How to use digital potentiometer as variable resistance

Discussion in 'The Projects Forum' started by watha2, Jun 27, 2013.

  1. watha2

    Thread Starter New Member

    Jun 1, 2013
    2
    0
    Hello, my name is Watha.
    I have a project to measure the output voltage of Photovoltaic using adjustable resistance in digital potentiometer AD5260.
    I need to use the digpot as variable resistance.
    In an mechanical potentiometer, if I need to use it as variable resistance, I could let the pin A floating, wire pin W (Wiper) to the source (photovoltaic) and pin B to the Ground. Or I could tie the pin A and pin W together.
    However, it didn't work in this digital potentiometer. If I did it that way, the output voltage is max and didn't change.
    But if I wire pin A to the source, pin B to the ground, and use pin W as output voltage, it just means that I use the digpot as voltage divider. The voltage output is linear because as I said, it used as voltage divider. I want it to be used as variable resistance and the output supposed to be logarithmic.
    Is my circuit is wrong?
    Or do I need to modified the program so it could be used as variable resistance?

    Please help.

    I attached the skematic circuit when I used it as voltage divider.

    http://www.google.com/url?sa=t&rct=...=8q25ToBsgWgVJ8-EN5Rf7g&bvm=bv.48340889,d.bmk
     
  2. shortbus

    AAC Fanatic!

    Sep 30, 2009
    4,013
    1,531
    You have the SHDN, pin #6, pulled high. This disables the pot, to get it to work, pin #6 needs to be pulled low. This is shown in the data sheet by the line above the SHDN marking of pin #6. With pin #6 pulled high it stays at the last resistance setting it was set at.
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Shutdown is active low. With the pin pulled low, pin A is open circuit. See Table 5 on page 7.

    Page 16 has info on using the device as a rheostat. I'm not sure why you are having a problem.
     
    Last edited: Jun 28, 2013
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