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I am building a totally overkill electronics repair bench for fun while in quarantine.

Part of this bench build involves wiring up electrical distribution (i.e., a lighting feature to illuminate the work surface, and a bunch of power receptacles). To provide additional safety, I have a GFCI inline with the circuitry, as well as a SPST wall switch wired up to interrupt the live side of the circuit so that I can power off the bench completely when not in use.

Switch product info here: https://www.legrand.us/~/media/products/resources/peps/pass-and-seymour/lgrp00021v0101en.aspx

At every step of wiring this circuit, I check to ensure there is no continuity between live, neutral, and ground--twice, of course, before plugging it in to live mains. So far, everything is working quite well, with no real issues. However, what puzzles me is the neon indicator lamp included within the switch, which I chose because I wanted the switch to be illuminated due to the circuitry sitting on the underside of the workbench.

The switch works as expected: when tested via multimeter continuity, I found continuity when the switch is closed, and no continuity when the switch is open, and no continuity at all between either switch terminal and the earth bonding screw.

The neon lamp powers on when the switch is opened, illuminating as expected. However, I got to thinking: what is actually powering the neon? I took measurements across live and neutral (see OutletBox1 in the schematic below) and I found

  1. When the switch is closed, the GFCI is closed, and the circuit is powered, there is 120VAC as expected.
  2. When the GFCI is tripped open, the circuit measures 0VAC regardless of the switch state.
  3. When the switch is opened and the GFCI is closed, however, there is 24VAC across live and neutral at the outlet box.

schematic

simulate this circuit – Schematic created using CircuitLab

My question is: If the live side is interrupted by the switch being opened, what is actually powering the neon lamp in the switch? Am I correct in assuming the 24VAC measured at the outlet box when the switch is open is a residual voltage from powering neon lamp?

The only two theories I could come up with:

  1. There's some kind of ballast that kickstarts the neon lamp, meaning the continuity test would read negative if the switch was being tested independent of mains (which is how I tested it, for safety), and by extension, this means the circuit isn't actually totally broken by the switch; creating an observable voltage at the receptacles that's too low unable to meaningfully power any mains devices on the other end, or...

  2. The neon lamp is shunting current from live to ground, and because this is happening before the GFCI in the circuit, the GFCI doesn't detect it. This particular option is terrifying!

I'm about 95% sure it's the first one, since it's the only way I can think of how 24 volts would end up across live and neutral. However, before attaching this to the constructed workbench and leaving it plugged in in my office 24/7, I'd rather be completely certain about what's going on under the hood.

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  • \$\begingroup\$ Post a picture of the switch, or preferably a link to datasheet or product page. It is difficult to follow the wall of text with little technical details. And yes, what happen before the GFCI, the GFCI won't know about it. And the picture is missing earth connections and wiring, surely you have earth? Also, neon lamps are extremely sensitive, even stray capacitance between wires in mains cabling can couple enough voltage for the lamp to light up. \$\endgroup\$
    – Justme
    Commented Sep 21, 2020 at 6:35
  • \$\begingroup\$ I do have earth, and I've verified everything in the circuit has continuity to earth through the bonding screws or ground prongs as appropriate. I didn't include it in the diagram because my question was about 24VAC measured across live and neutral, and also because I'm unfamiliar with Stack Exchange's circuit diagram editor. Also, I've added the datasheet. \$\endgroup\$ Commented Sep 21, 2020 at 6:48
  • \$\begingroup\$ That's not a real product datasheet with technical details, it just a brochure that boasts how ecological the product is. \$\endgroup\$
    – Justme
    Commented Sep 21, 2020 at 6:58
  • \$\begingroup\$ It's the best I could find, as I'm on mobile and searching by model number turns up mostly shopping results with no real datasheet. This is the product page. legrand.us/passandseymour/light-switches/toggle/660islg.aspx \$\endgroup\$ Commented Sep 21, 2020 at 7:12

2 Answers 2

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If the live side is interrupted by the switch being opened, what is actually powering the neon lamp in the switch? Am I correct in assuming the 24VAC measured at the outlet box when the switch is open is a residual voltage from powering neon lamp?

It sounds as though the neon indicator is wired as a switch-open indicator.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Probable wiring.

A neon lamp won't conduct until the test voltage exceeds the neon's breakdown voltage - typically about 60 V.

If the indicator is wired like this the indication will be ambiguous. It won't light when the the power is on or if the GFCI is tripped or if nothing is plugged in and switched on so that the neon can have a neutral return path.

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  • \$\begingroup\$ Test-tripping the GFCI confirms that the lamp does not illuminate in either switch state, which means this is most likely the correct answer. Thanks. \$\endgroup\$ Commented Sep 21, 2020 at 7:13
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It appears that the neon indicator is wired across the switch and would glow only when it's in the 'off' position.

In that condition, with the indicator in series with the GFCI circuitry, the 24 V ~ voltage drop across it would show up at the output.

enter image description here

The right way to wire the indicator would be as shown below. It would glow only when the switch is on and there would be no voltage at the output when the switch is off.

enter image description here

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