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Topic: Replace Potentiometer (Read 801 times) previous topic - next topic

johnricco

Hi,

I am sure similar topics have been created on this subject and i have read a lot of them. They seem to back up my initial thinking but thought id also make a post to just be sure.

Basically i am attempting it replace a potentiometer within a pre-existing circuit. It is my understanding that I should be able to remove this component and use a PWM pin on the arduino board. Then send this through a low pass filter.

Am I right in thinking the outer pins on the pot are the Vin and the Ground with the center pin being the Vout?

These are the values taken from the potentiometer on its original 6v circuit.
full left position values, Sensor 739 with output 184
center position values, Sensor 495 with output 122
full right position values, Sensor 276 with output 68

Thanks

Ian_Lang

Yes you are right in thinking that the centre is the output; it's commonly known as the wiper. But there is a caveat- potentiometers come with a logarithmic OR a linear track. If it's a volume control, it's probably the latter.

johnricco

yeah so as I turn the dial I change the values of the voltage divider.

Do you think using the arduino with the PWM and a low pass filter is a suitable substitute?

Ian_Lang

It's worth a try, is what I say. The caveat is going to be if your original circuit was six volts,  and the PWM only goes up to five. I'd say if you attach the PWM pin to where the wiper used to be, there's a good chance it'll work.

cr0sh


Yes you are right in thinking that the centre is the output; it's commonly known as the wiper. But there is a caveat- potentiometers come with a logarithmic OR a linear track. If it's a volume control, it's probably the latter.


You mean "the former"...

Log potentiometers are also known as "audio taper" pots (vs "linear taper" pots); thus, any time you are dealing with audio signals (ie, a volume control, for example), it is likely to be a logarithmic/audio taper potentiometer.
I will not respond to Arduino help PM's from random forum users; if you have such a question, start a new topic thread.

Ian_Lang

I do mean the former- thanks for that correction :)

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