CH-5-Relationship Between Field Theory and Circuit Theory
Transcript of CH-5-Relationship Between Field Theory and Circuit Theory
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5 Relationship between Field Theoryand Circuit Theory
(ref: Ramo et al.)At lower frequencies where physical circuit dimensions are small compared
to the wavelength1
of electromagnetic waves, the behaviour of circuits is ac-curately modelled usinglumped elementcomponent models, together withKirchhoffs laws. At higher frequencies where the distances between compo-nents are a significant fraction of a wavelength and greater, the signals car-rying information or power from one place in a circuit to another are treatedas waves. Signals must be routed from one point to another using trans-mission lines, modelled using transmission line theory. If the componentdimensions be comparable to the wavelength then accurate understanding
and prediction of behaviour may require modelling using electromagneticfield and wave theory.
In this section we examine the relationship between Maxwells equationsand circuit theory. Both Kirchhoffs voltage law, relating to voltage dropsVi around a loop
Ni=1
Vi= 0
and Kirchhoffs current law, relating relating currentsIi leaving a node
Ni=1
Ii= 0
can be explained in terms of field theory.Consider for example, the circuit shown below shown firstly using standard
circuit symbols for R, L and C, and secondly as a physical representation
1 f 50 Hz 100 kHz 1 MHz 10 MHz 100 MHz 1 GHz 10 GHz 100 GHz= c/f 6000 km 3 km 300 m 30 m 3 m 30 cm 3 cm 3 mm
5-1
with wires of finite width, a resistor as a rod of carbon, an inductor as acoiled wire, and the capacitor as a pair of metal plates. We shall examinebriefly each lumped component model from a field theory perspective.
V(t)
C
R
L
I(t)
2
0 3
1
0
1 2
3
Coil
R
L
C
V(t)
0 3
3
2
0
21
1
I(t)
Figure 5.1: Circuit diagram showing component symbols (above) and a more physical de-piction of the components (below).
5.1 Resistors
A resistor can be constructed from a resistive material of conductivity ,lengthl and cross-sectional area A, as depicted below.
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0
0
0
0
0
1
1
1
1
1
e1 2
Area A
J= E
l
E
Figure 5.2: Resistor made from a cylinder of carbon. A current flows as a consequence ofthe (axial component) of the electric field.
If the material is subjected to an electric field orientated along the lengthof the cylinder, a current will flow, explained as follows:
Electrons move under the influence of the electric field to reach anaverage drift velocity.
A classical model explains this as follows:Electrons initially accelerate under the influence of the field, but repeat-edly collide with bound atoms, and bounce off, resulting in deceler-ation. The net result is a constantaveragevelocity for the electrons.This has some analogy to the terminal velocity reached by a falling
object as a results of the resistance from the air molecules.
0000011111
Area A
e
electron, which collides
with atoms.
Path of an accelerating
Average drive velocity
of the electrons
J= EE
Figure 5.3: The electrons accelerate, but are impeded by the atomic structure, hence reach-ing a finite (average) terminal velocity. An imagined path of a single electronis shown.
Because of the high density of electrons, the average drift speed is sur-prisingly slow. For example, Halliday, Resnick & Walker 6th Ed, do anexample calculation (Problem27.3) in which the drift velocity with a cop-per wire or radius 0.9 mm, carrying a current of 17 mA, is calculated to be
4.9 107
m/s or 1.8 mm/hr.
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The average current density in Am2 is proportional to the strength ofthe electric field, i.e.
J= E
where has units [Sm1] and is a property of the medium.
It is noted that any particular electron continually accelerates and de-celerates with each collision as illustrated.
Energy is dissipated as a result of the collisions (i.e. in the form ofheat).
The voltage developed across the resistor is found by integrating the electric
field through it from node 1 to node 2:
V21 = V2V1=
21
Edl=
21
Edl =
21
J
dl=
21
I/A
dl= I
l
A
The constant lA is identified as the resistance of the rod, i.e.
