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Topic: Using capacitor to stabilize 2-pin rotary potentiometer? (Read 1 time) previous topic - next topic

aetenos

I am trying to use a 2-pin rotary potentiometer but there is so much noise in the signal, it is impossible to read. I have found that decoupling can help with this kind of problem, but haven't been able to figure out how to wire it effectively. I read that putting a capacitor between the input and ground helps, but this potentiometer seems to only have an input and a read pin. I have also tried placing resistors before the input and between the read pin and AnalogIn, to no avail...

So basically, how do I smooth the signal of a 2-pin pot?

Thanks!

Grumpy_Mike

First off if it only has 2 pins it is not a pot but a variable resistor. You will not get the full range out of this. Put a 1K from +5 to the analogue input, then put your variable resistor between input and ground. Finally put the capacitor from input to ground.
However I suspect you will not need a capacitor once you wire it up correctly because without that top resistor you will not get stable results.

aetenos

#2
Aug 13, 2011, 07:28 am Last Edit: Aug 13, 2011, 07:23 pm by aetenos Reason: 1
I'm still confused...I imagine 'analog input' is the destination of the current on the arduino board and 'input' is the +5? what does the current going out of the variable resistor connect to before going to analog input? I'm fresh to electronics, so it may help if I could understand it in terms of: the current goes from a to b to c....

Edit:

Ok, I put a resistor between +5 and the var res' input. Then I connected the wiper to analogIn. It only works when I also connect an LED from the wiper to ground (and, as you said, the capacitor on +5 and ground does nothing). The significance of the LED I do not understand at all. And, like you said, I do not get the full range, only about 3450-7109 (at the very end of the var res' range it just jumps rapidly haha).
> Can anything be done to get the full range? If not, I'll try to map the logarithmic output to a linear scale and see if that works....

Grumpy_Mike

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Can anything be done to get the full range?

No not with a variable resistor, you really need a pot.

Quote
I imagine 'analog input' is the destination of the current on the arduino board and 'input' is the +5?

No, input is the analogue input. You need a resistor between that and +5V.
THEN your two wire thing goes between the analogue input and the GROUND.
Finally the capacitor is wired between the two connections of your two wire thing.

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The significance of the LED I do not understand at all

The LED is acting as a pull down, if you wire it as I say you will not need it.

justjed

Does this help? What this is, is making a variable voltage divider. Any reason why you can't just use a potentiometer?

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