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Reviving a dead battery pack

badreg
Go to solution Solved by akio123008,
On 4/6/2021 at 9:21 PM, badreg said:

the product was sitting on the shelf for too long

If a battery like that has discharged that much, there's more going on than it sitting on a shelf long.

 

On 4/6/2021 at 9:21 PM, badreg said:

Is the voltage too low to be revived?

Yes, I think so. I once tried to revive a 2.0V LiPo cell and it didn't work out, the damage was already done. In your case, you're at 0.43V, good luck with that. You can give it a shot, but don't expect much from it.

 

On 4/6/2021 at 9:21 PM, badreg said:

ould I be able to parallel charge with a working pack without disassembling the pack?

In theory you could but you need to keep in mind three things:

- do this outside

- use a power resistor in between the batteries to limit the current! Do not connect them in parallel directly.1

- you can't be sure about the balancing of the cells, so unless the pack has a balance lead, or you are able to open it up, it's guesswork what the cell voltages are.

 

Li-ion cells have a very low internal resistance. Connecting a lower voltage cell to a higher voltage cell in parallel would therefore cause huge amounts of current. For instance, if I try to charge a 3.5V cell using a 3.6V cell, I get an effective voltage of 3.6-3.5=0.1V. That may not seem much, but if the cell has a 0.001 ohm resistance (which is not unrealistic) you'd get 0.1/0.001 = 100A of current (!). You're at 0.43V, so you'd have a voltage of say 3.7-0.4=3.3V (across some tiny resistance). So you'd be generating an insane current through an already damaged cell. A recipe for disaster. This is why you need to add a resistor to limit the current (or use a proper charger, but I assume proper chargers refuse to charge that pack haha)

I received 14.4V Li-ion battery pack that seems to be DOA. The retailer is sending me a replacement and said that I don't need to return the dead battery. It seems that the product was sitting on the shelf for too long, and the voltage dropped too low. I have not disassembled the pack, but the size and shape suggests that it contains four 18650 cells. The pack reads 1.7V, which is around 0.43V per cell.

 

Is the voltage too low to be revived? If the pack is salvageable, would I be able to parallel charge with a working pack without disassembling the pack?

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I am not an electronics or electrician, but I think if you can attempt a force charge (inserting the correct voltage and amperage) and you can revive it but it will not be as new.

 

Typically from my limited knowledge when something sits for a long time one or more cells deplete to 0 and thus become useless.

 

Opening the battery pack and checking, replacing and testing would be your best attempt at cheaply repairing.

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There are videos out there.. but I don't know man.. maybe do it outside 😄

 

If you think you can rebuild the pack with fresh cells I would probably go that route..

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On 4/6/2021 at 9:21 PM, badreg said:

the product was sitting on the shelf for too long

If a battery like that has discharged that much, there's more going on than it sitting on a shelf long.

 

On 4/6/2021 at 9:21 PM, badreg said:

Is the voltage too low to be revived?

Yes, I think so. I once tried to revive a 2.0V LiPo cell and it didn't work out, the damage was already done. In your case, you're at 0.43V, good luck with that. You can give it a shot, but don't expect much from it.

 

On 4/6/2021 at 9:21 PM, badreg said:

ould I be able to parallel charge with a working pack without disassembling the pack?

In theory you could but you need to keep in mind three things:

- do this outside

- use a power resistor in between the batteries to limit the current! Do not connect them in parallel directly.1

- you can't be sure about the balancing of the cells, so unless the pack has a balance lead, or you are able to open it up, it's guesswork what the cell voltages are.

 

Li-ion cells have a very low internal resistance. Connecting a lower voltage cell to a higher voltage cell in parallel would therefore cause huge amounts of current. For instance, if I try to charge a 3.5V cell using a 3.6V cell, I get an effective voltage of 3.6-3.5=0.1V. That may not seem much, but if the cell has a 0.001 ohm resistance (which is not unrealistic) you'd get 0.1/0.001 = 100A of current (!). You're at 0.43V, so you'd have a voltage of say 3.7-0.4=3.3V (across some tiny resistance). So you'd be generating an insane current through an already damaged cell. A recipe for disaster. This is why you need to add a resistor to limit the current (or use a proper charger, but I assume proper chargers refuse to charge that pack haha)

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4 hours ago, akio123008 said:

 

Thank you so much for the helpful information! I figured that it was fully dead, but wanted to avoid throwing away batteries unnecessarily. But I guess it's not really worth the effort to try and salvage this one.

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