### Categorized |General Basics, Power Electronics Basics

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Posted on 25 September 2020

# The Skin Effect

If a DC current I flows through a conductor with diameter d, then the current density is $J = \frac{4I}{\pi d^2}$ (A/mm2) and is constant throughout the cross-sectional area. In the case of an AC current this is not so. We can visualise this by imagining a conductor (fig. 1) as being composed of n concentric cylinders in which a part of the total current I flows. The current that flows in each cylinder produces a magnetic field as shown for example in fig. 2 which is drawn for cylinders 1 and 2.

Figure 1. Solid current carrying conductor                Figure 2. Cylinders one and two from figure 1

The smaller the follow-on number (1, 2, ... n) of the cylinders in fig. 1, the more magnetic field lines that are linked and therefore the larger the cell inductance of these sub cylinders. And in the case of an AC current, the impedance of one cylinder would therefore be greater than that of a greater sized. In other words the impedance rises as the cylinder numbers decrease. The current density will be greatest near the circumference of the conductor: we call this the skin effect.

The higher the frequency, the more pronounced this effect. The useful cross-sectional area of current carrying conductors decreases with an increase of the frequency of the current. It may be necessary for example with switch mode power supplies that operate at higher frequencies (50 to 100kHz) to use (instead of solid conductors) a number of thinner conductors in parallel for the windings of coils and transformers. Above 100kHz a so-called Litze wire is used to wind coils. This Litze wire is in fact a cable, composed of a large number of thin wires switched in parallel. The currents carrying cross-sectional area of this Litze wire is obviously much greater than a solid conductor of the same outside diameter.

With extremely high frequencies the current flows only in a thin layer on the outside edge of the conductor. We may just as well use a hollow conductor under such circumstances (waveguides in radar applications: 3 to 10 GHz!).

### Pronounced and non-pronounced skin effect

Fig. 3 shows the current density respectively for non-pronounced and pronounced skin effect. Figure 3. Current density in conductor

We can determine the factor u which indicates if the skin effect is pronounced or not:

$d =$ diameter of conductor (m)
$\mu_0 = 4\pi \cdot 10^{-7}$ (H/m)
$\mu_r =$ relative permeability
$\rho =$ specific resistance (Ω m)
$f =$ frequency (Hz)

If u > 6 there is pronounced skin effect (S.E.)

For copper wire ( ρ = 1.72 x 10-8 Ωm ) and with d  in mm and f  in kHz, gives : $u = 0.33868\cdot d\sqrt{f}$

The ohmic AC-resistance is calculated with: $R_{AC} = R_{DC}\cdot K$

u > 6 gives: $K = \frac{u}{2\sqrt{2}} + 0.25$

u < 6: K is given in table 1 Table 1. K values for non-pronounced skin effect

### Skin depth figure

Fig. 4 shows the current distribution inside a conductor in relation to the current density on the circumference of the conductor. Figure 4. Current density with pronounced skin effect

An interpretation for the skin depth is the distance to the circumference whereby the current density is reduced to 1/e times the current density of the circumference. Another approach is the replacement of the distributed current density by a fictive constant current density (amplitude and phase!) in a layer Sdi at the circumference of the conductor. We consider then the useful current carrying part of the conductor, a cross-sectional area given by $S_{di}\pi d$.

The value for the skin depth is given by:

In the case of copper conductors and with f  in kHz then Sdi  in mm: $S_{di} = \frac{2.08}{\sqrt{f}}$

Numeric example:

1. Given:        f = 100kHz; respective cross-sectional area of 0.76mm (AWG21) and 0.56mm (AWG24)
Question:   pronounced or non-pronounced skin effect?

• $R_{AC} =$ ?
• $S_{di} =$ ?

Solution:

a) wire 0.76 mm:
$u = 0.33868\cdot 0.76\sqrt{100} = 3.01$; u < 6 → non-pronounced S.E.

Table 1: $K = 1.318;\; R_{AC} = 1.318\cdot R_{DC}:$ the AC- resistance is 31.5% larger than the DC-resistance of the wire.

$S_{di} = \frac{0.208}{\sqrt{100}} = 0.208 \text{ mm}$

b) wire 0.56 mm:
$u = 0.33868\cdot 0.56\sqrt{100} = 1.89$; u < 6 → non-pronounced S.E.

Table 1: $K = 1.067;\; R_{AC} = 1.067\cdot R_{DC}:$

$S_{di} = 0.208 \text{ mm}$

2. Given:        a copper wire with d =1.35 mm (AWG16). We use frequencies of 100kHz and 1MHz
Question:   determine for both frequencies if there is pronounced S. E.

Solution:

a) f1 = 100kHz:
$u = 0.33868\cdot 1.35\sqrt{100} = 4.572$ → non-pronounced S.E.

Table 1: $K = 1.86;\; R_{AC} = 1.86\cdot R_{DC}:$

$S_{di} = \frac{0.208}{\sqrt{100}} = 0.208 \text{ mm}$

b) f2 = 1MHz:
$u = 0.33868\cdot 0.56\sqrt{10^4} = 45.72$; u > 6 → pronounced S.E.

$K = \frac{u}{1\sqrt{2}} + 0.25= 16.41;\; R_{AC} = 16.41\cdot R_{DC}:$

$S_{di} = \frac{2.08}{\sqrt{10^4}} = 0.0208 \text{ mm}$

If we use the model for the skin depth with constant current division, then we find a cross-sectional area for current conduction:

$S_{di}\cdot \pi \cdot d = 0.0208\cdot \pi \cdot 1.35 = 0.08821$ mm2

The cross-sectional area for DC current would be: $\frac{\pi d^2}{4} = \frac{\pi \cdot 1.35^2}{4} = 1.4314$ mm2

The ratio $\frac{1.4314}{0.08821} = 16.23$ approaches the value for K !

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### This post was written by:

- who has written 7 posts on PowerGuru - Power Electronics Information Portal.

Professor Dr. Jean Pollefliet is the author of several best-selling textbooks in Flanders and the Netherlands

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