How to use digital potentiometer as variable resistance

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watha2

Joined Jun 1, 2013
2
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
 

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shortbus

Joined Sep 30, 2009
10,049
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.
 

Ron H

Joined Apr 14, 2005
7,063
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.
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.
 
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