50k to 25k pot - which method?

[Beedoola]

its my understanding you can make a 50k pot 25k by either using a 50k resistor across lugs 1&3 or 1&2 - or is it 2&3?

which method works best/replicates the sweep/function of a normal 25k pot?

[shawnee]

Just put a 50k across lugs 1 and 3 and you end up with 25K.

[jimilee]

So if you take a 50k pot and put a 50k resistor across the lugs, It only takes it down by 25k? Can you explain it to me and a very short way?

[RobA]

It'll take it down to 25kΩ, but it won't be a linear law pot response. You can see that it'll a max of 25k just by doing the parallel resistor rule with the pot all the way clockwise. But, if you breadboard it and measure the result, the pin 1 to wiper resistance will be around 19kΩ instead of 12.5kΩ at 12:00.

As far as I know, there is no way to change the value of a pot without changing the law as well. If you use two resistors, you can do some different laws that might be closer to what you want. This link talks about some approaches.
http://sound.westhost.com/pots.htm#chg-law

[shawnee]

It can change the taper a little but that doesn't bother me. You could always switch to a Log taper pot if it was a linear to start with and it will correct some of that if your sweet spot was towards the middle. You probably won't get much change at all on the tail end of the turn though.

What DOES bother me is if you don't get close enough to the MEASURED value of the pot. What happens then is your max change when turning the pot may be at about 9 or 9 1/2 on the rotation of the pot and 10 is going backwards just a little. It hate that when maxing the pot isn't the maximum audible change. Measure your 50K pot from lug 1 to 3. Pots have a pretty loose tolerance. It may be something like 47K. If so, use a 47K resistor and you will get close to 23.5K. Just match the measure value as closely as you can.

A cool trick on gain pots is to switch the resistor in or out. That gives you an instant gain increase when you flip the switch and take the resistor out of the circuit because it's like doubling you gain pot value (kind of like an Fulltone OCD).

[RobA]

Assuming my math is right and you actually can match the resistor to the pot, the law you'll end up with is (2 x - x^2)V/2 where V is the pot value and x is the fraction of rotation CW. This is measuring the resistance from pin 1 to pin 2. So, it acts quite a bit like a reverse log pot. So, using a log pot to start with would straighten it back out some. But since none of the log pots are really log, I guess it would really be bent twice. But, that has me wondering if you can rig a way to start with a 100k linear and do the trick to make a 50k log and then use that to make a 25k linear.

Edit: And the answer is no. Basically because the two different law changes only work in two different settings. The resistor in parallel with the 1 & 2 pins only converts to a log law for a voltage divider and the resistor in parallel with the 1 & 3 pins only works in the case of a series resistor.

[RobA]

its my understanding you can make a 50k pot 25k by either using a 50k resistor across lugs 1&3 or 1&2 - or is it 2&3?

which method works best/replicates the sweep/function of a normal 25k pot?

So, my last edit brings up the question of what setting you are going to use the pot in? What's the circuit?

[kothoma]

So if you take a 50k pot and put a 50k resistor across the lugs, It only takes it down by 25k? Can you explain it to me and a very short way?

Not sure if this was already addressed.

Short answer is: two resistors in parallel provide two ways for the current so the overall resistance gets lower.
The two values here are the same so twice as much current gets through the two paths, which is the same as saying the resistance is halfed.

The general formula for this is: 1/Rpar = 1/R1 + 1/R2 + .... If you turn it around for two resistors you get Rpar = R1 R2 / (R1 + R2).
As R2=R1 here you have Rpar = R1 R1 / (2 R1) = R1 / 2.

Edit: As to

[...] It only takes it down by 25k? [...]

No, not only, the taper is changed too, but that's already discussed here. (Sorry it this was the heart of your question in the first place.)

[Gledison]

Assuming my math is right and you actually can match the resistor to the pot, the law you'll end up with is (2 x - x^2)V/2 where V is the pot value and x is the fraction of rotation CW. This is measuring the resistance from pin 1 to pin 2. So, it acts quite a bit like a reverse log pot. So, using a log pot to start with would straighten it back out some. But since none of the log pots are really log, I guess it would really be bent twice. But, that has me wondering if you can rig a way to start with a 100k linear and do the trick to make a 50k log and then use that to make a 25k linear.

Edit: And the answer is no. Basically because the two different law changes only work in two different settings. The resistor in parallel with the 1 & 2 pins only converts to a log law for a voltage divider and the resistor in parallel with the 1 & 3 pins only works in the case of a series resistor.

Yeah! Talk dirty to me! 😛
Sorry could miss the joke! Love this kind of discussions where i can learn a lot from your experiences. Seems like going back to the physics lesson in the Uni! The weird thing is that in that time i didnt like so much and now im pretty interested about it! Teachers should make students build pedals to fall in love with electronics!:)

[RobA]

[...]
Yeah! Talk dirty to me! 😛
Sorry could miss the joke! Love this kind of discussions where i can learn a lot from your experiences. Seems like going back to the physics lesson in the Uni! The wierd thung is that in that time i disnt kike so much and now im pretty interested about it! Teachers should make students build pedals to fall in live with electronics!:)

Funny thing is, when I was taking physics in Uni, building pedals wouldn't have interested me much. I only played classical guitar back then. The thing I really wanted in my physics courses was a good acoustics class. In our electronics lab course, the guy who was by far the best at it was an auto mechanic. He had, by far, the best intuition for what was going to happen. It's hard to tell what's going to motivate people. It couldn't hurt though to include some synth/effects building in a course.

[Gledison]

[...]
Yeah! Talk dirty to me! 😛
Sorry could miss the joke! Love this kind of discussions where i can learn a lot from your experiences. Seems like going back to the physics lesson in the Uni! The wierd thung is that in that time i disnt kike so much and now im pretty interested about it! Teachers should make students build pedals to fall in live with electronics!:)

Funny thing is, when I was taking physics in Uni, building pedals wouldn't have interested me much. I only played classical guitar back then. The thing I really wanted in my physics courses was a good acoustics class. In our electronics lab course, the guy who was by far the best at it was an auto mechanic. He had, by far, the best intuition for what was going to happen. It's hard to tell what's going to motivate people. It couldn't hurt though to include some synth/effects building in a course.

Definatly!
I thing if the teachers were more open minded to ask the students what they are interested on, would be much easy to diggest some physics theories:)
As a chemist i was more interested in how to create electricity than aplying it!