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I am currently looking at some multi-conductor connectors that will handle reasonably high currents (approximately 30 A at 24 v). When reading datasheets, I see that the connectors have both a maximum current and a maximum voltage. For example,

Voltage Rating: 600 VAC / Current Rating: 9 Amps Max. in 2-position applications.

I am having a hard time interpreting this. My understanding is that the maximum current is dictated by the resistance of the pins. My intuition is that this means that it would be safe to use the conector for any application that draws less than (9 A)(600 V) = 5.4 kW of power as long as the voltage does not exceed 600 VAC.

Is this true? If so, why isn't there a single "maximum power" rating? If not, can you explain how to interpet the rating at different voltages?

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This is a common misconception, and this sort of question has been asked before. I just can't find the question, which probably means that it's good that it's being asked again. – Kevin Vermeer Feb 18 '12 at 18:39
30A is a lot of current. You could parallel pins in your multipin connectors, but the best option is probably Anderson Powerpole or similar connectors. At least if you are connecting to wires and not PCB<->PCB. – markrages♦ Feb 18 '12 at 21:12
power-handling capability is not an inherent property of any electrical component; it is only an indirect consequence of current/voltage limits. – Jason S Feb 18 '12 at 23:07

3 Answers

up vote 10 down vote accepted

The voltage rating is related to the breakdown voltage of the plastic between the pins. You shouldn't exceed the breakdown voltage even of theres no current at all.

The current rating is, as you say, relates to the pin resistance and how much the connector will heat up. You shouldn't exceed the current rating even at very low voltages.

Edit:

As KellenJB says in another answer, it looks like the key thing you're missing is that the power consumed in the connector (and so the self-heating which could damage the connector) is not related to the voltage between the pins, but to the current through the pin. This current, combined with the (very small) resistance of the pin or contact, generates a small voltage between one end of the pin and the other (or between one pin and the socket its mated to). This voltage, multiplied by the current, gives the heat generated in the connector.

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up vote 4 down vote

Everything The Photon has said is correct, but let me give you an actual example.

enter image description here

Power=resistance*current^2

So 9 amps going through the pin would result in power dissipation in the pin of P=x*9^2. As I am sure you can see by now, the voltage doesn't come into play here at all.

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Just came back to add this to my answer, it looks like OP was not totally clear on how to apply Ohm's law or how to tell whether power is consumed in the interconnect or the load, so this should help to clarify. – The Photon Feb 19 '12 at 2:56
Yes, you're absolutely right! I was considering the power consumed by the load---which is a function of the voltage between the pins---instead of the power lost in the connector. @ThePhoton's reply makes perfect sense now. – Michael Koval Feb 19 '12 at 3:29
up vote 1 down vote

You're answering the question yourself: "as long as the voltage does not exceed 600 VAC". The 5.4kW isn't the only limitation. Otherwise you could connect 5.4kV @ 1A or 540A @ 10V; both are also 5.4kW. The 9A limit is for the dissipation in the contacts. Higher currents may melt the connector or weld the contacts. The 600V limitation is about the insulation between the contacts, or between contact and connector housing if the latter is metal.

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