You say it breaks the circuit, but parasitic inductance says no.
Parasitics are a bitch, they are giving me headaches in pcb design right now.
When it breaks, it makes a spark, a fairly large spark to say the least, depending on what your pack is made of, that can very well result in a fire anyway.
Proper fuses are designed to minimize and contain this spark.
Here with the hot tip… aluminium wire doesn’t play well with regular solder and flux.
Apparently the oxide on the face of the aluminium requires mega awesome flux to break through it.
This is a no go without serious effort.
MysticalDork,
With my pack (20s7p of 18650PF cells), I used 24AWG tinned copper wire, 10mm long.
Resistance of pack in Ohms equals number of serial cells divided by the number of parallel cells multiplied by the average resistance of the cells in Ohms
Rp=(xS/xV)Rc
with about 190mm ((xS-1)*10mm) of 24AWG per parallel run which is
R=ρL/A =1.86E-8Ω.m * 0.19m / 2.05E-4m^2
=1.7E-5Ω
=.000017Ω
=0.017mΩ
That is the resistance for one leg of 20S7P using 10mm for the length of the fuse wire.
Rp=(xS/xP)Rc
=1/7*0.000017Ω
=2.43N-6Ω
=0.00000243Ω
=0.00243mΩ
or the long way.
1/Rt=1/R1+1/R2+1/R3+1/R4+1/R5+1/R6+1/R7
Rt=1/(1/R1+1/R2+1/R3+1/R4+1/R5+1/R6+1/R7)
=1/(58823+58823+58823+58823+58823+58823+58823)
=1/411764
=2.42E-6
=0.00000242Ω
=0.00242mΩ
So 7 paralleled lines of 190mm of 24AWG copper only adds 0.00242mΩ to the battery pack?
I guess I’m just skeptical of the added resistance from the fuse wires and the (small) voltage drop.
I think you misplaced a decimal somewhere in there, but you’ve got the concept.
24AWG wire has a resistance of about 84 mΩ per meter. 1cm=0.01m, so 1cm of wire has a resistance of 0.84mΩ.
0.84 x 20 / 7 = 2.4mΩ added resistance overall.
I think you misplaced a decimal somewhere in there, but you’ve got the concept.
Ya my bad. Error Corrected A of 24AWG is 2.05E-7m^2 not 2.05E-5m^2
How does it look now.
Resistance of pack in Ohms equals number of serial cells divided by the number of parallel cells multiplied by the average resistance of the cells in Ohms
Rp=(xS/xV)Rc
with about 190mm ((xS-1)*10mm) of 24AWG per parallel run which is
R=ρL/A
=1.86E-8Ω.m * 0.19m / 2.05E-7m^2
=0.017Ω
=17mΩ
That is the resistance for one leg of 20S7P using 10mm for the length of the fuse wire.
Rp=(xS/xP)Rc
=1/7*0.017Ω
=0.00242Ω
=02.43mΩ
or the long way.
1/Rt=1/R1+1/R2+1/R3+1/R4+1/R5+1/R6+1/R7
Rt=1/(1/R1+1/R2+1/R3+1/R4+1/R5+1/R6+1/R7)
=1/(59+59+59+59+59+59+59)
=1/411.76
=2.42E-6
=0.00242Ω
=02.42mΩ
So 7 paralleled lines of 190mm of 24AWG copper only adds 2.42mΩ to the battery pack?
It would be 7 paralleled lines of 200mm not 190mm, because I have cell level fuses on each cell in each P group, not just the first 19.
But yes, the amount of extra resistance it adds to the system is negligible compared to the internal resistances of the cells themselves.
Anyone have a good source for tinned copper wire? Preferably in N. America?
I’ve been using the @akhlut p-group calculator to decide on the gauge for a pack I’m building- been super helpful in wrapping my head around the concept as is the excellent write up (thx homie!).
my cheapo spot welder just bit the dust - can you tell me more about this method? are you doing CLF on both sides of the cells?
I am interested in this method as well.
Is it better to use copper or brass for the bus bars / p-pack strips?
You should not solder directly on the cells. The heat damages them, and yes , even if you do it real fast
Buying a spot welder on Amazon would be better than soldering. Batteries are dangerous, if a $100 investment in the correct tools is out of the question then so is building a battery.
I don’t do it that way anymore. I have the spot welder from flipsky now and use the precut duck battery nickel. 
