How is incoming charge distributed amongst multiple batteries wired in parallel?
July 27, 2009 4:03 PM   Subscribe

Say I have a three batteries (all 12V SLA of the same Ah capacity) wired in parallel to form a bank. I connect a dumb charging source (let's say 12V, 5A). How will the incoming charge be distributed amongst the three batteries, and what does it depend on?

I'm keen to understand the principlesat work here. I've built several battery banks (largest was eight deep cycle marine batts), so I've done a fair bit of just plain trying it out. I'm not trying to enhance my theoretical understanding, so I can figure out what will happen in a simple situation like:

Battery 1: 100% charged
Battery 2: 50% charged

...Battery 2 will receive more of the incoming current. But how much more? Will it receive 100% of the current, with Battery 1 receiving 0%? That's my guess - and in fact I'm guessing that Battery 1 will be charging Battery2 in this situation, and they'll end up equalizing then charging in parallel.

If I'm right about that, how do I generalize to:

B1: 100% charged
B2: 75%
B3: 50%

...I'm guessing B1 and B2 will charge B3 until B2 and B3 equalize, then B1 will charge B2and B3 until all three equalize?

What are the key factors at play here? State of charge of the batteries has to be the main one, but... well any help appreciated. I've googled for info, and read the basic battery faqs, but none have helped me understand this.
posted by gribbly to Science & Nature (16 answers total) 1 user marked this as a favorite
I think that the model of the battery is an voltage source in series with a resistance that is related to the state of charge.

Current is going to go mostly into the lowest opposing voltage node. Its source will be the charger output and anything else that can provide current, which in this case means a battery with a higher voltage and lower series resistance. How much depends on the various batteries, their internal conditions, their state of charge, and the specifics of the charger. Charger specifics vary significantly depending on the type of charger, so there is no way to say for sure in your case.

To test your specifics, you can put a .1 (or .01) Ohm resistor in series with all the individual batteries and you can measure the input current division by looking at the voltages across all of them. Observe polarity to determine if a battery is sinking or sourcing current.

If you connected all the batteries in parallel and walked away for a day, they would equalize in voltage. They would also accept different amounts of current from the charger if you then connected it, since the individual characteristics of each would be slightly differenct.
posted by FauxScot at 4:27 PM on July 27, 2009

Best answer: Let's start with a simplifying assumption: the wiring connecting the charger and batteries together is heavy enough that the charger voltage and battery terminal voltages are all the same as each other. In reality this will never quite be true, but it's close enough to understand the system correctly.

The key to this whole thing is the terminal voltage, specifically:
  1. What is the terminal voltage?
  2. How do the batteries respond to it?
From the point of view of the charger, it doesn't care how many batteries are connected in the bank, nor what their states of charge are. It just sees that it is connected to something that will draw current that increases with voltage.

The charger will output whatever voltage causes the battery bank as a whole to draw 5 amps, up to whatever its maximum voltage is (which for a nominally 12V "dumb" charger could be 15 volts or more). In other words, the battery terminal voltage will be whatever voltage causes the entire bank to draw 5 amps from the charger.

Now that we know the terminal voltage, we can look at how each individual battery behaves with that terminal voltage applied to it. Depending on each battery's type, size, state of charge, temperature, and even its history, it could be charging, discharging, or neither. But it will do whatever it would have done by itself, if it was connected to a constant-voltage charger maintaining the same terminal voltage.

For example, the fully-charged battery might still accept some charge (say, 1 amp) while the partly discharged battery accepts more (say, 4 amps). Or if these were very large batteries, one could imagine the fully charged battery discharging at 100 amps, and the partly discharged battery charging at 105 amps.

To summarize, the characteristics of the battery bank as a whole, and the charger, will determine the terminal voltage, and each battery's response to that terminal voltage determines what each battery does, independently of all the other batteries.
posted by FishBike at 4:46 PM on July 27, 2009

Parallel connected batteries will indeed try to charge each other to equalize their terminal voltages. If there's a significant difference in the state of charge of the batteries in a parallel connected array, then these inter-battery currents can be high enough to cost quite a bit of service life.

Instead of a straight parallel connection, you can use diodes or FETs configured as effective diodes to isolate batteries from each other while still allowing them to charge from a common source. If you do this, the battery with the lowest terminal voltage will receive all the charge current and the others will receive none. Once the terminal voltages have equalized, each battery will receive a charging current inversely proportional to its internal resistance.

