Delta CT2000 Series User Manual page 627

Chapter 19 EMC Standard Installation GuideCT2000 Series
80
60
40
20
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0.05
19-4.2 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.
Wall of
shielded
Greater leakage
enclosure
form bigger
apertures
( ap er tu re d im en sio n)
Shield in g ef fectiveness
( distan ce that fields
( SE)in dB
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
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
Electromagnetic fields
"Waveguide below cut-off"
doesn't leak very much
G=gap
(does not have to be a tube)
d=depth
have to travel)
GHz
F<0.5Fcutoff SE is approximately 27d/g
2 5
19-14
d
(depth)
g
(gap)