| SIGNAL INTERFERENCE |
| When a particular installation is prone to
EMI/RFI/ESI interference from either internal or external sources, some form of cable
shielding will be required. The types of interference, or noise cables are exposed to can
determine the type of cable shielding required. There are basically four types of noise
which will affect the wiring or cabling of an instrument or control circuit: static,
magnetic, common mode and Crosstalk noise. |
| Static noise. This refers to
signal distortion due to the electrical field radiated by a voltage source, which has
coupled into the signal-bearing circuit. Simple shielding of the full circuit is a typical
means of mitigating this Electrostatic
type of interference. Foil shields, which offer
100% shielding efficiency, have proven most effective against this type of interference.
It is critical that the shield be continued to, and completely encompass, the transmitting
and receiving ends of the circuit if high levels of noise reduction are required.
Effective Ground of the shield is also required; "floating" or non-Grounded
shields only partially reduce the effects of noise. |
| Magnetic Noise. Magnetic
fields, radiated by power wiring found in large AC motors, transformers and knife
switches, can set up current flows in opposition to the instrument circuit field. The
result is the superimposing of a noise current on the signal current. The simplest and
best means of mitigating the effects of such magnetic interference is by simple twisting
of the cable elements. |
| Common mode noise. Common
mode interference is the result of currents flowing between different potential Grounds
located at various points within a system. Receivers with very high common mode rejection
ratios minimize this type of interference. |
| Crosstalk. This refers to
the superimposing of either pulsed DC or standard AC signals carried on one wire pair to
another wire pair in close proximity. Although pair twist tends to reduce Crosstalk
levels, the most effective means of mitigation is individual cable pair shielding coupled
to pair twist. |
| Noise levels. Once it has
been determined that noise currents are going to pose a system problem, it becomes
necessary to determine if the noise is of a low, medium or high level. The table below
gives general guidelines as to the areas which are subject to these generalized noise
levels: |
| NOISE
LEVEL CHART |
| Noise
Level |
Sources
of Noises |
Typical
Locations |
| High |
Electrolytic
processes, heavy motors, generators, transformers, induction
heating, relay controls, power lines and control wire in close proximity. |
Heavy processing
plants such as steel mills and foundries. |
| Medium |
Wiring near
medium-sized motors, control relays. |
Average manufacturing
plants. |
| Low |
Wiring located
far from power lines, motors; motors < 5 hp; no induction heating,
arcs, control or power relays nearby. |
Storage areas,
labs, offices and light assembly operations |
|
| SHIELDING
PERFORMANCE |
| The shielding
of electronic interconnect cables can play a critical role in overall
system performance. System configuration, type of signals transmitted
and proximity to noise generating sources all must be considered.
These factors plus the type of interference,
whether Electromagnetic
(EMI),
Electrostatic
(ESI) or radio Frequency
(RFI),
will determine the necessity and type of shielding required. Alpha's
XTRA•GUARD® Cables are available in two shielded
constructions providing protection for the majority of installation
needs. |
|
| SHIELD
COVERAGE --- BRAIDED SHIELD |
| The effectiveness
of a braided shield depends upon the percent coverage afforded by
the shield. Leakage in a braided shield is due to air spaces which
exist between the weave. The following equation can be used to determine
the percent coverage of a braided shield. |
2 P (D + 2d) P
Tan a = ___________
C |
Where:
D = Diameter under shield, inches
d = Diameter of 1 strand, inches
P =Picks per inch
N = Number of strands per carrier
a = Shield
angle, degrees
K = Percent coverage
C = Number of carriers |
When:
K = (2F - F2) x 100
and:
NdP
F = _____
Sin a
|
| Shield
Resistance |
| The D.C. resistance
for braided shields can be calculated using the following equation: |
dR
R = _________
Cos a (NC) |
 |
Where:
R = D.C. resistance, ohms/unit length
dR = D.C. resistance of 1 strand end,
ohms/unit length
a = Braid Angle, degrees
N = Number of strand ends in one carrier
C = Number of carriers |
|
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