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

Shielded Wires

Let us say that the voltages measured at the terminal of the receptorcircuit are beyond the desired level. What can we do?

Two common solutions to reduce the level of interference are1) 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

University of Illinois at Chicago

ECE 423, Dr. D. Erricolo, Lecture 22



The reference conductor can be either another wire

or a common ground plane

Figure 3

Figure 2





In both cases the transmission line contains 4 conductors, hence the results previously found are not applicable. Solutions for a multi-conductor transmission line can be computed using programs thatsolve the corresponding equations, which are, in the frequency domain:

)(ˆ)(ˆ

)(ˆ)()(ˆ

zVCjdz

zId

zILjRdz

zVd

where

R

S

G

R

S

G

I

I

I

zI

V

V

V

zV

ˆ

ˆ

ˆ

)(ˆ,

ˆ

ˆ

ˆ

)(ˆ

Note that losses in the medium have been neglected ).0( G

(1)

(2)

(3)





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 University of Illinois at Chicago




RSRS

RSRSGSSGS

GSGSG

R

S

G

R

S

G

CC

CCCCC

CCC

C

llll

llll

llll

L

rrrr

rrrr

rrrr

R

0

0

000

000

000

000

000

000

The per-unit length matrices CLR ,, are:

two elements are missing in the capacitance matrix!

(4)





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 thepresence of the shield.





Computation of the per-unit length parameters

Generator and receptor wires: and are computed in the usual mannerGr Rr

Resistance

Shield: its resistance depends on the construction technique.

braided-wire shieldW

bS BW

rr

cos

where resistance of one braid wire number of belts number of braid wires/belt weave angle

brBWW

solid shieldshsh

S trr

2

1

computed assuming well-developed skin-effect;:shr:sht

shield interior radiusshield thickness

(5)

(6)









InductanceWe will only consider computation of inductance parameters for thecase of a ground plane reference conductor.

generator:

shield:

receptor:

generator-shield:

The equality holds only if shield and generator wireare widely separated.

WG

GG r

hl

2ln

20

shsh

GS tr

hl

2ln

20

WR

GR r

hl

2ln

20

GRrG

GS ls

hhl

20 4

1ln4

GRGS ll

Receptor-shield: Sshsh

GRS l

tr

hl

2ln

20

(7)

(8)

(9)

(10)

(11)





The fact that is very important since it explains how inductivecoupling is eliminated. Consider the following circuit for the computation of the mutual inductance between shield and receptor:

SRS ll

Figure 6

The mutual inductance may be computed by placing a current on theshield 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

SRS ll





CapacitancePer-unit length cpacitances are obtained using the relation

2ILCCL The medium inside the shield may have . The capacitance betweenthe receptor wire and the shield is the same as for a coaxial cable:

1r

WRsh

rRS rrc

/ln

2 0

The other capacitances may be obtained applying (12) to the subsystemconstituted by the generator wire and the shield so that:

SGS

GSG

GSSGS

GSGSG

ll

ll

CCC

CCC00

It can be shown that:

100

10

001

00 rrCL

(12)

(13)

(14)

(15)





Capacitive couplingThe notion of cross-talk voltages being composed of capacitive and inductive contributions holds for coupled, electrically short lines. Let ususe some results for small frequency. Consider the following circuit:

Figure 7

Capacitive coupling through and leads to:GSC RSC

GSRS

GSRS

FENE

FENEG

CAPFE

CAPNE CC

CCC

RR

RRRV

RCj

RCjVV

,,

1ˆˆ

(16)





When the frequency is sufficiently small, the cross-talk voltage may beapproximated by:

DCGRSGS

RSGS

FENE

FENECAPFE

CAPNE V

CC

CC

RR

RRjVV ˆˆˆ

whereSL

LGG RR

RVV

DC

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 bothends, so that its voltage is reduced to zero and the capacitive couplingcontribution is removed.

When the line is not electrically short, one has to connect the shield tothe reference conductor at many points spaced by an amount of toremove the capacitive coupling.

(17)

(18)

10/





Inductive couplingWe 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.

The current generates a magnetic flux that is picked up by theshield-reference circuit. By Faraday’s law, the resulting emf induces a current in the shield. The magnetic flux associated with tendsto cancel .

GI

SI

G

G SIS

Figure 8





Let us consider in more detail the mechanism of inductive coupling.Our considerations will be based upon the following circuit for theshield receptor wire:

Figure 9

From this circuit we easily obtain that:

SRSGGRFENE

NEINDNE ILILj

RR

RV ˆˆˆ

whereG

SHSH

GSG I

LjR

LjI ˆˆ

(19)

(20)





Substitution of (20) into (19) yields:

GSHSH

SHGRRSGSGRSH

FENE

NEINDNE I

LjR

LLLLLRj

RR

RV ˆ)(ˆ

2

(21)

Now we use the relations:

SHRSGSGR LLLL ,

which give

effectshieldshieldwithout

ˆˆSHSH

SHGGR

NEFE

NEINDNE LjR

RILj

RR

RV

effectshieldshieldwithout

ˆˆSHSH

SHGGR

NEFE

FEINDFE LjR

RILj

RR

RV

(22)

(23)

(24)





So the results previously found without shield are still valid providedthat they are multiplied by the factor.

SH

SHs

sSHSH

SH

R

L

jLjR

RSF

,

1

1(25)

The shield factor is approximated by

sSHSH

s

LRSF

1,

1,1

so that, overall, the behavior of the inductive cross-talk contribution asa function of the frequency is:

Figure 10

(26)





From a qualitative viewpoint there are two different situations:

1) : The lowest impedance path goes through the ground plane so the flux due to threads the entire receptor circuit.

2) : The lowest impedance path goes through the shield, instead of the ground plane, resulting in which causes no magnetic flux threading the receptor circuit.

s 1

s 1

GI

GS II ˆˆ

To summarize, when the shield is grounded at both ends the inductive coupling contributions are given by the following transfer functions.

LSSHSH

SHGR

NEFE

NEINDNE RRLjR

RLj

RR

RV

1ˆ

LSSHSH

SHGR

NEFE

FEINDFE RRLjR

RLj

RR

RV

1ˆ

(27)

(28)





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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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