# Different batteries in-parallel - possible problems



## ltiu (Dec 19, 2009)

I did a google search of cpf to see if anyone has this already listed but I found none.

I found a lot of issues with different batteries in-series. Specially those that go  and the ones that leak.

But now, I want to know what problems are possible with different batteries in-parallel.

Can you guys help me by listing possible issues with putting different batteries in parallel.

By different I mean, different voltages, brands, types, sizes, chemistry, primaries and rechargeables all mixed-in ... 

Thanks.


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## italianboy (Dec 19, 2009)

Basically, you can only put in parallel batteries with the same voltage and chemistry, otherwise it won't work, or it might be dangerous.


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## ltiu (Dec 19, 2009)

italianboy said:


> Basically, you can only put in parallel batteries with the same voltage and chemistry, otherwise it won't work, or it might be dangerous.



Thanks. But you did not answer my question. 

I know it works since I have tried it briefly.

I would like to know exactly what dangerous things can happen.


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## Mugrunty (Dec 19, 2009)

I think people on the forum have discussed this before, but charging Nickel based batteries in parallel doesn't work to well (or safely). Specifically when you try to charge the pack quickly. I found out the hard way when I was working on my senior project. It used two 7.2V Ni-Cd packs in parallel. They were quite old (1984ish sintered plate technology 450mAh). 

Anyway, I decided to charge a few of them quickly in parallel using Delta-V...and well...one of them started to leak. Turns out...when one of the series string of the battery reaches full, it starts to dip in voltage. BUT, since it is in parallel with another series string...the dip in voltage cannot be seen, causing the charger to continue charging. Eventually if not caught in time, one or more of the series strings will vent.

My senior project cheats and charges so slowly that no problems would ever arise. It was somewhere around 1/10C. Probably more like 1/8C or something. Plus it slows the charge to a crawl when the voltage reaches approximately 1.43V/cell.

The other bad thing was I was charging those parallel packs at 4C...lol.
Normally they can handle it...if done independently. They stay cool to the touch right up till the end of charge. Even then, they are only luke warm. Good old endothermic Ni-cds! But that's another topic...


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## Mr Happy (Dec 19, 2009)

ltiu said:


> Can you guys help me by listing possible issues with putting different batteries in parallel.
> 
> By different I mean, different voltages, brands, types, sizes, chemistry, primaries and rechargeables all mixed-in ...



Whoah! What are you thinking?

By asking for possible issues it means you are thinking of doing it. But don't even think it. If you break this rule you will get poor performance, excess heat, damaged batteries, or worse. Just don't.

For discharging you can put two or more identical batteries in parallel.

For charging you can put two or more identical batteries in parallel if they are Li-ion or lead-acid, but not if they are nickel chemistry as Mugrunty mentioned.


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## Benson (Dec 19, 2009)

Mr Happy said:


> For discharging you can put two or more identical batteries in parallel.
> 
> For charging you can put two or more identical batteries in parallel if they are Li-ion or lead-acid, but not if they are nickel chemistry as Mugrunty mentioned.


True, but "identical" is unnecessarily strict. Same chemistry is necessary, but they can be different capacity with no problems.


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## ltiu (Dec 19, 2009)

Thanks for the feedback so far. 

I am thinking of *Discharging Only*.

I am thinking of NiMH - different brands, different capacities.

I am thinking of alkalines - different sizes, different brands.

I am *NOT* using Lithiums here (primary or rechargeables).

AND I am *NOT* mixing alkys and NiMH. Just alkys with alkys and NiMH with NiMH. 

I wish to mix different NiMH brands/capacities together. 

I am also thinking of mixing different alkys together.

All in parallel.

Can you guys specifically tell me what could happen?


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## rmteo (Dec 19, 2009)

If you don't mind, can I ask you what is the point of this exercise?


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## Mr Happy (Dec 19, 2009)

Benson said:


> True, but "identical" is unnecessarily strict. Same chemistry is necessary, but they can be different capacity with no problems.


In theory, maybe, as long as the capacities are not not too dissimilar. But for good design you would always match cells like for like.



ltiu said:


> Thanks for the feedback so far.
> 
> I am thinking of *Discharging Only*.
> 
> ...


In short, nothing terrible will happen. But it doesn't make sense and normally nobody would do this.


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## ltiu (Dec 19, 2009)

rmteo said:


> If you don't mind, can I ask you what is the point of this exercise?



For emergency situations when you need high capacity and have many different battery types lying around.


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## rmteo (Dec 19, 2009)

ltiu said:


> For emergency situations when you need high capacity and have many different battery types lying around.



OK, thanks.


