Changing Potentiometer Value
in [Design]
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From: Martin Brown on 30 May 2010 06:49 On 29/05/2010 01:25, Joel Koltner wrote: > "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.) 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<= x <= 1 (only half the motion of the wiper giving unique values) > >> 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. :-) Well there is after a fashion. The mismatch comes from having a 10k paralleled with the wiper to ground on a 2M pot. > > 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. 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 |