TX to Ae Ch 2. Contents << Part 1 Part 3 >>

2. Components and Materials: Part 2.
by David Knight
2-6. Inductance of a wire:
As should be apparent from all of the foregoing discussion, every electrical conductor exhibits inductance when it carries a current, and it is useful to have an idea of what that inductance will be. Many radio textbooks address this issue by quoting a formula for the inductance of a straight cylindrical wire, derived by Edward B. Rosa, which appeared in the Bulletin of the American Bureau of Standards in 1908 [Rosa 1908][see also "Inductance of a Straight Wire", by Tim Healy]. We will look at this formula in a somewhat more detailed way than usual; so that we can try to find out why it is that such a widely quoted and used (and misquoted, and misused) equation tends to lead people to the wrong answers (even though it is correct). Rosa's formula can be written:
 L / l = Li + (μ(e)/2π) [ln(4l/d) - 1]     Henrys / metre,     l>>d
(Note: The notation used here differs from that used by Rosa)
The quantity on the left-hand side L/l is the inductance per unit length; which means that the number given by calculating the right hand side must be multiplied by the length of the wire l in order to obtain an inductance in Henrys. The right hand side of the equation is separated into two parts: The term "Li" is the so-called internal inductance of the wire, i.e., the inductance per metre due to the magnetic fields inside the wire; and the term after it is the external inductance, which we might also call "Le", i.e., the inductance per metre due to the fields outside the wire. "ln" means "take the natural logarithm (log to the base e, or Loge) of the quantity in the brackets directly after it, and "d" is the diameter of the wire. The quantity inside the logarithm brackets must be dimensionless, and so d must be in the same units as l.
μ(e) is the magnetic permeability of the medium external to the wire. It is of course composed of a relative permeability multiplied by the permeability of free space, i.e.:
μ(e) = μ0 μr(e)
Thus if the wire is surrounded by air, which has effectively the same permeability as free space (i.e., μr=1), then μ(e)0. If, on the other hand, the length of wire for which we intend to calculate the inductance is surrounded by a tight-fitting ferrite bead of (say) μr=850, then the external inductance will be 850 times greater than it would have been for the same wire with no magnetic materials in its immediate vicinity.
At very low frequencies (DC), the internal inductance per unit length is given by:
Li(DC) = μ(i) / 8π     Henrys / metre
(see, for example., ref [9a], section 5.17) where μ(i) is the permeability of the material from which the wire is made. If the wire is non-ferromagnetic (e.g., copper), then we know from previous discussion that we may assume, to an excellent approximation, that μ(i)0.
Now, if we assume that the wire is non-magnetic, and it is surrounded by air, we may rewrite Rosa's equation:
L(DC) / l = (μ0 /8π) + (μ0/2π)[ln(4l/d) - 1]     Henrys / metre
and observing that the internal and external inductances have a common factor μ0/2π, we obtain:
L(DC) / l = (μ0/2π)[ln(4l/d) - 1 + ¼]     Henrys / metre.
Now, since μ0=4π×10-7H/m, we can reduce the equation further to:
 L(DC) = 200 l [ln(4l/d) - 0.75]     nano Henrys
This, apart from possible conversion to non-SI units, is the form in which Rosa's formula appears in handbooks. The subscript "(DC)" however is likely to be omitted, and herein lies the first misuse. It is normal to state that the internal inductance term disappears at "very high frequencies" due to the skin-effect, in which case the 0.75 term becomes 1. "Very high frequencies" may also be carelessly changed to "VHF", but the formula was derived in 1908, when "great frequency" (Rosa's actual phrase) meant: "frequencies beyond the capability of mechanical alternators". In any case, we may be sure that the internal inductance has gone at 'infinite' frequency and so may write:
 L∞ = l Le = 200 l [ln(4l/d) - 1]     nano Henrys
(where Le is the external inductance per metre). At intermediate frequencies, where internal inductance must be taken into account, we may correct the formula thus:
L = l ( Le + Li )

Internal Impedance
To find the internal inductance at a particular frequency, we may start with the internal impedance of a wire; which is given (for example) by Ramo, Whinnery and Van Duzer (ref [9a], section 5.17) for the 'high frequency' case as:
Zhf / l = Rs (1 + j ) / (π d)      [Ohms / metre]
Where Rs is the surface resistivity of the wire, and is given by the expression:
Rs = √(π f μ(i) ρ )     [Ohms]
μ(i) being the permeability of the wire (=μ0 for non-ferromagnetic), and ρ being the ordinary volume resistivity (in Ohm metres). Alternatively:
Rs = √(π f μ(i) / σ )     [Ohms]
where σ is the conductivity (in Siemens / metre).
