Quick question for any battery expert

[Monkman]

Hello All,

I just signed up here because it seems that there are a lot of very knowledgable people here who can answer a battery question. I had a friend build me a 52v 14s 9p Lithium battery with Samsung 35e cells. It was running great until the BMS decided not to balance the number 9 pack. So as I read through on here I purchase a Neptune 15 which I installed in place of the cheap one I had. I got it all wired up and it is working as it should. I then proceeded to solder the positive wire to the positive of the battery pack but got burned by the iron which dropped, melted to the side of one of the cells which caused it to start leaking. I quickly removed the cell from the bank to prevent any further damage. I ordered a replacement cell but finally, my question is:

What would happen if I ran the battery with one missing cell? Effectively it would be a14s 9p with one of the packs as an 8p.

Just something Id like to know, sorry for the long story lol.

Thanks, Rob

 

[Sunder]

As long as the BMS is still working, you'd just lose 11% of the capacity.

When in use, the 8P string will empty first, then the BMS will cut off discharge. When recharging, it should still charge up reasonably equally, but may spend more time balancing.

If the BMS isn't working properly, a fireball is a possibility. I'd definitely watch it with a multimeter on your first few recharges until you get it fixed.

 

[Monkman]

Man,

That is exactly the most informative answer I needed. I really appreciate you taking the time to pass on some knowledge.

Thanks Robert

 

[tony rodriguez]

As long as the BMS is still working, you'd just lose 11% of the capacity.

When in use, the 8P string will empty first, then the BMS will cut off discharge. When recharging, it should still charge up reasonably equally, but may spend more time balancing.

If the BMS isn't working properly, a fireball is a possibility. I'd definitely watch it with a multimeter on your first few recharges until you get it fixed.

im new here and im just looking around to see the builds and knowledge . i was amazed on the projects and experience you have. if you dont mind me picking your brain and asking questions? ill be posting info details and pix soon. thanks again.

 

[Hwy89]

While Sunder is technically correct that if the BMS is functional the battery could still be used but with a lower capacity, the amount of time required by most BMS boards to correct the imbalance would probably make it impractical. To make matters worse, unless you run sense wires to an external connector you won’t be able to check if the battery is being balanced at all.
My opinion is that there are two options: 1) replace the damaged cell, or 2) remove one cell from each of the other groups.

 

[Monkman]

I'm going to replace the damaged cell this week. I do have a question though, all the cells in the pack are around 4.13 - 4.15 volts. If I add a new cell back in since it's new it should be 4.2v. Do I need to charge the pack to 4.2 or is that a no no since I have a missing cell?

Thanks Robert

 

[Monkman]

While Sunder is technically correct that if the BMS is functional the battery could still be used but with a lower capacity, the amount of time required by most BMS boards to correct the imbalance would probably make it impractical. To make matters worse, unless you run sense wires to an external connector you won’t be able to check if the battery is being balanced at all.
My opinion is that there are two options: 1) replace the damaged cell, or 2) remove one cell from each of the other groups.

I'm going to replace the damaged cell this week. I do have a question though, all the cells in the pack are around 4.13 - 4.15 volts. If I add a new cell back in since it's new it should be 4.2v. Do I need to charge the pack to 4.2 or is that a no no since I have a missing cell?

Thanks Robert

 

[Hwy89]

The replacement cell will most likely be shipped at around 3.6 volts. You will need to rig up a single cell charger and bring it up to match the rest of the cells in its group. Try to have them within .1v of each other or you may see sparks when you make the connection.
An old cell phone charger or similar “wall wart” type charger can be used as a single cell charger as long as it has an output of 4.2 volts. Be sure to pay attention to the polarity when you attach it and monitor the process closely. If you happen to overshoot the target voltage just discharge it with an automotive bulb ( filament type not LED).

 

[Hillhater]

Much safer to discharge both the new cell and the damaged pack to a minimum voltage (<3 v) before you do the repair.
Minimum energy in the components incase of another “accident” !

 

[999zip999]

Watch watch when charging the cell to match and never overcharge so watch watch and make sure if you go to the bathroom unplug it.

