From: FyberOptic on
Hey folks. I've been trying to find out how to modify the value of a
potentiometer, but so far I haven't had much luck. Things I found on
the web only seem to refer to if I'm using it as a rheostat, but I
need it as a voltage divider.

Basically, I have a microcontroller with an analog-to-digital
converter, and I'm using a 2 megohm pot. The input impedance of the
mcu seems to be 10k however, and since I'm having a really jittery
input, I thought reducing the pot's overall value might help some. I
don't fully understand impedance matching though to be honest, so it's
still a guess. But I've tried putting equal value resistors from vcc
to wiper, and another from wiper to ground, thinking this might do the
trick. I ended up with a large dead zone in the center, and the rest
ends up being too sensitive, so that's obviously not the correct
method.

I know that just replacing the pot is probably just the easiest
solution, but that would take several days to order one and more money
wasted to ship it. Surely there's a way to do this with just some
resistors or something, I hope! Thanks in advance.

From: Joel Koltner on
"FyberOptic" <fyberoptic(a)gmail.com> wrote in message
news:6c91e649-12cb-4c34-a2aa-52341b2b446b(a)a16g2000vbr.googlegroups.com...
> Basically, I have a microcontroller with an analog-to-digital
> converter, and I'm using a 2 megohm pot. The input impedance of the
> mcu seems to be 10k however, and since I'm having a really jittery
> input, I thought reducing the pot's overall value might help some.

Have you tried placing a 100nF (or bigger) capacitor from the
microcontroller's ADC input pin to ground?

With a 10k input impedance, your pot isn't going to be all that horribly
linear, but you should still be able to use the pot to indicate between Vcc
and ground. (Well, actually... 2M is enough larger than 10k that even with
the pot all the way to one end, instead of getting close to zero ohms from the
wiper to the end, you might get something big enough to matter, so maybe you
won't quite get Vcc at one end, but it'll be close.)

> I
> don't fully understand impedance matching though to be honest, so it's
> still a guess.

That's OK, for what your doing there's really nothing to be impedance matched.
:-)

If you have an op-amp kicking around, you could buffer the signal (Google can
find schematics, e.g., "op-amp unity gain buffer") -- the op-amp's input will
be much larger than 2M and the output will be much smaller than 10k, so you'll
get a linear voltage vs. pot position output back if that matters to you.

> I know that just replacing the pot is probably just the easiest
> solution, but that would take several days to order one and more money
> wasted to ship it.

When you do end up ordering parts anyway, you might get yourself an assortment
of pots from, say, 100ohms up to 10megs in a 1-3-10 sequence (e.g., 1k, 3k,
10k, 30k, 100k, etc.) or 1-2-5-10 if you want to get a few more. The surplus
places like All Electronics will often let you do this very inexpensively.

---Joel

From: Grant on
On Fri, 28 May 2010 17:16:49 -0700 (PDT), FyberOptic <fyberoptic(a)gmail.com> wrote:

>Hey folks. I've been trying to find out how to modify the value of a
>potentiometer, but so far I haven't had much luck. Things I found on
>the web only seem to refer to if I'm using it as a rheostat, but I
>need it as a voltage divider.
>
>Basically, I have a microcontroller with an analog-to-digital
>converter, and I'm using a 2 megohm pot. The input impedance of the
>mcu seems to be 10k however, and since I'm having a really jittery
>input, I thought reducing the pot's overall value might help some.

Just put a capacitor there. The ADC input is sampled data and wants
low impedance for a short time in order to charge the sample/hold cap
inside the chip. A much larger cap outside the chip will allow the
sample/hold to acquire the signal with less jitter.

Grant.

> I
>don't fully understand impedance matching though to be honest, so it's
>still a guess. But I've tried putting equal value resistors from vcc
>to wiper, and another from wiper to ground, thinking this might do the
>trick. I ended up with a large dead zone in the center, and the rest
>ends up being too sensitive, so that's obviously not the correct
>method.
>
>I know that just replacing the pot is probably just the easiest
>solution, but that would take several days to order one and more money
>wasted to ship it. Surely there's a way to do this with just some
>resistors or something, I hope! Thanks in advance.
--
http://bugs.id.au/
From: Jan Panteltje on
On a sunny day (Fri, 28 May 2010 17:16:49 -0700 (PDT)) it happened FyberOptic
<fyberoptic(a)gmail.com> wrote in
<6c91e649-12cb-4c34-a2aa-52341b2b446b(a)a16g2000vbr.googlegroups.com>:

>Hey folks. I've been trying to find out how to modify the value of a
>potentiometer, but so far I haven't had much luck. Things I found on
>the web only seem to refer to if I'm using it as a rheostat, but I
>need it as a voltage divider.
>
>Basically, I have a microcontroller with an analog-to-digital
>converter, and I'm using a 2 megohm pot. The input impedance of the
>mcu seems to be 10k however, and since I'm having a really jittery
>input, I thought reducing the pot's overall value might help some. I
>don't fully understand impedance matching though to be honest, so it's
>still a guess. But I've tried putting equal value resistors from vcc
>to wiper, and another from wiper to ground, thinking this might do the
>trick. I ended up with a large dead zone in the center, and the rest
>ends up being too sensitive, so that's obviously not the correct
>method.
>
>I know that just replacing the pot is probably just the easiest
>solution, but that would take several days to order one and more money
>wasted to ship it. Surely there's a way to do this with just some
>resistors or something, I hope! Thanks in advance.

No there is not, but if speed is no issue you could try acapacitor from
ADC input to ground to quiet it down.
A capacitor PLUS a lower value pot is the best solution perhaps.
From: Bob Eld on
"FyberOptic" <fyberoptic(a)gmail.com> wrote in message
news:6c91e649-12cb-4c34-a2aa-52341b2b446b(a)a16g2000vbr.googlegroups.com...
> Hey folks. I've been trying to find out how to modify the value of a
> potentiometer, but so far I haven't had much luck. Things I found on
> the web only seem to refer to if I'm using it as a rheostat, but I
> need it as a voltage divider.
>
> Basically, I have a microcontroller with an analog-to-digital
> converter, and I'm using a 2 megohm pot. The input impedance of the
> mcu seems to be 10k however, and since I'm having a really jittery
> input, I thought reducing the pot's overall value might help some. I
> don't fully understand impedance matching though to be honest, so it's
> still a guess. But I've tried putting equal value resistors from vcc
> to wiper, and another from wiper to ground, thinking this might do the
> trick. I ended up with a large dead zone in the center, and the rest
> ends up being too sensitive, so that's obviously not the correct
> method.
>
> I know that just replacing the pot is probably just the easiest
> solution, but that would take several days to order one and more money
> wasted to ship it. Surely there's a way to do this with just some
> resistors or something, I hope! Thanks in advance.
>

Two Meg pot is way too high of a resistance for the 10 k input impedance. It
works but is very nonlinear and subject to noise. A cap can help quiet it
down.

The obvious solution is to get a proper pot, 1k, 2k, or 5k, so the driving
impedance is nice and low, more like constant voltage.

But, you can add an NPN emitter follower, base to the pot wiper, emitter to
the input with a 10k resistor to ground, collector to +V. The follower gives
a low dive point impedance. This won't give you full range but might be
sufficient for your application. The range and offset can be adjusted with
appropriate values in software. This is a stop-gap hobby solution, not for
production.