ADDITIONAL TOPICS SKIN EFFECT: HF RESISTANCE OF A CONDUCTOR
Consider the cylindrical conductor shown in Figure 2.30a, which is carrying a current / into the paper (x). The magnetic field B of / is clockwise. Consider two magnetic field values B and B2, which are shown in Figure 2.30a. B is inside the core and B2 is just outside the conductor.
Assume that the conductor is divided into two conductors. The hypothetical cut is taken just outside of B. The conductor in Figure 2.30a is now cut into a hollow cylinder and a smaller solid cylinder, as shown in Figure 2.30b and c, respectively. The currents 11 and I2 in the solid and hollow cylinders sum to /. We can arrange things and choose B such that our cut gives I = I2 = l. Obviously, I flowing in the inner conductor is threaded (or linked) by both B and B2. (Remember that B is just inside the conductor in Figure 2.30b, so it threads at least 99% of I.) On the other hand, the outer conductor is only threaded by B2, simply because I2 Jflows in the hollow cylinder and there is no current in the hollow, which means that B is not threaded by I2. Clearly, 7i threads more magnetic field than I2 and thus conductor (c) has a higher inductance than (b). Recall that inductance is defined as the total magnetic flux threaded per unit current. Consequently, an ac current will prefer paths near the surface where the inductive impedance is smaller. As the frequency increases, the current is confined more and more to the surface region.
For a given conductor, we can assume that most of the current flows in a surface region of depth 8, called the skin depth, as indicated in Figure 2.31. In the central region,
(b) Current in hollow outer cylinder is 1/2.
(c) Current in solid inner cylinder is 1/2.
Figure 2.30 Illustration of the skin effect.
A hypothetical cut produces a hollow outer cylinder and a solid inner cylinder. Cut is placed where it would give equal current in each section. The two sections are in parallel so that the currents in (b) and (c) sum to that in (a).
Figure 2.31 At high frequencies, the core region exhibits more inductive impedance than the surface region, and the current flows in the surface region of a conductor defined approximately by the skin depth, <5.
The current will be negligibly small. The skin depth will obviously depend on the frequency co. To find 8, we must solve Maxwell’s equations in a conductive medium, a tedious task that, fortunately, has been done by others. We can therefore simply take the result that the skin depth 8 is given by
Skin depth for conduction
8 = ■■—}= [2.53]
Where co is the angular frequency of the current, a is the conductivity (a is constant from dc up to ~ 1014 Hz in metals), and ii is the magnetic permeability of the medium, which is the product of the absolute (free space) permeability /jlq and the relative permeability /zr.
We can imagine the central conductor as a resistance R in series with an inductance L. Intuitively, those factors that enhance the inductive impedance coL over the resistance R will also tend to emphasize the skin effect and will hence tend to decrease the skin depth. For example, the greater the permeability of the conducting medium, the stronger the magnetic field inside the conductor, and hence the larger the inductance of the central region. The higher the frequency of the current, the greater the inductive impedance coL compared with R and the more significant is the skin effect. The greater is the conductivity a, the smaller is R compared with coL and hence the more important is the skin effect. All these dependences are accounted for in Equation 2.53.
With the skin depth known, the effective cross-sectional area is given approximately by
A — na1 — n(a — <5)2 % 2ixa8
HF resistance per unit length due to skin effect
Where 82 is neglected (8 < a). The ac resistance rac of the conductor per unit length is therefore
Where p is the ac resistivity at the frequency of interest, which for all practical pur
Poses is equal to the dc resistivity of the metal. Equation 2.54 clearly shows that as co increases, 8 decreases, by virtue of 8 a and, as a result, rac increases.
From this discussion, it is obvious that the skin effect arises because the magnetic field of the ac current in the conductor restricts the current flow to the surface region within a depth of 8 < a. Since the current can only flow in the surface region, there is an effective increase in the resistance due to a decrease in the cross-sectional area for current flow. Taking this effective area for current flow as 2na8 leads to Equation 2.54.
The skin effect plays an important role in electronic engineering because it limits the use of solid-core conductors in high-frequency applications. As the signal frequencies reach and surpass the gigahertz (10 Hz) range, the transmission of the signal over a long distance becomes almost impossible through an ordinary, solid-metal conductor. We must then resort to pipes (or waveguides).
SKIN EFFECT FROM DIMENSIONAL ANALYSIS Using dimensional analysis, obtain the general form of the equation for the skin depth 8 in terms of the angular frequency of the current co, conductivity cr, and permeability /x.
The skin effect depends on the angular frequency co of the current, the conductivity cr, and the magnetic permeability /x of the conducting medium. In the most general way, we can group these effects as
Where the indices x, y, and z are to be determined. We then substitute the dimensions of each quantity in this expression. The dimensions of each, in terms of the fundamental units, are as follows:
Quantity Units Fundamental Units Comment
TOC o "1-5" h z 6 m m
Co s-1 s”1
S2“1 m“1 C2 skg“1 m“3 Q = VA~i = (JC-‘XCs“1)“1
= Nms C~2 = (kg m s~2)(m s C~2)
H Wb A”1 m~1 kg m C~2 Wb = T m2 = (N A-1 m-1 )(m2)
= (kg m s~2)(C~! s)(m)
Clearly, x = y = z = is the only possibility. Then, 8 a [coa. It should be reemphasized that the dimensional analysis is not a proof of the skin depth expression, but a consistency check that assures confidence in the equation.
SKIN EFFECT IN AN INDUCTOR What is the change in the dc resistance of a copper wire of radius 1 mm for an ac signal at 10 MHz? What is the change in the dc resistance at 1 GHz? Copper has pdc = 1.70 x 10“8 Ј2 m or crdc = 5.9 x 107 Q~ m-1 and a relative permeability near unity.
Per unit length, rdc = pdc/na2 and at high frequencies, from Equation 2.54, rac = pdc/2na8. Therefore, rac/rdc = a/28.
We need to find 8. From Equation 2.53, at 10 MHz we have
= [^coadcfji] 1/2 = [| x 2n x 10 x 106 x 5.9 x 107 x 1.257 x 10“6] 1/2
2.07 x 10“5 m = 20.7 /xm
= — = <io~3 m> _ 2413 rdc 2S (2 x 2.07 x 10"5 m) ‘
The resistance has increased by 24 times. At 1 GHz, the increase is 240 times. Furthermore, the current is confined to a surface region of about ^2 x 10~5(20 |im) at 10 MHz and ~2 x 10~6 m (2 |im) at 1 GHz, so most of the material is wasted. This is exactly the reason why solid conductors would not be used for high-frequency work. As very high frequencies, in the gigahertz range and above, are reached, the best bet would be to use pipes (waveguides).
One final comment is appropriate. An inductor wound from a copper wire would have a certain Q (quality factor) value9 that depends inversely on its resistance. At high frequencies, Q would drop, because the current would be limited to the surface of the wire. One way to overcome this problem is to use a thick conductor that has a surface coating of higher-conductivity metal, such as silver. This is what the early radio engineers practiced. In fact, tank circuits of high-power radio transmitters often have coils made from copper tubes with a coolant flowing inside.