Does a high-frequency signal affect the skin depth of another low-frequency signal?

E.g.: On a solid 6mm2 copper conductor, there are three currents running through it at the same time:

  • 15 A DC

  • 15 A rms @ 50Hz

  • 5 A rms @ 15kHz (edited)

Can I calculate the skin effect individually for each signal and sum the losses? Or do the currents influence the skin effect of one another?

  • \$\begingroup\$ out of curiosity: where does your 5A @ 15 kHz current come from? That sounds like an interesting application! (also, do consider the inductivity and conductive effects if that cable is long enough) \$\endgroup\$ Commented Aug 1, 2018 at 14:36
  • \$\begingroup\$ The three currents are only for easier explanation. The real current has no DC component. It is for a GTI, and the 15kHz is the switching frequency. I was just wandering if Litz was necessary for the chokes, given I did not understand the interaction between the frequencies. \$\endgroup\$ Commented Aug 1, 2018 at 14:48

1 Answer 1


Short answer

no. Skin Effect is fully explained by the linear model of Maxwell's equation, so different frequencies can be considered independently.


at 50 Hz, your skin depth is about 9mm; far thicker than your conductor is (makes sense, right? Otherwise we wouldn't be using massive copper for power distribution!).

Long answer

Skin depth being non-zero is due to non-ideality of your conductor. Of course, if you heat up a metal, it changes conductivity / resistance.

In your case, 6mm² carrying a maximum sum current of 35 A: Ignore. Your cable has about 2.4 mΩ resistance per 1m of length; P=I²·R~=10³ A² · 2.4·10⁻³ Ω = 2.4 W. Getting rid of 2.4 W of heat over 1 m of length: will happen by itself.

With the three currents at the right frequencies, we can even be specific:

  • 5A @ 0 Hz: "infinite" skin depth. No significant heating due to this current.
  • 15 A @ 50 Hz: skin depth >> radius. No significant heating due to this current.
  • 15 @ 15 kHz: skin depth ca 0.5mm. Resulting area of conducting cross section is \$\pi\$·(outer diameter² - (outer diameter-skin depth)²)=\$\pi\$·(0.78² - 0.28²) ~= 6mm² - 0.88mm² ~= 5mm². No significant heating through this current.

Things get worse in nonlinear materials, but I'm pretty optimistic that your copper conductor is linear enough.

  • \$\begingroup\$ thank you! I meant 15kHz on the last current, already edited. Hadn't thought about the thermal influence between eachother, interesting. \$\endgroup\$ Commented Aug 1, 2018 at 14:03
  • \$\begingroup\$ Another interesting example are parallel Grid Tied Inverters (GTI) where a massive Solar Farm with hundreds of thousands of inverters. Most producing pure sine wave currents but some producing higher switching noise & harmonics can have say 50 A of current going out at line freq. thru a 60A fuse but if harmonics from outside get shunted inside from a filter the current adds and if the skin effect causes the fuse to be a higher resistance, it heats up easier or gives a false fuse rating. so that GTI fuse might be affected from grid harmonic noise currents if the GTI shunts the high f current. \$\endgroup\$ Commented Aug 1, 2018 at 14:30

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