Post on 29-Jan-2016
ELEC 3105 Basic EM and Power Engineering
Conductivity / Resistivity
Current Flow
Resistance
Capacitance
Boundary conditions
Conductivity and resistivityThe relaxation time model for conductivity works for most metals and
semiconductors.
In a conductor at room temperature, electrons are in random thermal motion, with mean time between collisions.
Random motion of the electron in the metal. Electron undergoes collisions then moves off in different direction.
0E
electron
collision
Conductivity and resistivityThe relaxation time model for conductivity works for most metals and semiconductors.
In a conductor at room temperature, electrons are in random thermal motion, with mean time between collisions.
Electrons acquire a small systematic velocity v* component in response to applied electric field
E
electron
collision
Conductivity and resistivityFor a weak electric field v* can be obtained.
m = mass of electron = carrier mobility (ELEC 2507)
E
electron
collision
EEm
q
m
Eq
m
Fv
*
{} units of
Vs
m2
Conductivity and resistivity
FOR A STRONG ELECTRIC FIELD
E
for low fields v* proportional to E
For strong electric fields, electrons acquire so much energy between collision that mean time between collisions is reduced.
v*
E
Conductivity and resistivityAs long as we stay in the weak electric field regime, i.e. the linear region of the curve in the previous slide, then the current density can be defined as:
E
v*
E
This regionENqJ
EJ
Nq Conductivity
1
Resistivity
Conductivity of elements
Current flowThe total amount of charge moving through a given cross section per unit time is the current, usually denoted by I.
Conductor ???
vdt
dq v
dt
dqI CURRENT
Current flowIf we consider the current per unit cross-sectional area, we get a value which can be defined any point in space
as a vector, typically denoted
vdt
dq v
J
cross-sectional area A
N charged particles per unit volume moving at v meters per second
dq = N q vdt A Charge moving through cross-sectional area A in time dt
Current flowThe charge density is simply this quantity divided by the unit time and area. The current density is:
vdt
dq v
vNqJ
cross-sectional area A
N charged particles per unit volume moving at v meters per second
dq = N q vdt A Charge moving through cross-sectional area A in time dt
dq = N q vdt A
Current flowThe total current through the end face can be obtained from the current density as an integration over the cross-sectional area of the conducting medium.
vdt
dq v
davNqdaJIAA
cross-sectional area A
TOTAL CURRENT
Current flow
The total charge passing through the cross-sectional area A over a time interval from t1 to t2 can be obtained from:
vdt
Q v
2
1
2
1
t
t A
t
t
dtdaJIdtQ
cross-sectional area A
TOTAL CHARGE
MOSFET
Resistance of conductors: any shape
ab
abab I
VR
b
a
ab dEV
AA
ab dAEdAJI
RESISTANCE
Resistance of conductors: any shape
ab
abab I
VR
b
a
ab dEV
A uniform rectangular bar
ELVab
Electric field is uniform and in the direction of a bar length L.
Resistance of conductors: any shape
ab
abab I
VR
A uniform rectangular bar
Electric field is normal to the cross-sectional area A.
AA
ab dAEdAJI
EAIab
Resistance of conductors: any shape
ab
abab I
VR
A uniform rectangular bar
EAIab ELVab
A
LRab
A
LRab
SUPERCONDUCTORS
Capacitance• Capacitance is a property of a geometric configuration, usually
two conducting objects separated by an insulating medium.
• Capacitance is a measure of how much charge a particular configuration is able to retain when a battery of V volts is
connected and then removed.
• The amount of charge Q deposited on each conductor will be proportional to the voltage V of the battery and some constant C,
called the capacitance.
V
QC Capacitance {C/V}
Parallel plate capacitor
Free space between plates o
z
z dEV0
o
sE
+Q
-Q
V = 0 volts
V = V voltsz
zVo
sz
Plate area APlate separation D
Between plates
A
Qs
At z = D DA
QV
o
DA
QV
o Rearrange
D
AC o
Capacitance of parallel plate capacitor
V
QC
CAPACITORS IN SERIES/ PARALLEL/ DECOMPOSITION
PARALLEL
C1C2
Ceq
21 CCCeq
C1
C2
SERIES
Ceq
21
111
CCCeq
CAPACITORS IN SERIES/ PARALLEL/ DECOMPOSITION
DECOMPOSITION
3
21
111
C
CC
Ceq
Ceq
C1
C2
C3
CAPACITANCE OF A COAXIAL TRANSMISSION LINE
a
bVV L
ba ln2 Prove this result as part of
next assignment.
