#1

OK, I got a 100K pot, and I heard it is possible to change it's value to 50K.

If so, what resistor must I use and to which terminals to solder it?

If so, what resistor must I use and to which terminals to solder it?

#2

Solder a 100k resistor between the outside lugs.

#3

what are you doing??? what is it for???

#4

what are you doing??? what is it for???

Expression pedal, got the thing running, but still experimenting.

#5

you cant magically make a 50k pot with a 100k pot and a 100k resistor. you wont be able to control the entire range of resistance with the pot.

#6

you cant magically make a 50k pot with a 100k pot and a 100k resistor. you wont be able to control the entire range of resistance with the pot.

I already cant, with my 100K pot.

#7

I dissagree, it's entirely possible.you cant magically make a 50k pot with a 100k pot and a 100k resistor. you wont be able to control the entire range of resistance with the pot.

#8

What Jim means is it screws up the taper of the pot. Yes, at full value, it will be the same as a 50K pot, but throughout its range it'll be a set of weird values with an odd taper.

#9

What Jim means is it screws up the taper of the pot. Yes, at full value, it will be the same as a 50K pot, but throughout its range it'll be a set of weird values with an odd taper.

OK, I kinda didnt udnerstand what you said.

#10

Time for a lesson in potentiometers and parallel resistance. The unit for resistance is ohms and its symbol is the Greek letter omega (Ω.

This is a potentiometer. It consists of two lugs connected to both ends of a resistive strip (usually carbon) with a middle lug connected to a wiper that travels across the resistive strip. End lug 1 is usually called lug 1, the wiper is called lug 2, and end lug 2 is called lug 3. For our purposes let's assume this is a 100KΩ pot with a linear taper for ease of calculations. Right now the resistance between lugs 1 and 2 is, for all practical purposes, 0Ω, the resistance between lugs 2 and 3 is 100KΩ, and the resistance between lugs 1 and 3 is also 100KΩ.

We've now rotated the pot to 50% of its travel. The resistance between lugs 1 and 2 is now 50KΩ, lugs 2 and 3 50KΩ, and lugs 1 and 3 100KΩ. Now let's slap a 100KΩ resistor in parallel with the carbon strip by soldering it across lugs 1 and 3 and see what happens.

1/Req=1/R1+1/R2+1/R3+...+1/Rn. That reads as "The inverse of the equivalent resistance of resistors in parallel is equal to the inverse of resistor one plus the inverse of resistor two and so on and so forth for n number of resistors in parallel." Therefore, the resistance if turned all the way to the left as in the original picture would now be 0Ω between lugs 1 and 2, 50KΩ between lugs 2 and 3, and 50KΩ between lugs 1 and 3. Sounds good, right? We've just made a 50kΩ potentiometer. Not quite. If you rotate to 12 o' clock as in the above diagram, the resistance between lugs 1 and 2 is 37.5kΩ, lugs 2 and 3 is 37.5kΩ, and the resistance between lugs 1 and 3 is 50kΩ. This happens because you provide an alternate path for current to travel from lug 2 to lugs 1 and 3, which is the carbon strip back to lug 1 or 3 in series with the resistor across lugs 1 and 3. Resistance is added in series, so lug 2 sees a 150kΩ resistor in parallel with a 50kΩ resistor between it and both lugs 1 and 3. So, although the resistance is correct between lugs 1 and 3, it is lower than intended between both lugs 1 and 2 and lugs 2 and 3.

This is still acceptable for most uses, but things get weird in the positions between half and full rotation. If you turn the pot 75% of the way there are 25KΩ of resistance between lugs 2 and 3 in parallel with 175KΩ, whereas there are 75KΩ of resistance between lugs 1 and 2 in parallel with 125KΩ. This means the resistance between lugs 1 and 2 is about 47KΩ, whereas the resistance between lugs 2 and 3 is only about 22KΩ. That's a 2.1:1 ratio between the two resistances, as opposed to the 3:1 ratio expected with a linear taper potentiometer where, in a 50KΩ resistor, the values would be 37.5KΩ and 12.5KΩ, respectively.

Hopefully that cleared things up for you.

This is a potentiometer. It consists of two lugs connected to both ends of a resistive strip (usually carbon) with a middle lug connected to a wiper that travels across the resistive strip. End lug 1 is usually called lug 1, the wiper is called lug 2, and end lug 2 is called lug 3. For our purposes let's assume this is a 100KΩ pot with a linear taper for ease of calculations. Right now the resistance between lugs 1 and 2 is, for all practical purposes, 0Ω, the resistance between lugs 2 and 3 is 100KΩ, and the resistance between lugs 1 and 3 is also 100KΩ.

We've now rotated the pot to 50% of its travel. The resistance between lugs 1 and 2 is now 50KΩ, lugs 2 and 3 50KΩ, and lugs 1 and 3 100KΩ. Now let's slap a 100KΩ resistor in parallel with the carbon strip by soldering it across lugs 1 and 3 and see what happens.

