3.12: Skin Depth

The electric and magnetic fields of a wave are diminished as the wave propagates through lossy media. The magnitude of these fields is proportional to

$$e^{-\alpha l} \nonumber$$

where $\alpha\triangleq\mbox{Re}\left\{\gamma\right\}$ is the attenuation constant (SI base units of m$^{-1}$), $\gamma$ is the propagation constant, and $l$ is the distance traveled. Although the rate at which magnitude is reduced is completely described by $\alpha$, particular values of $\alpha$ typically do not necessarily provide an intuitive sense of this rate.

An alternative way to characterize attenuation is in terms of skin depth $\delta_s$, which is defined as the distance at which the magnitude of the electric and magnetic fields is reduced by a factor of $1/e$. In other words:

$$e^{-\alpha\delta_s} = e^{-1} \cong 0.368 \label{m0158_esddef}$$

Skin depth $\delta_s$ is the distance over which the magnitude of the electric or magnetic field is reduced by a factor of $1/e \cong 0.368$.

Since power is proportional to the square of the field magnitude, $\delta_s$ may also be interpreted as the distance at which the power in the wave is reduced by a factor of $(1/e)^2\cong 0.135$. In yet other words: $\delta_s$ is the distance at which $\cong 86.5$% of the power in the wave is lost.

This definition for skin depth makes $\delta_s$ easy to compute: From Equation $\ref{m0158_esddef}$, it is simply

$$\boxed{ \delta_s = \frac{1}{\alpha} } \nonumber$$

The concept of skin depth is most commonly applied to good conductors. For a good conductor, $\alpha\approx \sqrt{\pi f\mu\sigma}$ (Section 3.11), so

$$\delta_s \approx \frac{1}{\sqrt{\pi f\mu\sigma}} ~~~\mbox{(good conductors)} \label{m0158_eSDGC}$$

Example $\PageIndex{1}$: Skin depth of aluminum

Aluminum, a good conductor, exhibits $\sigma\approx 3.7 \times 10^7$ S/m and $\mu\approx\mu_0$ over a broad range of radio frequencies. Using Equation $\ref{m0158_eSDGC}$, we find $\delta_s \sim 26~\mu$m at 10 MHz. Aluminum sheet which is 1/16-in ($\cong 1.59$ mm) thick can also be said to have a thickness of $\sim 61\delta_s$ at 10 MHz. The reduction in the power density of an electromagnetic wave after traveling through this sheet will be

$$\sim\left(e^{-\alpha\left(61\delta_s\right)}\right)^2 = \left(e^{-\alpha\left(61/\alpha\right)}\right)^2 = e^{-122} \nonumber$$

which is effectively zero from a practical engineering perspective. Therefore, 1/16-in aluminum sheet provides excellent shielding from electromagnetic waves at 10 MHz.

At 1 kHz, the situation is significantly different. At this frequency, $\delta_s\sim 2.6$ mm, so 1/16-in aluminum is only $\sim 0.6\delta_s$ thick. In this case the power density is reduced by only

$$\sim\left(e^{-\alpha\left(0.6\delta_s\right)}\right)^2 = \left(e^{-\alpha\left(0.6/\alpha\right)}\right)^2 = e^{-1.2} \approx 0.3 \nonumber$$

This is a reduction of only $\sim70$% in power density. Therefore, 1/16-in aluminum sheet provides very little shielding at 1 kHz.