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I have a digital potentiometer that I want to calibrate by sending in my commands and then measuring what the actual resistance is. However, the resistance from the digital pot is (obviously) connected to other bits of the circuit and I'd need to isolate it to get an accurate measurement.

My question is, if I use this analog mux or something similar, can I feed my "resistance" through it ?

I do believe that the analog mux is really just a "switch" and there isn't anything fancy about them and that this would work but I'd like to make sure before I go ahead any further with it.

I appreciate your time and look forward to your opinions.

Thanks!

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  • \$\begingroup\$ If you need the accuracy, and can spare four mux channels instead of two, and you measure with a current source, a Kelvin (four-wire) measurement lets you measure resistance in a way that minimizes errors from resistance of your connections. That's lots of ifs, though.. \$\endgroup\$ – tariksbl Jun 20 '15 at 22:57
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The data sheet says that it has a resistance of 0.2 ohms with an analog signal of -5.5V to +5.5V. If the output of your variable potentiometer falls between that voltage range I see no reason why it wouldn't work. You are not really feeding "resistance" through it but current, no matter how small, but as long as your voltage and current are within the specs of the device it should work.

When you are "calibrating" the device you are sending a current through it and measuring the voltage drop to get at the resistance.

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