Shielding: absorption, reflection and apertures
Shielding effectiveness, absorption 8.69 t/δ and reflection proved with copper examples, low-frequency magnetic fields, slots and cables passing through the shield.
A shield is a conductive enclosure separating the source from the victim: a device’s metal case, a cable’s braid, a foil over a sensitive circuit. It works in two ways, reflecting the wave and absorbing it, and the calculations show that even a very thin foil would suffice for most fields. The real problem is apertures: slots, seams and cables passing through the shield. In this lesson we look at shielding effectiveness, absorption and reflection with their proofs and examples, the difficult case of low-frequency magnetic fields, and apertures.
Shielding effectiveness
Shielding effectiveness is the ratio, in decibels, between the field without the shield and with it:
![\[ SE = 20\log_{10}\frac{E_{without}}{E_{with}} \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-bf9feb59df28f7e8b60af778003a42ae_l3.png "Rendered by QuickLaTeX.com")
According to Schelkunoff’s theory it splits into three terms: 
, reflection, absorption and a correction term for multiple reflections inside the shield, negligible when absorption exceeds about ten decibels.
Absorption
Inside a good conductor the field decays exponentially with depth, with a characteristic length called the skin depth (Electromagnetic Fields lesson on the skin effect):
![\[ \delta = \frac{1}{\sqrt{\pi f\mu\sigma}} \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-fda23b659b3682f60abac82f18f567bf_l3.png "Rendered by QuickLaTeX.com")
Theorem. A shield of thickness 
introduces an absorption of
![\[ A = 8.69\,\frac{t}{\delta}\ \text{dB} \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-604590357d4ecb4b7020698b5792d11f_l3.png "Rendered by QuickLaTeX.com")
Proof. Crossing the thickness , the field amplitude is reduced by the factor 
. In decibels: 
. ∎
Example. Copper, 
S/m: at 1 MHz 
µm, and a 0.1 mm foil absorbs about 13 dB; at 100 MHz 
µm and the same foil absorbs about 130 dB. Absorption grows with the square root of frequency: at high frequencies even a thin metal layer deposited on plastic is enough.
Reflection
A wave reaching a surface with an impedance different from its own is largely reflected (lesson on plane waves). The wave in free space has impedance 
Ω, the metal a very small impedance, 
.
Theorem. For 
the reflection is
![\[ R = 20\log_{10}\frac{|Z_w|}{4|Z_s|} \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-1040436901503f754c833e4cc4c25155_l3.png "Rendered by QuickLaTeX.com")
Proof. The transmission coefficient of the electric field from a medium of impedance 
to one of impedance 
is 
. Entering the metal it is 
; leaving it, 
. The product is 
, and its inverse in decibels gives the formula. ∎
Example. Copper at 1 MHz: 
Ω, 
dB. With absorption, even a thin copper shield gives more than 120 dB against a plane wave: far more than is needed.
The difficult case: near magnetic fields
Close to a magnetic source (a loop, a transformer) the wave impedance is much lower than 377 Ω, and reflection is reduced accordingly; at low frequency absorption is also small, because 
is large (at 50 Hz in copper it is about 9 mm). A mains transformer cannot be shielded with a copper foil. What is needed is:
- high-permeability materials (iron-nickel alloys such as mu-metal), which offer the field lines a preferential path and divert them around the area to be protected;
- or, better, reducing the source and the victim loop (lesson on coupling).
Near electric fields, on the contrary, with high impedance, are easily shielded: a grounded conductive surface is enough.
Apertures
In practice a shield is never closed: there are ventilation slots, seams between panels, displays, connectors. A slot behaves like a slot antenna (lesson on aperture antennas), and what matters is its maximum dimension, not its area. For a slot of length 
, less than half a wavelength:
![\[ SE \approx 20\log_{10}\frac{\lambda}{2L} \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-cfd79c7c09d92d4b352a526e0c86fcb1_l3.png "Rendered by QuickLaTeX.com")
Example. A 10 cm slot at 1 GHz (
cm) leaves only 3.5 dB of shielding; at 100 MHz, 23.5 dB. The thickness and material of the shield become irrelevant: the slot is what counts. Many small holes shield much better than a slot of the same area, because each has a small maximum dimension.
The resulting rules:
- seams with frequent contacts (closely spaced screws, conductive gaskets), at a spacing much less than
at the highest frequency of interest; - ventilation with grids of small holes or honeycombs, which behave like waveguides below cut-off;
- every cable passing through the shield must be filtered at the point of entry, or its braid must be bonded to the shield around its whole circumference (360°). An unfiltered cable carries the disturbance outside and radiates it, defeating the shield: it is the most common mistake. A braid connected with a short wire (pigtail) loses much of its effectiveness at high frequencies, because the pigtail has inductance.
