EM wave shielding & skin depth

[tj50]

If I use 10mm (thickness) of a metal for the sides of a Faraday cage, with a 5mm skin depth at the frequency to be shielded, & increase the v/m of the EM wave to be blocked gradually, will there be a stage where the skin depth will increase to 6mm, or is 5mm the maximum limit?

 

[tech99]

The radiation does not stop abruptly at the skin depth but continues to penetrate the material, tapering off exponentially. In other words, at a fixed number of dB per mm.

 

[tj50]

So with a high enough voltage you can get EM waves through every metal at any frequency so what use is a skin depth no.?

 

[tech99]

So with a high enough voltage you can get EM waves through every metal at any frequency so what use is a skin depth no.?

Yes. The skin depth tells you the depth at which the electric field has fallen by 1/e.

 

[tj50]

Thanks tech99, at 100Hz the skin depth for aluminium is around 8.2mm.

Does this tell us that at approx. 8.2mm the E.field will drop by 1/e, and then after a further 8.2mm the field once again drops by 1/e and so on?

 

[sophiecentaur]

Yes. That's the best you can do, I'm afraid because there is no thickness for 'absolute extinction'. All Fat=raday screens have finite limits to their screening performance.
Once you get to, say 60dB of isolation, there are other mechanisms for the EM to get through. Seams and door seals can leak like a seive if they are not make well aned, of course, the holes in the side, through which the power and signal leads may be brought in and out. (Plus the ventilation holes, when people are inside, making measurements) You have to specify what isolation you require and the conditions and then open your wallet and say "help yourself".

 

[tech99]

The attenuation is 8.7dB for every skin depth below the surface.

 

[sophiecentaur]

The attenuation is 8.7dB for every skin depth below the surface.

Yes - that's 1/e Volts in dB.

 

[tj50]

How can the electric field decreases of an em wave in free space be calculated? For example calculating what e new electric field 5m from a transmitter or the initial electric field is?

 

[jtbell]

Conservation of energy requires that the power density decrease as 1/r² from a pointlike source. The energy and power densities in an EM wave are proportional to the square of the amplitude of the wave's electric field. Therefore... (I'll let you fill in the remaining step. )

 

[tj50]

@ sophiecentaur & tech99:

Thanks both, that was very helpful & informative.

 

[tj50]

Great, that's cleared things up, thanks jtbell.

 

[tech99]

But if you are very near the antenna, bear in mind the radiated power does not start to fall as 1/r^2 straight away, but remains constant out to a distance of about lambda/5 from the antenna.
A simple formula for the electric field strength from a dipole antenna beyond this distance is : E = (7 sqrt p) / d^2 (volts/metre, watts and metres).

 

[tech99]

Sorry, correction, E = (7 sqrt p )/d

 

[tj50]

Thanks tech99, that should help me check my recent dipole antenna work.

 

[tech99]

If you are working very close to the antenna, ask me again.

 

[tj50]

Ok, thanks again tech99