Potentiometer Question

[RobA]

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.

[DutchMF]

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

[RobA]

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.

[Gledison]

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

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!!

[davent]

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