Skin effect calculator

Calculate the penetration depth for various materials, and find wire diameter for a given current density.

frequency: Hz
conductivity: \(\mathrm{\frac{S}{m}}\)
Current density: \(\mathrm{\frac{A}{mm^2}}\)
Current: A

Conductivity of typical materials:

Copper 58 \(\mathrm{\frac{MS}{m}}\)
Brass 16 \(\mathrm{\frac{MS}{m}}\)
Aluminium 37 \(\mathrm{\frac{MS}{m}}\)
Silver 62 \(\mathrm{\frac{MS}{m}}\)
Gold 44 \(\mathrm{\frac{MS}{m}}\)
skin depth \(\delta\): \(\mu\mathrm{m}\)
Wire diameter D: mm

Remarks, and: How does it work?

The calculator assumes nonmagnetic conductors, and in this case, the skin depth \(\delta\) is calculated as follows:

\delta = \sqrt{\frac{2}{\omega\,\mu_0\,\sigma}} = \sqrt{\frac{1}{\pi\,f\,\mu_0\,\sigma}}

At a depth of one skin depth, the current density has decreased by 63%; therefore, after approximately 5 skin depths, the current has decreased almost to zero. This is important to find the diameter of the conductor. Assume the conductor has a radius \(r\). Then, there are two distinct cases:

Case 1

\(5\,\delta \geq r\): The current flows in the whole conductor, and therefore, the current density can be found by \(J = \frac{I}{A}\).

Case 2

\(5\,\delta \lt r\): The conductor diameter is large compared to the skin depth, and therefore, the current only flows in an annular ring. The outer diameter \(D\) is the conductor diameter, while the inner diameter \(d = D – 2\cdot 5\,\delta\). The surface area of the annular ring is therefore
\(A_\mathrm{ring} = \frac{\pi}{4} \left(D^2 – \left(D-10\,\delta\right)^2\right)\) and the current density is then \(J=\frac{I}{A_\mathrm{ring}}\).