From my understanding of a Xenon flash circuit, a storage capacitor and a trigger capacitor are charged up to about 240V (both with similar RC values, but less capacitance is needed for the trigger capacitor), and then a switch allows the trigger capacitor to discharge through a transformer, producing a very high, approximately 4KV pulse which ionises the xenon gas allowing the storage capacitor (charged to 240V) to discharge through the bulb. This keeps the gas ionised until the voltage drops too low.

That last sentence is what confuses me. Why do you need a very high voltage ≈4KV for the original ionisation, but then less than 240V for it to stay ionised? From my understanding of ionisation, an electric field applied over a certain distance (a voltage) will cause the electrons to be separated from the positively charged ions when the dielectric strength of the gas is exceeded. I imagine this is something to do with the potential energy of a point charge in an electric field, which I cant find an equation for, and the ionisation energy required to move the electrons from the valence to the conduction band.

So if a certain energy is required to ionise the gas, why does it stay ionised when this energy level is decreased?

Here is a simple diagram and description of a flash circuit:

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    $\begingroup$ Once you have free electrons accelerating with the (lower) voltage you easily get more ionization and more free electrons. The hard part is getting the first one... $\endgroup$ – Jon Custer Aug 28 '15 at 23:15
  • $\begingroup$ @JonCuster did you mean to say "Once you have free electrons accelerating with the (higher) voltage...". And do you know why this is? It seems like there is some positive feedback happening, but what is the mechanism that causes this? $\endgroup$ – Blue7 Aug 29 '15 at 18:09
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    $\begingroup$ Because a free electron will accelerate in the field, collide with a neutral atom, knock an electron off of that, and now two electrons are accelerating to make two more electrons. Avalanche breakdown it is called. $\endgroup$ – Jon Custer Aug 29 '15 at 18:58
  • $\begingroup$ In addition to Jon Custer said: Usually the breakdown voltage is defined as the voltage applied to a gap that induces a breakdown within a certain amount of time; I want to say I remember 30 minutes but the reference I need is at work. Regardless, it is a lengthy time. No one wants to wait around that long to light a lamp (or other glow discharge) so you over voltage; i.e. apply a bias higher than the minimum breakdown voltage. This reduces the statistical time lag until breakdown. The higher the over voltage, the shorter the lag until it's almost instantaneous. $\endgroup$ – hmode Nov 10 '17 at 20:43

It takes less voltage to maintain the lamp discharge because once the high voltage kicks the gas into being a conductor, current begins to flow through the ionized gas and since the gas has electrical resistance, power will be dissipated in the gas (which causes the gas to emit light). That power dissipation also serves to help keep the gas hot & ionized- so that once the current flow begins, less voltage is needed to maintain it than it took to start it.


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