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I've become fairly familiar with the clearance and creepage requirements of UL/IEC/etc 60950, 60730 and derived/related standards. I've just realised, though, that they only refer to distances through air (clearance) or along a surface(creepage). In fact, solid insulation is specifically used as a barrier to increase the distances. Does this mean, then, that the PCB itself is considered solid insulation, and completely removed from the clearance rules? Can I safely have 240VAC on one side of the PCB and user accessible low-voltage circuits on the other, provided there aren't any vias in the vicinity?

Extending the question, what role does the thickness, dielectric strength, etc play? If a thing is considered "solid insulation" is the only requirement that it passes the hipot testing, and if so, at what voltages?

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Can I safely have 240VAC on one side of the PCB and user accessible low-voltage circuits on the other, provided there aren't any vias in the vicinity?

It's best to separate high voltage with other secondary voltage components. If you need them on the same board, ensure the trace is at least 5mm (forgot the actual number) away. Also the components nearby when bend should not get close to the primary voltage.

That being said, board thickness is around 1.6mm. So no, you can't mount them on reverse side.

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    \$\begingroup\$ Consider, for example, a transformer. Does it have 5mm between the primary and secondary windings? No, but it does have insulation between them - enough to pass the hipot test for whatever its isolation requirements are. \$\endgroup\$ – UStralian Aug 4 '17 at 23:52
  • \$\begingroup\$ PCB safety design is not component design. The rules are there for it to pass safety regulations. Without passing safety, the product can't sell in many countries that require them. \$\endgroup\$ – Jason Han Aug 5 '17 at 0:10
  • \$\begingroup\$ Yes, I know. I'm not asking if I can break the rules, I'm asking for help understanding the rules. The clearance rules say clearance through air, and around solid insulation. Is the pcb solid insulation? \$\endgroup\$ – UStralian Aug 6 '17 at 1:33
  • \$\begingroup\$ You can read more here: forum.mysensors.org/topic/4175/… From what I understand in this article, you may use your PCB as an insulator to increase your creepage distance. However the insulator must be 2.5mm. \$\endgroup\$ – Jason Han Aug 6 '17 at 2:27
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Extending the risks to AC_coupling through the FR-4. 1mm trace width, 10cm (4") long in the low-level side of the PCB, has

$$ Capacitance = Eo * Er * Area/Distance $$

C = 9 e-12 Farad/meter * (5 = FR-4 Er) * 1mm * 100mm / 1.5mm (1/16th ")

C = 9e-12Farad/meter * 5 * 67mm * [1meter/1000mm] = 2.5 e-12 = 2.5pF

Using I = C * dV/dT, and knowing the slewrate of a CLEAN 60Hz at 120vac is 377 * (1.414 * 120) = 64,000 volts per second, we find the injected current is

$$ Iinjected = 64,000 * 2.5pF = 160,000 picoamps or 160 nanoamps or 0.16uA.

If the 120vac is FREE of any spikes.

Microwave ovens will up that slewrate by 100X, if not more, in which case you and 16 uA of spikes, occurring 120 time per second.

Please reconsider the casual placement of any circuitry (or logic) under or near the 120VAC power.

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  • \$\begingroup\$ Try pressing alt gr + shift + * and tell me what you get. You should get ×. And tell me what you get if you use \\$\frac{1}{2}\\$. You should get \$\frac{1}{2}\$ \$\endgroup\$ – Harry Svensson Aug 4 '17 at 4:56
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    \$\begingroup\$ Sorry, I understand the capacitive coupling concerns, and can easily mitigate them in the application. I'm not suggesting anything like 100mm overlap, just a couple of sensing components. I'm asking specifically about compliance - if I can make the circuit correctly do what I need, will it fail 60950 solely based on only 1mm clearance, or does the fact that it's solid insulation solve that problem? \$\endgroup\$ – UStralian Aug 4 '17 at 16:15

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