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## ltiu (Dec 19, 2009)

Mr Happy said:


> But it doesn't make sense and normally nobody would do this.



I'm a flashaholic. I can't be normal.


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## Benson (Dec 19, 2009)

ltiu said:


> I wish to mix different NiMH brands/capacities together.


Probably perfectly fine.



> I am also thinking of mixing different alkys together.


This configuration is likely to leak and ruin your light.

But then, a single alkaline is likely to do that, too. :laughing:


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## Lynx_Arc (Dec 19, 2009)

I don't see any issue for mixing different nimh batteries in parallel if you are running at low to moderate current levels. I would think though when the current draw started pushing the limit of the the battery with the least internal resistance it could overheat it because it would try to provide whatever current it could and the other batteries with higher internal resistances would not be helping equally as well. One battery under high drains could become discharged very quickly even possibly damaging it. As it discharged greatly and then the current load was removed then the other batteries would *try* to recharge it up to their voltage level as fast as their internal resistance would permit.... more heating.


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## 45/70 (Dec 19, 2009)

Lynx, I'm with you on what you're saying, but with cells in parallel, the voltage of all of the cells will remain the same. No cell will be discharged lower (as far as voltage goes) than any other. That is not suggesting that _all_ of the cells couldn't be over discharged, just that all of the cells will be at the same voltage level (which doesn't necessarily mean the same SOC). That is a property of cells in parallel. I think you're thinking of what happens when cells are in series.

Dave


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## Lynx_Arc (Dec 19, 2009)

sure the voltages will remain the same but think about it. If each cell were all resistors in parallel the path of least resistance would be through the cell with the least resistance and it would discharge first and fastest while the other cells would still have power in them it would quickly be drained dry compared to them.


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## lumenal (Dec 20, 2009)

ltiu said:


> I would like to know exactly what dangerous things can happen.


 
Why don't you try rigging something up, putting it in an ammo box in your driveway, and report back?


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## Benson (Dec 20, 2009)

Lynx_Arc said:


> sure the voltages will remain the same but think about it. If each cell were all resistors in parallel the path of least resistance would be through the cell with the least resistance and it would discharge first and fastest while the other cells would still have power in them it would quickly be drained dry compared to them.


I'd say, not that it would drain first and fastest, but that it would drain fastest _at first_ -- then it reaches a steady state where it's somewhat discharged, showing a similar external voltage (despite its low resistance) to the sagged voltage of the other cells, then they're all running the same current (more-or-less). Suppose they take it the whole way down to empty (say 0.8V) -- that means they're sagging to 0.8V under serious current while it's internally at 0.8V (empty) connected through its internal resistance (whatever value) to an external 0.8V -- V = 0 --> I = 0 neither charging nor discharging.

Basically, I see differences of internal resistance mainly showing up as short surges where one battery takes more than it's fair share for a little while following a load change -- if loads change gradually (not the case for startup, but definitely true for, say, an incan as the batteries run down) there will be no such effect visible, and if loads are switched suddenly, such effects will be visible, but generally not harmful.

Of course, if the "strong" cell drained itself to 0.8V like this, and the weak cells weren't fully discharged, when the load was removed abruptly, they'd shunt-charge the strong cell. But since NiMH are safe to charge rather fast _as long as you don't keep charging them when they're full_, and it's impossible for the other, partially-depleted cells to overcharge this one, I don't see it being a problem. Then again, I'm not a Ni** guru, so I don't know this stuff...

Besides, worst-case, doesn't abusing a NiMH just lead to higher internal resistance? So if this _is_ hard on the battery, it'll just catch up to the group, then they'll all pull evenly. (Although maybe a single "weak" battery maintaining a higher SOC and trying to recharge the rest of the pack would be a problem...)


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## Lynx_Arc (Dec 20, 2009)

I think it depends on the internal resistance of the cells. I have some 2500 energizers that after about 2 days they are under 1v if they were put with good cells that don't discharge to 1v even in 6 months the high self discharge cells would drain them dry in just a few days more. I don't know the internal resistance differences so I wouldn't be able to just guess at what could possibly happen in use.


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## ltiu (Dec 20, 2009)

Benson said:


> This configuration is likely to leak and ruin your light.



I did not say it was for a light


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## ltiu (Dec 20, 2009)

lumenal said:


> Why don't you try rigging something up, putting it in an ammo box in your driveway, and report back?



No. Plans are to keep it in open air with a fan blowing through it to keep things cool. Won't happen until the next hurricane. As I mentioned, for emergency situations.


