ACR Dual Batteries and Charging Current

I PM'd RRRAAAYYY2 with this question, but figured we could all benefit from knowing the answer.

I've been looking at dual battery systems for my boat and found a basic principle using an automatic charging relay(ACR).

http://bluesea.com/category/2/productline/overview/387

I've devised what the circuit should look like, but am curious what the current is between the batteries when the relay switches from isolated to parallel, particularly when one battery is already drained.





best I can tell, this is close to how the painless dual battery setup is designed.

Anywhere from 0 to the charged battery's CA rating (several hundred), depending on how good it is, and how DIScharged the other battery is. You need a winch relay there.



IDK what you want to accomplish, but that circuit is unnecessarily complicated just to have a backup battery. This is the way Ford did it in the early 80s:

The reason for the complication is two fold.

1. The first relay is there to isolate the house battery during cranking. The boating crowd seems to think there are possible transients present during cranking that can damage electronics if the batteries were no isolated.

2. A typical boat ignition switch has OFF, ON & START. There is no HOT-IN-RUN from which I would have picked up the signal to relay otherwise. again, this is for isolation during cranking.

Regarding current, I must assume it is not so great because jumper cables do not regularly melt. Which is to say, that when you attach the cables from a running vehicle to a dead one, the current is not at the CA of the good battery. If you were to dump 750 amps through 12' of 2ga cable, I suspect it would more than heat up. So I must assume that the current is far less than a basic 12' jumper cable is rated for.

I see so many posts on the web where people think it is only the available current provided by the alternator that jumps the car, and they completely ignore the massive storage capacitor(battery).

Which relay do you use? I've seen the Warn 68379 from ~$30.

The current that charges a battery is dependant on two things, the voltage pushing that current (alternator or second battery) and the resistance of the battery (and cabling) itself. When a battery discharges it's internal resistance decreases. As it charges back up it's internal resistance increases. That's why a discharged battery will take a large charge current initially and then slowly decrease the charge current over time.

How much current will it take? Well batteries are very good at providing massive amounts of current when they are fully charged (higher voltage) if that current is going into a very low resistance. However when discharged, they are not (should not) exhibit a zero ohm resistance. If they did there would be no way to recharge them. Since a discharged battery has some degree of resistance it will not consume every bit of current that is capable from the alternator or second battery. Think of the battery as a current self-regulator, as long as it is not damaged. Damaged batteries can approach zero ohms and will destroy alternators also. I'd say it's safe to assume the battery charge current can be anywhere from zero to possibly 200 amps for a short time if everything is in good condition (other than discharge).

If you use that ACR relay just wire it up like it shows in the instructions here. Run your alternator output directly to the battery you want for starting and the ACR will take care of charging the second battery when it needs it.

Disclaimer: In no way should you ever attempt to use an ohmmeter on a battery. Your ohmmeter will melt and kittens will die.

Excellent explanation. I devised the circuit because I'm cheap and already have some high current relay's at the house. The ACR simply adds additional logic that may be overkill, and the Ford design should suffice.

Is there any truth to transients due to starting that would justify isolation during starting?

lonesouth said:

Is there any truth to transients due to starting that would justify isolation during starting?

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It's possible but probably not very likely with any automotive grade electronics. Use the full time charging battery as the starting battery and you shouldn't see any transients during starting.

lonesouth said:

Regarding current, I must assume it is not so great because jumper cables do not regularly melt. Which is to say, that when you attach the cables from a running vehicle to a dead one, the current is not at the CA of the good battery.

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I didn't say it would ALWAYS be that high; I said:

Steve83 said:

Anywhere from 0 to the charged battery's CA rating (several hundred), depending on how good it is, and how DIScharged the other battery is.

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It would only spike that high if one is totally dead AND good AND has very high reserve minutes, and the other was fully charged & in new condition.

BTW
A battery stores & releases electric energy in a totally different physical process (electrochemistry) from a capacitor (electric field).

Steve83 said:

BTW
A battery stores & releases electric energy in a totally different physical process (electrochemistry) from a capacitor (electric field).

Click to expand...

I understand that, but the concept from a cave man is that the little dohicky makes the light bulb go on. Lots of people seems to think that a battery only provides power when the engine is off and the alternator does the rest when running. They don't seem to understand that the battery is always there and is a source of tremendous current.