How to use a 20K potentiometer with a recommended 10K interface?

P

pavjayt

Member
I have a driver that get a voltage input between 0-5VDC that is reocmmened to be across a 10K trimmer. I am using a digital pot which only strats from 20K and up. How can I use this 20K to interface with the above voltage input.

I am reading **broken link removed** article inorder to get some understanding on it can be done properly, but couldnt understand it properly.

I could use an op-amp, but trying to find if there is any other simple way to do this other than an op-amp.

Any suggestions?

thanks
 
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I would think the 20k pot would work okay, but perhaps with some non-linearity in the output voltage versus pot position. How critical is pot position (setting) versus voltage in this application?

Do you know what the input impedance of the driver is?
 
If you can put up with a non-linear output-voltage versus input-number response then the simplest way would be to use two 10k resistors with the 20k pot. One R goes from the 'wiper' to one end terminal; the other R goes from the 'wiper' to the other end terminal.
 

The input impedance of the driver is 10K. Since I had the only option of 20K digital pot in my design, I ended up inputting 10VDC across the pot to get 5VDC at about 3/4 poistion when connected to the driver which works for now but not ideal. I would like to use 5VDC across the pot too, this way users wont go above 5VDC which is not recommened for the driver.
 

So, that would be similar to Fig 9 in the article above. Does this give 10VDC full scale if I have 10VDC across the pot when connected to the driver?
 
Here's a simulation comparing the outputs from a simple 10k pot and a 20k pot + resistors combination, assuming Rin (the driver input impedance) is 10k.


Edit: Note the supply voltage V+ is 5V here, NOT 10V.
 
Last edited:
the OP said:
digital pot which only strats from 20K
Click to expand...

Please be more specific: digital pot, like which chip
strats, I assume to me "starts". Is this like 64 segments of aprox. 20K each?

Or am I barking up the wrong tree.
 
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