This article explains the technique used at
Introduction
If we consider a two wire transmission line, we can define currents I1 and I2 flowing in the same direction in each conductor.
These two currents can be decomposed into differential and common mode components:
- Id=(I1-I2)/2; and
- Ic=(I1+I2)/2.
The diagram shows the relationship of the various example phasors, their sums and phase relationships.
Rearranging these, we can write:
- I1=Ic+Id; and
- I2=Ic-Id.
So the component currents Ic and Id fully account for the current at each terminal, I1 and I2.
I1 and I2 can be measured directly by placing a current probe around each of the wires, and I1+I2 (I12 or 2Ic) can be measured directly by placing a current probe around both of the wires.
Measurement of the magnitudes of these three currents I1, I2 and I12 can be resolved into components Id and Ic.
Calculation
The calculation is not very difficult, it used no more than high school maths.
The measured values I1, I2 and I12 are the magnitudes of three phasors, and for some of the calculator results, we need to find the magnitude of the phase angle between I1 and I2. The Law of Cosines provides the solution, \(|\theta_{12}|=acos\frac{(I_{12})^2-I_{1}^2-I_2^2}{-2 I_1 I_2}\).
We can then calculate using the Law of Cosines \(2I_d=I_1^2+I_2^2-2 I_1 I_2 cos \alpha\) and since \(\alpha=\pi-\theta_{12}\) we can write \(2I_d=I_1^2+I_2^2-2 I_1 I_2 cos(\pi-\theta_{12})\).
Now to find magnitude of the phase angle between Ic and Id, |θdc| using the Law of Cosines \(|\theta_{dc}|=acos\frac{I_2^2-I_c^2- I_d^2}{-2 I_c I_d}\).
Example
Example 1
Using the lengths of the phasors in the figure above, we can calculate the components.
Above, the example to scale.
Above, the calculated results.
My own 40m wire antenna
A graph is not presented, it will be impractical to read.
I12 varies from 10-30mA depending on recent weather, seasonal vegetation changes etc.
Low Ic is achievable with good design, good symmetry, and an effective common mode choke.