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## ltiu (Dec 20, 2009)

Lynx_Arc said:


> I don't see any issue for mixing different nimh batteries in parallel if you are running at low to moderate current levels. I would think though when the current draw started pushing the limit of the the battery



That's why I want to run things in-parallel, to distribute the current load across many batteries so each bank's current draw is small.


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## 45/70 (Dec 20, 2009)

Lynx_Arc said:


> .....in parallel the path of least resistance would be through the cell with the least resistance and it would discharge first and fastest.....



True, until the voltages of the cells in parallel became equal (in a very short time). At this point, all cells will have equal voltage potential and thus, at least appear to the load, to all supply an equal amount of current.




Lynx_Arc said:


> .....while the other cells would still have power in them it would quickly be drained dry compared to them.



This can't happen, for the reason stated above.



Lynx_Arc said:


> I think it depends on the internal resistance of the cells.



The load sees the cells that are paralleled as one cell, not individual cells. The internal resistance seen by the load, would be the average IR of all the cells in parallel. I might not have this correct, but I think it is. 



Lynx_Arc said:


> I have some 2500 energizers that after about 2 days they are under 1v if they were put with good cells that don't discharge to 1v even in 6 months the high self discharge cells would drain them dry in just a few days more.



This is partially correct. The high self discharge cells would suck the power out of the good cells (in a relatively short time). However, again due to the cells being in parallel, no individual cell in the pack could be drained any "dryer" than any other cell. They would all be at the same voltage potential.



Lynx_Arc said:


> I don't know the internal resistance differences so I wouldn't be able to just guess at what could possibly happen in use.



Again, the IR differences between the cells in parallel, would be averaged between the cells, and the load would see this as a single IR value.

I hope this isn't really off topic, it does relate to the OP. I'm pretty sure I'm correct here, but I certainly wouldn't mind being corrected if I'm wrong. I recently started a thread asking a somewhat similar question about charging Li-Ion cells in parallel, so I'm no expert. I'm just stating what I believe to be correct. 

Dave


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## Mr Happy (Dec 20, 2009)

45/70 said:


> True, until the voltages of the cells in parallel became equal (in a very short time). At this point, all cells will have equal voltage potential and thus, at least appear to the load, to all supply an equal amount of current.


It's not a question of how the cells appear to the load, but how the load appears to the cells.

In fact, the cell with the lowest internal resistance will supply more current, and the cell with the highest internal resistance will supply less current. For it to be otherwise would imply a logical contradiction. Consider, as a thought experiment, that one cell had a very high internal resistance: in that case for it to supply any significant current it would have a very high voltage difference across it (V = IR). Since we know the voltage difference across all cells is equal and low, it follows that the cell with the high resistance must supply only a low current.

By analogy, consider the case of pure resistances in parallel. When viewed from the external circuit these appear as a single resistance. But internally, the current through each resistor is given by I = V/R. Since V is the same for all resistors, we see that the current through any resistor is inversely proportional to its resistance.

Therefore if you put two cells of equal capacity but differing internal resistance in parallel, the cell with the lowest internal resistance will drain faster and will become empty before the other one.


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## 45/70 (Dec 20, 2009)

OK Mr H. I can follow your explanation. The point I was trying to make was that when cells are in parallel, it is not possible to discharge one cell lower than the others. Yes, the cells with lower IR _will_ deliver more current, but in the end, the cells will all have the same voltage and no individual cell will be discharged "dry", unless they all are. Is this correct?

*EDIT: *By "lower" I'm not referring to capacity, but rather voltage, as in discharging an individual cell in the parallel "pack" to a voltage that would be damaging to the cell.

Dave


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## rmteo (Dec 20, 2009)

ltiu said:


> I'm a flashaholic. *I can't be normal.*



Yes, it shows.


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## Mr Happy (Dec 20, 2009)

45/70 said:


> OK Mr H. I can follow your explanation. The point I was trying to make was that when cells are in parallel, it is not possible to discharge one cell lower than the others. Yes, the cells with lower IR _will_ deliver more current, but in the end, the cells will all have the same voltage and no individual cell will be discharged "dry", unless they all are. Is this correct?


No, it cannot be correct. The amount of discharge of a cell is defined by current x time (mA x h = mAh). Therefore the cell supplying more current will discharge faster and will become empty first, while other cells still have charge remaining.

Where you appear to be coming unstuck is in considering voltage alone. Voltage is only a comparable indicator of charge remaining in a cell when the cell is isolated and no current is flowing. When you place various cells in parallel there are loops between cells and current can flow in these circuits. The existence of these extra loops in the circuit changes things and the voltage can no longer be compared to the open circuit voltage.


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## 45/70 (Dec 20, 2009)

OK, Mr H. I guess I'm going to have to rethink this situation. 