R= l
A
In the labelled circuit loop,
V21 = V2 V1= IR
5.1.1 Calculation of average drift velocity
To calculate the average drift velocity of the electrons in a cylindrical wireconductor, one needs to know the current I, the charge on one electron
(qe = 1.6 10
19
coulombs), the thickness of the wire, and the electrondensitye [m3] (not to be confused with charge density). Copper contains
e= 8.5 1028 electrons per m3 - a very large number!
Imagine a slug of electrons moving along a wire rod in the x direction.The electron drift velocity in metres per second (in the x direction) can beexpressed in terms of the current as
vdrift=dx
dt =
dx
dQ
dQ
dt =
1dQ
dx
(I)
where
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dxdt is the velocity of the leading edge of the slug as it passes some pointx0,
dQ
dx is the ratio of charge passing the point x0 per distance dx movedin thex direction. This is identical the charge density in coulombspermetreof wire.
dQdt is the charge per second passing pointx0per second. If the conven-tional current is I amperes moving in the negative x direction, thendQdt = I.
To determine dQdx , consider a wire segment of lengthdx, and of thickness 2r.
The volume of the segment is r2dx.The number of coulombs per metre is
dQ
dx =
(charge/vol) vol
length =
(eqe)(r2dx)
dx =eqer
2
and hence
vdrift= I
eqer2
For copper wire of 1 mm thickness,dQ
dx =eqer
2 = 8.5 1028 (1.6 1019) 3.14 (0.5 103)2 = 10676 Cm1
If a current ofI= 1 ampere flows in the wire, the electron drift velocity is
vdrift= I
eqer2=
1[Cs1]
10676 [Cm1]= 9.37 105ms1 = 34 cm/hr.
The number of electrons making up one coulomb is 1/ |qe| = 6.25 1018.Thus for a 1 ampere current, 6.25 1018 electrons pass per second.
5.2 Capacitors
Consider a parallel plate capacitor. As the current flows through the wires,a surface charge builds up on the inner sides of the capacitors plates.
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Note that all excess charge will sit on the surface of the capacitorplates, in a thin layer (not inside the metal). Recall that E= 0 insidea perfect conductor, and since div D = or divE = this implies,
= 0 inside the conductor. All excess charge musttherefore lie on thesurface, described by a surfacecharge density s in Cm2.
The charge Q on the plate onto which the conventional current flows isfound by integrating the current flowing onto the plate, i.e.
Q(t) =
tt0
I(t)dt+Q0
whereQ0is the initial charge at some starting time t0. The other plate willhave a charge ofQ(t).
A potential difference builds up between the places. The potential differ-ence can be shown by careful argument [Griffiths] to be proportional to thechargeQ on the plates, i.e.
Vc(t) = 1
CQ(t)
where Cis the constant known as the capacitance.Substituting forQ(t),we get
Vc(t) = 1
C
tt0
I(t)dt+Q0
C
In the circuit loop, V03 = V0 V3= Vc(t).
5.3 InductorsInductors are made by winding several turns of wire either in air, or aroundsome high permeability material (which boosts the inductance, requiringfewer turns).
We shall explain the operation of an inductor by considering first a singleturn, and then a coil of several turns, in the context of the series circuitunder analysis.
As already discussed, we are interested in applying Faradays law aroundthe dashed loop shown in the physical circuit. For the inductor, we are
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interested in the integral of the electric field through the air gap betweenthe terminals as indicated by the dotted line between nodes 2 and 3 in thecircuit.
5.3.1 Single-turn Inductor
Consider a single turn inductor that forms part of the series circuit underanalysis.
+
contour C
Integration
B
V
I
B
Figure 5.4: Single turn inductor.
We can apply Faradays law locally to a closed contourCthat goes clockwise
around the insideof the wire and then across the air gap (in the shape ofthe dotted path),
C
E dl=
(air)
E dl +
(wire)
E dl= d
dt
S
B dS=M
t
where and dS points into the page, and M is the flux threading the inte-gration loop (and cutting a chosen surface S, bounded by C).
