I am controlling a high voltage power supply to generate corona discharge. To switch it on and off I am using an AC relay controlled by a microcontroller as shown in the figure.

enter image description here The controller switches the power supply on for one minute and than shuts it down for 10 seconds.

Now the problem which I am facing is that sometimes during corona discharge there are sparks and when ever it sparks the microcontroller restarts(I know this because when it boots the LEDs blink on the board) and the HV power supply turns off before that one minute duration.

I tried putting the micro and the relay inside a Faraday shield and grounded everything but it does not seems to work. Also, other thing I noticed is that whenever I removed the relay and drive the HV power supply without the microcontroller the sparks have no effect on the controller.

Any suggestions on how to use the HV power supply with the microcontroller and protect it from the sparks?


It looks like your problem is the wire that runs relay to HVPSU. That wire conducts the interference straight in to the supposedly screened region.

If you must have a galvanic connection through the Faraday screen, then take it in through a feedthrough filter.

It's not clear from your diagram what the relay is doing to the HVPSU. It has only a single wire drawn. Where does the return current flow, and is it amps, to power the PSU, or mA, to trigger it? If you want the Faraday screen to work, then it should flow back to the screen at the filter, in a small loop that doesn't magnetically couple to the MCU.

You might be better off simply coupling the MCU to the relay with an opto-coupler. That might be sufficient. If it's not, then reinstate the Faraday screen, with the hole in it bridged by the opto.

  • \$\begingroup\$ I am using the relay with an optocoupler but I am still facing this issue. What do you mean by "If it's not, then reinstate the Faraday screen, with the hole in it bridged by the opto.". Also, is there a way to make DIY feed through filter? PS. I have a limited background of electrical engineering. \$\endgroup\$
    – Kadyan
    Nov 7 '17 at 10:47
  • \$\begingroup\$ sounds like you have a lot to learn before you can even understand the answers. The fact that you've drawn your diagram the way you have indicates that you don't understand what a Faraday screen does, or the precautions you need to take in order to make it work. How does this sound, there must be no conductors penetrating the Faraday screen, and any that do must be grounded to the screen by capacitors. An opto has no conductors from one side to the isolated other side. A feedthrough filter means short connections to the shunt capacitors. \$\endgroup\$
    – Neil_UK
    Nov 7 '17 at 11:37

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