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I'm trying to understand how this AD8402 digital potentiometer works, I need it to be decoupled from the circuit it's driving, that is Vdd, GND and all logic inputs not connected in any way to A, B and W pins.

According to the figure 45 (page 20) of the datasheet I think I can safely connect a 5V power supply to the IC power pins and connect a completely different system (also 5V) on the A, B and W pins.

On page 11 however it seems that VA, VB and VW must be between VDD and GND...

Also on page 24:

Certain boundary conditions must be satisfied for proper AD8400/AD8402/AD8403 operation. First, all analog signals must remain within the GND to VDD range used to operate the single-supply AD8400/AD8402/AD8403.

My question is: can I connect the VDD and GND of my AD8402 to my 5V supply and regulate a potentiometer input on another device that supplies its own 5V and GND to the A and B pins, but without the GND of the AD8402 and the device being connected?

Solenoid
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  • Are you seriously asking whether you can use a single wire to pass a voltage signal between two devices, with no conductor to provide for return currents and a common voltage reference? – Kaz Apr 23 '13 at 15:46
  • @Solenoid It's recommended not to immediately accept answers to give incentive for multiple viewpoints and to stimulate discussion on the site. 24 hours is a reasonable grace period (upvote away in the interim, of course) – Adam Lawrence Apr 23 '13 at 18:29
  • @Kaz: for my defence I didn't know what the bilateral CMOS switch exactly did, I searched Google image to find its name. Since in my circuit I provide the secondary circuit VCC and GND signals I think I'm in the right to assume that if those little switches acted as small relays then the system is decoupled. – Solenoid Apr 25 '13 at 11:59

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No, you can't do that without using a separate isolation circuit between the logic circuit and the digital pot. I've never seen a digital pot that also had that type of isolation built in.

Dave Tweed
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