I've seen tiny solenoid electromagnet (that is the solenoid used to open contacts inside a GFCI receptacle) with very fine wire used in the winding.. won't it burn if exposed directly to 120v ac power at each end?

How is the winding computed?

  • Cable capabilities are calculated thanks to cross section and current. Do you know what kind of current would go through (in amps) and roughly the cable cross section ?
    – Jules R
    Jan 15, 2019 at 10:41
  • The wire is heated by the current flowing through it. If there is very little current, there will be little heating. With only a little heating, the wire will not melt. Jan 15, 2019 at 10:45
  • By just plugging it directly to 120v source. The load will be the solenoid only as electromagnet
    – Jtl
    Jan 15, 2019 at 10:59
  • Don't coils have high self inductance which limits ac current? But if the same coil would be connected to a dc power source of the same voltage, the current would rise to levels that would cause overheating. Jan 15, 2019 at 11:41

3 Answers 3


Wire thickness doesn't relate to voltage. It relates to current. So a coil of that nature would either need to be externally current-limited (resistor in series), or internally limited due to its own practical impedance, being a coil and all. In that impedance, frequency would be a factor.

Insulation thickness relates to voltage. And coil wires don't need enamel strong enough to block the entire voltage, only the worst-case voltage between adjacent windings. If each winding layer has a separator with higher dielectric strength, that may be all that is needed.


The current to operate the solenoid is tiny, so the current draw is small, and I believe it just operates a trigger that disconnects power, so the loading is very brief. So the insulation and the wire gauge can be much lighter than what's usually see for 120V devices.

  • But if you plug the solenoid continuously.. would it burn or would it only draw very small current making it still safe for continuous operation?
    – Jtl
    Jan 15, 2019 at 13:32
  • 1
    A tiny solenoid would only draw a tiny current. I couldn't say for sure, but I would bet that it's designed around how it actually works, and the wires are not sized for continuous duty. Jan 15, 2019 at 13:42

The 'tiny' wire will have a high resistance per length, yes, and it will likely melt with a constant 120 Vrms applied to it. But the solenoid only engages for half a cycle (17 ms for 60 Hz, 20 ms for 50 Hz) before the contacts open, so the heating is minimal and designed for.

For more info, see US patent US8760824B2 and related GFCI controller datasheet (FAN4149). Especially interesting may be the Function Description in the latter document. Further questions about the design process, however, belong in the EE StackExchange.

GFCI circuit

  • When the gate to the scr is triggered, does the scr remain on and the solenoid remain on? I read an scr would remain on even after the gate trigger is turned off.
    – Jtl
    Jan 16, 2019 at 1:13
  • 1
    It remains on until the current through it ('holding current') is low enough. Since mains voltage alternates between positive and negative current through it, the SCR will only conduct for a maximum of one half cycle. Since the gate trigger is removed, it won't conduct on the next cycle. Make sense? Jan 16, 2019 at 14:07
  • Ok. But are SCR designed to be continuously triggered? When the circuit continuously trigger the gate, the SCR just burns out (from almost continuous) AC.
    – Jtl
    Jan 16, 2019 at 22:23
  • 1
    @Jtl It depends on the SCR rating for the application, but here the SCR isn't continuously triggered. Once the circuit detects the imbalance in current the SCR is triggered, current flows through it (and the solenoid), and the contacts open. Current is only conducted for half a cycle, maximum. In such a short amount of time the SCR -- if it isn't rated for continuous operation -- won't be damaged. Jan 17, 2019 at 2:07

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