SinceE= 0 in the wire, the potential different is then
V =
(air)
E dl= d
dt
S
B dS
The magnetic field Bis linearly proportional to the currentIflowing in thewire2, i.e. B I, and so is
SB dS I. The constant of proportionality
2The magnetic field vector owing to a short current segment can be computed using the Biot-Savart
law (reviewed in Section 2.5). The total field is found by integration of all contributions from currentelements in the wire.
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is known as the inductance, i.e.S
B dS= LI
or
L=
SB(t) dS
I(t) =
MI
The units of inductance3 are henrys [H], The voltage across the inductor is
V = d
dt
S
B dS=d[LI]
dt =L
dI(t)
dt
In the labelled series circuit,
V32 = V3 V2= LdI(t)dt
5.3.2 Multi-turn Inductor
A multi-turn inductor is constructed by winding a coil of wire as depictedin the figure below.
+
V
V3
V2
V1
I
Figure 5.5: Multiturn inductor (N= 3 here).
The voltage drop across the terminals is the line integral along the dashedline:
V =
+
E dl
= [
gap1
E dl +
gap2
E dl +...
gapN
E dl]
=V1+V2+...VN
3The units of inductance are equivalently [H] = [Wb A1] = [VA1s1] = [VCs2] = [NC1mCs2] =[Nms2].
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If we further assume that the flux linking each turn is the same, then
V1=d
dt =V2= ... = VN
and
V =N V1= Nd
dtBecause there are N turns, the flux theading the coil will be N timesstronger than the contribution from a single turn, i.e.
=N 1turn
where 1turnis the flux contribution from a single turn. Substituting, weobtain
V =N V1= Nd
dt =N2
d1turndt
=N2d(L1turnI)
dt =N2L1turn
dI
dt
Thus the inductance for a tightly woundN-turn coil is
L= N2L1turn
where L1turn is the inductance of a single turn.
It is worth remembering that the inductance increases as a function ofN2.If one doubles the number of turns, the inductances increases by a factor offour.
Alternative Explanation
Analysis of a multi-turn coil is similar to the case of a single turn coil, withthe added complication that the integration contour C is not a circle, butrather made up of a spiral that follows the wire (and a short section in theair gap between its terminals). As argued for a single turn case, the voltageacross the terminals is
V =
(air)
E dl= d
dt
S
B dS
where surface S is now a complicated-to-visualise surface that is boundedby the contourC. It helps to imagine the surface within the coil as a smooth
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spiral staircase winding around an imaginary centre line. The total fluxM passing through S is proportional to the current I in the wire, and isgiven by
M= S
B dS= LI
where L is the inductance of the multi-turn coil. Thus
L=M
I =
SB(t) dS
I(t)
where M must be carefully evaluated for the particular coil structure. Fora short, compact coil ofNturns, it is not difficult to show that
L= N2L1turn
whereL1turnis the inductance of a single turn coil of the same radius. Tosee this, one must grasp two points:
the spiral surfaceSthrough with the flux lines pass consists of a stackofNidentical contributing flatish discs (the total surface area is ap-
proximatelyNtimes larger than for a single turn)
the flux density on each componentdiscis Ntimes stronger than theflux density generated by a single turn (superposition of contributionsfromN turns, each carrying currentI)
Thus the total flux threading the surface of the multi-turn contour is
M=
S
B dS N
N 1turn
where 1turn is equal to the flux generated by a singleturn coil carryingcurrentI .
The inductance of the multi-turn coil is then
L=M
I =
N21turnI
=N2L1turn
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5.4 Formulas for Practical Coils
In order to calculate the inductance accurately, we need to consider both the
field inside, and outside of the wire making up the coil. The total inductanceis usually computed in two parts:
L= Linternal+Lexternal
whereLinternal is the contribution arising from the magnetic field within thewire, andLexternal is the contribution from the field outside the wire.
5.4.1 Inductance of a Circular Wire Ring
Internal field
in the surrounding
air.
wire 2a
inside the metal
External field
2ar
wire
diameter
radius
of circle
Figure 5.6: Circular wire ring, and its cross section.
