RF shields do not need to be "air-tight"
RF shields do not need to be "air-tight"
Posted Nov 22, 2005 2:39 UTC (Tue) by stevenj (guest, #421)In reply to: RF shields do not need to be "air-tight" by hamjudo
Parent article: Richard Stallman's Tin-Foil Hat (Bruce Perens' Journal)
- 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.
Posted Nov 22, 2005 2:57 UTC (Tue)
by stevenj (guest, #421)
[Link]
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.
RF shields do not need to be "air-tight"
...the amount of radiation that gets through is proportional to the fraction of the surface area occupied by the holes