Transcript ECE423Lect22 - University of Illinois at Chicago

Shielded Wires
Let us say that the voltages measured at the terminal of the receptor
circuit are beyond the desired level. What can we do?
Two common solutions to reduce the level of interference are
1) Replace the generator/receptor wire with a shielded wire;
2) Replace the generator/receptor wire with a twisted wire.
Let us consider the use of a shielded wire. The equivalent electrical
circuit is:
Figure 1
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ECE 423, Dr. D. Erricolo, Lecture 22
The reference conductor can be either another wire
Figure 2
or a common ground plane
Figure 3
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 22
In both cases the transmission line contains 4 conductors, hence the
results previously found are not applicable. Solutions for a multiconductor transmission line can be computed using programs that
solve the corresponding equations, which are, in the frequency domain:
dVˆ ( z )
(1)
 ( R  j L) Iˆ( z )
dz
d Iˆ( z )
(2)
  j C Vˆ ( z )
dz
where
 VˆG 
 IˆG 
 
 
ˆ
ˆ
ˆ
V ( z )   VS  ,
I ( z )   IˆS 
(3)
 ˆ 
ˆ 
 VR 
 I R 
Note that losses in the medium have been neglected (G  0).
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ECE 423, Dr. D. Erricolo, Lecture 22
We will not solve the multi-conductor transmission line equations here,
but we will simply concentrate on the computations of the
per-unit-length parameters. The equivalent circuit for a length dz of the
transmission line, assuming TEM propagation mode, is:
Figure 4
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ECE 423, Dr. D. Erricolo, Lecture 22
The per-unit length matrices R, L, C are:
r0
r0 
 rG  r0
R   r0 rS  r0 r0 


 r0
r0 rR  r0 
l0
l0 
 lG  l0
L   l0 lS  l0 l0 


 l0
l0 lR  l0 
 CGS
0 
 CG  CGS
C    CGS CS  CGS  CRS  CRS 


 0
 CRS
CRS 















(4)
two elements are missing in the capacitance matrix!
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 22
Referring to the following figure
Figure 5
The mutual capacitances between 1) the generator and the receptor
wires and 2) the receptor and reference wire are missing because of the
presence of the shield.
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 22
Computation of the per-unit length parameters
Resistance
Generator and receptor wires:rG andrR are computed in the usual manner
Shield: its resistance depends on the construction technique.
rb

r

(5)
 braided-wire shield
S

BW
cos

W












where rb  resistance of one braid wire
B  number of belts
W  number of braid wires/belt
W  weave angle
1

r

solid shield
S

 2 rsht sh


computed assuming well-developed skin-effect;



rsh : shield interior radius



t sh : shield thickness



(6)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 22
Inductance
We will only consider computation of inductance parameters for the
case of a ground plane reference conductor.

2h
(7)
generator:
lG  0 ln G
2
rW G
shield:
lS 
0
2hG
ln
2 rsh  tsh
(8)
receptor:
lR 
0 2hG
ln
2
rW R
(9)
generator-shield: lGS
0  4hG hr 

ln 1  2   lGR
4 
s 
(10)
The equality lGS  lGR holds only if shield and generator wire
are widely separated.
0
2hG
l

ln
 lS
Receptor-shield: RS
(11)
2 rsh  tsh
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 22
The fact that lRS  lS is very important since it explains how inductive
coupling is eliminated. Consider the following circuit for the
computation of the mutual inductance between shield and receptor:
Figure 6
The mutual inductance may be computed by placing a current on the
shield and determining the magnetic flux through the receptor circuit or,
conversely, by placing a current on the receptor wire and determining
the magnetic flux through the shield circuit. The second option is
equivalent to placing all the current on the shield, which explains why
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lRS  lS
ECE 423, Dr. D. Erricolo, Lecture 22
Capacitance
Per-unit length cpacitances are obtained using the relation
(12)
LC  C L   I 2
The medium inside the shield may have  r  1. The capacitance between
the receptor wire and the shield is the same as for a coaxial cable:
20 r
cRS 
(13)
ln rsh / rW R 
The other capacitances may be obtained applying (12) to the subsystem
constituted by the generator wire and the shield so that:
CG  CGS  CGS 
 lG lGS 
 0 0 
 C