The inductance of a wire ring can be found from
L=
SB dS
I
=M
Iwhere B can be found by integrating the field contributions (using the Biot-Savart law, reviewed in Section 2.5) from each elemental current segmentaround the ring.
Considering only the contribution from the flux outside the wire, it canbe shown, by integration, that M Ir[ln
8ra 2] and hence the external
inductance (Ramo et al.) is
Lext= MI 0r[ln8r
a 2]
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for the case where a
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5.4.2 Inductance of a Short Coil (short length to radius ratio)
r
Radius
Figure 5.7: Short inductor.
For a short length to radius ratio, the external inductance of an Nturn coilis N2 times that of a single turn, i.e.
L N20r[ln
8r
a
2]
When winding a coil, it is useful to remember that the inductance isproportional to the square of the number of turns. The inductance may be
increased by winding the coil on an iron or ferrite rod or on a toroid, whichhas a relative permeability of hundreds or thousands that of air.
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5.4.3 Inductance of a Long Multi-turn Coil (long length to radiusratio)
length l
Integration contour
B ~ 0 outside
I
I
rRadius
B field
Figure 5.8: Long inductor.
For a long coil of length l, containing Nturns carrying current I, the coilcan be treated as a wrapped current sheet (row of dots in the illustra-tion), from whichHcan easily be obtained by application of Amperes lawH dl=
SJ dS. Consider the integration contour shown in the sketch.
The vertical side contributions to the integral are negligibly small becausethe flux lines are perpendicular to the contour. Outside the coil, the fluxlines spread out, and H becomes negligibly small compared to inside thecoil. Consequently, we can also ignore the horizontal segment of the integral
on the outside of the coil. Thus H dl H l whereHis the magnetic fieldinside the coil. The total current passing through the integration contourisSJ dS= N I.
Thus Amperes law impliesH l NIand from which
H NI/l
and B 0NI/l
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An additional point to note is that this result is independent of the exactposition of the the horizontal segment of the contour within the coil. Thisimplies that the field intensity is uniform in a cross section of a long coil,
and hence the flux through the cross section of coil is = Br2 =0Hr2(i.e. linking one turn of the coil)
The inductance is the ratio of the total flux linking the coil to the current,being
L=
SB dS
I =
N
I =
N Br2
I
N(0NI/l)r2
I =
0r2N2
l
5.4.4 Inductance of an Intermediate Length Multi-turn Coil
In cases where the coil can neither be considered very long or very short,the following approximate formula is commonly used:
L 0r2N2
l+ 0.9r
The formula incorporates an empirical correction factor (+0.9r) in the de-
nominator.
5.5 Mutual Inductance
If two wire coils are close to one another, flux resulting from current flowingin one coil will thread the coil of the other. Thus a changing current in onecoil, will result in a changing flux in the other and hence induce a voltage
across its terminals. This is the basis of a transformer.
I2(t)I1(t)
V1(t) V2(t)
1+1from2 2+2from1
Figure 5.9: Coupled coils.
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Consider two coils in close proximity, one containing N1 turns, and theother containingN2 turns. Although not shown in the sketch, these coilsare parts of circuits and carry currents.
Let 1(t) be the component of the flux threading coil 1, resulting purelyfrom the currentI1(t) flowing in coil 1. Let the 1 from 2be the flux threadingcoil 1 arising from the currentI2(t) in coil 2. From Faradays law, the voltageacross the terminals of coil 1 is
V1(t) =d (N11+N11 from 2)
dt =N1
d1dt
+N1d1 from 2
dt
where N11 is the total flux threading the multi-turn coiled surface. Since
1 is proportional to I1 and 1 from 2 is proportional to I2, we have,
V1(t) =L1dI1dt
+M12dI2dt
where L1 is the self inductance constant of coil 1, and M12 is another con-stant. Since N1
d1 from2dt =M12
dI2dt ,
M12=
N11 from 2
I2
Similarly, the voltage across the terminals of coil 2 is
V2(t) =d (N22+N22 from 1)
dt =N2
d2dt
+N2d2 from 1
dt
and
V2(t) =L2dI2dt
+M21dI1dt
where L2 is the self inductance constant of coil 2, andM21 is a constant:
M21=N22 from 1
I1
It can be shown (consult more detailed texts), that regardless of the ge-
ometry, M12 = M21. The constant M = M12 = M21 must have the same
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units ofL1 and L2 being henrys, and is known as the mutual inductancebetween the coils5.
