I am working on a design that requires Two Chips - Energy measuring chip (for analogue measurement) and a MCU (for connecting to a communication peripherals, and connected to a PC.) Both Chips is to be powered by a 3.3V DC supply.

But the ground for both chip has to be different. How can I possibly go about that? Please refer to page 3 of ADE7878 Eval Board Power supplies. I intend to use a SMPS for this design.

Thanks a lot.


In a production design, you would use an isolated DC-DC converter to transfer power from the microprocessor domain to the metering domain. You can purchase these as pre-built modules from any number of vendors. Make sure that the isolation rating of the converter is sufficient for the mains voltages you're dealing with.

For prototyping, two separate AC-input power supplies can be used.

  • 1
    \$\begingroup\$ And don't forget the optoisolators for signals that cross between the different power domains. \$\endgroup\$ – Joe Hass Sep 13 '12 at 15:06
  • \$\begingroup\$ Optoisolators aren't the only option here either, ADI makes a line of magnetic isolators that work well for logic level signals as well. \$\endgroup\$ – Ryan Sep 13 '12 at 15:12
  • \$\begingroup\$ Thanks @dave. Please, let me understand you better on this - "transferring of power from the MCU domain to the metering domain". I am considering using the ADuM5000 from Analog Devices on this. I hope its okay for a mains voltage of 220V - 230V. \$\endgroup\$ – Paul A. Sep 13 '12 at 15:14
  • \$\begingroup\$ Yes, the ADuM5000 looks ideal for this application. \$\endgroup\$ – Dave Tweed Sep 13 '12 at 15:28
  • \$\begingroup\$ @DaveTweed thanks for your suggestions. One more thing. I checked the ADum5000, and I want to know if it really matters which side of the domain (MCU or AFE) is connected to the Primary/secondary side of the DC-DC converter. This is as regards the statement in your answer " use an isolated DC-DC converter to transfer power from the MCU domain to the Metering Domain". Thanks. \$\endgroup\$ – Paul A. Sep 14 '12 at 12:33

One way is to use two optocoupler. Connect your chip to input optocoupler and output to MCU. So, your MCU ground and chip's ground will be differ. But it depends on your chips behavior.


In normal configurations, you maintain the analogue and digital grounds as separate grounds, joining them together at a single point.

Depending on your circuit, you may include coupling capacitors or an inductor at the connection.

However, as observed by @Dave (and I'll quote the datasheet) this device needs to maintain different, isolated grounds:

The ground of the ADE7878 power domain is determined by the ground of the phase voltages, VAP, VBP, VCP, and VN, and must be different from the ground of the micro-controller’s power domain.

So please refer to Dave's answer...

  • 2
    \$\begingroup\$ Bzzt! Thank you for playing. You didn't read the data sheet, did you? We're talking about galvanic isolation of two separate power domains here. You do NOT connect the grounds together! \$\endgroup\$ – Dave Tweed Sep 13 '12 at 14:49
  • \$\begingroup\$ @Andrew Thanks for your comment. I have actually seen such isolation using an inductor with the ADE7755 reference design. But from Dave Tweed's comment, i guess this does not apply here. \$\endgroup\$ – Paul A. Sep 13 '12 at 14:54
  • \$\begingroup\$ @DaveTweed can you throw more light to you comment. Please give a suggestion! Thanks \$\endgroup\$ – Paul A. Sep 13 '12 at 14:55
  • \$\begingroup\$ @DaveTweed - humble pie being eaten... and answer revised \$\endgroup\$ – Andrew Sep 13 '12 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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