4
\$\begingroup\$

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?

\$\endgroup\$
3
  • \$\begingroup\$ 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? \$\endgroup\$
    – Kaz
    Commented Apr 23, 2013 at 15:46
  • \$\begingroup\$ @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) \$\endgroup\$
    – Adam Lawrence
    Commented Apr 23, 2013 at 18:29
  • \$\begingroup\$ @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. \$\endgroup\$
    – Solenoid
    Commented Apr 25, 2013 at 11:59

1 Answer 1

Reset to default
3
\$\begingroup\$

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.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.