The Wikipedia article on ground plane says

In addition, a ground plane under printed circuit traces [the paths that the circuit currents take] can reduce crosstalk between adjacent traces. When two traces run parallel, an electrical signal in one can be coupled into the other through electromagnetic induction by magnetic field lines from one linking the other; this is called crosstalk. When a ground plane layer is present underneath, it forms a transmission line with the trace. The oppositely-directed return currents flow through the ground plane directly beneath the trace. This confines most of the electromagnetic fields to the area near the trace and consequently reduces crosstalk.

but unfortunately there are no citations. Page 5 of this document says the same thing ("Return Current Flow is directly below the signal trace") and gives a formula that the magnitude of the return current at point $x$ on the ground plane is proportional to $\cos(\theta)/r$, where $r$ is the distance between point $x$ and the circuit trace and $\theta$ is the angle that point $x$ makes relative to the vertical. But the document gives no explanation justifying this formula.

My naive intuition is that the return current would want to avoid the circuit trace, since opposite currents repel, and so it would spread out and increase crosstalk. Why does the return current instead concentrate below the circuit trace?


2 Answers 2


Remember that the story we tell children about electric currents --- that energy in electric circuits is carried by moving electric charges --- is somewhere between an oversimplification and a fiction. Energy flow in electric circuits is carried by electromagnetic fields. That's true even for DC circuits, like the one below. You can think of the transmission line between the voltage source and the load as a capacitor, with an electric field $\color{red}{\vec E}$ pointing from the upper wire down to the lower, and as an inductor with a magnetic field $\color{green}{\vec H}$ pointing into the page in the loop and returning outside. The magnitude and direction of power flow is given by the Poynting vector $\color{blue}{\vec S} \propto \color{red}{\vec E}\times\color{green}{\vec H} $, which points out from the battery, down the transmission line, and into the resistor. If you do the hard, three-dimensional integral to find out how much of the power flow in this sort of circuit is described by the Poynting vector, you find ... all of it.

Poynting vectors of DC circuit.svg
By Chetvorno - Own work, CC0, Link

To find the distribution of charges on the ground plane, we have to satisfy the boundary conditions for the electromagnetic field at the surfaces of the conductors. Let's do a static charge on the conducting trace (for example: a constant-voltage signal to some FET with very large input impedance, so that it draws no current) first, to warm up. For a point charge above an infinite conducting plane at $z=0$, the method of image charges gives charge density on the ground plane

$$ \sigma = -\epsilon_0 \frac{\partial V}{\partial z} \Bigg|_{z=0} = \frac{-q a}{2 \pi \left(\rho^2 + a^2\right)^{3/2} } \tag 1 $$

where $\rho$ is the cylindrical distance from the point charge. Note that the total distance between a point $(\rho,\phi,0)$ on the ground plane and the point charge at $(0,\text{anything},a)$ is $r=(\rho^2+a^2)^{1/2}$. If $\theta$ is the angle between $\vec r$ and the vertical, we have $\sigma \propto r^{-3}a \propto r^{-2}\cos\theta$ for a point charge.

We got that result using the potential for a point charge (and its mirror image) of $V\propto r^{-1}$ for a distance $r$ from the charge. The potential due to a line charge goes like $V\propto \ln r$ for a (perpendicular) distance $r$ from the charge. Doing the same derivative as in (1) ought to then give us $\sigma \propto r^{-1}\cos\theta$, but I'll leave the details and $2\pi$s and so on to you.

So there's your derivation for a static line charge. How does the charge distribution on the ground plane change if there's a current? The answer is that it doesn't. The boundary condition on $\vec E$ at the ground plane is still that $\vec E$ must be normal to the plane. The current introduces everywhere a magnetic field $\vec B$ that's perpendicular to the direction of the line current. Consider the DC/zero-frequency case first: there's no $\partial\vec B/\partial t$ to change the electric field, so the charge density must be the same as in the static case. This rebuts the (reasonable) intuition you state in your question, that the ground-plane current should be magnetically repelled by the line current: doing so would set up a transverse electric field on the surface of the ground plane conductor.