I edited my previous post with a better explanation of what I meant by "lower". From what you say, though, I've misunderstood what actually happens to cells in parallel. :sigh:

Thanks,

Dave


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## Lynx_Arc (Dec 20, 2009)

45/70 said:


> True, until the voltages of the cells in parallel became equal (in a very short time). At this point, all cells will have equal voltage potential and thus, at least appear to the load, to all supply an equal amount of current.


the cells will equalize in a time frame determined by the internal resistances and power left in them and whatever load is also in the circuit


> The load sees the cells that are paralleled as one cell, not individual cells. The internal resistance seen by the load, would be the average IR of all the cells in parallel. I might not have this correct, but I think it is.


this is wrong, there is many things going on here because batteries are not just a voltage source that is fixed but a power source that supplies current based upon the capacity of the cells and their internal resistance and the load of the circuit acting upon them. Under lighter loads much of this is marginalized and has less effect but as the load increases a batteries internal resistance starts to figure in greatly and can allow one battery to drain more than the other. If you go look up threads on how much some nimh cells can provide currentwise you will find some nimh are capable of 10 amps output while others cannot supply that much without damaging themselves. If you were to think you could get 20 amps from two batteries in parallel but only one can do 10 amps and the other less then that could put 12 amps on the one that can do 10 and 8 amps on the other and the excessive load could drain it dry fast while the other has capacity left. If a nimh runs out of capacity (chemical reaction producing voltage) it is easier for it to be overdischarged under a load while one with some capacity left may not be damaged. Think of it as to having two men carrying something heavy close together and one man can carry 200 lbs and another man only one hundred. If they are carrying 200 lbs between them then it is split 100/each but if they are carrying 300 lbs between them you start having problems because one man can only carry 100 and he either has to resist carrying more or fail and if the resistance of the man that carries 200 is greater than the reistance of the man carrying 100 because he is stronger to begin with then the man carrying 100 will be stuck with more than he can manage and be overloaded a lot more tire a lot more and fail before the other fails.... even be damaged. This is an oversimplification of things and not really 1:1 accurate with the two cells but it show how problematic things can be about balancing. If the two men were carrying the 300 lb item up the stairs it would be ok if the stronger man was on the lower end but if it were the other way around and the weaker man was on the lower end carrying most of the load (due to internal resistance) then he would tire quickly and the other man would be unable to drag the load up the stairs from his weaker position (of higher internal resistance).


> This is partially correct. The high self discharge cells would suck the power out of the good cells (in a relatively short time). However, again due to the cells being in parallel, no individual cell in the pack could be drained any "dryer" than any other cell. They would all be at the same voltage potential.


the issue I was bringing up is if you have good cells that do not suffer from HSD (high self discharge) putting them in with cells that do causes them to be discharged completely when they would alone not do so.


> Again, the IR differences between the cells in parallel, would be averaged between the cells, and the load would see this as a single IR value.


you cannot average things, ohms law is not a law of averages and is the only correct formula. You must take into account the resistance of each cell (R), the power it can supply (IxE), and also each cell has a voltage that it maintains at the capacity it has left. In other words you are oversimplying something that can be rather complex in nature. Under very light loads the internal resistances of the cells do not figure into the equation as much and can be ignored more than as the load increases.


> I hope this isn't really off topic, it does relate to the OP. I'm pretty sure I'm correct here, but I certainly wouldn't mind being corrected if I'm wrong. I recently started a thread asking a somewhat similar question about charging Li-Ion cells in parallel, so I'm no expert. I'm just stating what I believe to be correct.


charging cells has the same problems but in reverse. The cell that has the least capacity CAN fill up faster, internal resistances will figure into which cell charges faster. when one cell is full, it will be overcharging which is a big problem with lithium ion cells. Typically if the current supplied to cells is lower the internal resistance of the cells charging factors less into things but at higher rates you could have problems just as with higher discharge rates of cells in parallel.
I am by no means an expert in this area, there are many that are a lot more qualified and have more research but my point is you cannot consider a battery as just a voltage source, but rather a resistive power force so to speak with the resistive part figuring in more and more under larger current conditions whether charging or discharging. If you check out some of Silverfox's and others battery reviews that have charts you will see the discharge curves of many batteries under all sorts of conditions and the curves differ all over the place for different batteries.


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## ltiu (Dec 20, 2009)

I'm the OP and I say to all of you: Keep it coming. This is good learning.


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## 45/70 (Dec 20, 2009)

Thanks for the lengthy explanation Lynx. :thumbsup:

First let me say that I've never charged or discharged cells in parallel that were different capacities, brands, chemistry, etc. and they have always been very close in voltage as well. I've just never thought about such a situation seriously, until reading this thread.

