Ok all, first time poster. I did some searching and found what I think might work, but wanted to make sure.
I have just installed a new car alarm in my s-10 pickup. the alarm is wired into the main battery right now. When all is said and done I will have a second battery in the bed (for my sound system). it will be isolated from the main.
I would like to setup a relay so that if someone gets under the hood to disconect the main battery the second battery in the bed will take over and the alarm will continue to go off. This is the relay layout I found,
Main battery + to pins 86, 87a
Main battery - to pin 85
Second battery to pin 87
12 volt output (alarm) pin 30
Please let me know if this is correct. Thanks for the help.
u need 87a to the second batt. and not the main...everything else is right...also pin 87a and 87 go to the postive of the second batt.
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Where theres is a wire there's a way.
Thats a big 10-4, thanks for the help.
Keep in mind that if you hook everything up this way you will have a constant current draw on the battery that can somewhat significant. It won't kill your battery over night, but it will reduce the 'sitting around' time of the battery.
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Kevin Pierson
This shouldn;t be any more of a current draw then the alarm has hooked up normally right? I mean right now it is just hooked directly to the battery anyway.
DG
Consider this.. once someone cuts the primary battery lead, *all* power requirements are going to be feeding through that relay from the second battery. All that's got to happen is for current draw to exceed fusing and your second battery power feed is gone.
Perhaps use diodes to feed both batteries to the alarm system.. you'd have to watch where you feed your annunicators (lights/siren) from, but that shouldn't be too difficult to figure out.
Jim
The coil to the relay you are adding will be connected directly to the main battery of the vehicle. The resistance of the coil will determine how much current is continously drawn. If the relay has a 100 ohm coil then Ohm's Law states that at 12vdc (typical car off voltage) the current draw will be I = V/R or I = 12/100. This results in a 120mA continous load on you battery. Like I said, not enough to kill the battery overnight most likely, but I wouldn't leave it sit for a week straight.
If you look around at comprable systems (ie DEI's back up battery) you will find that they have electronic control modules that switch back and forth from battery to battery with out the use of relays.
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Kevin Pierson