# Skin effect with superimposed frequencies

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?

• 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) Commented Aug 1, 2018 at 14:36
• 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. Commented Aug 1, 2018 at 14:48

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

## Also

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!).

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.

• thank you! I meant 15kHz on the last current, already edited. Hadn't thought about the thermal influence between eachother, interesting. Commented Aug 1, 2018 at 14:03
• 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. Commented Aug 1, 2018 at 14:30