I have a room aka cuboid with 4 x 3 x 2.5meters. In the 3D center is a device under test. To shield against 36kHz electric fiels I have built a faraday cage by sticking 0,012mm thin and 30cm broad aluminium foil at the edges of the cuboid. Overlap on the corners is 15cm.

Resulting attentuation is not so great (4dB).

I want to investigate why, because my aim is 20dB. My faraday cage has several issues:

  1. My aluminium foil is thinner than skin depth but should be sufficent after this formula (https://physics.stackexchange.com/a/160195/271766) for 70dB
  2. Overlap is small, but should be sufficent because I sticked it together at some points so it has lots of capacitance (https://electronics.stackexchange.com/a/423112/245958)
  3. The holes, in fact the entire cuboid faces, are small with respect to wavelength and the device under test is in the middle

I wonder if 1. is the main problem and I need better shielding material or if I should work on 2. and 3.?

  • 2
    \$\begingroup\$ A Faraday cage only works if it is considerably thicker than the skin depth. The EM field is shielded if and only if it cannot propagate away. The skin depth figure (the Helmholtz equation developed from Maxwell's equations while looking for the Cartesian component of the field trying to penetrate the metal shield) gives \$\frac1{e}\$ protection. Call the depth, \$\tau\$. You'll need several \$\tau\$ to get useful shielding. So this means more thickness than you imagined and that doesn't mean separate, stacked layers, either. The electrons need to have all 3 degrees of freedom. \$\endgroup\$
    – jonk
    Jun 27, 2021 at 19:20
  • 1
    \$\begingroup\$ Also, I get over 432 microns for 36 kHz. Your thickness is significantly less. Your figure of 4 dB is about right. \$e^{{-}\frac{0.012\;\text{in}}{432.2\; \mu\text{m}}}=0.494\$. Pick a value you want to achieve and solve it for the thickness. Then build that and try again. \$\endgroup\$
    – jonk
    Jun 27, 2021 at 19:24
  • \$\begingroup\$ Makes sense, would accept it as answer. Is a thick wire at the edges a better solution then? \$\endgroup\$ Jun 27, 2021 at 19:30
  • \$\begingroup\$ You second comment confuses me, because you said it will only start working at skin depth and greater. \$\endgroup\$ Jun 27, 2021 at 19:33
  • \$\begingroup\$ I don't understand your confusion. I gave you an equation where you can compute the attenuation. What more could be desired? \$\endgroup\$
    – jonk
    Jun 27, 2021 at 19:58

2 Answers 2


Aluminum foil makes a poor shield because the surface is covered with Al2O3, aluminum oxide, a very effective insulator. It is also thin, and will not stop the magnetic component of the 36 kHz signal.

Use copper screening or foil. Admittedly, expensive for a whole room -- might cost hundreds of US$ or Euros.

Not as effective, but less expensive, is metal lathe or "stucco netting". A 70 x 240 cm sheet is ~US$14. Whether that would provide sufficient isolation for your use would have to be determined.

It is very important to ensure that all edges are thoroughly bonded. Three or four conductive latches might be needed at the doors, and edges should be tack-soldered or tightly wired together.

BTW, if any wires penetrate the Faraday cage, use effective L-C RFI filters on each.


Aluminium foil works. When I measure resistance I get values < 0.5Ohm, even when the connection between two is made with tape.

To enhance your faraday cage take the 4 walls of the cuboid and make a vertical lane at each. Take the floor and ceiling and make a cross intersecting in the middle. Then you will get 10dB in the 3D middle of your cuboid.

If you attach aluminium foil at the entire room so tight that the maximum gap is 42cm, then you can get more than 20dB at 36kHz and more than 30dB on 6MHz in the middle. If you still want daylight in the room you can use chicken wire instead of aluminium foil which you get for 2€/m^2.

The reasons why you have to have such great area covered is explained here: https://people.maths.ox.ac.uk/trefethen/chapman_hewett_trefethen.pdf


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