Field Precision title

How to calculate inductance (1)

Finding inductance is one of the main applications of our magnetic field codes PerMag (2D) and Magnum (3D). In this note I'll discuss how to use the static-field codes to determine inductance at low and high frequency. I'll concentrate on the inductance between two wires. Another article covers the energy method for high-accuracy calculations of mutual inductances in multi-wire systems. As an illustration I will use a PerMag calculation of the inductance per length of a two-wire transmission line (e.g., the lead wire on an old-time TV antenna).

The basic idea is to create a field solution with a drive current I = 1.0 A and then to calculate the integral of field energy U. The inductance is related to the field energy by

U = L*I^2/2.

This approach is well-suited to numerical calculations. It is accurate because it involves a global integral rather than a surface integral over a faceted boundary.

The tranmission line consists of two wires of diameter D carrying current in opposite directions. The wire centers are separated by distance S. Checking S. Ramo, J. Whinnery and T. Van Duzer, Fields and Waves in Communications Electronics (Wiley, New York, 1965), Table 8.09, the inductance per unit length is:

l = (μ0/π)*acosh(S/D).

(The reference contains a clue to when I bought my textbooks.) For the example, we shall use S = 6.0 cm and D = 2.0 cm. The expected result is 7.0510E-7 H/m.

Here are the input files if you want to run the example

two_wire.min
two_wire.pin
two_wire.scr

There are some interesting features in the setup:

  • The calculation has planar symmetry (x,y) with infinite extent in z (out of the page). The code determines field energy per length u from which we can determine inductance per length l.
  • By symmetry, magnetic field lines are parallel to the midplane between the wires. Therefore, we need only model half the system and apply the Dirichlet condition Az = 0.0 along the midplane. We must remember to multiply the code result for u by 2.0.
  • Numerical models of familiar analytic benchmarks are often difficult. In this case, the ideal solution extends to infinity. We include a large surrounding volume at low element resolution with Dirichlet boundaries (conducting box). We must realize that the answer will not be perfect. It will get closer to the analytic value as we make the solution volume larger.

The solution has three regions: 1) the main volume with μr = 1.0, 2) the wire with a drive current of 1.0 A and 3) the Dirichlet boundary with Az = 0.0. The figure below marked "Low frequency" shows a detail of field lines when μr = 1.0 in the wire. By default, PerMag has distributed the current density uniformly over the wire cross section. This condition is value for static solutions or at very low frequency. There is significant field energy (i.e., self-inductance) inside the wire. We can estimate the valid frequency range from the formula:

f ≪ ρ/π μ0 D^2,

where ρ is the resistivity of the wire. For copper wire with diameter D = 0.02 m and resistivity ρ = 1.678E-8 ohm-m, the low frequency solution is valid when f < 500 Hz.

Usually, we are interested in the inductance at much higher frequencies. In this case, the current lies close to surface of the wire. The conundrum is how do we set the current distribution? The layer is quite thin and the current density varies with position around the wire perimeter. The answer is to let the code do the work. What we really want to do is to ensure that the normal component of B equals zero around the wire boundary. We can achieve this condition by setting μ ≪ 1.0 inside the wire. Figure 1 shows the field distribution with μ = 0.0005. Here are some considerations and results:

  • The low value of μ impedes solution convergence. We must set a large value of MaxCycle to ensure convergence. If you pick too low a value for the magnetic permeability, you may get an invalid solution with strange-looking field lines.
  • To check the physical validity of the solution, we can perform an ampere integral around the wire to confirm that the enclosed current is close to 1.0 A. The integral in two_wire.scr gives a value of 0.9995 A.
  • Volume integrals of the field energy u appear at the end of the listing file two_wire.pls. The value for a solution volume with sides of about 30 cm is u = 1.72203E-07 J/m, impying that l = 6.8881E-7 H/m (about 2.3% below the ideal value). The difference depends on the solution size. If we reduce the solution volume length to about 15 cm, the code gives l = 6.4931 H/m (7.9% difference).

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Figure 1. Two wire example, high frequency.


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