Electrical – Considerations for mismatched 18650’s in series

battery-charginglithium ion

I'm building a large battery pack. I'm in the planning stage, collecting parts and such, but my goal is to have something like a 4s9p or 4s10p arrangement. The cells I have vary though. Not by a ton, but a hundred or so mAh here and there. All of the parallel sets would be identical cells that were all originally used together, but each set in series would have slightly different capacities. What considerations do I need to keep in mind with this sort of setup? I know it's not advised to use mismatched cells, but I might not have a lot of choice in the matter. I know I'll at least have to have good balancing, anything else to keep in mind?

Best Answer

All of the parallel sets would be identical cells that were all originally used together, but each set in series would have slightly different capacities.

You want to do exactly the opposite. A 1 Ah cell in parallel with a 2 Ah cell behaves as a single 3 Ah cell, but a 1 Ah cell in series with a 2 Ah cell yields a 2S battery that can only be discharged 1 Ah, wasting the mismatching capacity of the better cell. If you discharge the hypothetical battery further in order to benefit from the capacity of the better 2 Ah cell, you will overdischarge and ruin the 1Ah cell.

If you insist on reusing salvaged batteries (I wouldn't), you want to combine the mismatching cells into parallel sets of equal capacity, and then place those in series. You also must add overcharge protection, undercharge protection and balancing to your battery in some form or another.

As pointed out by Robherc, if the internal resistances of parallel cells are vastly different (and not in proportion of their differing capacities), you can exceed the maximum charge and discharge rates of individual cells in a parallel set. For example:

You have a battery of two parallel 1 Ah cells rated for 3 A discharge, but one cell has an internal resistance of 0.1 Ω and the other has degraded to 1 Ω. The battery is fully charged to 4.2 V, and you begin a 4 A discharge. The battery voltage will immediately fall to† 3.84 V due to the internal resistance, but the current is not shared equally. Initially the 0.1 Ω cell will be discharging at 3.64 A, exceeding the rating, while the 1 Ω cell will only provide 0.36 A. However as the 0.1 Ω discharges much faster, the 1 Ω cell will eventually catch up during the discharge and start to share the load (assuming that it doesn't combust first). Also, the 0.1 Ω cell will wear much faster with subsequent charge-discharge cycles, slowly equalizing itself with the more degraded cell in the pack.

While this is a rather extreme example, you should still throw out cells that differ greatly in the product of cell capacity and internal resistance, and de-rate the charge and discharge rates from the theoretical values accordingly.


† Internal resistance of the battery as a whole:
1/(1/0.1 Ω + 1/1 Ω) ≈ 0.091 Ω
Voltage loss over the internal resistance:
4 A * 0.091 Ω = 0.364 V
Remaining (true) battery voltage:
4.2 V - 0.364 V = 3.836 V ≈ 3.84 V