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I am looking at this formula from wiki for breakdown voltage in gas discharge lamps, and I see its linear by length (d) (oops, I see it's divided by $ln$)

$$\frac{Bpd}{\mathrm{ln}\ Apd-\mathrm{ln}\ (1+\frac{1}{\gamma_\text{se}})}$$

Does that mean that I can make tiny neon lamp with just 0.1 to 0.05mm spacing between electrodes and it will glow as low as 5V for example?

Are there any gases which have lower minimal breakdown voltage than neon?

I guess using niddles instead of parallel plates would also lower breakdown voltage, as field is stronger?

Waffle's Crazy Peanut
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BarsMonster
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1 Answers1

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As $A$, $B$ and $\gamma_{se}$ are parameters that do not depend on the distance of the electrodes try to visualize $V=V(d)$ (all in arbitrary units): enter image description here

The breakdown voltage increases dramatically for smaller electrode distances. In a normal neon lamp the electrons are accelerated by the electric field and excite the neon atoms. This can only happen if the electrons have enough kinetic energy when they collide with the neon atoms. If you reduce the distance between the electrodes and keep the other parameters constant the electrons will not have enough distance to accelerate before the hit the other electrode and you will not get a glow. To still achieve a breakdown you have to increase the voltage quite a bit. For even smaller distances the formula does not work anymore. Such high fields will probably lead to destruction of your electrodes from sparks between them.

Alexander
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