



Question
Connectors, amp ratings
I am not sure of the following. lets take a XT60 as an example. It is rated at a constant 60 A. But at what voltage? This is not specified. Or does voltage make no difference? Will the strain on the connector be the same under the following two conditions?
1. When you have a 3S (11.1V) battery with a 30 A drain 2. When you use a 400V source with a device pulling 30A at that voltage. In case number two the watts is 36 times more. Will the XT60 be operating at the same temperature under both scenarious? The reason for my question is just out of a matter of interest. Here is a very practical example. It is written on a particular household plug '' 125 VAC at 5 amps''. Now if you want to use that same plug in a 220V household. Is it still good for up to 5 Amps? Sorry if this does not seem boating related, but it is. My question basically: At what voltage do manufacturers give the particular amp ratings of any specific connector? 





In general the amp rating stays the same when working with different voltages. How ever a switch rated for 125v probably will not be suitable for 220v sure to other factors like contact separation and insulation.
In short, the current rating is based on contact resistance. In your example when the switch its closed out wool see the same voltage in both scenarios as according to ohms law v=I*r. I=30 in both cases and the resistance it's also the same so the voltage drop will be the same too. When the switch opens how ever the full voltage will be across it so must be rated for 400v 





No I agree. But I was always wondering when manufacturers quote the max continious amprating. It is never said at what voltage this was achieved.
Looking at normal bedlamp wires, they are seperated by very little plastic between them. With XT60's for example, there are a lot more plastix seperating the two connectors. I'm sure they can handle a lot of voltage. But I'm merely interested at what voltage are the continious amprating achieved. 





The amp rating is probably based on the connector resistance and power dissipation.
If you have a look at ohms law you'll see that supply voltage has no effect on the current capacity of the connector. In this case Power = Current squared * Resistance There is no voltage in that equation. If you don't apply the heatshink correctly to an XT60, there is bare copper exposed even when connected. It appears it's only a couple of millimetres between the positive and negative. Not taking into account any moisture on the plastic, a surge of just a few KV would be enough to start an arc, and being DC going through it, the arc could be kept going, effectively shorting your battery through the plasma. You don't get this problem with AC (if an arc is created, the voltage soon drops to zero, stopping the arc), which is why you may see a switch or relay with contacts rated at a particular current with a voltage of 125VAC/30VDC 





Yeah... For the most part (within some reasonable range) the voltage does not matter.
As viperdae mentioned it would seem they are rating the connector using the equation: Power = Amps ^ 2 * Resistance The equations for heat are very similar to power, so the more power you generate the more heat you generate over time. I am going to assume that when they rate the something for 60amps they are probably thinking they have a given resistance and at over 60 amps the heat buildup becomes a problem  like softening or melting the plastic. The resistance of a connector is probably somewhere in the 0.0009 ohm range which means something around 4 watts of power dissapated in the connector is a problem. You could use any voltage you like, but if nothing changes in the overall circuit and you crank up the voltage you will also increase the number of amps. It would seem that you work mostly with voltage and the motor to determine RPM and an amp draw. Then given the amp draw you can determine if a given connector will work. I don't think arcing is a problem until the voltage gets really high. I am pretty sure it takes about 3000 volts to jump a 1mm gap in air. 





Ahhh. Ok. Thanks Viper.



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