So, you're saying that if I discharge three cells in parallel, that are identical in every way, but two are fully charged and one is fully discharged, that the discharged cell will drop to near zero Volts? I am presently discharging an AAA NiMH to 0.8 Volts under load, from my recycle bin. I am going to fully charge two other identical AAA cells. Then I'm going to discharge the three cells in parallel on my CBA II, just to see what happens.

I never thought about it before, but is there any way to read individual cell voltages while the discharge is taking place? I suppose not, as the voltmeter will be in parallel as well, and read the 3 cell average, correct? My big mistake was figuring that the voltage of "the pack" was the voltage of the cells in parallel and that the more fully charged cells would bring up the voltage of the discharged cell, until they were more or less even (voltage wise, not capacity wise). I guess that's where my understanding went awry.

Thanks again Lynx,

Dave

*EDIT:*


ltiu said:


> I'm the OP and I say to all of you: Keep it coming. This is good learning.



That's good ltiu, I was beginning to worry.


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## SilverFox (Dec 20, 2009)

Hello Dave,

While you are "playing..." you may find this paper an interesting read.

Tom


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## ltiu (Dec 20, 2009)

SilverFox said:


> ... you may find this paper an interesting read.



Wow, now I have indigestion. Too much information.


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## Lynx_Arc (Dec 20, 2009)

45/70 said:


> Thanks for the lengthy explanation Lynx. :thumbsup:
> So, you're saying that if I discharge three cells in parallel, that are identical in every way, but two are fully charged and one is fully discharged, that the discharged cell will drop to near zero Volts? I am presently discharging an AAA NiMH to 0.8 Volts under load, from my recycle bin. I am going to fully charge two other identical AAA cells. Then I'm going to discharge the three cells in parallel on my CBA II, just to see what happens.
> I never thought about it before, but is there any way to read individual cell voltages while the discharge is taking place? I suppose not, as the voltmeter will be in parallel as well, and read the 3 cell average, correct? My big mistake was figuring that the voltage of "the pack" was the voltage of the cells in parallel and that the more fully charged cells would bring up the voltage of the discharged cell, until they were more or less even (voltage wise, not capacity wise). I guess that's where my understanding went awry.



the minute you connect cells in parallel they will try to equalize in voltage, and if given enough time the voltage will equal. If the cells are identical in every way except charge state they should be equal given time and no external loads or power introduced upon them. However if you put a load on the cells or start charging them the balancing will be affected by that and depending on the current drain on the cells those that are fully charged may not be able to charge the dead cell but instead would be trying to power a heavy load. Without specifics and calcuations there is no way to predict what would happen. If two cells had 2000mah left in them full charged hooked to a third dead cell then all three cells would share 4000mah minus whatever is lost in internal resistance given off in heat etc 


> Thanks again Lynx,
> 
> Dave


I sort of wish someone else with an EE degree would have explained it to you as my explanation is probably less accurate and crude but should give you an idea that more is going on here than realized. I would say at ratings of half an amp or less the effects of internal resistance would probably be of little problem with the cells balancing between themselves in both discharge and charging but as you increase to several amps then you could find any differences in cells coming into play as I would also suggest it is quite possible as cells discharge/charge the internal resistance in circuits is affected by the state of charge some... in other words internal resistance could actually vary to some extent over the range of charge which would partially explain why nimh cells on their own self discharge at differing rates to some extent down to certain voltages slowing down when they hit them.


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## 45/70 (Dec 20, 2009)

Thanks, Tom, Lynx, and especially, ltiu, for letting us highjack his thread. 

I did skim through "_The Paper'_ quickly. I was impressed with the general uncertainty presented. It seems to me that involving multiple strings in parallel and the enormous amounts of voltage and amperage that are involved, would tend to complicate things a bit at the individual cell level. Still, the principle is the same, I guess. I'll have to read it over again tomorrow.

I have an important meeting that has come up for tomorrow, that I am totally unprepared for, so while my AAA has discharged, my 3 cell experiment will have to wait a day or so. I still may check in to see if there are any more words of wisdom, however. 

Dave


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## Lynx_Arc (Dec 20, 2009)

I always learn stuff, even when I try and explain things to others I get to thinking and realize I understand more than I thought from reading all the stuff here. There are many here considerably more knowledgeable but I was just the first one to speak up this time. You can find plenty of threads talking about batteries and charging to keep you busy for a long time.


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## ltiu (Dec 21, 2009)

45/70 said:


> Thanks, Tom, Lynx, and especially, ltiu, for letting us highjack his thread.


 
Oh please do hijack. 

