Suppose there is a conductor above a ground plane. Current flows from a source through the conductor to a load on the other side. Depending on the frequency of the current the return path through the ground plane can take a variety of paths. At DC the return current takes a straight line. As the frequency starts to increase the current starts following the path under the conductor. Why is that?

Initially I thought that the current path under the conductor minimizes the current loop area. But why does the current path change depending on the frequency? I guess that has something to do with the impedance at different frequencies.

Edit Here is a diagram of what is going on: return path

I'm looking for an explanation why this is happening.

  • $\begingroup$ In a superconducting plane the DC path wouldn't take a "straight path", either, because the magnetic field could not penetrate the superconductor, i.e. the field could not spread out beyond the conductor about the plane. Only in the case that the penetration depth into the metal is large can there be significant field components far away from the conductor and only in that case can a current be induced in the plane far away from the conductor. $\endgroup$
    – CuriousOne
    Jun 15 '16 at 17:32
  • $\begingroup$ It is not clear to me what the scenario is which you are asking about. Perhaps a diagram would make it clear. Are you claiming that this effect does happen, what is your authority? Where is this stated? If this is what you think will happen, please can you explain why you think so? Have you seen it happen? $\endgroup$ Jun 16 '16 at 1:56
  • $\begingroup$ @sammygerbil I'm saying it does happen and is widely known in the electrical engineering community. Pretty much all of digital PCB layout is based on this principle in order to meet regulatory compliance. $\endgroup$
    – user110971
    Jun 16 '16 at 17:54
  • $\begingroup$ Thanks for the diagram but personally I'm still having trouble visualising it. Are you describing a PCB? If this is well known in the EE community, have you tried asking on Elec Eng Stack Exchange? $\endgroup$ Jun 16 '16 at 20:16
  • $\begingroup$ @sammygerbil Yes, it is a PCB. It's a track on the top layer with a ground plane on the bottom layer. I am looking for an explanation why this is happening from an EM perspective i.e. Maxwell's equations. That is why I asked it here. $\endgroup$
    – user110971
    Jun 16 '16 at 20:20

This is a really common effect observed in high-frequency circuits, and it's often the difference between a good one and a noisy one.

There are two relevant length scales in this problem: the first is the distance between the plane and the conductor, and the second is the wavelength of the electromagnetic waves being generated by the circuit. If you're coming at this from an electrical engineering perspective (which is where this kind of effect is most often studied) it's sometimes easy to forget that your circuits are all still just electric and magnetic fields and radiation. Whenever your wavelength is much longer than your circuit elements and wires, you can simply look for a low resistance straight-line path. Once your wavelength starts to be comparable or even shorter than your components (3 GHz corresponds to 10 cm wavelength in air) you have to start thinking in terms of EM waves and waveguides. At high frequencies, your conductors stop being equipotentials, but instead start to see charge bunching up dynamically and oscillating back and forth. The current path through the ground plane in your question changes because you cross over between these two regimes.

The clearest physical picture for the high frequency case is that power is being transferred by an EM wave. The conductors are basically sources of mobile charges which slosh around in response to the E and B fields and act to localize the wave in space. In the case of a thin conductor above a ground plane, this is what happens--the wave is localized between the two and follows along the wire. As the wave travels, it perturbs the electrons directly beneath the conductor more than anywhere else on the ground plane, which leads us to say that the return current runs directly beneath the top conductor. Your system becomes a type of what's called a "waveguide" at higher frequencies.

  • $\begingroup$ This still does not explain why the return current follows the conductor at frequencies where transmission line effects are not usually present. At 1 Mhz, which has a wavelength of 146 m in FR4, the return current follows the conductor. I am looking for a quantitative description of this effect. $\endgroup$
    – user110971
    Jun 16 '16 at 19:18

I did a bit of research and will give my own answer to my question for those that are interested. The return current follows the path of least impedance in the ground plane. There are two sources of impedance: the resistivity of the conductor, and the coupling between the trace and the ground plane: $$Z = R + jX\omega$$ At low frequencies the resistivity dominates. The path of least impedance is a straight line. As frequency increases the inductive coupling starts affecting the path, because it is dependent on the frequency.

The coupling is dependent on the loop area of the current, as given by Faraday's law. When the frequency increases enough for the resistivity to become insignificant, the current flows below the conductor in order to minimize the loop area, minimizing the impedance, as seen from the source.


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