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I can not find any information about the type of load of neon lamps, for example neon probe or wall contact plug or extension cord with switch with small red lamp. I think this is a passive capacitive load, but in the workplace we measured an extension cord and the meter showed it is a purely active load. How does it work?

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    \$\begingroup\$ The physics of gas discharge requires at least 2 spatial and one time dimension of PDEs, coupled to at least 6-dim ODEs to apprehend well. (That doesn't include radiation transport and atomic interactions.) A simpler version will use global rate equations that assume spatially averaged densities for the charged particles and neutral atoms and molecules, but then it still needs to deal with excited and metastable states. There is a modest, intermediate level text by Lieberman and Lichtenberg called "Principles of Plasma Discharges and Materials Processing." (From John Wiley and Sons.) \$\endgroup\$ – jonk Sep 28 '20 at 9:48
  • \$\begingroup\$ How does what work? You haven't adequately defined what you are talking about or what you want specific answers on. \$\endgroup\$ – Andy aka Sep 28 '20 at 10:34
  • \$\begingroup\$ What makes you think it is capacitive? It works by allowing a current to form in ionised gas. That is a real current consuming active power. In the little lamps you are describing, the current is limited by a resistor (usually 180 kilohms in the ones I've seen). \$\endgroup\$ – user_1818839 Sep 28 '20 at 12:41
  • \$\begingroup\$ @Szymon Here you will find minimal information about how it works. And if you want to model the operation, there will also be a model of the neon lamp. \$\endgroup\$ – csabahu Sep 29 '20 at 12:15
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Electrically look at neon lamp as a bi-directional zener of 55V. The voltage relates to the distance between electrodes.

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A neon indicator consists of a small neon lamp in series with a suitable resistor. The resistor will be around 330k ohms for 230VAC, and lower for 120V.

When fed with an AC waveform, between 0V and 90V, the neon will not strike, and is effectively open circuit. Above that voltage, the neon strikes, and its resistance rapidly drops. At that point, the resistor limits the current. As the AC voltage drops again, the neon will remain lit down to quite a low voltage, at which point it goes out.

Overall, it will look like a resistive load, dominated by the resistor chosen.

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