Conductors at Radio Frequency
"Mathematics . . . a kind of lazy-tongs by means of which conclusions may be reached without straining the intellect."
Alfred M. Still
I have a problem. What is the better form of conductor to use at radio frequency? Is it a tube or is it a flat strip? The problem is not made any easier by the fact that one contender expresses his views non-mathematically in a way which is at once easy to grasp, appeals to the reason and exercises one's appreciation of some fundamental physical realities, whilst his opponents state theirs in terms of partial derivatives and Bessel functions. But let me introduce the contenders.
In the red corner, the late, the great, Frederick Emmons Terman, ScD, Professor of Electrical Engineering and Dean of the School of Engineering, Stanford University; Past President of the Institute of Radio Engineers, author of "Radio Engineering", "Radio Engineers' Handbook" etc etc.
In the blue corner, the almost-unheard-of Mr. A.G. Warren, late of the Armament Research Department, Ministry of Supply, ably aided and abetted by the late Professor Alexander Russell, Fellow of the Royal Society, the late Professors C.L. Fortescue and G.W.O. Howe, the internationally renowned Mr. S. Butterworth designer of filters extraordinaire and that Victorian bastion of the scientific community, His Lordship the late Lord Rayleigh, whose collective achievements and honours would fill (and do fill) several books. It should be noted that Messrs. Russell, Fortescue, Howe, Butterworth and Rayleigh each contributed a significant portion to our knowledge of the flow of alternating current in metallic conductors at high frequencies. Also sticking their oar in for the blue corner (a rowing blue no doubt; somewhat unsportingly in my view) we have apparently the assistance of the Bureau of Standards, Washington.
Right now it doesn't look too good for Professor Terman, but it has been known for more hopeless cases to rise triumphant.
Let me state the cause of the dispute.
Professor Terman says (p. 20, "Radio Engineering") that a flat strip conductor is not a good choice for radio frequencies since the interaction of the current with the magnetic flux it produces will, by virtue of producing the skin effect, force the current to flow down the two opposite edges of the strip and hence the majority of the surface is unused as a conductor. The suitability of a conductor for radio frequencies is not a simple matter of its surface area, but the disposition of the conductor in space is a vital factor. He implies that a tubular conductor of circumference equal to the width of the flat strip is always to be preferred.
Dr. Cadd (that's me) adds by way of comment that it is easy to see Professor Terman's point by considering a cross-sectional view through a circular wire carrying radio frequency current. By the skin effect, the radial current distribution decreases exponentially from the outside, tending thus to be concentrated in an annulus around the circumference of the wire. A flat strip is simply a thin slice taken across the diameter of the wire, in which the current flows down the two edges, exactly as Professor Terman states.
So far, so good. Now it's Mr. Warren's turn. Mr. Warren says . . . actually, Mr. Warren doesn't say much at all, leastways not in English, and therein lies half my problem. Mr. Warren mutters various obscure incantations, inscribes mystical symbols and makes peculiar signs with his learned pen which only the truly wise and enlightened may comprehend the significance of. However, I do not hold that against him, nor do I see any great reason to curtail his views simply because they are expressed in a language other than my own. Unfortunately, he does go on a bit; succinctness is not a strong point of his. Mr. Warren, the webpage (and half of cyberspace) is yours .....
The wide parallel strip is represented as in the diagram where it is shown in cross-section. Its width is b , its thickness 2a ; b >>2a. The current flows in a direction represented by the normal to the paper (or screen - Ed.). At a distance x from the mid plane OO' the current density is g , the intensity of the magnetic field is H (directed along BC, or CB), the flux per unit length within the conductor, and to the right of BC, is f; the external flux to the right of the conductor is f'; the current within the section OBCO' is Ix ; the potential difference per unit length of the conductor is e . Proceeding round the path ABCD the current enclosed is 2Ix and therefore the magnetomotive force is 8pIx . This is equal to H x 2b . (In what follows, m is the magnetic permeability of the conductor, w is 2pf, r is the resistivity of the conductor and j is the square root of -1. Ed.) Hence
From equation 1
From equations 4 and 5
Differentiating this with respect to x gives
Assuming that g varies sinusoidally and interpreting equation 6 as a vector equation one obtains
the solution to which is
Since g is the same for ±x , A1 = A2 and the solution may be written
The current per centimetre width of the conductor is
then at the surface
and therefore the impedance per centimetre of 1cm width of the conductor is
(This impedance does not include the "external" reactance; this does not matter since Z1 is being determined to find R1, the resistance of 1cm width of the strip.)
