Block description of my board

Hi all, so I am creating a board to acquire data from industrial sensors. As you can see in the image, the board has a logic part (where the uC, memory... are connected and driven by the 3.3V voltage), and then the MAX22190 IC, which takes the input from these sensors and converts to a 3.3V level and sends the data through SPI to the uC, and lastly the MAX14483, which is basically an IC that creates an isolation between the logic part and the MAX22190. As you can see in the picture, the MAX14483 also creates an isolation between grounds.

The problem is the next, the MAX22190 is supplied with the 24V, but those 24V are referenced to SIDE A's ground, therefore, to ensure that there's not any potential difference between SIDE A's and SIDE B's ground I need to connect them, but at the same time I want to avoid any potential harm to my logic side (for example, if a huge discharge comes from the industrial sensor to my board, "I'm fine" with the industrial chip getting burned, but I don't want to get into my logic part, or I want to protect it as much as possible).

If you wonder why I'm not connecting the 24V directly to the SIDE B's ground, the 24V will come from another board (a carrier board), so those 24V and 3.3V are attached to that SIDE A's ground, that's a design constrain and cannot be changed.

I thought that maybe using ferrite beads to connect both grounds could at least give some protection in terms of intensity spikes, but any idea/improvement that you could come with would be very much appreciated.

  • 1
    \$\begingroup\$ This question seems very similar to your previous: electronics.stackexchange.com/questions/643487/… <-- what is the difference? Why don't you just implement a 24 volt isolator and use it or not use it (following testing)? \$\endgroup\$
    – Andy aka
    Nov 29, 2022 at 14:20
  • \$\begingroup\$ The safest way, would be another isolated DC-DC-converter from side A's 24V to a new 24V ref'd to side B. But even with your current setup, if you bond the GNDs together, I doubt a large discharge into side B GND would do any harm. It would flow through side A ground to Earth, but as your GND should have low impedance to Earth, this shouldn't create a lot of trouble unless you are speaking of huge energy discharges. \$\endgroup\$
    – tobalt
    Nov 29, 2022 at 14:28
  • \$\begingroup\$ @tobalt yeah, totally agree with you on this. The thing is that I'm the only hardware designer in my company (they hired me for a new project) and I don't know what kind of hazards can be found in the industry. I assume it would have to be a really big discharge, but I'm not sure if that's so unlikely. Regarding to the transformer, that's the first thing I thought about, but as I wrote in another comment, I was only able to find 1 in digikey that would work with 24V and it was too big and expensive to fit in the desing. \$\endgroup\$
    – Kike Rueda
    Nov 29, 2022 at 15:41

1 Answer 1


A couple of things here; 24V isolation is fairly straightforward. Isolated DC/DC converters can be bought off the shelf (i get roughly 700 hits on digikey with some rather broad filtering) , or trough transformer drivers and appropriate 1:1 transformers. I would recommend an off-the-shelf solution - building your own can get complicated.

However: While isolation is a good practice for protection (sometimes even mandatory by law) , there are other, perhaps even additional considerations you can make. Have you considered TVS diodes on your sensor inputs?

  • \$\begingroup\$ And could you help me finding an SMD (if possible) transformer that could do the deal? I thought about it, but only found one transformer that worked with 24V, and it was to big in size and expensive (that's important since the idea is to control up to 16 modules like this one from the carrier). To be completely fair, my knowledge into transformers it's close to none. \$\endgroup\$
    – Kike Rueda
    Nov 29, 2022 at 15:35
  • \$\begingroup\$ Btw, TVS diodes (and fuses) are included in the input. This would be just an extra layer of protection. \$\endgroup\$
    – Kike Rueda
    Nov 29, 2022 at 15:37

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.