Mutual Inductance for Tightly Coupled Coils
A special case is when the coils are tightly coupled, e.g. stacked on topof one another such that 1 from 2 = 2 and 2 from 1 = 1 (or coils wouldaround a common torroid). For this case,
M12dI2dt
=N1d1 from 2
dt =N1
d2dt
=N1
L2N2
dI2dt
and hence M12= N1N2L2. Similarly,M21= N2N1
L1.
SinceM=M12= M21, the product M12M21 yields
M2 =N1N2
L2N2N1
L1= L1L2
orM= L1L2
for tightly coupled coils.The voltage ratio is
V1V2
=L1
dI1dt +M
dI2dt
L2dI2dt +M
dI1dt
=N21 L1turn
dI1dt +
N21 L1turnN
22 L1turn
dI2dt
N22 L1turndI2dt +
N21 L1turnN
22 L1turn
dI1dt
=N1N2
which is a well known result for tightly coupled coils.
5One could in principle, calculate M12 = N11 from 2
I2for a given coil geometry by doing a surface inte-
gration ofB(i2) dS to obtain 1 from2(I2) = S1 B(i2) dS. B(i2) can be obtained directly from theBiot-Savart law (which requires a contour integration along coil 2). There is a better way to do it(consult other texts for details).
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5.6 Kirchhoffs Voltage Law
The relationship between Kirchhoffs law for a lumped element circuit model
and the physical component layout, is established by application of Fara-days law.Consider applying Faradays law to the closed contour indicated by the
dotted line in the following physical circuit representation. DefinedSpoint-ing into the page, which implies the integration direction is clockwise.
Coil
L
C
V(t)
R
0 3
3
2
0
1
1
I(t)
V10
2
V21
V03
V32
Figure 5.10: Series circuit loop - Faradays law is applied along the dotted line to deriveKirchhoffs voltage law.
Since E 0 in the wires, the voltage drops around the circuit occuracross the components. Thus, we can write
Edl=
10
Edl 2
1
Edl 3
2
Edl 0
3
Edl=S
B
t dS=
dMdt
or
V10+V21+V32+V03 =dM
dtThe flux threading the loop can be split into three contributions:
M= applied+ self+ mutual
where
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appliedrefers to any flux imposed on the circuit e.g. wave a bar magnetpast the circuit.
selfrefers to the flux generated by the current flowing in the circuitloop itself (the circuit can be thought of as a single turn inductor).self = LselfI where Lself is the self inductance of the loop, whichcarries currentI.
mutual refers any leakage flux from other parts of the circuit (notablythe inductive element) that threads the loop.
Substituting the lumped element relationships derived above,
V(t) IR 1
C
tt0
I(t)dt LdI
dt =
dMdt
Kirchhoffs lawN
i=1 Vi= 0 describes the circuit model, and hence wemust introduce additional model component(s) into the circuit modelto account for the term dMdt .
It is noted that the term dMdt
will modify the current flowing in thecircuit, and should be included for accurate prediction of the behaviourof the circuit.
In practice however, this term is usually small compared to the otherterms, and is often neglected in practical circuit design.
For the complete model shown below, the equation in the form of Kirchhoffslaw is written as
V(t) IR 1
C
tt0
I(t)dt LdI
dt
dapplieddt
dself
dt
dmutualdt
= 0
orNi=1
Vi= 0
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(the direction of positive flux)
R
L
I
C
V(t)
dSpoints into the page
applied
t
Lself
Figure 5.11: Circuit modified to incorporate an additional series inductor Lselfwhich mod-els the series inductance of the loop, and an additional voltage source whichmodels unwanted external signals.
