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Posted
Hello, I have a really long 24VDC guard door circuit, and I went and calculated how much votlage drop there would be and whether there is still enough voltage at the end of the run to pull in a relay. Could someone look at the why I calculate this to see if it's right? Run: 1900ft of 18 guage wire: <0.00751ohm/feet * 1900ft = 14.3ohms> Relay: 24VDC coil, rated at 6.5W power consumption, so that's about 0.28A. So coil has 24V/.28A=86ohms Total resistance: 24V/ (14.3ohm+86ohm) = .24 amps. So I can expect .24amps on the wire, which is slightly less than the .28A needed to pull in the coil. Does that mean I wouldn't be able to use this relay, or might have to go 120VAC? How much wiggle room is there on these ratings? The relay doesn't list a minimum, simply the rated coil energy. Thanks,
Posted
You should have taken your math one step further, the amount of voltage across the relay will be a tad over 20v, check the relay rating for the minimum pull-in voltage, if this is <= 20v then you should be OK
Posted
Another possible solution: Most of the newer switching power supplies, can be adjusted up to 28 VDC. Buy a dedicated power supply just for this circuit, and adjust the voltage up until the relay functions as desired.
Posted
Thanks for the replies: 1. I can not up the voltage of the power supply as it is being used for other things too. 2. The relay is a AB 700S relay. It has the smallest power consumption compared to other industrial relays. 3. I'm using a cable, so the gauge is not easily changed. 4. Weird, but the spec sheet on the relay doesn't list minimum pull in voltage - only power consumption. Is there anything wrong with my math?
Posted (edited)
Just to make sure I've got the right spec sheet... Are you using a 700S-CF? If so, the Operating Voltage for 24VDC is: Pick-up = 16.8 to 30 VDC (see note 1 - 0.7...1.25 x Us) Drop Out = 14.4 to 2.4 VDC (0.1... 0.6 Us) I found this info right above the Coil Consumption rating. HTH! Susan Edited by ssommers

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