Changing Potentiometer Value

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.

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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.)

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.

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.


Reply to
Joel Koltner

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.


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On a sunny day (Fri, 28 May 2010 17:16:49 -0700 (PDT)) it happened FyberOptic wrote in :

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.

Reply to
Jan Panteltje

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.

Reply to
Bob Eld

No one has suggested the crudest solution that may get the OP something roughly like what he wants with the limited resources available.

Put a fixed 5k or 10k from the ADC input to ground and put the pot in series with the input line to the signal. You then have a potential divider but with variable input impedance and a linear divider ratio determined only by the pot. Likely to be rather jerky but better than nothing. A quick an dirty trick to get the 2M pot to behave as a 500k non-linear variable resistor is to connect the two ends together and use the wiper as the output. R = Ro.x(1-x) where 0> I

Well there is after a fashion. The mismatch comes from having a 10k paralleled with the wiper to ground on a 2M pot.

Buffering the signal with variable gain would be one way of using the original pot without having to worry too much about its rather high and so noisy resistance.

Regards, Martin Brown

Reply to
Martin Brown

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