Re: Modelling pot performance


"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

.

Relevant Pages

  • Re: Modelling pot performance
    ... Tony Done wrote: ... was in the last small bit of knob rotation. ... I've used log pots for tone controls in a handful of my guitars. ... the resistance at the center of the rotation being 1/10 of the full ...
    (alt.guitar)
  • Re: Modelling pot performance
    ... was in the last small bit of knob rotation. ... behaviour of a typical log pot to compare - my maths ain't ... Say expected resistance as a function of knob ...
    (alt.guitar)
  • Re: Modelling pot performance
    ... controls, ... last small bit of knob rotation. ... present about 1/10th to 1/3rd of its full resistance between those ...
    (alt.guitar)
  • Re: A rotating disk PARADOX??
    ... Why don't you think reversing the rotation should reverse the ... Because resistance changes with temperature and everybody knows that the ... > If you spin a metal disk, you put a charge on its rim, as a result of the ...
    (sci.physics.relativity)
  • Re: Modelling pot performance
    ... last small bit of knob rotation. ... I'm guessing that you mean "in the -first- bit of knob ... present about 1/10th to 1/3rd of its full resistance between those ... an algorithm that models the behaviour of a typical log pot to ...
    (alt.guitar)
Loading