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Given the below pictured 1S parallel battery configuration that draws 20 amps (for example), with the pictured connections, what would the approximate current flow among the various spots along the bus-lattice be (not 'how many amps does each cell provide')?

Could the current distribution be further diluted by adding additional diagonal bars between cells?

I realize that if/when this parallel group gets put in series with other groups, the series connections will all carry the full 20A, but I'm specifically curious about intra-parallel current flow.

enter image description here

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  • \$\begingroup\$ Each battery ideally supplies 20A/12, since you correctly take power off opposite ends. The current distribution follows from this. \$\endgroup\$ – Brian Drummond Oct 4 at 22:46
  • \$\begingroup\$ The question should not be how many amps does each cell supply, but rather what is the current along the bus bar. I.e. does it mean the bus bar must only withstand 20A/12 e.g. 1.7Amps ? \$\endgroup\$ – user1142433 Oct 4 at 22:49
  • \$\begingroup\$ Current in Series stays the same or is added but in a parallel connection current is split using current division. So what's happening is in series all the current is being pulled at once whereas in parallel the current is being "split" while being pulled hence it will also make the battery last longer. \$\endgroup\$ – Mauz Oct 4 at 22:59
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    \$\begingroup\$ No it does not. Each segment carries the sum of currents of batteries upstream of it. \$\endgroup\$ – Brian Drummond Oct 4 at 22:59
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    \$\begingroup\$ @RussellMcMahon it's unclear what your point is. The "series connections" referenced in the question are not depicted, but rather come about only when the asker takes the parallel grouping they are asking about and joins it in series with other copies of itself. The asker is basically saying they are setting that aside; the question covers only the internal bussing of the parallel assembly. It's also true that cell matching is important for such an assembly. \$\endgroup\$ – Chris Stratton Oct 10 at 5:10
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By symmetry, the current through each cell is the same at 20/12 = 1.66A per cell.

There would be no current through the lateral connections (assuming all cells are matched).

The current through each of the lengthwise connections would be the same and each would contribute half of the current.

The current through each successive leg of the interconnect would go down by 1.66A as it goes past each cell. That is ~8.33A after the first cell, ~6.67A after the first two cells down to 1.67A for the last link.

Assuming the connections on the bottom are the same the current pattern would be the same although starting from the opposite end.

It is significant that the black wire connects to the opposite end of the pack from the red wire, if it didn't the connections would not be symmetric and there would be higher currents through the cells near the wire connections.

Additional lateral connections wouldn't help but adding extra thickness of conductor in the lengthwise direction would.

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  • \$\begingroup\$ "There would be no current through the lateral connections". So, the "rungs" of the ladder structure here serve no purpose? Could they be omitted without effect? \$\endgroup\$ – user1142433 Oct 5 at 1:19
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    \$\begingroup\$ @user1142433 yes, theoretically. But if one cell fails in one string, then those cross bars will enable the other cells in that string to still contribute via two cross bars. \$\endgroup\$ – Solar Mike Oct 5 at 4:15
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In series cells, the weakest has the largest voltage mismatch error prone to over/under voltage damage during CV charge or low SOC.

In parallel cells the “strongest” battery has the largest current mismatch due to $$ΔI = Vc / ΔESR [mΩ]$$

Therefore all cells must be perfectly tested and matched before any strings series or shunt or both. Balancing extends the life of series cells. But for parallel cells, the strongest (highest C[kF] lowest ESR) is stressed more to become as weak as the shared one then the pair is compared to the rest of the series string for mismatch voltages for aging effect margin.

Shunting two strong cells of different chemistry voltage can lead to thermal problems due to circulating currents in formula above until balanced voltage and null circulating current. Then the current sharing depends on ESR mismatch. The “strength” of a battery can be computed by the set of ESR*C=T time constants (more than one). Lower is better for short term surge power, which is often rated by C rating * Ah (20h) where higher T is better for long term T’s for duration of Ah.

Other info

These are different capacitors inside the complex Battery equivalent circuit.(beyond the scope of this query.)

BTW your 1S pack is actually 1S12P. The bus bar is <<1 mΩ while each cell may be 50 mΩ +/- 50% range but ought to be matched <<1% in that range. You may test each with pulse load tests for ESR1 and steady avg I Amps to get ESR1 and C2 values, where C2=I dt/dV measured for typical duration expected (0.5h~tbd h) in a ESR1C1//ESR2C2 simplified 2 cap equivalent cct. This is based on the approx. asymptotic 2 slope V vs t curves.

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