How to wire digital pot without direct short.

Forum 2005-2010 (read only) Hardware Interfacing
1

I have an external device that needs an analog voltage input. I tried using analogWrite PWM mode on Arduino, but suspected that the external device was picking up the off/on. I researched, and came up with two options -- an RC Filter or a digital pot.

I chose digital pot (AD5206 -- 10K) , partly to keep the external device on a completely separate circuit. I used this tutorial http://www.arduino.cc/en/Tutorial/SPIDigitalPot

I connected the external device negative to A1, external 5V to B1, and the analog input of external device to W1. I wired it up, and it works.

However, I am sure I am doing something wrong. If I put all of the digPot resistance on W1, then I will have a direct short from A1 to B1.

Do I add a resistor between B1 and 5V? How big? If I do this, wont I get skewed results. (If DigPot is at midpoint, wont W1 see 5K of resistance and B1 see 5K plus new resistor -- say 6K total)? Or am I overthinking it.

My objective is to use the DigPot to divide the 5V and put an analog signalling voltage on W1, without creating a direct short. How do I do this?

2

If your device's analog input is high impedance, then you don't have a problem. More accurately, if no current is flowing through the pot to the device, the addition or omission of an inline resistor will not really make a difference.

You may want to make sure your potentiometer ground connected on the Arduino and the external device. Most digital potentiometers can't handle more than VCC or less than GND on any terminal.

3

I believe the analog input is high impedance, and so no problem on W1 -- that is what goes to the device analog input.

My problem is with the A1 - B1 circuit. If all resistance is on W1, then there is no resistance between A1 and B1. (Maybe I am thinking about pots incorrectly -- I think the 10K is "split" between W1 and B1).

Do I just make B1 high impedance by putting a resistor between B1 and 5V?

4

Or am I overthinking it.

Yes the resistance between A1 and B1 is fixed there is nothing you can do about it, it's 10K.
The software command determines where along the 10K you tap off. The tap is the W connection. There should not be a problem unless you apply a voltage source to the W connector, do not do this. Normally W will be connected to an input.

So if you set W to half way there will be 5K between W and A and 5K between W and B. BUT still 10K between A and B.