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For some relays they may give a maximum switching capacity graph like below

enter image description here

From the graph can I assume when the current is 0, the maximum switching voltage is also 200V? That is the contact can only withstand a voltage 200V when open?

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  • \$\begingroup\$ no, when switching. That's why the axis is called switching current. But if no current is flowing through the relay when it's closed, then there can be no voltage across itn when it opens, either. Ohm's law. \$\endgroup\$ – Marcus Müller Dec 10 '16 at 13:26
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The vertical line at 200V means you can switch up to 300mA, at up to 200V, as long as the load is resistive.

A straight line on a log-log graph like this one means a constant product, in this case 60 watts.

The horizontal line at 2A says you can switch up to 30V at that current.

So the relay can be used for resistive loads, as long as it stays within all three constraints of 2A, 200V, and 60W.

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Yes. The graph shows you that the basic worst case insulation the manufacturer guarantees when the relay is open is a bit over 200 V. That's when the current being switched is 300 mA or less.

The rest of the graph shows how the maximum voltage spec must be derated for currents above 300 mA. At 2 A, it can only switch 30 V.

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The relay can always withstand 200V when open, just not always when opening, these graphs deal with the small arc that occurs when the contacts open under load.

Higher voltages and currents can sustain bigger arcs and the relay might not open wide enough to safely switch 200V @ 2A, but can handle 200V @ <0.3A.

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  • \$\begingroup\$ Sam, I think your answer is different from Olin Lathrop at the point "The relay can always withstand 200V when open", can you explain more? \$\endgroup\$ – diverger Dec 11 '16 at 1:18
  • \$\begingroup\$ @diverger When the relay is already open, the only limiting factor is dielectric breakdown, there is no current as the contacts are open. However, when the contacts are closed, the limiting factor is I2R losses and is dependent only on current as there is very little voltage drop across closed contacts. The biggest problems crop up when you have closed contacts and then you open them. This creates a little arc, more power flowing means the arc is harder to extinguish. It's the relay's (limited) ability to deal with the arc that results in a graph like the one shown in the question. \$\endgroup\$ – Sam Dec 11 '16 at 21:37

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