If one of the diode-isolated batteries has a collapsed cell, it will hog all the charge current (because its terminal voltage will never rise as high as that of the others) but at least it won't actually discharge any of the other batteries.
posted by flabdablet at 5:37 PM on July 27, 2009 [1 favorite]

It is not the case that once any of the batteries hits "full" that it will stop drawing current from the charger. There will still be current flowing through it, and it'll turn that into heat.

If they are 12V batteries and your charger is trying to backflow 12V into them, then "fully charged" is an asymptote you'll never reach. I suspect the charger is producing 13 or 14 volts. In either case, there will still be current flow through the most-charged battery as long as the charger is running.
posted by Chocolate Pickle at 7:39 PM on July 27, 2009

Response by poster: @FauxScot - thanks, that's helpful. Adding a resistor so I can "see" the ratios is a good idea, and something I can actually do (I have three small 5Ah SLA batteries that I can connect to a charger).

@FishBike - when you say "terminal voltage", which terminals do you mean? The terminals of each battery? I have to admit I found your answer a little confusing, probably because my knowledge on this topic is limited! When you say "the battery terminal voltage will be whatever voltage causes the entire bank to draw 5 amps from the charger" I don't really know what you mean. If you can spare the time, can you try and unpack that for me a bit more?

@flabdablet - thanks for the link, interesting possibility. As for collapsed cells, I've experienced that... my eight battery bank was constantly dumping into one dud battery before I figured out what was going on :-/ If only I'd had isolators!

@Chocolate Pickle - yep, good point. My "dumb" charger will require me to disconnect it when all batteries are close enough to charged. I should point out, though, that this is all entirely hypothetical - I'm not talking about a real battery bank, I'm trying to increase my theoretical knowledge!
posted by gribbly at 8:02 PM on July 27, 2009

Best answer: Analyzing networks of devices whose behavior is dominated by DC effects is made much, much easier when you grasp the principles of Thevenin and Norton equivalent circuits, and Kirchoff's current and voltage laws. Once you grok those, large numbers of electrical problems become much less confusing.
posted by flabdablet at 8:50 PM on July 27, 2009

Yeah, by "terminal voltage" I mean the voltage across the terminals of each battery... and because all the batteries are connected in parallel with thick wire, the terminal voltage will be the same for all the batteries in the pack, regardless of their state of charge or anything else.

So let's say we have a battery pack that, just sitting there with the charger switched off, has a terminal voltage of 12.6 volts. All the batteries in that pack will read 12.6 volts across their terminals if you measure them with a voltmeter. There's no current flowing between the charger and the pack because the charger is off.

Now switch on the charger. Two things will happen: the terminal voltage will start to rise, and current will start to flow from the charger into the battery pack. The higher the terminal voltage, the more current will flow. The terminal voltage will rise very quickly until the charger is putting 5 amps into the pack, at which point the voltage will stay almost constant. Let's say that with our hypothetical battery pack, the terminal voltage rises to 13 volts to draw 5 amps from the charger.

At this point we can now look at what each individual battery in the pack is going to do. They've all got a terminal voltage of 13 volts, but depending on their state of charge (and all those other factors I mentioned), 13 volts might be enough to cause a battery to charge, do nothing, or even discharge. Some possibilities:
  • The fully charged battery just sits there, neither charging nor discharging, while the partly discharged battery charges at 5 amps.
  • The fully charged battery actually discharges at 1 amp (or "charges" at -1 amp if you like), while the partly discharged battery charges at 6 amps.
  • The fully charged battery charges at 1 amp, while the partly discharged battery charges at 4 amps.
In all the above examples, if you add up the charging current for all the batteries together, it adds up to 5 amps--exactly what the charger is putting out. But each individual battery is doing its own thing that depends only on the terminal voltage of the entire pack.

I'm not sure if this explanation is making things clearer or just confusing things further. I'm hoping for the former, though. ;)
posted by FishBike at 6:21 AM on July 28, 2009

This probably doesn't help at all, but I have a "smart" charger that is overly sensitive. Won't charge a battery that isn't already mostly charged. I have been able to charge batteries that it wouldn't charge by hooking it up in parallel with another.