C

C
l
l
GS
S
GS 

 GS S 
It can be shown that:
0 
1 0
LC  0 0  0  r  r  1


 0 0
1 
(14)
(15)
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ECE 423, Dr. D. Erricolo, Lecture 22
Capacitive coupling
The notion of cross-talk voltages being composed of capacitive and
inductive contributions holds for coupled, electrically short lines. Let us
use some results for small frequency. Consider the following circuit:
Figure 7
Capacitive coupling through CGS and C RS leads to:
j RC
RNE RFE
CRSCGS
CAP
CAP
ˆ
ˆ
VNE  VFE 
VG , R 
, C
1  j RC
RNE  RFE
CRS  CGS
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ECE 423, Dr. D. Erricolo, Lecture 22
(16)
When the frequency is sufficiently small, the cross-talk voltage may be
approximated by:
RNE RFE CGS CRS ˆ
CAP
CAP
ˆ
ˆ
(17)
VNE  VFE  j
VGDC
RNE  RFE CGS  CRS
where VGDC
RL
 VG
RL  RS
(18)
is the low frequency value of the voltage along the generator wire.
Observe that the form of (17) is equivalent to the one previously
considered for the coupling between two wires without shield.
In practice the shield is connected to the reference conductor at both
ends, so that its voltage is reduced to zero and the capacitive coupling
contribution is removed.
When the line is not electrically short, one has to connect the shield to
the reference conductor at many points spaced by an amount of  / 10 to
remove the capacitive coupling.
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ECE 423, Dr. D. Erricolo, Lecture 22
Inductive coupling
We have seen that shield must be grounded at both ends to remove
capacitive coupling: the same must be done in order to remove
inductive coupling. To understand this concept, let us consider the
effect of the shield around the receptor wire.
Figure 8
The current IˆG generates a magnetic flux  G that is picked up by the
shield-reference circuit. By Faraday’s law, the resulting emf induces a
current IˆS in the shield. The magnetic flux  S associated with IˆS tends
to cancel  G .
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 22
Let us consider in more detail the mechanism of inductive coupling.
Our considerations will be based upon the following circuit for the
shield receptor wire:
Figure 9
From this circuit we easily obtain that:
RNE
IND
VˆNE

j LGR IˆG  LRS IˆS
RNE  RFE

where
IˆG 
j LGS
IˆG
RSH  j LSH

(19)
(20)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 22
Substitution of (20) into (19) yields:
2
R
j

R
L


( LGS LRS  LGR LSH ) ˆ
IND
NE
SH GR
ˆ
VNE 
IG
RNE  RFE
RSH  j LSH
(21)
Now we use the relations:
LGR  LGS , LRS  LSH
(22)
which give
IND
VˆNE
IND
VˆFE
without shield
shield effect

 


RNE
RSH
ˆ

j LGR I G
RFE  RNE
RSH  j LSH
without shield
shield effect

 


 RFE
RSH
ˆ

j LGR I G
RFE  RNE
RSH  j LSH
(23)
(24)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 22
So the results previously found without shield are still valid provided
that they are multiplied by the factor.
RSH
1
LSH
SF 

, s 
(25)
RSH  j LSH 1  j  s
RSH
The shield factor is approximated by
  1 s
 1,
SF  
(26)
R
L
,


1

s
 SH SH
so that, overall, the behavior of the inductive cross-talk contribution as
a function of the frequency is:
Figure 10
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ECE 423, Dr. D. Erricolo, Lecture 22
From a qualitative viewpoint there are two different situations:
1)  
1
: The lowest impedance path goes through the ground plane
 s so the flux due to Iˆ threads the entire receptor circuit.
G
1
2)  
: The lowest impedance path goes through the shield, instead
 s of the ground plane, resulting in ˆ ˆ which causes
I S  IG
no magnetic flux threading the receptor circuit.
To summarize, when the shield is grounded at both ends the inductive
coupling contributions are given by the following transfer functions.
IND
VˆNE

RNE
RSH
1
j LGR
RFE  RNE
RSH  j LSH RS  RL
(27)
IND
VˆFE

RFE
RSH
1
j LGR
RFE  RNE
RSH  j LSH RS  RL
(28)
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ECE 423, Dr. D. Erricolo, Lecture 22
Effect of pigtails
“Pigtails” refer to a break in a shield required to terminate it to a
grounding point. The interior shielded wire is exposed to direct
radiation from the pigtail section.
As a result, the shielding effectiveness is reduced
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ECE 423, Dr. D. Erricolo, Lecture 22