width of nickel tabs
- Thread starter jonny
- Start date
dogman dan
1 PW
Since nobody has replied, I'll take my guess.
Calculate the wire size you'd need, 12g, 10g, whatever you need. Then figure out the math to convert the nickel strips into the same thickness if it was round shaped like wire.
Calculate the wire size you'd need, 12g, 10g, whatever you need. Then figure out the math to convert the nickel strips into the same thickness if it was round shaped like wire.
John in CR
100 TW
That's some pretty small tab material. Assuming 7 strips going all the way through the series connection, that's only a cross sectional area of only 4.2mm2, which is only the equivalent of about 13 gauge wire.
In Makita tool packs when they went with the thinner tab material it was about 30mm wide to connect just 2 p. Before that they were using a thicker material that was 10mm wide to make the series connection for 2p.
What kind of max current were you looking to draw through your pack, and how did you plan to connect the battery leads (at 1 place on the tab or multiple?
John
In Makita tool packs when they went with the thinner tab material it was about 30mm wide to connect just 2 p. Before that they were using a thicker material that was 10mm wide to make the series connection for 2p.
What kind of max current were you looking to draw through your pack, and how did you plan to connect the battery leads (at 1 place on the tab or multiple?
John
hi, i have calculated that:
we take as an example a tab with a 40mm(L) and 5mm (W) with a thickness of 0.12 mm, its surface will be in mm2 of 410.8 mm2.
If we take a wire with a diameter of 1.5 mm, a lenght of 40mm (L), its surface will be of 191.92 mm2.
So to have the same value of the tab of nickel should I use an electric wire of nearly 3.5 mm2 ( AWG8)
191.92:1.5=410.8:X
X= (410.8x1.5)/191.2= 3,21 (awg8)
It's right or I'm wrong??
Current 20A work current max current 40A...
we take as an example a tab with a 40mm(L) and 5mm (W) with a thickness of 0.12 mm, its surface will be in mm2 of 410.8 mm2.
If we take a wire with a diameter of 1.5 mm, a lenght of 40mm (L), its surface will be of 191.92 mm2.
So to have the same value of the tab of nickel should I use an electric wire of nearly 3.5 mm2 ( AWG8)
191.92:1.5=410.8:X
X= (410.8x1.5)/191.2= 3,21 (awg8)
It's right or I'm wrong??
Current 20A work current max current 40A...
sorry don't understand
John in CR
100 TW
Not surface area, cross sectional area of your series connections 5mm x .12mm = .6mm2 x 7 = 4.mm2 only equal to 13 gauge wire, and that's if you have the series connection with tab on every cell.
You were probably thinking making 7p with a tab which is fine, but then how are you going to connect the 10 blocks of 7p into series?....certainly not with one piece of that thin tab. That tab material is only the equivalent of 21awg wire. To me even 7 strips of that tab isn't enough to carry the current.
This is how I'd build a 7p10s pack of 18650 cells using tab of that thickness, but probably about 25mm wide. Note the blue is the tab on one end of the cells and green is the tab on the other.
You were probably thinking making 7p with a tab which is fine, but then how are you going to connect the 10 blocks of 7p into series?....certainly not with one piece of that thin tab. That tab material is only the equivalent of 21awg wire. To me even 7 strips of that tab isn't enough to carry the current.
This is how I'd build a 7p10s pack of 18650 cells using tab of that thickness, but probably about 25mm wide. Note the blue is the tab on one end of the cells and green is the tab on the other.
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