For an AC current, you can think of the current trace as a line charge whose charge density varies sinusoidally along its length. The method of image charges suggests that the surface charge under the ground plane should still vary like $\sigma\propto r^{-1}\cos\theta$ for any "transverse slice" under the current trace, giving the same electric fields as if there were an anti-trace below the ground plane with the opposite charge distribution. That charge distribution does have nonzero $\vec E_\parallel$ at the ground plane, in the direction of the line current; those are the fields that, together with the changing $\vec B$, move the ground-plane charges in the direction parallel to the current flow. The job of proving that the ground-plane component of $\vec E$ transverse to the circuit trace vanishes everywhere I'll leave to you as well.

Here's another argument that the static and dynamic charge distributions transverse to the circuit trace should be the same, based on symmetry. (I think I learned about this problem from Griffiths's E&M textbook.) Imagine you have two infinite, parallel, opposite-sign line charges at rest relative to each other: they'll experience an electrostatic attraction, but no magnetic interaction. However, another experimenter walking past you sees the two line charges, in her boosted reference frame, sees the line charges as antiparallel currents, which have a magnetic repulsion in addition to their electrostatic attraction; the magnetic repulsion becomes stronger, in her reference frame, as she passed you more quickly and the apparent current becomes larger. It seems plausible that, if the other experimenter walks past you sufficiently rapidly, she might see the magnetic repulsion become larger than the electrostatic attraction. Then you would disagree about whether the two line charges were attracted or repelled from each other. It is instructive to compute the speed of the boost at which the magnetic and electric forces balance each other exactly --- though if you've studied special relativity you might be able to guess the answer.

  • $\begingroup$ @tparker First paragraph: think about fields first, charges second. (Even in your comment you make this mistake: in a circuit it's fields that run the show, dictating the arrangement of charges and currents.) Second paragraph: a DC signal with negligible current is like a charged capacitor. I was guessing you had computed charge distributions on capacitors somewhere in your E&M background and would recall the mirror-charge method. Third paragraph: to figure out where, in space, the power is transmitted, use the transmission-line formalism to find the fields. Does that clarify things any? $\endgroup$
    – rob
    Commented Dec 26, 2016 at 20:27
  • $\begingroup$ I agree that "a DC signal with negligible current is like a charged capacitor." But in practice, does a DC signal in a circuit board have negligible current? I though that circuit traces have neutral net charge so they produce a negligible electric field, and the magnetic field produced by the current flow is much more important. $\endgroup$
    – tparker
    Commented Dec 26, 2016 at 21:01
  • $\begingroup$ @tparker (two comments up): For a zero-current circuit board signal, imagine a digital "ON/HIGH" going to the input of a FET with effectively infinite input impedance. For any circuit, if there's a potential difference between the signal trace and the ground, an electric field will point from the signal trace to the ground. (previous comment): The location of the return current is determined by the fields which transmit the signal. $\endgroup$
    – rob
    Commented Dec 26, 2016 at 21:20
  • $\begingroup$ It seems to me that your claims boil down to this: the circuit trace emits both an electric and magnetic field. The electric field tends to induce charge onto the ground plane directly below the circuit trace. The magnetic field (coming from the current flowing through the circuit) tends to repel current away from the circuit trace. But in practice, for most circuit elements the impedance is high enough that the effect of the electric field dominates the effect of the magnetic field, so most of the return current is below the circuit trace despite the magnetostatic repulsion. Is ... $\endgroup$
    – tparker
    Commented Dec 27, 2016 at 15:00
  • 1
    $\begingroup$ Okay wow this answer pruned SO many loose ends I've had in my understanding of energy in circuits for a long time $\endgroup$
    – D. W.
    Commented Jan 13, 2017 at 19:12

I just asked a professor this question, and he thinks that the concentration of return current below the circuit trace is actually just an AC effect. The rapidly oscillating electric and magnetic fields induce a fictitious image current on the other side of the conductor, and the real and image wires together act as a transmission line. But this only works if the fields oscillate fast enough - in the DC limit, the magnetostatic repulsion between the trace and the return current does indeed cause the return current to spread out far away from the circuit trace.


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