2
\$\begingroup\$

I was thinking about shielding and its effectiveness. An ungrounded shield doesn't work well for either inductive or capacitive coupling.

However, if you think about plane wave theory, a PEC surface completely reflects the EM wave (reflection coefficient is -1), so you don't have any power density at the other side of the PEC surface. That seems to be the ideal shield for any EM wave/interference. This should lead to the conclusion that if I wrap around a cable with a PEC shield, no EM waves will be able to either go out or into the shield, be it grounded or ungrounded. Therefore, it should be immune to any interference or coupling.

But it seems not to be the case, why?

I suppose it has to do with the fact that coupling is a near field phenomenon, while plane wave theory usually considers the far field. But how exactly it is different, that I don't know.

\$\endgroup\$
0

3 Answers 3

Reset to default
1
\$\begingroup\$

A perfectly electrically conducting (PEC) surface would reflect plane waves if it were of infinite extent. If it's got holes, or if it's got edges, the waves will go around it, and even refract around to the back surface depending on the relative size of the plate and the waves.

If you wrap a cable with a PEC shield and do not ground it at one end, that lack of ground connection constitutes a hole in the shielding. If you do not ground it at either end, that constitutes two holes in the shielding. Either way, to one extent or another, EM will radiate.

You can think of this in electrical terms if the wavelength is long with respect to the cable length: in that case, particularly with shielding that's not grounded at either end, the entire shield will be capacitively coupled to the cable innards, and will to some extent follow it. Then the outside of the shield will radiate.

\$\endgroup\$
2
  • \$\begingroup\$ Thanks for the answer, that makes sense! However, if the hole is a lot smaller than the wavelength, there should be - almost - no refraction from the hole. Also, when crosstalk is explained in books, it's never mentioned (as far as I know) the effects of "how ungrounded" the cable is. That is, how large the hole that the un-grounding leaves. Could you please add something more about this? \$\endgroup\$
    – Tripola
    May 3, 2019 at 20:57
  • \$\begingroup\$ I don't know the answers to that well enough to say. I'm certain that any hole will let some radiation through, and that it'll be proportional to something that shrinks faster than \$(\mathrm{hole\ size})^1\$ as the hole size goes down. Searching on "how big can the holes in my shielding be?" may help. I do know that coax cable is rated for how well it shields, and that the tighter the braid the better the shielding, and that the holes in the shielding in cheap coax is a tiny fraction of the wavelength (although there's a lot of them). \$\endgroup\$
    – TimWescott
    May 3, 2019 at 21:07
1
\$\begingroup\$

High frequency noise currents flowing in the shielded conductor capacitively couple to the shield and use the shield as a return wire. If you leave the shield floating, it must flow through the capactive coupling between shield and ground to circulate and complete the current loop. The electromagnetic energy energy traveling through the air via the capacitive coupling is the EMI itself.

But if you ground the shield properly so that flowing in the solid conductors is an easier path (lower impedance) than through the capacitive coupling in the air, the electromagnetic energy stays in the solid conductors and out of the air resulting in less radiated EMI.

Similarly, if an external EM wave induces noise currents to flow in an ungrounded shield, they will capacitively couple into the cable and use the cable to flow to ground. This manifests as noise on the cable. But if you ground the shield, then the induced currents in the shield stay in the shield, and out of the shielded conductor, on their way to ground.

Note in the diagrams below that these are high frequency currents subject to the skin effect so they cannot travel through the shield. They must travel to the end of the shield, and around the edge in order to get to the shield's outer surface.

Also note that if you provide a grounding connection that is low resistance, but high impedance, the noise currents won't flow through it since the impedance dominates. So even if you ground the shield with a piece of wire of low resistance, the high frequency noise currents may still not flow through if the impedance through the wire is higher than the impedance of capacitively coupling through the air. That's why shields should have a continuous connection to ground along their entire open edge. These are high frequencies subject to the skin effect so they care more about distance they need to travel and perimeter of connecting surfaces more than the "bulk" of conductive material in the connection.

enter image description here

From EMC Engineering, Henry Ott 2009

\$\endgroup\$
4
  • \$\begingroup\$ Hi Toor, thank you for your detailed answer! About your first paragraph: If I had two infinite wires close to each other separated by an infinite PEC plane, would I still have capacitive coupling between them? Also, could you add an explanation of how there can be coupling when the EM wave should be completely reflected by the shield? \$\endgroup\$
    – Tripola
    May 3, 2019 at 20:47
  • \$\begingroup\$ @Elia Does a PEC still have the skin effect? How thick is the plane? If there is a skin effect, higher frequencies that cannot penetrate the thickness of the plane due to the skin effect will not be coupled because there is no edge of the plane that they can wrap around to cross from one side of the plane to the other. But frequencies that are low enough where they can penetrate the thickness of the plane would still couple. \$\endgroup\$
    – DKNguyen
    May 3, 2019 at 21:19
  • \$\begingroup\$ @Elia But if a PEC does not have a skin effect, then any frequency, no matter how high, would still be able to penetrate the plane, but I feel like there still would be no coupling since the PEC infinite plane is a perfect infinite sink. But at the same time, energy can only propagate in the plane at the speed of light so that might cause localized coupling still. It feels like a paradox which is the happens when you have idealized objects. Things don't make much sense which is why they don't exist to begin with. Sort of like unstoppable force and immovable object. \$\endgroup\$
    – DKNguyen
    May 3, 2019 at 21:26
  • \$\begingroup\$ @Elia it seems it might be the case that a perfect conductor suffers from the skin effect the most so no frequencies can penetrate its thickness it at all and all currents must travel along the surface. Again, it starts to not make sense here since if electrons can only travel on one layer on the surface, there must be a limit to how much current it can carry since you can't cram an infinite number of electrons into a space. You see this threshold in superconductors when the current density gets too high and it stops superconducting. \$\endgroup\$
    – DKNguyen
    May 3, 2019 at 21:49
0
\$\begingroup\$

But it seems not to be the case, why?

Because PEC's are not practical when building conductive surfaces in the real world, they exist in the ideal academic relm. While one can make materials that are near perfectly reflective for select wavelengths, for all frequencies this is not possible. When selecting materials, for most designs they need to be economically practical and readily available.

In addition, we have no materials that can readily reflect magnetic fields, only attenuate them. Maybe a superconductor would closely approximate a PEC and be perfectly reflective, and you could dip your device/cable in liquid nitrogen, again not practical.

A shield is a metallic partition placed between two regions of space. It is used to control the propagation of electromagnetic fields from one region to the other. Shields may be used to contain electromagnetic fields Electromagnetic Compatibility Engineering pg 238

So when we build shields, we mainly use cheap metal conductors. These reflect, and absorb the E-field component of the wave. This is mainly what we care about because the magnetic field attenuates by the inverse of the distance cubed, so it doesn't travel far.

An ungrounded shield does work as a shield in blocking the E-field component of a EM wave as long as the shield is continuous. But it does so by means of attenuation (the conductor depth must be greater than the skin depth) and redirecting the wave as current around the volume that the shield encloses.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.