 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 . 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 , section 5.24, or ref  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
 Now notice that if the length of the wire l becomes very long, the term ln(4l/d) will become considerably larger than either 1 or 2.45, in which case it will make little difference which formula is used. Hence it appears that we can define an inductance per unit length for infinitely large loops (except that the concept of lumped inductance presumes an infinite speed of light). In realistic situations, the loop formula always gives a lower inductance, and Rosa himself would not have expected anything else. The truth of the matter is that Rosa's formula is not a formula for the inductance of a wire, and he would no doubt have been horrified to think that it would come to be used in such a way. It is instead the starting point for the calculation of inductance, and it is perfectly correct when so interpreted. It sets up the magnetic field around the wire; and can be used in conjunction with other formulae (for mutual inductance), which establish the extent to which that field is cancelled or augmented by the rest of the circuit (All of the relevant techniques were developed by E B Rosa and F W Grover during the early part of the 20th Century, and were collected by Grover in his book "Inductance Calculations: Working Formulas and Tables" ). If the wire is formed into a loop, the magnetic field due to the outgoing conductor is partially cancelled by the field due to the return conductor (and vice versa) and the inductance is reduced. For a single-turn loop, the largest inductance is obtained when the area enclosed by the loop is at its maximum, i.e., when the loop is circular, because this is the configuration that gives the least cancellation of the magnetic field. This maximum possible value of inductance is given by the single-turn loop formula (6.3). The maximum inductance can only be exceeded by stacking conductors in such a way that their fields add together, i.e., by coiling the loop into several turns; in which case, if the turns are stacked closely, the inductance compared to that of a single turn is increased by a factor of N², where N is the number of turns. If a single turn loop is not circular, its inductance will be less than that predicted by equation (6.3), the extreme (but impossible) limit being reached when the outgoing and returning conductors are superimposed, the enclosed area is zero, and the inductance is zero.      So, after all of that, what is the inductance of a piece of wire? The correct, but unhelpful answer, is that wires only have a very small internal inductance, which disappears at moderate frequencies; but they exhibit a much larger external inductance due to the current flowing in the circuit of which they are part. Ian Hickman (among others) tells us that we should keep in mind that the inductance of a wire is about 10 nH/cm, and this is certainly a useful figure to remember, but it is obtained by taking a much longer length of wire and dividing the total inductance by the length.The bottom line is therefore; that the inductance of a small segment of a circuit does not exist in isolation from the rest of the circuit, and there is no such thing as inductance per unit length. We suggest therefore, that if you want to know the inductance of a piece of wire, then reformulate the question: take a look at the current-loop in which the wire is involved and estimate the inductance of that. Equation (6.3) will gave a fair approximation to the inductance if the loop is circular, and the inductance for other shapes can be estimated by multiplying this value by the ratio of the enclosed area of the actual loop to the enclosed area of a circular loop of the same length. Then, if you want the inductance of part of the loop, you can express it as a fraction of the total; but it won't tell you anything useful.       As a practical example, let us estimate the inductance of a wire-ended ½Watt metal-film resistor. Such a resistor might typically have a body-length of 9 mm, a body diameter of 2.5 mm, and lead diameter of 0.8 mm. To measure its inductance we might bend the leads over, trim them, and solder them to a miniature coaxial socket. This, without bending the leads dangerously close to the body, results in a total loop length of about 32 mm; and we may make the reasonable assumption that everything on the other side of the connector is magnetically isolated from this loop and will not modify its inductance. Thus we have 23 mm of 0.8 mm diameter conductor, and 9 mm of 2.5 mm diameter conductor. The average conductor diameter is therefore: (2.5 × 9 / 32) + (0.8 × 23 / 32) = 1.28 mm. If we want the inductance at several MHz, then the internal inductance can be neglected; and the circular loop formula (6.3) gives: L = 200 × 32 × 10-3 × [ln(4 × 32 / 1.28) - 2.4516] nH = 13.8 nH (within a few %) which is the upper limit for the inductance that will be found in practice. Misapplication of Rosa's formula (6.2) to this system gives L=23 nH which, given that the loop will not be truly circular, is too large by about a factor of 2.      One possible misleading implication of the foregoing discussion is that the stray inductance introduced by circuit wiring can be minimised by squashing the wires together to minimise the enclosed area. This is indeed true, but that does not mean that it is a good idea to do so. The problem is that by bringing conductors into proximity, the stray inductance will be minimised, but the stray capacitance will be maximised. We should instead be mindful of the two extremes: Maximum enclosed area: Maximum stray inductance, minimum stray capacitance. Minimum enclosed area: Minimum stray inductance, maximum stray capacitance. In the next section, for example, we will see that the high frequency characteristics of resistors are usually dominated by capacitance, in which case it is best to keep the connecting leads well apart.      One final caveat for this discussion, is that the concept of a lumped inductance to represent the inductance of a conductor breaks down without the assumption that the speed of light is infinite. Consequently, all of the preceding arguments assume that the length of any wires is very short in comparison to the wavelength at the operating frequency. If the length of a conductor becomes comparable to the wavelength, we must abandon lumped-component concepts and resort to transmission-line theory in order to understand the circuit behaviour. Part 3 >>

© D W Knight 2007, 2014
David Knight asserts the right to be recognised as the author of this work.  TX to Ae Ch 2. Contents << Part 1 Part 3 >>