Bill, uhhh, that puts a lot of pressure on me, if you wanna follow along. (don’t blame me, if it’s not working out )
by the way, your calculation:
is correct. however the 315A is the “acceptable” current, not the “optimum” … this could still cause some hazard if you’re going full throttle for a long time … I am not sure, just thinking.
Any my problem is, that I have already started with obviously too thin strips… I should have done the calculations earlier
I am not an electrical engineer myself, I just have some basic knowledge about electronics, a good common sense, and I bought that book “DIY Lithium Batteries” that I mentioned before here in this forum (it’s a really good read).
The cutting and positioning is very very easy! You just cut the strip using a wire cutter and the positioning is done with your fingers. The plastic thing that keeps the cells in place also helps you positioning the strips.
So for my project, I am short of 204 A current carrying capacity between each series connection.
This is, see above, because my first and second layer of Nickel Strips can carry 96A so far only.
This product (22.54€ |0,2x8x100mm 100 pcs/lot Reine Nickel Platte Band Streifen Blätter 99.96% für 18650 power batterie pack spot schweißen spot schweißer|sheets|sheet strapssheet plate - AliExpress ) 100 pieces of 100mm length 8x 0.2mm pure Nickel strip (I need 25mm length, so each 100mm piece can be cut apart into 4x 25mm pieces) can carry a little over 6,4A per strip. (this is the table in an earlier post talks about 0.2x7mm, whereas this product is 0.2x8mm)
To be safe, let me calculate with 6,4A:
so 204A divided by 6,4A = 31,875 pieces. that’s roundabout two layers per parallel group (15 cells per parallel group = 15 connections, 2 Layers = 30 connections).
The above link is a set of 100 strips, each cut into 4 pieces makes a total of 400. For an additional double layer to make 180 connections I need 360 pieces, so I’d be good to go if I order one package.
… and just thinking again, If I do all the calculations with the “acceptable” current instead of the “optimum” - then one additional layer of the 0.2x8mm would be enough