electromagnetism - Faraday cage in real life

In electrical engineering we talk about using a "Faraday Cage" all the time. In general we mean putting the circuit in a metal box and grounding it, or putting a EMI shield over the top of a chip. My question is how perfect is a real faraday cage, can electro magnetic signals escape it or enter it if there's enough power? I don't mean how perfect is a case I might make with holes, but rather a perfectly enclosed case.

For example if I made a perfect sphere of tinfoil and put a powerful transmitter inside can no signal ever escape or because the material is so thin is there a power level where it starts to?

I vaguely remember someone explaining this to me but I can't recall how it works, or the theory behind it.

Answer

The theory for a tinfoil screen is fairly straightforward.

There are two things going on: first, a lot (most) of the signal is reflected. Second, what penetrates into the foil is attenuated and dissipated by currents.

The impedance of the aluminium "tinfoil" is given by $\eta_{\rm Al} = (\mu_r \mu_0 \sigma / \omega)^{1/2}$, where $\omega$ is the angular signal frequency, and $\mu_r=1$ and conductivity $\sigma= 3.5 \times 10^{7}$ S/m are reasonable values for Al. Therefore $\eta_{\rm Al}= 44\omega^{-1/2}$ $\Omega$.

The transmitted E-field fraction is given by the following equation $$\frac{E_t}{E_i} = \frac{2 \eta_{\rm Al}}{\eta_0 + \eta_{\rm Al}} \simeq 2\frac{\eta_{\rm Al}}{\eta_0}\, ,$$ where the impedance of free space (or air), $\eta_0 \simeq 377$ $\Omega$

Once the field gets into the foil it is exponentially attenuated according to the skin depth $\delta = (2/\mu_r \mu_0 \sigma \omega)^{1/2} = 0.045 \omega^{-1/2}\,$m.

Let's now make the assumption that we ignore reflection from the foil/air interface on the way out. In that case the transmission factor at that interface is given by $2 \eta_{0}/(\eta_{\rm Al} + \eta_{0}) \simeq 2$$.

Putting this all together we get a final transmission fraction of $$\frac{E_t}{E_i} \simeq 4 \frac{\eta_{\rm Al}}{\eta_0} \exp(-t/\delta) = 0.47 \omega^{-1/2} \exp(-22 \omega^{1/2} t),$$ where $t$ is the foil thickness. The transmitted power fraction would be the square of this.

Take an example: Typical domestic Al foil has $t= 3\times10^{-5}$ m and let's use a low radio frequency of 1 MHz or $\omega = 6.3 \times 10^{6}$ rad/s. The foil is only just over a skin depth thick at this frequency, but most of the signal is reflected and $E_t/E_i \simeq 3.6\times 10^{-5}$. Even if the foil was much thinner, only $2\times 10^{-4}$ of the field would be transmitted.

At a frequency of 1 GHz the foil is many ($\sim 50$) skin depths thick and is more reflective, so the attenuation of the field is more than 20 orders of magnitude.