Why are the voltage ratings of switch contacts so much lower than breakdown voltage?

Question

I need a switch for a high voltage application, say 5 kV, and I'm looking at relays. From a breakdown voltage standpoint, a contact with a 5 mm gap should be able to withstand 15 kV, which is 3x what I need, so I figure it should be fine. Yet the rated voltage of the contacts is far lower, in the 100's of volts.

Are there rules that are not directly technical in nature that mandate a manufacturer must rate a switch lower? For instance, does law or UL have guidelines that spell out how much a switch can be rated for?

I figure (perhaps naively) a factor of 5 ought to be enough to shut current off, but is that true? Does it need to be more like a factor of 20? What am I missing here?

Answer 1

When contacts that are carrying a current are opened there will always be an arc due to the inductance of the circuit. The switch cannot be considered in isolation.

If the contacts are opened then the voltage is applied, the breakdown voltage is not necessarily of the air but of the insulating material used to mount the contacts. Depending on the mounting, the most likely path from one contact to another is along the surface of the insulating material.
Again the contacts cannot be considered in isolation.

Both the circuit and the mounting hardware must be included in the analysis of voltage breakdown.

Even though the breakdown rating is indicated for the contacts, it is really for the insulating material that has the lowest breakdown voltage: the air between the contacts or the material holding the contacts.

Update: An HV electric field can compromise the lattice structure of the insulating material throughout the volume. AC HV fields also cause heating which can vaporize chemicals near the surface leaving behind a carbon rich track. This is not caused by current, but the electric field. Jagged tracks are left on the surface of materials like plastic. If these tracks reach from contact to contact then a high current may flow. This is a voltage breakdown effect that will lower the rating

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This breakdown voltage can be considerably lower than the breakdown in the air.

Answer 2

The problem isn't the breakdown voltage. It's correct that a 5mm gap can withstand at least a couple kilovolts, but only as long as no arc has ignited yet. The actual voltage needed to sustain an arc over a distance of a few millimeters is much lower than the voltage needed to ignite it. This paper shows some typical arc voltages at high currents - note that they're on the order of 50V, not kilovolts, even with fairly large arc distances. Ionized air is a pretty good conductor.

When the relay opens, the distance between the contacts is initially very small and an arc can ignite. As the contacts depart from each other, the already-burning arc is just drawn out to become longer. Extinguishing this arc can be quite difficult, especially when higher voltages are available to sustain the arc. Most DC relays have magnetic arc extinguishing circuits for that reason.

AC relays often rely on the zero-crossing of the AC waveform to extinguish the arc. For that to work, the air between the contacts has to cool down enough around the zero-crossing point so that the arc can't continue burning during the next half-cycle of the waveform.

Answer 3

Relay contacts are not just open or closed. When opening, the distance is first small and it grows to final value.

So at the voltage they are already rated for, they will still spark and arc while trying to break the current, and the distance and voltage must be such that the arc extinquishes.

Even if the relay had an off distance enough for 5kV arc to not form by itself, it would likely spark and sustain the arc.