I like where this thread is going and it is really answering a lot of the questions in my head about batteries in-parallel.


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## 45/70 (Dec 22, 2009)

OK, I did get to run the experiment tonight.

These cells are from my recycle bag. They are La Crosse 700mAh NiMH AAA cells that came with my BC-900 in early 2005. The cells never were very impressive, and now have been in the recycle bag for months, at least. Their last duty was use in LCD thermometers etc. in which they worked relatively well, but were replaced with similar condition cells that are newer. They all hold their charge reasonably well, but suffer serious voltage depression, high IR, and lack of capacity. Probably not the best choice for an accurate scientific study, but that's what I used. 

All three cells out of the bag were at 1.20 Volts, or higher. Two of the cells were charged up on a BC-900 @ 700mA rate. The third cell was discharged on a CBA II several times, down to 0.8 Volts, the last discharge was at a 10mA rate.

Before the parallel discharge the cells OC voltages were,

# 1 1.38 Volt
# 2 1.38 Volt
# 3 1.22 Volt

The cells were then discharged in parallel on the CBA at a 900mA rate down to 0.9 Volts. I stopped the discharge a couple times and lost track of the total capacity, thinking it wasn't really important anyway, as these cells lost their capacity long ago. I'd guess they turned out in the neighborhood of 500mAh. Right after the discharge to 0.9 Volts, the cell readings were,

# 1 1.23 Volt
# 2 1.23 Volt
# 3 1.21 Volt

At least at an approximate discharge rate of 0.4C, there does not appear to be any damage done to the discharged cell, or the others. Of course, as I am learning, there is more going on here than I previously knew about.

What I should have done, is do a "Test" cycle on all three cells before doing this. Really though, I just wanted to see what happened with the discharge. I will run a "Test" cycle tomorrow to see if there is a significant difference between the cell that was discharged and the two cells that had been charged. As I remember, these cells were always relatively close in performance. Again though, this experiment lacks good scientific quality. More like a "spareaminute", than an experiment. 

I may try discharging two "dead" cells with one charged cell as well. We'll see.

Dave


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## Lynx_Arc (Dec 22, 2009)

45/70 said:


> OK, I did get to run the experiment tonight.
> 
> These cells are from my recycle bag. They are La Crosse 700mAh NiMH AAA cells that came with my BC-900 in early 2005. The cells never were very impressive, and now have been in the recycle bag for months, at least. Their last duty was use in LCD thermometers etc. in which they worked relatively well, but were replaced with similar condition cells that are newer. They all hold their charge reasonably well, but suffer serious voltage depression, high IR, and lack of capacity. Probably not the best choice for an accurate scientific study, but that's what I used.


 I have some of these... they were *ok* because they were free but if I could have gotten the charger cheaper without them I would have.


> All three cells out of the bag were at 1.20 Volts, or higher. Two of the cells were charged up on a BC-900 @ 700mA rate. The third cell was discharged on a CBA II several times, down to 0.8 Volts, the last discharge was at a 10mA rate.
> 
> Before the parallel discharge the cells OC voltages were,
> 
> ...


900ma over 3 cells is not much of a test that is averaging 300ma/cell which is a walk in the park.


> # 1 1.23 Volt
> # 2 1.23 Volt
> # 3 1.21 Volt
> 
> ...


a better test would be discharging on cell to 0.8v charging two cells to full and putting a 3+ amp load on them all till the voltage drops below 1v combined. Then quickly take off the leads and measure the cells starting with the dead cell to find the battery voltage. This most likely could damage one cell for good though.


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## Mr Happy (Dec 22, 2009)

There's a lot here to reply to in several posts above, so I won't try to pick and choose.

Firstly, if you take one discharged cell and one fully charged cell and put them in parallel, the voltages will be equal immediately. It won't take any time (well, a microsecond maybe...). This is because the point where the the positive (or negative) ends are connected is a single point. A single point has a single potential, by definition.

The way this is explained is that current will start flowing from the charged cell to the discharged cell, and this current will produce a potential difference through the internal resistances of the cells. The charged cell will experience a voltage decrease due to the current it is delivering, and the discharged cell will experience a voltage increase due to the charging current it is receiving. The current will adjust itself until these exactly balance each other.

Now if you attach an external load, some of the current will start flowing through the load instead of the discharged battery. This means the empty battery will be charging slower and the full battery will be discharging faster.

Depending on the size of the load (i.e. its resistance) the proportion of current that flows each way will vary.

One thing to avoid is the idea of "average voltages". There is no theory of voltage averaging, only Ohm's law. If you know the potentials and the resistances you can work out the currents, but sometimes this might involve solving simultaneous equations...


