# square wave skin depth

It is known that formula for skin depth is only applicable if waveform of the voltage is sinusoidal. So, i have an smps which works at 83khz frequency (square wave). What is skin depth for such frequency? I have heard something about equivalent sinusoidal frequency concept, but i dont understand how to apply it.

• Skin depth is related to the current density in a conductor, not type of waveform. Whether you have a sine or a pulse of the same frequency, the skin depth is the same. Mar 6 '18 at 7:18
• Your square wave is actually a series of sine waves, see Fourier analysis Mar 6 '18 at 7:30
• Thank you. I know that Fourier analysis is helpful and waveform can be presented as a series of sine waves. I am more interested in a concept of equivalent sinusoidal frequency. Mar 6 '18 at 8:09
• Don't forget proximity effect in your transformer. Mar 6 '18 at 11:08

Skin depth is defined only for sinusoidal fields and depends on frequency.

A rectangular signal can be seen as a series of sinusoidals signals with frequencies $f(2n +1)$ with $n = 0, 1, 2, ...$ where $f$ is the frequency of the rectangular signal (cf. TonyM's comment).

The higher frequency components will have lower skin depths. Higher frequencies will be attenuated more than lower frequencies. I.e. the waveform inside the conductor will not only get weaker it will also change its waveform.

So to answer your question you first have to define skin depth for non-sinusoidal fields (e.g.: is it where the amplitude of the new waveform is $1/e$ of the amplitude of the original waveform; or is it where the RMS of the new waveform is 1/e of the RMS of the original waveform; ...)

So only when you have a definiton of skin depth for non-sinusoidal signals your question can be answered.

The speed of propagation THRU (not along the surface, but thru) standard copper foil is about 150 nanoseconds. Such foil is 1 ounce/foot^2 or 1.4 mils thick or 35 microns thick. The Fskin is about 4MHz, depending on what numbers you use for conductivity of copper.

You won't see attenuation, just that 150 nanosecond delay.

What you will "see" on the far side, away from the incoming field, depends on what angle of sensor (normal, tilted, orthogonal) you use.

• Speed has dimensions of [distance]/[time] - is that some term-of-art you've used? Mar 6 '18 at 18:22