1/Req=1/R1+1/R2+1/R3+...+1/Rn. That reads as "The inverse of the equivalent resistance of resistors in parallel is equal to the inverse of resistor one plus the inverse of resistor two and so on and so forth for n number of resistors in parallel." Therefore, the resistance if turned all the way to the left as in the original picture would now be 0Ω between lugs 1 and 2, 50KΩ between lugs 2 and 3, and 50KΩ between lugs 1 and 3. Sounds good, right? We've just made a 50kΩ potentiometer. Not quite. If you rotate to 12 o' clock as in the above diagram, the resistance between lugs 1 and 2 is 37.5kΩ, lugs 2 and 3 is 37.5kΩ, and the resistance between lugs 1 and 3 is 50kΩ. This happens because you provide an alternate path for current to travel from lug 2 to lugs 1 and 3, which is the carbon strip back to lug 1 or 3 in series with the resistor across lugs 1 and 3. Resistance is added in series, so lug 2 sees a 150kΩ resistor in parallel with a 50kΩ resistor between it and both lugs 1 and 3. So, although the resistance is correct between lugs 1 and 3, it is lower than intended between both lugs 1 and 2 and lugs 2 and 3.

This is still acceptable for most uses, but things get weird in the positions between half and full rotation. If you turn the pot 75% of the way there are 25KΩ of resistance between lugs 2 and 3 in parallel with 175KΩ, whereas there are 75KΩ of resistance between lugs 1 and 2 in parallel with 125KΩ. This means the resistance between lugs 1 and 2 is about 47KΩ, whereas the resistance between lugs 2 and 3 is only about 22KΩ. That's a 2.1:1 ratio between the two resistances, as opposed to the 3:1 ratio expected with a linear taper potentiometer where, in a 50KΩ resistor, the values would be 37.5KΩ and 12.5KΩ, respectively.

Hopefully that cleared things up for you.

*Last edited by Mike-T93 at Jun 1, 2010,*

#11

u'd want to put a 50k resistor between lug 1 and 2, and lug 2 and 3 instead...

but 50k is hard to find, look for a 47k.

but 50k is hard to find, look for a 47k.

#12

u'd want to put a 50k resistor between lug 1 and 2, and lug 2 and 3 instead...

but 50k is hard to find, look for a 47k.

+1

I can't believe I explained the whole thing and forgot to tell him the solution

If it's a logarithmic (audio taper) pot then the values will need to be different. I haven't run the calculation, though, and I'm headed off to bed.

*Last edited by Mike-T93 at Jun 1, 2010,*

#13

yeah. it'll mess up the taper but i think that's the best solution?

#14

Eh what the hell, I googled it and a log taper looks to be about 18KΩ between lugs 1 and 2 and 82KΩ between lugs 2 and 3 at 50% rotation. If picking from standard values you'd need to solder 22KΩ across lugs 1 and 2 and 47KΩ in series with 22KΩ across lugs 2 and 3. Test your resistors with a multimeter to arrive at closer values to 18KΩ and 82KΩ. If you're really nitpicky replace the 22KΩ resistors with two 10KΩ resistors in series for both cases and you'll get a bit closer, but the difference won't be that important.

Edit: G'night folks!

Edit: G'night folks!

*Last edited by Mike-T93 at Jun 1, 2010,*

#15

It only gets worse from here I guess....

#16

yes it does, TS. sounds like you need to do some reading.

here: http://www.sound.westhost.com/pots.htm

and here: http://www.geofex.com/article_folders/potsecrets/potscret.htm

are good places to learn all about pots.

Mike: since we are only dealing with 2 resistors, a MUCH less confusing formula would be the product-over-sum version. you know it, im sure. plus its easyier for beginners to understand.

(R1*R2) / (R1+R2)

in fact i prefer it instead of that long confusing one you used. It can only calculate 2 resistors at a time but it's way easier to do mentally.

here: http://www.sound.westhost.com/pots.htm

and here: http://www.geofex.com/article_folders/potsecrets/potscret.htm

are good places to learn all about pots.

Mike: since we are only dealing with 2 resistors, a MUCH less confusing formula would be the product-over-sum version. you know it, im sure. plus its easyier for beginners to understand.

(R1*R2) / (R1+R2)

in fact i prefer it instead of that long confusing one you used. It can only calculate 2 resistors at a time but it's way easier to do mentally.

#17

OK, can you give simple orders of what to solder where?

Please?

Please?

#18

You need to tell us whether is linear taper or logarithmic (audio) taper.

#19

You need to tell us whether is linear taper or logarithmic (audio) taper.

How can I tell?

All I know is that it's a standard 100K pot that comes with a wah.

#20

does it have A or B anywhere on it

a= log taper, b=linear taper. you'd know that if you'd have read my links.

a= log taper, b=linear taper. you'd know that if you'd have read my links.

#21

I'll have to check.

#22

Apperantely It's a linear taper.

#23

So..... found this thread while googling changing pot values w/ resistors. I was wondering how you would calculate the two resistor values for an audio pot? Seems like this thread went kaput a few months ago, but I'm curious. Heck, I even created an account to find out! Thanks.

#25

Lots of good things being posted on here. Going to have to do some reading.

mothermuxer: What value is the pot?

mothermuxer: What value is the pot?

*Last edited by salgala2000 at Dec 9, 2010,*

#26

500k pot w/ audio taper. Would like it to be either 300k.

#27

doh! Hit post before finishing my thought.... 300 or 250k. I know I could purchase a 250k pot, but I've got 500k pots out the wazoo in my shop & would rather buy some resistors than new pots.

#28

^ i know how you feel. i have 50k pots out the ass but have little use for them. idk why 50k are so common.

anyway codemonk's link should be all you need.

anyway codemonk's link should be all you need.

kgraham0103

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