Electronic – Short-circuit xPyS LiFePO4 batteries

batterieslifepo4short-circuit

I have a LiFePO4 battery connected in 2P4S configuration. What happen if one of the batteries is short-circuited? The other batteries will also be short circuits, or this will be result in the venting of one battery?

update:

Thank you for the answer, but the batteries are connected in this way:

enter image description here

What happen to the other cells conneceted in series (BAT6, BAT7, BAT8) if we have an internal short circuit of the BAT5 ?

Best Answer

EDIT1: New answer to modified question (original see below)

First of all I need to tell you that this set-up is very risky. Very very very risky. This way of connecting the batteries means there's a much higher risk of damage over multiple charge/discharge cycles.

To your specific problem, if one battery in one chain becomes short circuited, the other three are "requested" to charge up to a 4cell voltage by whatever charges them, seriously over stressing them.

Put simply, if any battery becomes shorted in use, you can bet on it the other ones in the same chain are completely wasted as well, not to mention the severely increased risk of one cell blowing up or becoming shorted in the first place.

This is why battery packs are always built as in my original answer below by the factories that make them. Now, before you go and re-build a pack like that, let me warn you to always balance the cells to each other before you put them parallel, with a set up like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Because cells are never exactly charged the same, ever, so you let them charge each other through a resistor, making sure no idiotic current flow. How long you leave them like this before connecting them depends on whether they have been fully charged with the same charger. If they have, 10 to 15 minutes should be more than enough. If they have not, leave them for at least an hour, but probably just half a day or more. You can reduce the resistor to increase the current and reduce the time, but be sure the resistor can handle the power, usually 10 Ohm is safe enough with a 1/4W resistor, lower values you may have to go to 1/2W, 1W or even 3W. If you want to make sure they stay alive as long as possible, don't go over 1/3 their capacity for the current through the resistor.

Also, if you want to prevent all damage as good as possible, look into balance-chargers and how they work, as they try to keep all cells in your pack healthy during the charging, severely reducing the risks of damage over time.


ORIGINAL ANSWER:


If your setup is like this:

schematic

simulate this circuit

And one battery, say BAT6, becomes a short circuit you will get:

schematic

simulate this circuit

So BAT5 is then shorted between it's + and - terminals. Often when one cell of a parallel set-up goes bad the other one is very damaged as well, because the unbalance that causes premature cell-death affects both the cells, just the slightly weaker one will go first. BUT even if that wasn't the case, shorting BAT5 will cause it to get severely damaged soon enough.

So they only thing you should do to fix this is remove both BAT6 and BAT5 and create a lower voltage pack like this:

schematic

simulate this circuit

Or get a pack with half the capacity and two cells left over, like this:

schematic

simulate this circuit

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