Electromagnetic Fields; How To Reduce Emi By Shielding?; Metallic Shielding Effectiveness - Delta C2000 series User Manual

Shield in g ef fectiveness
80
60
40
20
0
0.05

4.2 How to reduce EMI by Shielding?

Iron and other metals are high conductivity materials that provide effective shielding at extremely low
frequencies. But conductivity will decrease as:
1. High frequency signals are applied to the conductor.
2. Equipment is located in a strong magnetic field
3. The shielding frame is forced into a specific form by machines.
It is difficult to select a suitable high-conductivity material for shielding without the help from a shielding
material supplier or a related EMI institution.

Metallic Shielding Effectiveness

Shielding Effectiveness (SE) is used to assess the applicability of the shielding shell. The formula is:
SEdB=A+R+B (Measures in dB)
The absorption loss refers to the amount of energy loss as the electromagnetic wave travels through the
shield. The formula is:
AdB=1.314(fσμ)1/2t
The reflection loss depends on the source of the electromagnetic wave and the distance from that source.
For a rod or straight wire antenna, the wave impedance increases as it moves closer to the source and
decreases as it moves away from the source until it reaches the plane wave impedance (377) and shows
no change. If the wave source is a small wire loop, the magnetic field is dominant and the wave
impedance decreases as it moves closer to the source and increases as it moves away from the source;
but it levels out at 377 when the distance exceeds one-sixth of the wavelength.

Electromagnetic fields

Wall of
shielded
Greater leakage
enclosure
form bigger
apertures
( ap er tu re d im en sio n)
( distan ce that fields
( SE)in dB
have to travel)
d=18"
g=6"
d=6"
g=2"
d=12"
g=6"
d=4"
g=2"
d=6"
d=2"
g=6"
g=2"
0.1 0.2 0.5 1 2
where A= Absorption loss (dB)
R= Reflection loss (dB)
B= Correction factor (dB) (for multiple reflections in thin
shields)
where f= frequency (MHz)
μ= permeability relative to copper
σ= conductivity relative to copper
t= thickness of the shield in centimetres
"Waveguide below cut-off"
doesn't leak very much
G=gap
(does not have to be a tube)
d=depth
GHz
F<0.5Fcutoff SE is approximately 27d/g
5
10
d
(depth)
g
(gap)