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## 45/70 (Dec 22, 2009)

Lynx_Arc said:


> 900ma over 3 cells is not much of a test that is averaging 300ma/cell which is a walk in the park......
> 
> ......a better test would be discharging on cell to 0.8v charging two cells to full and putting a 3+ amp load on them all till the voltage drops below 1v combined.



I agree Lynx, but these particular cells would fall flat on their faces at anything near that current level due to their poor condition. Also, for me anyway, a 3A load on a AAA cell is more than I've ever encountered, close to 2A, yes, but never 3A. I've always used standard NiMH's or Sanyo eneloops in good condition for this type of load. At this point in the recession, I'm unwilling to break up any of my sets of good cells, by potentially destroying part of the set, just for a better understanding of science. 



Mr Happy said:


> One thing to avoid is the idea of "average voltages". There is no theory of voltage averaging, only Ohm's law. If you know the potentials and the resistances you can work out the currents, but sometimes this might involve solving simultaneous equations...



I'm OK on your post up to this point Mr H, and have been since 4th grade (1963-1964) when I was introduced to basic electricity. I have no real electrical/electronic background, except as a hobby. My college years were focused more on physics, chemistry and mathematics, but not much specifically involving electricity or electronics.

I think I'm beginning to understand the problem with "average voltages". This and the "loops between cells" are a new area to me, something I never considered. My original question remains, however. Can a cell be damaged from undervolting (or over discharging) in a situation where it is parallel discharged, in an already discharged state, with two other cells that are fully charged? This is the reason for my experiment. I can understand, I think, how damage may occur if the load is high enough to potentially cause damage to a cell, even if it were discharged singly, at a phenomenal rate (4+C as Lynx suggested *EDIT: It would actually be a 1.4C load, which would be appropriate, sorry Lynx. An actual 4C could potentially be though*), but under normal (eg. 1C) rates, can damage actually occur to the previously discharged cell, in a parallel situation? My summation from the ideas you guy's have provided is, yes it can, but I really haven't seen a direct answer presented.

As I said before, I have never run into this situation in the real world. I never intend to either, but I do find it interesting, and you never know. 

Thanks for all the input. :thumbsup:

Dave


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## ltiu (Dec 22, 2009)

Mr Happy said:


> The way this is explained is that current will start flowing from the charged cell to the discharged cell, and this current will produce a potential difference through the internal resistances of the cells. The charged cell will experience a voltage decrease due to the current it is delivering, and the discharged cell will experience a voltage increase due to the charging current it is receiving. The current will adjust itself until these exactly balance each other.


 
Is this a way to charge a discharged NiMH by putting it in-parallel to another set of "fully charged" batteries?


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## Mr Happy (Dec 22, 2009)

45/70 said:


> Can a cell be damaged from undervolting (or over discharging) in a situation where it is parallel discharged, in an already discharged state, with two other cells that are fully charged?


The short answer is yes, maybe, depending on the circumstances. If the external load was heavy enough, it might pull the voltage of the charged cells down so much that the already empty cell was discharged below the safe level. But this load would have to be something like a short circuit, and the charged cells would not appreciate that kind of load very much either...


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## 45/70 (Dec 22, 2009)

ltiu said:


> Is this a way to charge a discharged NiMH by putting it in-parallel to another set of "fully charged" batteries?



Not really ltiu. It is possible to do this to get a NiMH cell to work in a charger that won't accept the cell because the voltage is too low, however. Just parallel it with a charged cell for a few seconds and most often, it will then charge in the charger.



Mr Happy said:


> The short answer is yes, maybe, depending on the circumstances. If the external load was heavy enough, it might pull the voltage of the charged cells down so much that the already empty cell was discharged below the safe level. But this load would have to be something like a short circuit, and the charged cells would not appreciate that kind of load very much either...



Thanks Mr H. That's what I wanted to know, and pretty much what I figured all along. As I said, I was thinking of more "normal" applied loads as opposed to something approaching a short circuit.

Thanks for all the info and opinions. The learning continues.  I may still run a few more experiments, however the really poor quality of the cells I'm using makes the results somewhat questionable.

Dave


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## SilverFox (Dec 22, 2009)

Hello Dave,

Keep in mind that you parallel cells in order to handle higher loads. A "normal" load becomes quite high if one of the cells is weak. I think the damage comes from the healthy cells being subjected to higher loads as they try to compensate for the weak cell.

For example if you have 2 Ah cells and a 4 A load, you could parallel two cells and end up with a 1C discharge rate. However, if one of the cells became weak, the other cell would have to handle a load higher than 1C. This can lead to increased heat which precedes damage to the cell.

