Electronic – Will forced air on PCB improve current capacity of trace

I'm going to have a stab at some maths :)

The DC resistance of a conductor - any conductor - is calculated as:

\$R_{DC} = \frac{{\rho}l}{A}\$

Where \$\rho\$ is the resistivity of the conductor in \$\Omega/m\$, \$l\$ is the length in meters, and \$A\$ is the cross-sectional area in m².

The thickness of 1oz copper is \$0.000034798m\$. Say you have a 3mm (or 0.003m) wide trace. The cross-sectional area is (approximately, assuming a perfectly rectilinear cross-section) \$0.000034798 × 0.003 = 0.000000104m^2\$. Resistivity of copper is \$1.68×10^{−8}\$ at 20C, and your trace is 100mm long (0.1m).

\$R_{DC} = \frac{1.68×10^{−8} × 0.1}{0.000000104} = 0.016153846\Omega\$ at 20C.

Ok, now for the tricky bit. The temperature co-efficient (\$\alpha\$) for copper is 0.003862.

\$R(T) = R(T_o)(1+\alpha{\Delta}T)\$

So for a temperature of 30C we have a \${\Delta}T\$ of 10C, or 10K (30 - 20 = 10, K = C + 272.15).

So \$R(30) = R(20)(1+0.003862×10) = 0.016153846×1.03862 = 0.016777708\Omega\$

So now solve Ohm's Law for voltage. Say you have 100mA flowing through the trace. That's \$V=RI\$, so \$0.016777708×0.1 = 0.001677771\$ or \$1.678mV\$ dropped across the trace at 30C.

Who says you need online calculators?

(Now, it's been about 20 years since I did this kind of thing at college, so I may be completely wrong ;) )

They are using power planes, maybe even with double-thickness copper layers too (70 microns / 2oz, rather than the usual 35 micron / 1oz Cu thickness), so the power path to the outputs is not going to be high resistance, and therefore not waste much power as heat loss, and therefore will not get hot and explode and otherwise be unable to "handle" the amount of current shown.

120A for 8 outputs is not too bad, each contact (which looks like just exposed mask on copper planes with holes for thermal relief) only needs to handle about 15A each which is entirely possible with nice fat wires going out to the rest of the set-up.

I would be worried though, first about the size of the battery needed, and second about your safety, if 120A loading was a common thing in your system.

You must also understand that using a trace width calculator makes certain assumptions, for example what the ambient temperature is, what the acceptable rise in temperature of the traces is, and in some cases how long the trace is, and finally the thickness of the trace which is usually measured in micrometers or ounces/square foot (oz). If you allow up to 30-40 degrees Celsius increase in temperature of the traces (which is usually fine for power related systems, which are often rated to 120-150 degrees) you can get away with very thin (2-3mm) tracks for carrying 15A, let alone how much you can do with a whole power plane (could be modeled as a 30-40mm wide trace!).