Electronic – maximum trace temperature on FR-4

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 ;) )

When two small wires are bifilar wound around a magnetic core, they exhibit the same inductance as one single wire. The emphasis in the first sentance is "small" meaning that any flux produced by one of the small wires is perfectly coupled to the 2nd wire.

In effect, each wire (in the pair) acquires an inductance that is twice the inductance of a single wire and, when there are two inductors are in parallel, the net inductance IS the same as if a single wire were used.

The magnetic core helps achieve flux coupling between those two wires in this little thought experiment.

Now, remove the core and coupling is not the same; current flowing in one wire (despite being closely wound with the other) does not 100% couple to that other wire and, as you spread the gap between those two wires, coupling gets even smaller and ultimately, the inductance of the pair starts to look like L/2.

It's the same effect with a wide copper conductor - the current is largely spread evenly across the width but flux generated by a part of the current along one edge does not 100% couple with the flux generated by a similar part of current flowing along the other edge.

So progressively coupling starts to reduce as the copper conductor gets wider and you start to move away from L to L/2.

The limit of L/2 does not take into account frequencies where the applied signal wavelength starts to be significant in terms of track thickness of course. That creates a whole bunch of other things to consider!