Another thing to keep in mind is that although voltages of NiMh and NiCd cells in parallel immediately equalize, their state of charge does not. If you take a full cell and parallel it with a discharged cell for 24 hours, you will end up with only a little over 10% charge transfer.

The problem during discharge is that if you have one cell with 2 Ah capacity and another cell with only 1 Ah of capacity paralleled together supplying 4 A of load for an estimated 1 hour of run time, your load on the good cell will exceed 1C, and the total run time will be less than the target of 1 hour. In this example you would have 3 Ah of capacity and a run time of 3/4 of an hour, or slightly less depending on how damaged the under charged cell is.

As you add more cells to the battery, the problem becomes more complex.

Tom


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## 45/70 (Dec 22, 2009)

SilverFox said:


> Keep in mind that you parallel cells in order to handle higher loads.



Not necessarily. In my own experience, the few times I have paralleled cells or batteries, has been to increase runtime. I'm talking, for example, HAM transceivers for "field day" operations. This usually involved gel cells, SLA's, or deep cycle marine batteries, but the idea was to increase "on air" time without power interruption. I remember assembling a rather large NiCd pack that was a parallel arraignment for running handhelds, as well. Again, the goal was to not have to change battery packs during communications, not to get higher load capability.

Here on CPF, the hotwire folks are probably the ones that use paralleled cells to achieve higher load capability the most. My own preference here, with my hotwires, smaller is better, runtime be damned, so I haven't gotten into that aspect with regard to flashlights.

Thanks for the explanation of the possible problems with paralleling cells. I was pretty much aware of that already. As I said in previous posts, my concern was more along the lines of lower discharge rates. What I'm using paralleling of cells for nowadays, is simply charging Li-Ion cells with a hobby charger and, since I have a 4 bay parallel Charge/discharge setup, I can also discharge all chemistries of cylindrical cells on the CBA II, much slower than with the hobby charger, or conventional battery charger/analyzer such as the Maha C-9000's I have.

And, one more time, I have and probably always will, only charge/discharge like cells, at a similar voltage/state of charge in parallel. But, you never know, and this thread discussion has opened up some areas where I have gained some knowledge that I didn't possess before.

As for my question "Can a cell be damaged from undervolting (or over discharging) in a situation where it is parallel discharged, in an already discharged state, with two other cells that are fully charged?". The scientific answer is "Yes". The field operations, or applied science answer is "Yes, but probably not, unless high discharge current is involved". Good enough for me. 

Dave


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## ltiu (Dec 23, 2009)

(Updated/Edited)

To Summarize, parallel batteries possible issues:

1) Possible deep discharging of the disproportionately "discharged" bank.

2) Possible overheating of the disproportionately "charged" bank.


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## 45/70 (Dec 23, 2009)

Actually litu, I think the danger of overheating the _charged_ cells sounds like the bigger problem. To over discharge, you'd have to run the whole "pack" too low. I suppose if you put a really huge current load to the pack, you could get a  situation.



ltiu said:


> How about the possibility of frying the electronics you have the batteries connected to ?



You couldn't really damage your electronics with a parallel setup, unless of course, the voltage of the pack was too high to start with. Unless your one of those folks that thinks a 3000mAh cell will make their light run brighter than a 1000mAh cell, you should know better.  The larger capacity cell will only allow the light to run longer, not brighter, as the voltage is the same with either cell.

Dave


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## HartonoSagala (Mar 11, 2015)

45/70 said:


> Actually litu, I think the danger of overheating the _charged_ cells sounds like the bigger problem. To over discharge, you'd have to run the whole "pack" too low. I suppose if you put a really huge current load to the pack, you could get a  situation.
> 
> 
> 
> ...



Hi Guys... is this conversation still on? I am looking for some information... currently i am doing some experiment with 3 batteries 18A 12V paralel with 18A 12V paralel with 60A 12Vdc.. all of them charged with solar panel. And my temporary conclusion the performance of paralelling 2 batteries of 18A is better compared with 3 batteries above in patalel.


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## alpg88 (Mar 11, 2015)

Mugrunty said:


> I think people on the forum have discussed this before, but charging Nickel based batteries in parallel doesn't work to well (or safely). Specifically when you try to charge the pack quickly.


actually it does work well and pretty safe, done it many times, the problem exists only with smart chargers that need to sence deltaV to terminate charge, with parallel cells it becomes problematic, with simple dumb charger no problems, as long as you calculate time correctly, and shut off in time.


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## alpg88 (Mar 11, 2015)

for emergencies anything goes, hooking different brand cells in parallel, I've done it many times, I have not experienced any problems.


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