# 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.

• Assuming that you don't have to worry about the holes kinds kills the "real life" aspect here. For real life examples taking care of the necessary penetrations is the hard part... – dmckee --- ex-moderator kitten Jan 18 '15 at 22:15
• I understand that, and I understand what happens when there are holes. I just don't understand what the baseline is for something with no holes. Is it perfect no matter what the power level? – confused Jan 18 '15 at 22:43
• I've been inside a closet-sized Faraday cage made with copper wire screen. The holes in the screen were maybe 0.5 mm in size, meaning that any radio wave with a wavelength much larger than that would not get through. The school was within about 1000 feet of the broadcast antenna of a rock&roll AM station, and inside the cage was the only place on the entire campus where you could not detect it. – Hot Licks Jan 19 '15 at 0:18

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.

• Rob thanks for the great answer :) This is exactly what I was looking to understand. – confused Jan 19 '15 at 17:02