For those who seek confirmation that Rs really is measured in Ohms, note that the reactance formula XL=2πfL tells us that Henrys (neglecting the dimensionless 2π term) are equivalent to Ohm seconds, and so the units of permeability (H/m) can be given as Ohm seconds / metre. The square root of [ / second]×[Ohm seconds / metre]×[Ohm metres] is Ohms.
The expression above for internal impedance is stipulated to apply only at 'high frequencies', because it uses the approximation that the current carrying layer is sufficiently thin that that it can be regarded as a sheet of width equal to the circumference of the wire (this being the πd term on the bottom of the fraction). We can therefore qualify the statement a little, by saying that the approximation is good for 'high frequencies' or 'large diameters'. It is incidentally, exactly the same approximation as was used earlier when calculating the RF resistance of wires, and we may gain an idea of when it will fail us by comparing the wire diameter with the table of conduction layer thicknesses for various metals at various frequencies given earlier. We may note, for example, that copper has a skin depth of 50 μm at 1.75 MHz, and so for a copper wire of 1 mm diameter the approximation will be very good, but for a wire of 0.2 mm (200μm) diameter, it will introduce a significant error.
Now we, of course, want the internal inductance of the wire. We may therefore take the internal reactance, which is the imaginary part of the internal impedance, and obtain the inductance from the reactance formula XL=2πfL, i.e.:
Xi(hf) / l = Rs / (π d)
= [ √(π f μ(i) ρ ) ] / (π d)     [Ohms / metre]
and
Li(hf) / l = Xi(hf) / 2πf
= [ √(π f μ(i) ρ ) ] / (2π² f d)    [Henrys / metre]
Now, recalling the trick for rearranging equations involving square roots, i.e., any number is the square of its own square root; and factoring out μ/2π (which, as you may recall, is the common factor for the two parts of Rosa's formula when the internal end external permeabilities are the same), we obtain:
Li(hf) / l = (μ(i)/2π) / [d √(π f μ(i) σ ) ]
Which takes us all the way back to equation (2.2), for the skin depth, i.e.,
δi = 1 / √( π f μ σ )
Thus:
 Li(hf) / l = (μ(i)/2π) (δi / d)    Henrys / metre 6.1
and so it transpires, that the internal inductance at hf is simply μ/2π times the ratio of the skin depth to the conductor diameter. We can even work out what 'hf' means, by recalling that:
Li(DC) / l = (μ(i)/2π) ¼
Since Li(hf) can never be greater than Li(DC), then our expression for Li(hf) can only be valid when d is considerably greater than 4δi. It should also be apparent that the internal inductance starts to diminish at frequencies a lot lower than "VHF" in the modern sense.
Using the result just obtained, we can rewrite Rosa's formula properly for the 'high frequency' (or large diameter) case. We will do so first on the assumption that the wire is not ferromagnetic, and the external medium is air; thus:
 L = (μ0/2π) l [ln(4l/d) - 1 + (δi/d)]     Henrys,    d >> 4δi or L = 200 l [ln(4l/d) - 1 + (δi/d)]     nano Henrys,    d >> 4δi where δi = √[ ρ / ( π f μ0 ) ] 6.2
We can also generalise the formula to wires and external media of any permeability by multiplying the internal and external inductance terms by their respective relative permeabilities, thus:
 L = 200 l { μr(e)[ln(4l/d) - 1] + μr(i)(δi/d) }     nano Henrys,    d >> 4δi where δi = √[ ρ / ( π f μ0 μr(i) ) ]
The question that remains, of course, is does the formula work? To resolve this issue we need to examine some data, and suitable measurements are given in an article by Ian Hickman in Electronics World [17]. The article discusses the matter of wire inductance from a completely different perspective to the one used here, the point being to instil the idea that the inductance of a wire is about 10 nH/cm (25 nH/inch), and is more-or-less exactly so for a wire of 0.4 mm diameter. Ian Hickman chose to illustrate this point by making three impedance measurements on 20 cm long loops of copper wire using a HP8753D Network Analyser. The measurement frequency was 5 MHz. The results were as follows:
 SWG Wire diam / mm Inductance / nH 16 1.63 142 28 0.376 201 36 0.193 235.5
 So how do these results compare to our modified version of Rosa's formula (equation 6.2)? To find out, we start by calculating the skin depth for copper (ρ=17.241 nΩm, μr=1) at 5 MHz, i.e.: δi = 1 / √( π f μ0 / ρ )     = 1 / √[ π × 5 × 106 × 4 × π × 10-7 / ( 17.241 × 10-9 ) ]    = 29.55 μm This quantity (before truncation to two decimal places) is used to calculate the δi/d internal inductance correction term for equation (6.2), and the results are shown in the table below:
Copper wire in air. Wire length=20 cm. Frequency=5 MHz.
