! 

[slashandburn]

That might actually be an idea for a cover version idea Ive had in head for way too long.

I cant play Banjo though man. I studied at Stow, I can barely tie my own shoelaces.

[MrStab]

i used to live near Stow, in Garnethill. think it's City of Glasgow College or something now.
sorry for hijacking btw! i'll shut ma hole.

[slashandburn]

Its sound. Just doing our bit to reinforce the misconception that Scotland is only about 25 folk, eh? Aye I was just round the corner, drank in the Halt and the Arlington bar.  Seen the Halt's shut down now. Arlington still claiming they've got the Holy Stone in the basement. I figure its true. The stone must have had its own gravitational pull. Would explain the sticky carpet, anyway.

Genuinely though. Aye the Halt was my local.  Hole of a pub. Some good nights though.