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I'm creating a 20s6p li-ion pack from 18650 cells. Here is the design enter image description here

I am are using 8 by 0.2mm nickel strips rated for 6.5A. With 6 parallel branches that's 6*6.5 = around 40Amps. But since we doubled nickel strips we get 80A flowing in and out of the battery.

My question is about the current accumulation on each terminal of the battery which contains 6 cells in parallel. I'm thinking that the 80A current going to the battery cable will have to pass through a single point on the battery which can currently only handle 13A (1 parallel branch)

So I'm not sure if we should have 6 wires connected to each parallel cell to distribute the current before connecting to the main battery cable. Or could we just triple or quadruple up on the nickel strips at the terminals.

Thanks!

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I'm thinking that the 80A current going to the battery cable will have to pass through a single point on the battery which can currently only handle 13A (1 parallel branch)

Depending on where you connect the output leads, that could be correct. Even if you connect it to one of the inner cells the current going through straps in adjoining cells will exceed their rating, since it will be the sum of cell currents in that leg.

So I'm not sure if we should have 6 wires connected to each parallel cell to distribute the current before connecting to the main battery cable.

Yes, that is the best solution. Each wire only has to handle ~13 A so it can be a thinner gauge, which is easier to handle than a single wire rated for 80 A. If the wires are all made the same length then they should distribute the current more evenly between cells.

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