Putting 2a = t , the thickness of the strip, gives
where R1 is the resistance of a strip 1cm wide.
At low frequencies (mains power frequencies) equation 15 may be used to determine the increase in resistance due to eddy currents. To the first order, when a is small,
For copper, k = 0.215f½ and therefore a = 0.152f½. At f = 50, a = 1.075. For a strip 1cm thick at = 1.075; sinh 1.075 = 1.2943; sin 1.075 = 0.8796; cosh 1.075 = 1.6356; cos 1.075 = 0.4757; and so
an increase of resistance of 0.74%.
An approximate expression for l when at is small, is
At high frequencies, when at becomes big, sinh at and cosh at become equally very great, while sin at and cos at remain finite so that the resistance of unit width of strip approximates asymptotically to ra/2. Thus at f = 106, a = 152 and a strip of considerable thickness would have a resistance of 76r, independently of the actual value of the thickness. It is clear, however, from equation 15, that at a constant frequency, as the thickness is increased, the resistance passes through a number of maxima and minima. A strip of finite thickness may have a lower resistance than one which is indefinitely thick. Using the same nomenclature as for equation 14, then if
Maximum and minimum values occur when
Since both S and s being zero corresponds to a conductor of no thickness this case cannot be considered. S <> 0, therefore s = 0 for a maximum or minimum value.
At maximum or minimum values at = p, 2p, 3p, 4p, etc. When at = 2np, cos at = +1 and
When at = (2n + 1)p, cos at = -1
Successive maximum values of R1 are
Successive minimum values of R1 are
The particular value of interest is tanh(p/2) = 0.917; all subsequent values differ from unity by only a negligible amount. For this case, when at = p, i.e. t = p/152 = 0.0206 cm, the resistance is 8.3% less than the resistance of a thick strip.
Thank you Mr. Warren. It appears then from your work that a copper strip just a shade over 0.2mm in thickness has a resistance at 1Mc/s (a = 0.152 x the square root of frequency) which is over 8% lower than any other thickness. Unfortunately, Mr. Warren does not see fit to extending the argument to tubular conductors, which presumably follow a similar pattern.
The difficulty I have with this is that effectively this is saying that flat strip conductors are good at radio frequency, which disagrees with Prof. Terman. The specific argument I will pick with Mr. Warren's argument is that he has chosen from the beginning the way in which the current distributes itself over the conductor cross-section. Is there any good reason for not doing the calculations with the geometry as shown here, and then working out how the balance of factors distributes the current over the strip?
I would also suggest reaching the desired answer by considering a gradual perturbation of the system, by drawing the circular wire shown beside Professor Terman's argument through progressively more oval dies of equal circumference and noting the redistribution of current at each stage, until the last die is rectangular and of thin section.
Presumably, depending on frequency, resistivity, dimensions and current, an equilibrium will be reached between the penetration of the current and flux into the strip, and the forces tending to move the electron flow to the surfaces and to the edges of the strip. That is the computation which needs to be done in order to decide the current distribution and hence the strip resistance.
So is Professor Terman right or wrong when he says that a flat strip is not a good choice at radio frequencies? Evidently the basis for the calculations is here.
But how to actually do it?! I have no idea, the mathematics are quite beyond me. Perhaps someone has measured the radio frequency resistance of a strip and a tube of the same thickness on a bridge and factored the resistances according to the width of the strip and the circumference of the tube. I just wish I could find the answer somewhere.
The next section concludes the journey from the generator to the load with a look at inductive coupling.