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schematic

simulate this circuit – Schematic created using CircuitLab

I'm designing a circuit which should leak some amount of current from AC line to PE. The question is regarding clearance-creepage calculation for this specific case.

The part (triac optocoupler) is rated for 600V peak on TRIAC terminals, but the pin spacing is below minimum required by safety standard (2mm instead of 3mm required from line to PE).

The question is if I can count clearance-creepage added by each resistor leads spacing?

For example: If I have 4 resistors of 1206 size it will give me ~4x1.5mm=6mm donated by resistors + 2mm pin spacing of triac optocoupler.

The most intuitive answer will be yes, but in case triac is not conducting we will have line voltage at its terminals and it can probably arc the leads.

Thank you in advance.

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  • \$\begingroup\$ The intuitive answer to me is no, but I'm not familiar with the details of the standards. I would suggest getting a different optocoupler. \$\endgroup\$ – Hearth Nov 6 '18 at 14:53
  • \$\begingroup\$ Functionally, it makes very little difference unless we are talking about Gohm range. If you are trying to fulfill IEC 60950 or similar, then no, you can not use the resistors pin spacing. \$\endgroup\$ – winny Nov 6 '18 at 14:59
  • \$\begingroup\$ Felthty, Winny, thanks for quick response. The problem there is only one manufacturer which has more than 3mm pin spacing. This is wierd that in components like TRIAC that are used for AC mains do not have sufficient clearence-creepage... So I asume that I misunderstood something and it is legal to use them somehow. \$\endgroup\$ – ILYA Shu. Nov 6 '18 at 15:01
  • \$\begingroup\$ You have several ways to cheat here. Ask manufacurer which material group or CTI the plastic is. If CTI is 600, your creepage requirements are relaxed and you can make a cutout in your PCB. If not, you can coat it with conformal coating to get down to pollution degree 1 locally across the optocoupler. \$\endgroup\$ – winny Nov 6 '18 at 15:24
  • \$\begingroup\$ @ILYAShu. You might have more success searching for it if you look for solid-state relays. TRIAC-type solid-state relays are basically the same thing as TRIAC-type optocouplers, just designed with more power in mind. I can easily find a few on digikey. \$\endgroup\$ – Hearth Nov 6 '18 at 15:29
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As already stated by Felthry and winny, the spacing within the resistor terminals on the high voltage side of the circuit cannot be counted as a supplementary clearance/creepage distance across the isolation barrier offered by the optotriac \$\mathrm{TRI}1\$. The reason was already guessed by you: resistors do not offer galvanic insulation between their terminals, but only a functional one, meaning that they are properly spaced in order to work with an applied voltage sufficient to make them reach their maximum power dissipation without being destroyed. As long as no current flows through them, the highest voltage on the controlled load will appear at the optotriac anodes, leaving it as the only barrier for safety/galvanic isolation. Therefore you have two choices for solving your creepage/clearance problem:

  1. If you need a clearance distance (i.e. across the air, "at sight") larger than 2mm, you should change the optotriac model you are using. No matter how you design your control and load circuit, there will be always a condition where the maximum voltage to be "blocked" will be present across the optotriac, and it will cause an arcing discharge through air to reach your input control circuit, usually with unwanted effects, to say the least.

  2. If instead the clearance required can be relaxed, and you instead need absolutely to have a creepage distance (i.e. along a path laying on the board/component surfaces) larger than 2mm, then you can also decide to make a cut of sufficient thickness on the board just below the body of the optotriac. The requirement will be fulfilled (it depends also on how crowded is your board), with a modest increase of board costs.

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  • 1
    \$\begingroup\$ Daniele, thank you for quick response. Regarding 1: if my understanding of how arcs work right - arc is equivalent to a short-circuit. If this is true then it will case current flow in the resistors which will emidiately decrease voltage to 0V on TRIAC terminals, which will cause arc to stop. So arc can't even start, because to start it must have "insuficient" amount of energy (and in our case it's limited by resistors). Am I missing something? \$\endgroup\$ – ILYA Shu. Nov 6 '18 at 16:50
  • \$\begingroup\$ @ILYAShu. You're right: the term "arcing" is not exact since the current cannot rise up to that levels (several amperes) needed to start a proper arc discharge. It is better to say that it will cause a black/glow discharge (you'll see the characteristic light spike or nothing at all) and few (perhaps tens of) milliamperes will pass from the load circuit part to the control circuit part, in spite of the attempt to create a safety barrier. \$\endgroup\$ – Daniele Tampieri Nov 6 '18 at 20:24
  • \$\begingroup\$ didn't now about black/glow discharges, thanks for pointing out. But the isolation input-output is ok, the problem is distance between terminals of triac on load side. I'm afraid of possible arc from AC Line to PE across triac. \$\endgroup\$ – ILYA Shu. Nov 7 '18 at 7:45
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I figured out that one can use parts which are rated for 600V in 230AC even if the pin spacing is below required with the following restriction:

  • If the part is removed(not assembled) you'll need to keep the required clearance-creepage. This can be achieved in my case by disassembling one or more resistors when oproisolator is not present.

The answer is weird but I recieved it from verified source :) so I tend to believe this is true...

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    \$\begingroup\$ +1. I have been busy so I had not the time to search a more accurate answer. Your verified source is right since, if your circuit is designed in order to leak some current from the AC power to the PE, you implicitly cannot keep the safety crepage/clearance distance from AC to PE. Therefore, when the optotriac is mounted and provides its "leakage" function, you have to respect only the functional isolation distances, lower than the safety ones: when it is not mounted, you have to remove also the resistors in order to reach the safety levels. \$\endgroup\$ – Daniele Tampieri Nov 11 '18 at 10:02
  • \$\begingroup\$ Thanks for confirmation and deeper explonation, Daniele! \$\endgroup\$ – ILYA Shu. Nov 12 '18 at 13:52

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