I am currently looking at a layout and I have a trace which has to sustain this current waveform (single pulse) according to the simulation:

enter image description here

As you can see, the current is passing through the trace for just a short time, but the trace is also very thin. I think it won't work.

  • Trace thickness: 17.5 µm
  • Trace width: 150 µm
  • Trace Length : 30 mm
  • Transient time: 10-240 µs
  • It is an internal layer

I found a site on the internet that calculates the maximum current as a function of the above parameters, and it seems that it is possible according to their formula, but I have my doubts. What do you think?

  • 1
    \$\begingroup\$ What does Length Width : 30 mm mean? Please link to the sites you used and maybe copy into your question the analysis from one of those sites. Did you get a trace resistance of 0.194 ohms? \$\endgroup\$
    – Andy aka
    Jan 12 at 14:14
  • \$\begingroup\$ If you can use thicker copper or a wider trace, that definitely wouldn't hurt. \$\endgroup\$
    – Hearth
    Jan 12 at 14:21
  • \$\begingroup\$ 18A for 10µs, ouch. Also consider thermal heating effects (worse for internal traces); one pulse per hour would be a much different scenario than one pulse per second. Calculate the resistance of that trace, then the power lost at 18A to get an idea of the heat generated per-pulse. Could find a similar board with similar internal trace and "abuse" it to see what that does. :) \$\endgroup\$
    – rdtsc
    Jan 12 at 14:22
  • \$\begingroup\$ look up adiabatic heating, to see whther it will handle a single pulse, then compute the average power and temperature rise over time, to see whether it will handle the repetition of them. \$\endgroup\$
    – Neil_UK
    Jan 12 at 14:24
  • \$\begingroup\$ Have you ruled out any possibility route it on an external layer instead? Have you tried Saturn PCB design toolkit for calculations? \$\endgroup\$
    – winny
    Jan 12 at 14:27

2 Answers 2


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 $$ R = \rho \frac{l}{A} $$


  • \$\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\$) $$ \Delta T = \frac{W}{m \cdot C_{Cu}} $$


  • \$W\$ is the dissipated energy
  • \$m\$ is the mass
  • \$C_{Cu}\$ is the specific heat of the copper, 0.383 Joules per gram-Kelvin.


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?

  • 4
    \$\begingroup\$ 100 K does not sound healthy in the long run. \$\endgroup\$
    – winny
    Jan 12 at 18:44
  • \$\begingroup\$ Well thank you for this :) \$\endgroup\$
    – Jess
    Jan 14 at 14:20
  • \$\begingroup\$ FWIW with normal FR4 materials with Tg of 130˚C, a variation of 100 K is not healthy even in the short run. \$\endgroup\$ Oct 18 at 11:51

The Gem Circuits calculator seems to be using Onderdonk’s equation which relates temperature rise as a function of time and current in a copper wire. The primary assumption is that the wire is thermally insulated (heat can’t escape), so you can use it for internal and external traces if you like. The following are two good references with derivations, assumptions, and some discussion of the shortcomings of the equation:



It's bad form to reference your own posts, but if you need a brief derivation one is provided here: https://physics.stackexchange.com/a/718543

Many Onderdonk’s equation calculators (such as the Saturn PCB toolkit) are used to determine if a current pulse will melt the copper (temperature rise = copper melt temperature - room temperature). The Gem Circuits calculator is using a temperature rise the user inputs (20 °C seem to be the default).

I agree that a 150 µm trace seems too narrow to carry this pulse, but Onderdonk's equation is not a bad starting point. I do worry about applying Onderdonk's equation to such a short current pulse when skin effects may be non-trivial. Unless you have access to some type of electro-thermal simulation, you might want to apply a large factor of safety to the trace width.


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