Re: Modelling pot performance

On Sat, 03 Oct 2009 05:26:47 GMT, "Tony Done" <tonydone@xxxxxxxxxxx>
wrote:


"RS" <RS@xxxxxxxxxxx> wrote in message
news:b1jdc554ndkcsreomiifiug69skef3pg2u@xxxxxxxxxx
On Fri, 02 Oct 2009 22:35:01 GMT, "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.

Tony, I'm guessing that you mean "in the -first- bit of knob
rotation", right? Log pots come up very slowly when turned clockwise.
IOW, at full counterclockwise (CCW), the wiper-to-'bottom'-contact
would present 0 ohms. The first half of the log pot's rotation will
present about 1/10th to 1/3rd of its full resistance between those
terminals. It's near the -top- of its rotation that the rise in
resistance accelerates.

I know that you can get them to behave in a
non-linear manner by putting a fixed resistor in parallel.

Nonlinear, but it's going to go in the opposite direction from what
you want. The paralleled resistor is not going to affect the lower
end of the resistance curve much, so the pot will still come up fast
within the first few degrees of rotation. Instead, the paralleled
resistor is going to limit the max resistance, so the lower range will
appear to come up even faster, as compared to the full resistance.

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.

The formula in English: Add the reciprocals of the two resistances.
Then take the reciprocal of the result.

The math: Rfinal = 1 / (1/Ra + 1/Rb)

That can be converted to:

Rfinal = (Ra * Rb) / (Ra + Rb)

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)

With a linear pot, you'd have a directly proportional (linear) curve
for one of the two R's above. Let's say the pot is Ra. Then at 1/2
rotation, the second equation becomes:

Rfinal = (Ra/2 * Rb) / (Ra/2 + Rb)

It's that simple. A graphing calculator should be able to do the curve
easily.

But, alas, though a paralleled resistor will help to linearize a log
pot with too severe a curve, it's not going to generate a forward log
response from a linear pot. You'll see that when you model it.

<sigh> You're right, I was measuring the resistance from the wrong lug! -
Between normal hot and wiper instead of between normal earth and wiper (OK,
my excuse is that I turned the pot knob-side-up to do the measurements and
neglected to switch lugs), and the curve has the opposite curvature to the
Ohm's law combinations.

It happens. You won't believe the number of initial reverse-wirings
done by good techs and engineers. I usually just look at the pot and
visualize where the wiper is going to be when it's turned down. Then I
hook it up, it's wrong, and I reverse it. <g>

However it still remains true that the 1M linear pot
had virtually no effect until it was in the 1 to 2 range on the knob. In
summary, the pickup is about 6.6K resistance, the pot is about 250KA, the
cap is .022mF, and the treble cut becomes noticeable to me when the pot is
in about the 0 to 100K range.

Tony D

That sounds logical as well. You were referring to treble -cut- rather
than the overall response, and by "last" you evidently meant on the
counter-clockwise end of the rotation.

Anyway, you'll want to get a log pot. You do enough electronics that
you should consider keeping a few extras in stock, Tony.
.

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