### Battery Management Systems for Large Lithium-Ion Battery Packs

#### Section 6.1.1.3.1 Addenda

#### Cell self-balance time

The time required to balance a set of Li-Ion cells by connecting them in parallel (before building a pack), without a power supply, depends on these factors, in order of importance:

- Average SOC level:
- At either end of the charge curve, the voltage vs SOC curve is much steeper than at mid SOC levels, which means that the higher voltage of a more charged cell encourages more of its charge to go to lower voltage cells
- On the other side, at either end of the charge curve the internal cell resistance is higher, reducing the resulting current

Chemistry: Cobalt and standard LiPo are faster than LiFePO4, because their charge curve is steeper

- Amount of max initial imbalance: less imbalance (with respect to the average) is faster; fewer imbalanced cells is faster
- Resistance in the interconnecting wires, though this one can easily be insignificant compared to the cell resistance, and can safely be ignored
- For example, 12 AWG (3 mm^2) wire can handle 40 A continuous, but in this case the current is a pulse, so I'd trust it at 100 A peak;
- Assuming 2" (5 cm) between cells, a 12 AWG wire has a resistance of 0.2 mOhm, which is on the same order of magnitude than the internal resistance of the typical 100 Ah cell
- So, typically, the resistance of wire that is larger than 12 AWG can be ignored

Depending on the above factors, the self-balance time in on the order of hours to days.

The best case is for LiFePo4 cells, when most cells are full, and just a few are just below 100 % SOC. In that case:

- The voltage difference between the full cells and the not so full ones is relatively large (on the order of 200 mV)
- The resistance of the full cells is higher, but there are lots of them in parallel, so the total resistance is low
- The resistance of the not so full cells is low, because they are not yet in the high resistance area at the end of the charge curve
- The delta in SOC is low, meaning that not much charge needs to be transferred.

Let's look at an example of that:

100 LiFePo4 cells

- 100 Ah, all equal
- 10 mOhm at mid SOC levels; 100 mOhm at 100% SOC
- 99 cells at 100 # SOC, 3.6 V; 1 cell at 90 % SOC, 3.4 V
- The long term RC time constant of the cells is 30 minutes
- We can accept an imbalance of 0.5 %

Then:

- A charge of 10 Ah needs to be transferred into the low cell
- The source looks like 9900 Ah, 3.6 V, 0.1 mOhm
- The load looks like 100 Ah, 3.4 V going up to 3.6 V, 10 mOhm going up to 100 mOhm
- The RC time constant for both load and source is 30 minutes

So:

- The initial current is (3.6 - 3.4) / 10 mOhm = 20A
- At that rate, that would fill the low cell in 1/5 hour = 12 minutes
- In reality, as the low cell gets full:
- Its resistance increases, and does so non-linearly
- The delta in voltage decreases, and does so non-linearly
- The RC equivalent circuit in the cell enters into the equation, as its time constant is on the same order of magnitude as the 12 minutes (above)

- So, the curve drops down worse than plain exponentially

So, what is the time, when you consider all these factors? I used a spreadsheet to calculate this, and the result was:

- 72 minutes.

What about the worse case?

- Mid SOC levels (on the order of 2 mV delta in the OCV)
- The resistance is low
- Some cells have been exposed to significant current within the last few minutes, and their voltage has not yet relaxed
- To get a balance within 0.5 %, the OCVs must be matched to within 500 uV
- The OCV is masked by various thermodynamic effects, so the cell voltages must be allowed to relax

Let's look at an example of that:

100 LiFePo4 cells

- 100 Ah, all equal
- 10 mOhm at mid SOC levels; 100 mOhm at 100% SOC
- Bell curve distribution in capacity, from 30 to 70 % SOC
- OCV is within 1 mV of 3.2 V, but terminal voltage is within 100 mV of that
- The long term RC time constant of the cells is 30 minutes
- We can accept a imbalance of 0.5 %

To analyze this case, we need to employ a Monte Carlo statistical analysis. I have not yet done so. Once I do, I will add those results here.

For now, let me say that my guess is that it will take a week to two weeks to balance these cells.

Compare that to how long a BMS would take to balance those cells once they are in a pack will all the cells in series (assuming 100 mA balance current:

- Time = delta charge / current = 50 Ah / 100 mA = 500 hours = 20 days