I guess the biggest problem is: Can your copper handle the energy introduced per pulse or will the track vaporize.
So lets see:
Cross-section, volume, mass, resistance:
Your copper cross-sectional area is
$$
A=w \cdot t \cdot k
$$
where
- \$0\leq k \leq 1\$ for derating and tolerances
- \$w\$ is trace width
- \$t\$ is copper thickness
So, \$\mathrm{A(k=1) = 150 \cdot 17.5 = 2625 \ \mu m^2=0.002625 \ mm^2}\$.
Your copper volume therefore is
$$
V=A \cdot l = w \cdot t \cdot k \cdot l
$$
where
- \$A\$ is the cross-sectional area (comes from above, so do \$w\$, \$t\$, and \$k\$).
- \$l\$ is the trace length
So, \$\mathrm{V \approx 78.75\ 10^6 \mu m^3 = 7.875 \ 10^{-5} \ cm^3}\$
Your copper mass is
$$
m = V \cdot d
$$
where
- \$V\$ is the volume of the copper trace (comes from above)
- \$d\$ is the density of copper, 8.96 grams per cubic centimetre.
So, \$m\approx 7.875 \ 10^{-5} \cdot 8.96 \approx 705 \ \mu g\$
Your resistance is
$$
R = \rho \frac{l}{A}
$$
where
- \$\rho\$ is the resistivity of copper, 1.786 Ohm-metre
- \$A\$ is the cross-sectional area
- \$l\$ is the trace length
So, \$R\approx 205 m\Omega\$.
Power, energy and dT:
The total energy that your signal \$i(t)\$ delivers to a resistance \$R\$ over a period \$T_s\$ is
$$
W=\int_{T_s} i(t)^2 \ R \ dt
$$
Please do your own approximation for \$W\$.
Your temperature rise is (from \$W=Q=m \ C \ \Delta T\$)
$$
\Delta T = \frac{W}{m \cdot C_{Cu}}
$$
where
- \$W\$ is the dissipated energy
- \$m\$ is the mass
- \$C_{Cu}\$ is the specific heat of the copper, 0.383 Joules per gram-Kelvin.
Analysis:
If you allow a \$\Delta T\$ of 100 °K (we are safe here, as energy is also conducted to the PCB, so the temperature rise calculated will be the worst case!) your max heat-energy per pulse becomes: 27 mJ.
Or: A 21 A DC-signal for 300 µs into 205 mΩ copper track with 705 µg of mass will lead to a temperature rise of 100 K.
Now the question is: Can you get rid of these 27 mJ/pulse every period, before it is introduced again?
