When matching li-ion cells in a battery pack how do you use both the cell's resistance AND capacity?

I've seen sources mentioning that each parallel group should have about the same capacity, and that cell internal resistances should be "close".

How do we match based on both criteria?

Say I want to build a 3S2P pack with the following li-ion cells (made up values):

  • A: 2800mAh 50mΩ
  • B: 2950mAh 80mΩ
  • C: 2950mAh 100mΩ
  • D: 3000mAh 90mΩ
  • E: 3050mAh 70mΩ
  • F: 3150mAh 20mΩ

Method 1, capacity matching:

Combination Capacity(mAh) Equivalent resistance (mΩ)
a+f 5950 14
b+e 6000 37
c+d 5950 47

Method 2, resistance matching:

Combination Capacity(mAh) Equivalent resistance (mΩ)
a+d 5800 32
b+e 6000 37
c+f 6100 17

What is the correct way to balance a pack considering both cell values, and how would you match the example cells?

  • 2
    \$\begingroup\$ I do not understand why you would even want to assemble a battery pack with cells that aren't very similar. Battery packs for cars, laptops, E-bikes etc. are all assembled from batteries that are very similar, preferably from the same batch. Then the voltages, capacities and series resistances should match well enough to assemble a battery pack without issues. Your question appears to suggest the combination of "random" cells which is never a good idea as you have no guarantee that the cells will share the work equally (which is what you want). \$\endgroup\$ Commented Apr 13, 2021 at 14:17
  • 1
    \$\begingroup\$ even in a very good situation, your whole battery back would only perform slightly better than an array of the "worst" cell in it... don't do this. \$\endgroup\$ Commented Apr 13, 2021 at 14:21
  • \$\begingroup\$ I'm using cells from a similar batch. But they are refurbished. They have been discarded based on sketchy spot weld so the specs are still close to sheet and are never used in production. The capacity is pretty close, but the internal resistances differ a tad. I'm wondering what's the optimal way to combine the cells. \$\endgroup\$
    – rrmoelker
    Commented Apr 13, 2021 at 15:28
  • 1
    \$\begingroup\$ "even in a very good situation, your whole battery back would only perform slightly better than an array of the "worst" cell in it... don't do this" Every battery pack has to match the cells somehow. Even the ones with same batch cells. I'm asking how to match these cells in general. As I understand for mass produced packs no matching is done because the benefits do not outweigh the costs. But for a DIY build they might. \$\endgroup\$
    – rrmoelker
    Commented Apr 13, 2021 at 15:42
  • \$\begingroup\$ Are the values you've provided actual experimental values? They vary quite widely. Higher internal resistance loosely correlates with lower capacity, but if these are your samples, likely the way you would match them would be to buy a huge number of them such that you started getting more similar groups. \$\endgroup\$
    – K H
    Commented Apr 14, 2021 at 9:16

1 Answer 1


For optimized cycle life I would choose "method 3":

Individual cell parallel AC resistance matching

This method is based up on Internal resistance matching for parallel-connected lithium-ion cells and impacts on battery pack cycle life.

Resistance matching with lowest difference for the 2 parallel cells.

  • c+d, avg parallel IR = 95mΩ, parallel IR diff ≅ ±5%
  • b+e, avg parallel IR = 75mΩ, parallel IR diff ≅ ±7%
  • a+f, avg parallel IR = 35mΩ, parallel IR diff ≅ ±43% (<- too much)

By the way, I would:

  1. match AC-IR at 1 kHz, not DC-IR.
  2. measure AC-IR at the same cell voltage (< 0.02V).
  3. measure AC-IR at the same cell/room temperature.
  4. then bin in 0.5 mΩ differences (for INR18650)

My 3 purchased reclaimed Samsung INR18650-35E Li-Ion cells were in two different 0.5 mΩ bins at 0.002V difference:

  • 20.5 mΩ ± 0.25 : 1 pc
  • 21.0 mΩ ± 0.25 : 2 pcs


The cell with the higher internal resistance in a parallel line might go under 2.5V if a significant current is drawn by the load and thus will be damaged. When the load is switched off, the other cells in the parallel line will pump their energy into the bad cell with to high current. Further damaging the higher resistance cell.


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