uneven discharge in 2-cell 123A - questions

beezaur

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Can someone explain this for an electronics retard?

I have a Surefire A2 and also an L1. The A2 is done with its 2 batteries when they are too low to power the incan bulb, but there is still a lot of energy left for running my 1-cell L1 LED. So, new batts (Surefire or Duracell, not mixed) go into the A2, then into a pile for use in my L1 when they fail to light the A2 incan.

I test all my batteries by measuring their short-circuit current before use. New batts might do ~10 A each. But when they come out of the A2, one batt will still do ~6-7 A while the other will only do ~1 A.

I have read that this happens, but I don't know the details. Could someone please enlighten me about what is going on?

Also, I could swear that there was not such a disparity when I first took them out of the A2. Is there a reason one would "fade" just sitting on my desk?

Is there a danger in drawing 2 batts way down in a light that can squeeze more out of them, like an L2?

Scott
 

KevinL

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Batteries do discharge unevenly, I've documented this phenomenon with every 2-cell light I've used, and quite a number of times I've pulled the cells hot off the light. Don't worry about it.

I don't anticipate any more danger than if you had continued to use those two same cells in the A2, but the L2 will not run at full power. Eventually one cell will become exhausted and drag down the output of the system. I have seen extreme cases where I've run a light till it completely went out, one cell is 0 flash amps, but the other has 4+ flash amps left, and is still useable.

You may want to discard the 1A cells and use the 6-7A cells in your L1. May not be worth the trouble to salvage the 1As.
 

Skibane

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Strange...

Since both batteries are in series, they will be supplying the same amount of current. If they both started out in identical states of charge, they should end up in identical states. The fact that they aren't says something.

Perhaps the incan's high current amplifies any small differences in their initial state of charge.
 

paulr

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But it's always the front one that craps out first, I thought. I figured it was because that one was closer to the bulb and got hotter.
 

beezaur

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"You may want to discard the 1A cells and use the 6-7A cells in your L1."

Thanks, KevinL. That's pretty much what I do. In fact, the "deader" cell often won't even light up the L1's high setting.

I'll have to keep better track of which one drains most.

I don't quite understand the business of batteries discharging in series. If you have one battery, electrons come out of one end, and come in the other, but none go *through* all the way; they jump from, and return to, the "magic goo" inside.

It seems to me that, when 2 are in series, one battery might be more electrochemically "active," giving and receiving more electrons without simply having them pass through. So then you have one battery that is doing more chemistry, and one that is being more of a spectator. The spectator would have more electrons passing through.

If that is true, It would seem like each battery would have a different voltage differential across its terminals, thus exposing each battery to a different load.

I dunno. I am doing well to remember how to use the words /ubbthreads/images/graemlins/wink.gif

And, is it "cell," or "battery?"

Scott
 

Skibane

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[ QUOTE ]
beezaur said:
It would seem like each battery would have a different voltage differential across its terminals, thus exposing each battery to a different load.

[/ QUOTE ]

Yep, since the current supplied by both batteries is the same, the only way they can differ is in their voltages. If one of the batteries has a higher voltage, it will supply more power to the load than the other (since power = voltage x current).
 

eluminator

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In my experience with NiMH cells, there will always be a small difference in capacity between even well matched cells. If you discharge them in series long enough, the difference will become magnified.

Most of the time matched NiMH cells will have voltages that are identical to the closest hundredth of a volt. But when the strongest is discharged to 1.0 volts, the weaker will have a voltage of around 0.0 volts, or even be negative. This damages the cell.

I suspect the same thing happens to all cells regardless of their chemistry. The voltage readings may be different, but their ability to produce useable electrical energy will follow this pattern.
 

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