RF shields do not need to be "air-tight" [LWN.net]

RF shields do not need to be "air-tight"

Posted Nov 22, 2005 1:20 UTC (Tue) by hamjudo (subscriber, #363)
In reply to: RF shields do not need to be "air-tight" by stevenj
Parent article: Richard Stallman's Tin-Foil Hat (Bruce Perens' Journal)

AM pocket radios don't work inside cars and steel frame buildings unless they are "close enough" to a window. The shortest wavelength in the AM broadcast band (in the US) is about 175 meters, yet the radios can sometimes get a signal through a hole that is less than half a meter across. So a mesh with millimeter scale holes will only work reliably, when the mesh is kept a safe distance away from the antenna. (A distance that I don't know how to calculate.)

I've got a couple other reservations and another experiment.

Bruce's card didn't indicate that it had an RFID built in to it at all, so it certainly didn't indicate the operating frequency. I'm afraid we won't get a memo when "they" upgrade RFID to 5.8 Ghz in a few years, or 10 Ghz a few years after that. So a pouch effectively blocks everything at 2.45 Ghz, may not do anything usefull at 5.8 Ghz.
The pouch only works if all sides are connected together through a low impedance path. Aluminum foil wears out after too much flexing.
I just tried putting my cell phone into a translucent anti-static bag. The signal strength went from 5 bars to 2 bars. So don't try lining your wallet with antistatic wrap, it's not good enough.

On the plus side, I have established myself as enough of a kook, that they won't feel the need to monitor me. So I don't need to shield my entire body, I will just want to protect my wallet.

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RF shields do not need to be "air-tight"

Posted Nov 22, 2005 2:39 UTC (Tue) by stevenj (guest, #421) [Link]

Regarding the AM radio inside the car, that's not relevant here: if you have a thin metal surface perforated by subwavelength holes, then the amount of radiation that gets through (barring resonant effects that only occur with specially arranged holes) is proportional to the fraction of the surface area occupied by the holes. For your car, a large proportion of the surface area is taken up by the windows, and hence a large proportion of the RF energy can get in (or out). For an object wrapped in metal foil, however, the ratio of hole/surface is tiny.
Regarding going from 2.45 to 10GHz in the future, that's still 3cm wavelength, which is still much larger than than any apertures in an object wrapped in foil.
Regarding needing a continuous low-impedance path all around the object, I don't think that's correct: even if you had "flakes" of aluminum foil pasted together with insulating paste, as long as they completely covered the object and the gaps between them were of negligible area, it should still block RF. The size of the flakes only needs to be much larger than the skin depth of the RF in the aluminum (which is tiny, see below).

The real problem you may have is for low frequency RFID, where the thickness of the foil (0.03mm according to Wikipedia) is less than the skin depth (the distance for the power to decay by exp(-2) ~ 90%). At microwave frequencies (GHz), the foil is many skin-depths thick and the transmission through it should be negligible. At the lowest frequency RFID, around 130KHz according to Wikipedia, the foil is only 1/10 of the skin depth in thickness, which is not enough to attenuate the power by much. On the other hand, you can simply wrap your object in 20-30 layers of the foil.

RF shields do not need to be "air-tight"

Posted Nov 22, 2005 2:57 UTC (Tue) by stevenj (guest, #421) [Link]

...the amount of radiation that gets through is proportional to the fraction of the surface area occupied by the holes

Whoops, sorry, it's worse than that: the transmitted radiation is proportional to the hole's fraction of the surface area multiplied by (d/λ)⁴, where d is the hole diameter and λ is the wavelength. (H. Bethe, Phys Rev. 66, p. 163, 1944.)

However, once the hole occupies a significant fraction of the solid angle around the antenna (as when you bring your radio close to a window), then the problem completely changes because you are in the near field and you can't treat the incident field as ~constant amplitude in the vicinity of the hole. This shouldn't be the case for an object wrapped in foil, however.