Converting a 4.7k Linear Pot into a 2.5k Linear ?

[steveyraff]

Hey guys,

Can I bridge the lugs of a B4.7k pot to turn it into a B2.5K pot? Whats the best way to go about it, and with what resistor value?

Many thanks,
Stevey.

[Kipper4]

You can Stevey.

This should help you out.


http://www.sengpielaudio.com/calculator-paralresist.htm

[deadlyshart]

I've thought about this before, and I don't really think it's possible.

(cue someone smarter than me coming in and showing exactly how it's possible!)

Because your numbers aren't as round, I'm gonna pretend you have a 10k pot, and you want to make it into a 5k pot, and I'm gonna omit the k's because they don't matter.

So the way a pot is, is essentially 3 pins, A,B,C, where B is the middle one, and you have R1 between A and B, and R2 between B and C, and when you turn the pot, it changes R1 and R2, but they're constrained by R1+R2=10.

What you want to do is somehow add something so that you now have the same scenario, but it's now R3 between A and B, and R4 between B and C, and R3+R4=5.

You can show it more mathematically (basically by showing that the system is underconstrained), but you can kind of show pretty easily using a couple counterarguments that adding constant resistors across the pins in any configuration won't work.

Pretend you add a resistor R5 across A and B. Now, when the pot is set so that R1=10,R2=0, R3=1/(1/10+1/R5). What you want at this point is, R3=5, because that would be the extreme value of you "modified" pot. So you can get this if you choose R5=10. But now, with the same R5 value you chose, set the pot so that R1=R2=5 (the mid point). Now R3=1/(1/5+1/R5), and we want R3=2.5 here. To do this, you'd need R5=5 though, meaning there's no constant R5 value that works...

That's one configuration that can't work, but if you do it out for putting a single resistor from A to C, or multiple resistors, you'll see the same thing.

The one thing that actually *does* work is if you had a dual pot, and soldered the corresponding pins together, so it was like two pots in parallel, that have to turn at the same time. But at that point... you probably have better options.

Someone please tell me if there's a cool way to do this!

[slacker]

If you're using it as a voltage divider, like a typical volume pot then you can connect a resistor across the outside lugs to lower the total resistance of the pot and it will still be linear. A 4k7 resistor would make it 2k35 which is close enough for rock and roll. A lot of the time though you can change the value of a pot used as voltage divider and the circuit won't care, so a 4k7 pot might work fine.
If you're using it as a variable resistor, using only two lugs then like deadlyshart you can't make it smaller and be linear.

Have a look here for more info http://www.geofex.com/Article_Folders/potsecrets/potscret.htm

[steveyraff]

I've asked this question before, but it was with a 5k Pot.

Govmnt_Lacky responded:
"Take a 5K linear pot (5KB) and put a 5.1K resistor between Lug 1 and Lug3...... VIOLA.... its now a 2.5KB pot". I've done this before and it sounded great.

So I was just presuming all I had to do was change the resistor value bridging lug 1 and 3 if I was converting a 4.7k pot. I would have imagined 5.1k resistor would have been near enough still ?

P.S Its for the mids control of a Dr.Boogie.

[deadlyshart]

I've asked this question before, but it was with a 5k Pot.

Govmnt_Lacky responded:
"Take a 5K linear pot (5KB) and put a 5.1K resistor between Lug 1 and Lug3...... VIOLA.... its now a 2.5KB pot". I've done this before and it sounded great.

So I was just presuming all I had to do was change the resistor value bridging lug 1 and 3 if I was converting a 4.7k pot. I would have imagined 5.1k resistor would have been near enough still ?

P.S Its for the mids control of a Dr.Boogie.

It's true that if you put a constant resistor of the total value of the pot across its pins 1 and 3, it will halve the total resistance of it... but it won't be a "2.5Kohm pot", because the center pin won't scale correctly. Honestly, it may be fine depending on your application, though.

[askwho69]

I've asked this question before, but it was with a 5k Pot.

Govmnt_Lacky responded:
"Take a 5K linear pot (5KB) and put a 5.1K resistor between Lug 1 and Lug3...... VIOLA.... its now a 2.5KB pot". I've done this before and it sounded great.

So I was just presuming all I had to do was change the resistor value bridging lug 1 and 3 if I was converting a 4.7k pot. I would have imagined 5.1k resistor would have been near enough still ?

P.S Its for the mids control of a Dr.Boogie.

What if you wire the lug 1 and 2? It doesn't affect the lug 2.


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[steveyraff]

I also have a 2k pot I might try to see if its close enough.

[askwho69]

2k is ok or add 470ohm on the end of the lug , so that you can get the highest 2.5 slop


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