Battery charging current?
#1
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Boost Pope
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Battery charging current?
A problem has come up, and web-searching isn't yielding a good answer.
I need to figure out what the typical charging current will be when a conventional flooded lead-acid battery, which is at nearly full discharge, is first connected to an alternator putting out whatever voltage the stock alternator in a 2021 Ford Transit puts out.
Or, put another way, what's the ESR of a discharged 12v flooded deep-cycle lead acid battery with a 101Ah / 690 MCA rating?
Doesn't have to be an exact number. Being off by 50% in either direction wouldn't hurt my calculations. I'm going for rough order of magnitude here. Clearly it won't suck down the whole output capacity of the alternator, or else the vehicle would stall. But is it going to take 5 amps or 50? Assume no restrictions on current- I'm trying to figure out how large of a wire & fuse will be needed to provide short-circuit protection without melting under normal conditions.
This is *not* the main starting battery. It's an aux battery which is disconnected while the starter is turning.
And, no, I do not have the vehicle (or battery) here in front of me to test with. I won't have access to either until I get on-site and need to make the hookup, which will be in a remote location outside of easy access to auto-parts stores. I'm trying to pre-purchase the appropriate parts.
I need to figure out what the typical charging current will be when a conventional flooded lead-acid battery, which is at nearly full discharge, is first connected to an alternator putting out whatever voltage the stock alternator in a 2021 Ford Transit puts out.
Or, put another way, what's the ESR of a discharged 12v flooded deep-cycle lead acid battery with a 101Ah / 690 MCA rating?
Doesn't have to be an exact number. Being off by 50% in either direction wouldn't hurt my calculations. I'm going for rough order of magnitude here. Clearly it won't suck down the whole output capacity of the alternator, or else the vehicle would stall. But is it going to take 5 amps or 50? Assume no restrictions on current- I'm trying to figure out how large of a wire & fuse will be needed to provide short-circuit protection without melting under normal conditions.
This is *not* the main starting battery. It's an aux battery which is disconnected while the starter is turning.
And, no, I do not have the vehicle (or battery) here in front of me to test with. I won't have access to either until I get on-site and need to make the hookup, which will be in a remote location outside of easy access to auto-parts stores. I'm trying to pre-purchase the appropriate parts.
#2
Closer to 50 for a brief period of time. I monitor alt current, and if the battery is discharged to say 11.8 volts it will draw upwards of 70A+ for a very short time (singular seconds) dropping to 20-30A for 10s of seconds then stabilize to something in the 10-15A range. This is with a tightly regulated CV source of 14VDC.
#4
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Boost Pope
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That's useful info.
70A @ 14v yields an ESR of 0.2Ω. Which is way lower than I was expecting.
70A is too much in this scenario. But it's a temporary application, and the series-resistor idea is interesting. I could pre-build that with a switch across the resistor to bypass it after a few minutes.
70A @ 14v yields an ESR of 0.2Ω. Which is way lower than I was expecting.
70A is too much in this scenario. But it's a temporary application, and the series-resistor idea is interesting. I could pre-build that with a switch across the resistor to bypass it after a few minutes.
#5
You probably don't even need to bypass it. As the battery voltage approaches the final charge voltage the current will drop, as will the voltage across the resistor. You do probably want to run the resistor directly between the source (alt) and the battery, with the load connected directly to the battery so that the load current doesn't result in a voltage drop across the resistor.
#6
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Boost Pope
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Load will be connected directly to battery. It's only the charge-side of the equation that concerns me.
And, seriously, thank you for the insight. I was so deep in this rabbit hole that "Dude, just stick a resistor in there" had legitimately not occurred to me.
And, seriously, thank you for the insight. I was so deep in this rabbit hole that "Dude, just stick a resistor in there" had legitimately not occurred to me.
#9
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Boost Pope
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From: Chicago. (The less-murder part.)
This way, the second aux battery can be charged by the primary aux circuit when the engine is running, and also supply power back into it when the engine is off.
Also, I should have checked these very cheap resistors before assembling everything to ensure that they are both what the etching on their cases says they are.