Another thing to consider. But I don't know how it works, I just know it's "a thing". The internal resistance of the battery. I think that a bad SLA battery will have higher internal resistance than a good one. So if they are connected in series, a bad battery will wreck a good one. Might be applicable to your scanario?
posted by gjc at 6:57 AM on July 28, 2009

This isn't a good idea. Batteries essentially expend expend energy to maintain a constant voltage across their terminals. So if you have one fully charged (say 13V) and one not so fully charged (say 10V) you now have a 3V difference across the battery to battery circuit: basically, 3V across a short. Which will draw a lot of current across that link which can lead to heat, smoke, fire and all that.

This is an idea battery model. In reality, this won't happen but you still risk bad things happening. Charge them one at a time or with a circuit like flabdablet mentions.
posted by chairface at 10:48 AM on July 28, 2009

It sounds like you have a mental model of a battery as being like a bottle into which you pour water, and when it's full then the flow of water stops. That's not how it works.

A battery is a resistor. If you're backflowing current through it, some of it is converted into heat, and some of it causes chemical changes which make the resistance rise. But it doesn't rise to infinity. It doesn't inherently shut off the current on its own.

Different kinds of rechargable batteries have different behaviors. NiMH is the most extreme: its resistance doesn't increase much as it nears full charge. Instead, an increasing percentage of the energy of the current becomes heat instead of stored charge. If you backflow NiMH constantly, eventually the battery will melt or explode. In order to prevent that, it's necessary to go into what battery charging guys call "topping off mode". That means you flow current for one second, and then wait four seconds for the battery to cool.

The biggest problem with the "pour water into it" conceptual model is that it doesn't take into account that a significant proportion of the "water" turns into waste heat instead of stored charge. In many cases it's well over half. In some cases it's virtually all of it.

In other words, most of the water is leaking away rather than going into the bottle.
posted by Chocolate Pickle at 11:52 AM on July 28, 2009


it keeps leaking away even after the bottle is full.
posted by Chocolate Pickle at 1:04 PM on July 28, 2009

Response by poster: @flabdablet - thanks for those links, that's exactly the nudge I needed. I know have much more targeted search terms for google to chew on! I just learned the difference between "resistance" and "impedance" (impedance is a more general term that factors in "reactance", which is only relevant in AC scenarios - right? =])

@Chocolate Pickle - thanks for all of that, helpful stuff. If you will indulge me a little further I have two followups:

1) What do you mean by "backflowing" current?
2) "Topping off mode" is the same as "float charging", right?
posted by gribbly at 11:26 PM on July 28, 2009

"Topping off mode" is the same as "float charging", right?

No. Topping off mode is something you do near the end of the charging cycle when quick-charging a NiCd or NiMH cell to stop it catching fire, involving a smart charger and pulsed current. Float charging is where you continuously apply a voltage very close to a lead-acid battery's its open-circuit voltage (usually 13.8V for a nominally 12V lead-acid battery) so that the battery continuously draws a small charging current (very close to its self-discharge current). This is safe for lead-acid batteries because their terminal voltage increases pretty steadily with their state of charge; other battery chemistries have a much flatter voltage vs. charge curve and need active charge current management.
posted by flabdablet at 11:51 PM on July 28, 2009

"Backflowing" current means to run current into the battery in the opposite direction that the battery wants to supply current.
posted by Chocolate Pickle at 7:19 PM on July 29, 2009

Response by poster: @FishBike - I only just fully understood what you were saying, and it is very helpful. I hadn't grasped the significance of terminal voltage, and how it could be used to simplify the mental model by allowing the bank to be decomposed into individual elements that are "seeing" that terminal voltage. Thanks for that insight.

I did have a follow up question on your second post. I'm able to follow your example well, but I just want to clarify that you sort of start out implying a battery bank with many batteries. Then in the bulleted list you speak as if there are just two batteries ("the fully charged", "the partly discharged). That's you switching battery counts unannounced, right? Not me missing something? =]

Anyway, thanks again for your help. In fact, everyone's answers were very helpful. I used this help to run a seven battery bank off three 130W solar panels at Burning Man this year. Ran camp (lights, sound, light toaster oven duty, battery charging) for ~9 people for a week - worked great.
posted by gribbly at 10:05 PM on September 28, 2009

I did have a follow up question on your second post. I'm able to follow your example well, but I just want to clarify that you sort of start out implying a battery bank with many batteries. Then in the bulleted list you speak as if there are just two batteries ("the fully charged", "the partly discharged). That's you switching battery counts unannounced, right? Not me missing something?

Yeah, it seemed simplest to look at a battery bank with just two batteries. But the same approach works for any number of batteries in parallel. I don't think you're missing anything.
posted by FishBike at 4:38 PM on September 29, 2009

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