BMS 12 AWG enough?

Hey Dudes,

I'm building a 20s pack that will be connected to a 45A controller.
The BMS I have (50A continues and 100A peak) has wires already soldered to it and the negative wire on it (since the positive is soldered by me to the battery pack itself) is a 12 AWG wire (about 10" long).
Will that be enough for my use or should I solder a 10 AWG?

10awg is ~1mOhm/foot, 12awg is ~1.6mOhm/foot. Since P=R*I^2, 45A will dissipate ~2W/foot for 10awg and ~3.2W/foot for 12awg. Compared to 3kW total, it's not likely to matter much to performance. Personally I have an 18S pack using 12awg, and no issues up to 80A peak.

Wow 80A peak through 12g? damn that's alot!
Thanks alot man, though I already put 10 AWG just in case.

My general rule is a conductor's rated 10 sec fusing current / 3 for peak current rating.
For 12awg this is 235A. 235A/3 = 78A max.

I'm more interested in the continues current it can handle than the peak.
Anyway to know that number? I know it depends on length and stuff...

As far as peak current, it doesn't depend on wire length, it's just a matter of power per unit length, and how well the wire can shed heat without getting too hot. The problem comes when you start either overheating something or wasting a significant portion of your total power in anything other than the motor.

Take my 80A peak example. I have ~4 feet of total wire between the battery and controller (2 feet each direction). At 80A, that's about 40W of power dissipated in the wire. It's a lot, but compared to the roughly 4500W of power being delivered by the battery it's only 1% of the total. Going to 10AWG would only save a few tens of watts, and since heat isn't a problem for 12AWG at 80A peak, there isn't really any reason to do so unless you happen to have 10awg available already or are going to be running at 80A for a long time. At that point the controller or battery will likely be the limiting factor.

But don't thinner wires able to transfer less amps since they have only limited amount of surface (thickness) to them to transfer the 80A (for that example)?

The length of the wire does matter in as much as there is often a maximum allowable voltage drop, which will be greater on a longer wire.

rg12 said:

But don't thinner wires able to transfer less amps since they have only limited amount of surface (thickness) to them to transfer the 80A (for that example)?

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There's no real limit to how much current the wire can carry, as long as it's not enough to damage the wire. You could send 1000A through 32awg wire as long as it's a short enough pulse and you have enough voltage to overcome the wire resistance. The limitation is thermal, not some physical limit above which the wire can't conduct.

Punx0r said:

The length of the wire does matter in as much as there is often a maximum allowable voltage drop, which will be greater on a longer wire.

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True, but for most cases the wire itself is not the biggest source of voltage drop. Take my 4ft (2ft each direction) of 12awg example. That's 4x1.6mOhm (0.0064 ohm). At 80A, you get ~.5V of voltage drop across the wires (80A x .0064ohm = 0.512V). I'd be willing to bet that 99% of battery packs used by people on ES have greater total internal resistance than 4ft of 12awg wire.

I'm certainly not suggesting using heavier than needed wire is a bad thing, just that after a certain point it's not going to make a noticeable difference. You need to consider total system resistance, which includes battery cells, interconnects, connectors, wire, etc.

About the current limit, if it's only thermal, why do people thicken their controller solder traces in order to deliver more amps?

rg12 said:

About the current limit, if it's only thermal, why do people thicken their controller solder traces in order to deliver more amps?

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Did you see where he said "short enough pulse"? If the trace is too thin, it will heat up, and eventually it will fail.. usually peeling off the board then breaking... then again, sometimes vaporizing... lol.
so people add braid/wire to the main traces on the board. If nothing else, less voltage drop.


You might also be confusing the traces with the shunt. adding wire/solder to a shunt will make the controller pass more current. cutting into it makes it pass less. This is because the shunt is the measurement device the controller uses to measure current.

Oh ok, so I missed the part with the shunt, so they thicken up the shunt to pass more current and then also thicken up the traces to handle the extra current the shunt will pass right?

rg12 said:

Oh ok, so I missed the part with the shunt, so they thicken up the shunt to pass more current and then also thicken up the traces to handle the extra current the shunt will pass right?

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The shunt is just a precise, low value resistor. The controller measures the voltage across the shunt to calculate current. If you reduce the resistance of the shunt by adding solder, extra wire, or whatever, then the amount of current it takes to get a certain voltage goes up (V=I*R). The controller has no idea that the resistance of the shunt has changed, so the result is more current.

Adding to the traces reduces power loss in them, although by how much is a matter of debate. When it's just solder, I doubt it helps much. The best idea I've seen is using solid core household wire and tacking it on along the length of the power traces. I don't know a lot about controller design, but my semi-educated guess is that the biggest source of wasted power in the controller is due to MOSFET on state resistance and during switching. Not trace heating. That's based on the fact that the FETs get warm, despite having the best thermal path to the ambient, while the traces don't dissipate a lot of power (if they did they would burn up).