One is 0.5 ohms. The other is reading 120k, which tells me it's effectively open.
Ah, well. Replacements on order, should be here in time. Wheels up on Friday.
#10
Just thinking out loud, but a dead battery might be say 10.5V. If you throw 50A into it, maybe it comes up to 11.8V while the alternator is at 14.8, so a 3 volt difference accross the resistor. Power is V*A, 3*50 = 150W. Those resistors are going to get really hot. I think they would need a really beefy heat sink and or fan to keep them from overheating and going open-circuit.
Also, if you really use a 0.5 ohm, then you won't get much charging/Amps flowing. V=IR, I = V/R = 3/.5 = 6 amps of charging current. So won't overheat the reistor, but will take forever to charge a 100AH battery.
Also, if you really use a 0.5 ohm, then you won't get much charging/Amps flowing. V=IR, I = V/R = 3/.5 = 6 amps of charging current. So won't overheat the reistor, but will take forever to charge a 100AH battery.
#11
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Boost Pope
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From: Chicago. (The less-murder part.)
Just thinking out loud, but a dead battery might be say 10.5V. If you throw 50A into it, maybe it comes up to 11.8V while the alternator is at 14.8, so a 3 volt difference accross the resistor. Power is V*A, 3*50 = 150W. Those resistors are going to get really hot. I think they would need a really beefy heat sink and or fan to keep them from overheating and going open-circuit.
The idea is to use the resistors only to limit the initial inrush current. You can see the shunt (left) and voltmeter / ammeter (right) in the photo. Once those instruments tell me that things are calming down, I can flip the two switches (one at a time, of course) to remove the resistors from the circuit.
#12
Retired Mech Design Engr
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Battery Combiner
Maybe this will do what you want. I am installing one on my motor home now that the supplied relay has stopped working.
EDIT: Automatic in either direction once the Master battery is at 13.6, or something. Either supply can be the Master. OR, can be forced to combine with the "Manual" terminal. Other brands are available.
DNM
Maybe this will do what you want. I am installing one on my motor home now that the supplied relay has stopped working.
EDIT: Automatic in either direction once the Master battery is at 13.6, or something. Either supply can be the Master. OR, can be forced to combine with the "Manual" terminal. Other brands are available.
DNM
#13
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Boost Pope
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From: Chicago. (The less-murder part.)
Battery Combiner
Maybe this will do what you want. I am installing one on my motor home now that the supplied relay has stopped working.
Maybe this will do what you want. I am installing one on my motor home now that the supplied relay has stopped working.
This is a temporary connection, and the limitation here is that I can't cut into the vehicle, so I don't have direct physical access to the house battery. I pretty much have to go in through the RV-style 12v aux fuse panel with one of these:
Hence the current restriction.
#14
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Boost Pope
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From: Chicago. (The less-murder part.)
If we assume the 0.2Ω ESR for the discharged battery which was calculated initially, and two 0.5Ω resistors in series, then with a 14v alternator output, inrush current will be 11.7A with 67 watts of heat going into each resistor. That's perfectly manageable in my application.
After a few minutes, when current starts to ramp down, I can close the first switch and remove one resistor from the equation, then wait a few more minutes and close the second switch, at which point the two batteries are fully in parallel. I have an ammeter in series with that part of the circuit so I can monitor it.
I'm deliberately erring on the side of caution here. I'll be carrying a pocketful of spare fuses of various ratings (mostly 30 and 40A, a few lower values for other circuits), and I don't wish to expend them all.
The defective resistor has been replaced, it passed functional test (I used the 6 year old battery which I pulled out of the Miata a few weeks ago as a test subject) and the whole thing is final-assembled. No photos as it's out in the garage with the spray-adhesive curing on the rubber shelf-liner which I attached to the underside. Intended to prevent it from sliding around and putting stress on the cable while the vehicle is in motion.
Pro-tip: cheap inverters are cheap. Eg: non-sinewave. The switching power supplies were ok with it, but the AC fan motor makes a weird noise when being powered by it. Wattage and power factor were within spec, though.
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