 SWG Wire diam d / mm δi / d Measured L / nH Calculated L / nH 16 1.63 0.0181 142 208.57 28 0.376 0.0786 201 269.66 36 0.193 0.1531 235.5 299.31
The δi/d figures in the table show that the internal inductance contribution to the calculated inductance is very small except for the case of the ultra-thin 36SWG wire; i.e., "VHF" has already happened at 5 MHz. The calculated inductances however are much too high, and so we need to find out what went wrong. The problem, as pointed out by Ian Hickman, and of which Rosa was certainly aware, is that it is impossible to make an unambiguous definition of inductance without considering the electrical circuit that gives rise to it. A formula for the inductance of an isolated length of wire begs the question: "how can the current (which gives rise to the inductance) get into the wire without creating fields that interact with the fields used to define the inductance?" The answer is that it can't. Ian Hickman made his measurements by forming the wire into a loop, so that he could connect it to the terminals of a network analyser; and although the curvature of such a loop is sufficiently low that we might consider a small segment of it to be reasonably straight; what he actually measured is the inductance of a circuit. So, can we take the magnetic fields due to the whole circuit into account?

Loop Inductance
The clue to how we might improve the accuracy of the calculation is given by the well known formula for the external inductance of a circular loop (see for example, ref [9], section 5.24, or ref [18] p146). The expression is:
Le = R μ(e) [ln(8R/r) - 2]     [Henrys].   R>>r
Where Le is the external inductance of the loop, R is the radius of the loop, and r is the radius of the wire.
Now, we would like to relate this expression to Rosa's formula, and so straight away we will equate: l=2πR, i.e., the length of the wire is the circumference of the loop. Hence:
Le = l(e)/2π) [ln(8l / 2πr) - 2]     [Henrys]
and we can replace the radius of the wire with its diameter using d=2r, i.e.:
Le = l(e)/2π) [ln(8l /πd) - 2]     [Henrys].
This looks very much like the external inductance part of equation(6.2), except that the term inside the logarithm bracket has changed from 4l/d to 8l/πd, and the -1 has become a -2.
Now recall that two numbers can be multiplied by adding their logarithms and taking the antilogarithm, i.e.:
Log(ab) = Log(a) + Log(b).
This works for natural logarithms, not just base 10, and so we can remove an additive term from the logarithm in the expression for external inductance. In this case the substitution we want is:
ln(8l /πd) = ln(4l/d) + ln(2/π)
i.e.,
ln(8l /πd) = ln(4l/d) -0.451582705
and so the formula for the external inductance of a wire loop can be written:
Le = l(e)/2π) [ln(4l/d) - 2.451583 ]     [Henrys].   l>>d
What we see here is that the -1 from Rosa's formula has changed to -2.45 by the act of forming the wire into a loop. All we have to do now is add the internal inductance, which is the same as before, to obtain (assuming a non-ferromagnetic wire in air):
 L = (μ0/2π) l [ln(4l/d) - 2.4516 + (δi/d)]     Henrys,   l >> d >> 4δi 6.3
We can now re-calculate the inductances of Ian Hickman's wire loops using this expression, and compare it with the results obtained from Rosa's formula.
Copper wire loop in air. Circumference=20 cm. Frequency=5 MHz.
 SWG Wire diam d / mm δi / d Measured L / nH Rosa L / nH Loop formula. L /nH 16 1.63 0.0181 142 208.57 150.50 28 0.376 0.0786 201 269.66 211.59 36 0.193 0.1531 235.5 299.31 241.25
 What we find is that the results from equation (6.3) have predicted the inductances within a few percent (and there would have been something badly wrong if they hadn't). So Rosa's formula is definitely wrong for this application. The two formulae are shown below placed next to each other:
 Rosa L = 200 l [ln(4l/d) - 1 + (δi/d)]     nH,   l >> d >> 4δi 6.2 Loop L = 200 l [ln(4l/d) - 2.4516 + (δi/d)]     nH,   l >> d >> 4δi 6.3