Re: Modelling pot performance

Tony Done <tonydone@xxxxxxxxxxx> wrote:
"RichL" <rpleavitt@xxxxxxxxx> wrote in message
news:SeidncdKH7kHjFHXnZ2dnUVZ_j6dnZ2d@xxxxxxxxxxxxxxxx
Tony Done <tonydone@xxxxxxxxxxx> wrote:
I recently bought a few high values linear (B taper) pots to use in
tone controls, and they didn't work out. All the noticeable effect
was in the last small bit of knob rotation. I know that you can get
them to behave in a non-linear manner by putting a fixed resistor in
parallel. I also know how to model this effect in MS Excel by
applying Ohm's law. What I don't know is an algorithm that models
the behaviour of a typical log (A taper) pot to compare - my maths
ain't that good. I know I can just go out and take some
measurements off a couple with the multimeter, but an equation
would be nice if anyone can suggest one.... Say expected resistance
as a function of knob value (1 to 10)

Thanks,

Tony D

Just got back from vacation and noticed this thread...

I've used log pots for tone controls in a handful of my guitars.
Like you, I don't like all the "action" occurring in that last
quarter of the pot's motion.

Commercial "log" pots actually have two or three linear segments and
aren't truly "log". A common variation is to have two segments with
the resistance at the center of the rotation being 1/10 of the full
pot value. You can use a simple set of equations in that case. If x
represents the degree of rotation (with x = 0 being full
counterclockwise and x = 1 being full clockwise), then

If x < 0.5:

R = R0*0.2*x

If x > 0.5:

R = R0*(9*x - 4)/5

Here R is the resistance between the ground lug and the wiper and R0
is the pot's resistance from hot to ground. You can verify that if
x =
0.5, both expressions give R = 0.1*R0.



Thanks Rich. I remember reading that inexpensive log pots weren't in
fact log, but I didn't know the details.

Tony D

If you want to use the "correct" form for the same situation (i.e., true
"log" taper spanning two decades with R = R0/10 at the center position,
use this:

R = R0*(81^x - 1)/80.

(here ^ means "raised to the power").

At the clockwise endpoint x = 1, and R = R0 as expected.
At the center point, x = 0.5, and R = R0(9 - 1)/80 = R0/10.
At the counterclockwise endpoint, x = 0, and R = 0.


.

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