The term dself
dt =LselfdIdt resulting from the current in the loop, is mod-
elled by a (small) series inductance Lself. A feeling for the magnitude ofthis self inductance can be gained by considering a circuit arranged in a
circular loop of radius 10 cm. We previously calculated the self inductanceof a wire ring of radius 10 cm and wire radius of 0.5 mm to be 0.7 H.At an operating frequency of say 1 kHz, the AC reactance of this termis XL = 2f Lself 4 10
3 ohms, which is usually small enough to beneglected from calculations. At higher frequencies, this term may becomesignificant.
The term dapplied
dt arises if the circuit is exposed to some externally gen-
erated AC field, e.g. a nearby transmitter like a cell phone, or perhaps amotor, or close-by transformer. Usually this term can be neglected. Of
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course radio waves are ever present, but their contribution is usually in-significant compared to the voltage levels in the circuit. Circuits can beshielded from external sources by placing them in a metal enclosure known
as aFaraday cage6.The term dmutualdt in this context arises from the leakage flux from the
inductor L, and is in practice usually small compared to the voltage dropacross the (multi-turn) inductor. The net effect may either be to increaseor decrease the current in the circuit, depending on the physical orientationof the inductor.
NOTE: The fluxSB dSrequires the direction ofdSto be defined. Since
the integral of E is taken clockwise around the loop, the right hand ruletells us that dSpoints into the page. The flux will be a positive quantity ifB(threading the loop) points into the page.
6A Faraday cage will provide good shielding from DC electric fields. DC magnetic fields however dopenetrate metal enclosures. e.g. the earths magnetic field is still detected by a magnetic compasswithin a Faraday cage. The degree of penetration of time-varying AC electromagnetic fields is afunction of a frequency dependent parameter of the metal known as the skin depth, which will be
studied later in this course. For good shielding at a particular frequency, the enclosure wall should beconsiderably thicker that the skin depth (at that frequency).
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5.7 Kirchhoffs Current Law at a Node
Consider the illustration in Figure 5.12 showing four wires connecting to a
node carrying currentsI1, I2, I3andI4. Kirchhoffs node current law statesthat the sum of all currents leaving the node equals zero, i.e.
Ni=1
Ii= I1+I2+I3+I4= 0
closed surface S2
Displacement current "flows" between plates
closed surface S1
I1
I2
I3
I4
ndS
Figure 5.12: The relationship between Kirchhoffs cuurent law at a node the continuity
equation.
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Consider now the continuity of charge relationshipS
J dS= d
dt
V
dV = dQ
dt
which states that the total conduction current leaving an arbitrary closedsurfaceSis equal to (minus) the rate of change of charge within the volumeV enclosed by S. One can re-express the continuity relationship in a formthat looks similar to Kirchhoffs law by moving the charge term to the lefthand side:
S
J dS + d
dt
V
dV = 0
The 2nd term can be expressed as a surfaceintegral overSby substitutingGauss law
dV =
SD dS,
S
J dS + d
dt
S
D dS= 0
Moving the time derivative within the integral, the continuity equation be-comes
SJ dS + SD
t dS= 0
which says that the sum of the conduction current Ic and the displacementcurrentId leaving an arbitrary closed surface
7 is zero. I.e. for any closedsurface S,
Ic+Id= 0
where
Ic= SJ dS
and
Id=
S
D
t dS
Thus we have derived a generalised form of Kirchhoffs current law, whichcan be applied to an arbitrary closed surface.
7It is worth noting that the total displacement current flowing out of a closed surface is equal to the timerate of change of charge enclosed by the surface, i.e. Id=
S
Dt dS =dQ/dt.
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For example, consider the closed surface S = S1 surrounding the node inFigure 5.12. There are N= 4 wires piercing the surface and joining at thenode.
For S1, the total conduction current leaving the surface is
Ic=
S1
J dS= I1+I2+I3+I4
The displacement current is typically insignificant (there is no significantcharge build up withinS1, i.e.
