Changing the value of a potentiometer

[Moustache_Bash]

If I wanted to change a 50K pot to 25K, would I just put a 50K resistor on lugs 1 and 3?

[eatyourguitar]

yes

[multi_s]

actually it depends.

do you have a schematic?

if its setup just to use 2 terminals, the taper of the pot will get pretty skewed.

[morange]

Cut it in half.

[Moustache_Bash]

It's for the volume. I only have 50K pots. So, yeah, it does only use two lugs....

[crazynoises]

Try this:
http://www.diystompboxes.com/analogalchemy/emh/emh.html

[multi_s]

ya ok well i didn't mean to sound like it wont work, just the taper or response of what happens as you turn will not be as is intended. I would even just try putting the 50k in as is and see how it goes.

[eatyourguitar]

ok lets take a look at what really happening. OP275 pinout is on the datasheet http://www.analog.com/static/imported-files/data_sheets/OP275.pdf

then we see that the volume pot 25k is the negative feedback loop of the opamp. this controls opamp gain. the higher the resistance, the more gain you get. pin 6 has audio coming in through a 22k so your feedback loop should be 22k for unity gain. with a 5.1k resistor and 25k pot you get 30k. 30k/22k gives us a gain of 1.5

so what if we use 50K? ----> (50k+5.1k)/22k = gain of 2.5

nothing bad will happen. its just louder. if you don't like it, you can just turn the volume down.

P.S. your lin pot is still lin if you put a 50k resistor on a 50k pot when it is working as a VR. resistor or not, your choice!

[multi_s]

P.S. your lin pot is still lin if you put a 50k resistor on a 50k pot when it is working as a VR. resistor or not, your choice!

I don't think it actaully would be linear. maybe close enough for rock and roll and ya may not matter to the op either way. but if you put say 50k in parallel with a 50k variable resistor, the total resistance is 0 ohms at one extreme and 25k at the other, which is nice and what we were ideally going for. What's not so nice is the total resistance as a function of the rotational position of the pot, call it Rt

Rt (theta) = 1/( 1/50k + 1/f(theta))

where f(theta) is some function of where the resistor is dialed too. If you started with a linear pot (ie f(theta) is a straight line) then Rt( theta) is not a straight line.



anyways i assumed the pot travels 270 degrees. that's probably not correct but its just an illustratin.

Since eats did all the work descrambling the vero, if he has it right you can just change the 22k he mentioned to 42k or seomthign close and that will make the 50k lin have roughtly the same gain range a 25k lin would have without cramping the tapers style. using waht he said, 55/42 ~ 30/22. The ideal value is closer to 40.3k. 33K will give you a bit more love, 47k a bit less.

[eatyourguitar]

such a great idea changing that 22k. I love it.

[eatyourguitar]

more wood for the fire. thanks to digi2t on diystompboxes.com

[Nocentelli]

I understood it (and those graphs seem to bear this out) that a parallel resistor across lug 2+3, whether using the pot as a variable resistor or a voltage divider, makes a linear pot more like anti log: The smaller the resistor compared to the pot size, the greater the anti-log-ness. Since it's a volume control, it's pretty much set and forget, so taper is pretty irrelevant, and as stated, leaving it as a 50 will just make your max volume louder.