 

[Sunder]

While Sunder is technically correct that if the BMS is functional the battery could still be used but with a lower capacity, the amount of time required by most BMS boards to correct the imbalance would probably make it impractical.

I'm wondering if you think it would be very slow, because you're assuming something like you have to burn off 11% of the battery pack, for 13 of the 14 strings. I.e. something like 3400mah, on a BMS that can balance using 100ma i.e. at least 34 hours to Balance.

In reality, it wouldn't be anywhere near as bad as you think, if you start off top balanced. Remember, the BMS doesn't care about the capacity remaining - It only cares about the voltage it can see.

Say each one of those 1Ah for simplicity. You start off with 14 cells being:

9 9 9 9 9 9 9 9 8 9 9 9 9 9 - However, from a voltage (hence BMS point of view) it will be 4.2 for all cells, and thus balanced

If you run it, and the BMS is doing it's job, when it cuts off from cell low voltage, it will be:

1 1 1 1 1 1 1 1 0 1 1 1 1 1 - From a BMS point of view say, 3.6, 3.6 ... 3.2, 3.6 etc.

When you recharge, and are putting 8Ah back in, when it's done charging but before balancing, it should be back to:

9 9 9 9 9 9 9 9 8 9 9 9 9 9 and thus 4.2 for all cells.

However, as the 9P series will have a lower internal resistance, it will receive ever slightly more current than the 8P series, so will hit 4.2v per cell before the 8P series. The BMS will then burn off energy from these 13S of cells, and charge again. Assuming all the cells were 10mOhm, then the difference between a 9P and 8P string is 1.1mOhm vs 1.25mOhm. That's well within the normal variation of even a high quality cell.

 

[Monkman]

The replacement cell will most likely be shipped at around 3.6 volts. You will need to rig up a single cell charger and bring it up to match the rest of the cells in its group. Try to have them within .1v of each other or you may see sparks when you make the connection.
An old cell phone charger or similar “wall wart” type charger can be used as a single cell charger as long as it has an output of 4.2 volts. Be sure to pay attention to the polarity when you attach it and monitor the process closely. If you happen to overshoot the target voltage just discharge it with an automotive bulb ( filament type not LED).

Awesome, I will do just that, got a 18650 double cell charger off of Amazon.

Thanks for the assistance,

Rob

 

[Monkman]

While Sunder is technically correct that if the BMS is functional the battery could still be used but with a lower capacity, the amount of time required by most BMS boards to correct the imbalance would probably make it impractical.

I'm wondering if you think it would be very slow, because you're assuming something like you have to burn off 11% of the battery pack, for 13 of the 14 strings. I.e. something like 3400mah, on a BMS that can balance using 100ma i.e. at least 34 hours to Balance.

In reality, it wouldn't be anywhere near as bad as you think, if you start off top balanced. Remember, the BMS doesn't care about the capacity remaining - It only cares about the voltage it can see.

Say each one of those 1Ah for simplicity. You start off with 14 cells being:

9 9 9 9 9 9 9 9 8 9 9 9 9 9 - However, from a voltage (hence BMS point of view) it will be 4.2 for all cells, and thus balanced

If you run it, and the BMS is doing it's job, when it cuts off from cell low voltage, it will be:

1 1 1 1 1 1 1 1 0 1 1 1 1 1 - From a BMS point of view say, 3.6, 3.6 ... 3.2, 3.6 etc.

When you recharge, and are putting 8Ah back in, when it's done charging but before balancing, it should be back to:

9 9 9 9 9 9 9 9 8 9 9 9 9 9 and thus 4.2 for all cells.

However, as the 9P series will have a lower internal resistance, it will receive ever slightly more current than the 8P series, so will hit 4.2v per cell before the 8P series. The BMS will then burn off energy from these 13S of cells, and charge again. Assuming all the cells were 10mOhm, then the difference between a 9P and 8P string is 1.1mOhm vs 1.25mOhm. That's well within the normal variation of even a high quality cell.

This was very informative, thanks for the explanation.

Rob