How do I turn a 500k linear pot into a 250k linear. Can i use a resistor.
Just add. A 500k in parallel u get the 250k! 1/Rt=1/R1+1/R2
Add the resistor between lugs 1 and 3!
Cheers
i connect the outer lugs with a resistor?
Yep! Sorry, ive just updated my answer
yes but will it still be a linear pot?
Yes!
I would sugest you to get the right pot , but if you cant, try it with the resistor and see how it goes! You can always change afterwards
yea it's just to test it out then i'll order the right pots
I hate to disagree with Gledison, but it's a bit more complicated than that. To start with, it depends on how the pot functions in the circuit, is it a voltage devider, variable resistance etc... This is some good reading: http://www.geofex.com/article_folders/potsecrets/potscret.htm (http://www.geofex.com/article_folders/potsecrets/potscret.htm)
Getting the right pot is always preferable, but to start: a linear pot with a resistor bootstrapped over it will never remain linear.
Paul
Quote from: DutchMF on December 03, 2013, 11:52:42 AM
a linear pot with a resistor bootstrapped over it will never remain linear.
Yeah I had a gut feeling about that when I asked... I really should take time and read "Life of pots"... ::)
It's really difficult to get your head around stuff like this, if I say so myself. I made an Excel sheet once, with some graphs linked to all the data, to visualize what happens when you start messing with pots and resistors, but I lost that one......... If I get really bored, I will give it another go, because it worked fantastic!
Paul
Quote from: DutchMF on December 03, 2013, 11:52:42 AM
Getting the right pot is always preferable, but to start: a linear pot with a resistor bootstrapped over it will never remain linear.
I agree Paul however, I would be pretty confident saying that it will get you where you want to be ;D
It would be close enough to a linear taper that it would work. If you want perfection.... then get the 250KB pot. If you want it to work NOW, use the 500KB pot with a 500K (or 510K) resistor strapped between Lugs 1 and 3.
A small sacrifice for having it completed now. 8)
Quote from: Govmnt_Lacky on December 03, 2013, 12:40:33 PM
Quote from: DutchMF on December 03, 2013, 11:52:42 AM
Getting the right pot is always preferable, but to start: a linear pot with a resistor bootstrapped over it will never remain linear.
I agree Paul however, I would be pretty confident saying that it will get you where you want to be ;D
It would be close enough to a linear taper that it would work. If you want perfection.... then get the 250KB pot. If you want it to work NOW, use the 500KB pot with a 500K (or 510K) resistor strapped between Lugs 1 and 3.
A small sacrifice for having it completed now. 8)
Yeah, you can certainly do this to see everything works, but when the pot is used as a variable resistor, like the gain control in the Egodriver, this won't work......... Sorry for being an ass, I really only try to help. :D
Paul
Quote from: DutchMF on December 03, 2013, 12:46:39 PM
Quote from: Govmnt_Lacky on December 03, 2013, 12:40:33 PM
Quote from: DutchMF on December 03, 2013, 11:52:42 AM
Getting the right pot is always preferable, but to start: a linear pot with a resistor bootstrapped over it will never remain linear.
I agree Paul however, I would be pretty confident saying that it will get you where you want to be ;D
It would be close enough to a linear taper that it would work. If you want perfection.... then get the 250KB pot. If you want it to work NOW, use the 500KB pot with a 500K (or 510K) resistor strapped between Lugs 1 and 3.
A small sacrifice for having it completed now. 8)
Yeah, you can certainly do this to see everything works, but when the pot is used as a variable resistor, like the gain control in the Egodriver, this won't work......... Sorry for being an ass, I really only try to help. :D
Paul
Why do you say that? Connect lugs 2 and 3 and it's two resistors of 500K = 250K.
If you have a variable resistor pot, putting a resistor across lugs 1 and 3 wont work, was what I meant........ Or am I completely mistaken? At work at the moment, so no time to double check. If I'm wrong about this, I'm sorry and will delete the mis-information.......
Paul
To see what it does, you have to know how it's going to be used in a circuit. Putting a resistor across 1 and 3 really depends on how you are going to use it. Connecting 1 and 3 with a resistor and using it as a variable resistor between 1 and 2 does one thing. Resistor across 1 and 3 with 3 and 2 connected as a variable resistor between 1 and 2 does another. Resistor across 1 and 3 with the arrangement used as a voltage divider does another (mostly useless) thing.
If you want to use the resistance between lugs 1 and 2, connecting a resistor between lugs 1 and 3 wouldn't do anything, right?
Paul
Quote from: DutchMF on December 03, 2013, 02:24:45 PM
If you want to use the resistance between lugs 1 and 2, connecting a resistor between lugs 1 and 3 wouldn't do anything, right?
Paul
What happens depends on if you connect pins 3 and 2 as well (in the way that is often done when using a pot as a variable resistor.
The pot has resistance V, the resistor is R, R' will be the total resistance between 1 and 2, x is the fraction of rotation ( x in [0, 1]).
If you don't connect pins 3 and 2:
The easiest way to visualize this is that the portion of the pot that is not between 1 and 2 wraps around and adds to the R to form the total resistance that is in parallel with the resistance between pins 1 and 2.
R' = [1/xV + 1/((1 - x)V + R)]^-1 = xV[(1-x)V + R]/(V + R). So, if V = R
R' = (x - 0.5 * x^2) R
If you do connect pins 3 and 2: R' = [1/xV + 1/R]^-1. So, if R = V, R' = R x/(1 + x).
Note that it is way too early for me to be doing math, so I probably made a mistake somewhere. But, testing on a breadboard with the pot set to half seems to verify these calcs.
Quote from: DutchMF on December 03, 2013, 11:52:42 AM
I hate to disagree with Gledison, but it's a bit more complicated than that. To start with, it depends on how the pot functions in the circuit, is it a voltage devider, variable resistance etc... This is some good reading: http://www.geofex.com/article_folders/potsecrets/potscret.htm (http://www.geofex.com/article_folders/potsecrets/potscret.htm)
Getting the right pot is always preferable, but to start: a linear pot with a resistor bootstrapped over it will never remain linear.
Paul
Hey Paul, no worries. im far away of being an expert and you are much more qualified to help here :) .
I just remembered to read about this problem, and ive read that by adding a resistor would solve the problem. I need to read your article as well...
thanks!!
Besides the Geofex pot discussion there's this one; http://sound.westhost.com/pots.htm.
Towards the end is the info on creative changing of the pot's "law".
dave