## Will Copper Serial Connection Add Something?

Batteries, Chargers, and Battery Management Systems.
rg12   100 kW Posts: 1255
Joined: Jul 26 2014 4:22pm

### Will Copper Serial Connection Add Something?

I'm building my packs with 0.15mm thick pure nickel strips that have no pattern (one piece of about 27mm wide nickel) so instead of having the typical 8mm wide serial connection, I have an area that is as wide as the cell (18mm).
So 18*0.15=2.7 which I devide that 2.7 by 4 in order to get the square area of the conductor in copper language which is 0.67mm that is about 19AWG copper wire per cell.
The strongest cell I use is 25R which I draw a max of 10A (half of the rated discharge rate).
So 10A through 19AWG.
I calculated that the length of the conductor between each two cells is about 10mm and if I multiply it for a 20S pack then I have 19 pieces of 10mm which is 19cm of total 19AWG length.
Can it be calculated like this?
If so, how do I calculate the resistance and voltage drop of 10A across a 19cm long 19AWG (0.65mm2) copper wire?

nieles   10 kW Posts: 682
Joined: Jul 14 2008 5:39pm
Location: The Netherlands

### Re: Will Copper Serial Connection Add Something?

R=Rho * L / A

rho of copper = 1.72 x 10^-8 at room temperature
L = 0.19 meter
A = 0.65 x 10^-6 meter^2

R = ( 1.72 * 10^-2 * 0.19 ) / 0.65 (i crossed off the 10^-6 on both sides of the divider)
R = 0.005 Ohm
R = 5 miliohm

voltage drop = I x R = 10 x 0.005 = 0.05V = 50mV

heat generated = Vdrop x I = 0.05 x 10 = 0.5W

rg12   100 kW Posts: 1255
Joined: Jul 26 2014 4:22pm

### Re: Will Copper Serial Connection Add Something?

nieles wrote:
Feb 11 2020 5:02am
R=Rho * L / A

rho of copper = 1.72 x 10^-8 at room temperature
L = 0.19 meter
A = 0.65 x 10^-6 meter^2

R = ( 1.72 * 10^-2 * 0.19 ) / 0.65 (i crossed off the 10^-6 on both sides of the divider)
R = 0.005 Ohm
R = 5 miliohm

voltage drop = I x R = 10 x 0.005 = 0.05V = 50mV

heat generated = Vdrop x I = 0.05 x 10 = 0.5W
Thank you!
So there is no point in making an effort to go full copper for the serials since the loss is only 0.5W, am I right?