For a project, I use an optocoupler to detect 230V/50Hz AC on a line - if a light switch connected to mains is pressed.

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For my calculation, I use 240V as a maximum AC voltage. It will limit the current to 1.2mA for the LEDs of the optocoupler.

My board has a minimum trace distance of 3mm for all traces, except the distance of the pads between the optocoupler:

enter image description here

The shown distance (yellow) is 1.4mm.

I assume, there is only a voltage of 1.4V across these pins, therefore the distance would be enough. Is this correct?

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    \$\begingroup\$ Why do you care about clearance between these 2 pads? They are both on the HVAC side, and are connected with diodes. \$\endgroup\$ – Lior Bilia Nov 6 at 16:51

Yes, there is only about 1.2V peak between the pins, what you need to worry about is creepage across the isolation barrier.

  • \$\begingroup\$ So, adding an air gap between these pins would put me on the safe side? \$\endgroup\$ – Flovdis Nov 6 at 16:55
  • \$\begingroup\$ Between input and output, yes it would be better, but only you know the requirements of your application. For many purposes 8mm creepage is adequate. \$\endgroup\$ – Spehro Pefhany Nov 6 at 16:56
  • \$\begingroup\$ 8mm between the traces? Do you mean the 3mm distance between my traces is to narrow? The PCB will be sealed in a case, so there is no risk of dust or dirt. \$\endgroup\$ – Flovdis Nov 6 at 16:59
  • \$\begingroup\$ Ah, sorry, now I understand. There is a 4mm air gap routed under the optocoupler. \$\endgroup\$ – Flovdis Nov 6 at 17:13

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