I an designing a digitally controlled 5 band analog equalizer. In order to change the gains of each band i am using a few AD8403. I have it hooked up to a stmf3discovery which controls the CLK,SDI,nCS,nRS pins on the ad8403. I have the circuit configure as shown below just without the second pot in for testing purpose. Also the supply voltage levels are from 0v to 5 volts the test circuit does not have a floating ground at -2.5v enter image description here

The problem / problems that im having is when I hook up a multimeter across A to B on any of the terminals I get a very large resistance in order of mega ohms and these pots are suppose to be 10k max. So I tried applying a 5v to the A terminal then grounded both the b and the wiper terminals and tried measuring the voltage across A to W and got a full 5 volts.

So the main question is do digital potentiometer not have a directly readable resistance? Do you have to measure it via Vout / Iout ?


It looks like you have the SHDN pin tied to ground. The bar over the SHDN label means it is active low. This means that tying it to ground activates that circuit. According to page 20 of the data sheet you linked above, that will open circuit the devices.

"Both parts have a power shutdown SHDN pin that places the VR in a zero-power-consumption state where Terminal Ax is open-circuited and the Wiper Wx is connected to Terminal Bx, resulting in the consumption of only the leakage current in the VR."

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  • \$\begingroup\$ Thank you for pointing that out I will review my design and make some changes and let you know if i get correct results \$\endgroup\$ – JWL May 29 '13 at 17:31
  • \$\begingroup\$ Any updates on this? \$\endgroup\$ – jwygralak67 Jun 4 '13 at 18:02
  • \$\begingroup\$ Yes turns out I had forgotten basic electronics. When I was testing the voltage drop from the A terminal to the be wi per I had the wiper wired to ground and i was shorting it to ground. Once I found these silly mistake all was good. Also with regards to the interface I also wired the SHWN to high to solve that problem. \$\endgroup\$ – JWL Jun 16 '13 at 17:25

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