Id=
S1
D
t dS=
dQ
dt 0
Thus we haveI1+I2+I3+I4 0
If one shrinks surface S1 to a tiny surface surrounding the node, the dis-placement current shrinks to zero and the relationship converges exactly toKirchhoffs law.
If however, we choose a surface S=S2in such a way as to pass between theplates of the capacitor as illustrated in Figure 5.12, then we have a slightlymore subtle situation.As there is one less wire cutting the surface, the total conduction current is
Ic=
S2
J dS= I1+I2+I3
There is however a significant charge build-up on the plate(s) of the capac-itor as a result of current I
4. The charge Q
plateon the plate (within S
2)
builds up at a rate of dQplate
dt =I4.
Thus the continuity relationshipS
J dS= dQ
dt
applied to surface S2 becomes
I1+I2+I3 dQplatedt
= I4
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does enter the metal, decays exponentially with a decay constant called theskin depth (covered later in the course).
5.10 Twisted Pair Cables
An interesting application of field theory concerns the understanding ofhow twin-wire transmission lines are influenced by electric and magneticfields. The following figure depicts a parallel wire (non twisted) transmissionline, which could be used to carry a signal from one location to another.Such parallel wire transmission lines are particularly susceptible to inductivecoupling of magnetic fields, especially when several signals need to be carriedin the same bundle. Changing magnetic flux d/dtwithin the circuit loopinduces an additional voltage which adds to the signal voltage in accordancewith Faradays law. A clever solution to minimising inductive coupling isto reduce the net flux, by twisting the pair of wires as illustrated. Theflux contributions B dS in adjacent twists are opposite in polarity andwill tend to cancel, resulting in reducedd/dtand hence reduced magneticfield interference. This type of cable is known as twisted pair and is very
commonly used for data networks.In addition to the minimization of magnetic coupling, the twisting also
improves the immunity to capacitive coupling. If, for example, the cablelies close to a conductor that is varying in potential relative to ground, likethe live wire wire 50 Hz mains supply, this 50 Hz signal will capacitivelycouple to the conductors (imagine small valued capacitors between the 50Hz conductor and cables two wires). Twisting the cable, creates a moresymmetrical coupling arrangement, independent of the of the orientation ofthe pair, hence causing the effect to be common to both wires. A differentialamplifier at the receiver with a high common-mode rejection ratio extractsthe desired differential signal, and removes the common capacitively coupledinterference.
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V(t)
V(t) dm/dt
netdm/dt 0
Figure 5.13:
Figure 5.14: Illustrations of capacitive coupling onto parallel wire transmission lines for thecase of balanced versus unbalanced driving circuitry.
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Sometimes twisted pairs are also shielded, offering increased immunity toelectromagnetic interference and noise. The shielding also further reducesradiation from the cable itself. Several twisted pairs are sometimes bundled
within the same cable. The use of twisted pairs offers significantly lowercross talk between data channels compared to non-twisted side-by-sidewires within a cable.
Unshielded twisted pair (UTP) cable is now used for connecting standardPCs in in-door local area networks (LANs). UTP network cables replacedpreviously used 50 ohm coaxial cables for LANs because UTP cables arecheaper to manufacture than coaxial cable, and offer adequate immunityto electromagnetic interference. UTP network cable is typically used for
distances up 100m.The Cat-5e series cable is the cable commonly used for PC LANs (for both
100 Mbit/s and gigabit ethernet networks), and is designed to carry frequen-cies up to 100 MHz. PC LAN network cables contain four unshielded twistedpairs, with RJ-45 connectors on each end. The characteristic impedance ofCat-5e is 100 ohms.
Mechanical arrangement within the cable can further reduce coupling be-
tween pairs. For example the Power Cat-6 four pair cable sold by RSElectronics contains four unshielded twisted pair (UTP) cables, with a cen-tral separator, and is designed to support high speed data transmissionsystems (frequencies up to 250 MHz).
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Cat-5e
Cat-6
RJ-45 connector
Figure 5.15: Photos from from RS website; connector from Intel website
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