If we consider as the charge per unit length on each of the two coaxial surface, then:
V
QC
ba
L
VVC
a
bC
ln
2
ab
C
ln
2
m
pF
ab
C r
ln
6.55
L
(ELEC 3909)
CHARGE CONSERVATION AND THE CONTINUITY EQUATION
v
vin dvQ Charge in volume v
Current through surface A A
dAJI
Also recall dt
dQI
The main ingredients to the pie
31
CHARGE CONSERVATION AND THE CONTINUITY EQUATION
v
v
v
dvt
dvJ
Then:
Current out of volume ist
QI in
From divergence theorem
v
v
A
dvt
dAJ
Using previous expressions
Av
dAJdvJ
tJ v
CHARGE CONSERVATION AND THE CONTINUITY EQUATION
Interpretation of equation: The amount of current diverging from am infinitesimal volume element is equal to the time rate of change decrease of charge contained in the volume. I.e. conservation of charge.
tJ v
In circuits: 0inI If no accumulation of charge at node.
33
CHARGE CONSERVATION AND THE CONTINUITY EQUATION
EJ
A charge is deposited in a medium.
tJ v
tE v
AlsofreeD
freeE
t
0
t Tt
oet
CHARGE CONSERVATION AND THE CONTINUITY EQUATION
A charge is deposited in a medium.
Tt
oet
If you place a charge in a volume v, the charge will redistribute itself in the medium (repulsion???). The rearrangement of charge is governed by the constant
T = REARRANGEMENT TIME CONSTANT
T
TCu, Ag=10-19 s Tmica=10 h
Boundary conditionsTangential Component of E
0c
dE
Around closed path (a, b, c, d, a)
0V
ELECTROSTATICS
1
2
Boundary
t̂
a
b
cd
2E
1E
2tE
1tE
Potential around closed path
Boundary conditionsTangential Component of E
1
2
Boundary
t̂
a
b
cd
2E
1E
2tE
1tE
0lim 120,0,
a
d
c
bc
cdabdEdEdE
ELECTROSTATICS
Boundary conditionsTangential Component of E
t̂
a
b
cd
0lim 11220,0,
a
d
c
bc
cdabdEdEdE
012
dEE
012 dEE tt
012 tt EE
12 tt EE The tangential components of the electric field across a boundary separating two media are continuous.
ELECTROSTATICS
Boundary conditionsTangential Component of E
012 tt EE
12 tt EE
At the surface of a metal the electric field can have only a normal component since the tangential component is zero through the boundary condition.
ELECTROSTATICS
01 tEt̂
a
b
cd
metal
02 E
1 nEE ˆ11
Boundary conditions Normal Component ofE
enclosed
c
qdAD
Gauss’s law over pill box surface
ELECTROSTATICS
12
Boundary
n̂A
2E
1E
2nE
1nE
n̂
n
dAdADDdADs
cn
210
lim
ELECTROSTATICS
1
2
Boundary
n̂A
2E
1E
2nE
1nE
n̂
n
Boundary conditions Normal Component ofE
41
dAdADDdADs
cn
210
lim
1
2A
2nE
1nE
dAdADD snn 21
021 dADD snn
snn DD 21
snnEE
2211
The normal components of the electric flux density are discontinuous by the surface charge density.
Boundary conditions Normal Component ofE
ELECTROSTATICS
42
snn DD 21
snD 1
022 nn DE
1nEE ˆ11
11
snE
Boundary conditions Normal Component ofE
ELECTROSTATICS
metal
02 E
At the surface of a metal the electric field magnitude is given by En1 and is directly related to the surface charge density.
Boundary conditionsNormal Component of D
snn DD 21
ELECTROSTATICS
Gaussian Surface
Air Dielectric
Gaussian surface on metal interface encloses a real net charge s.
Gaussian surface on dielectric interface encloses a bound surface charge sp , but also encloses the other half of the dipole as well. As a result Gaussian surface encloses no net surface charge.
snD 1
021 nn DD
21 nn DD
ELEC 3105 Basic EM and Power Engineering
Extra extra read all about it!
44
45
Electric fields in metals
Electric fields in metals
(a) no current Einside = 0 (b) with current Einside 0
Inhomogeneous dielectrics
We can consider an inhomogeneous dielectric as being made up of small homogeneous pieces, at the interfaces of which bound charge will accumulate.
x
D
Suppose that we have a dielectric whose permittivity is a function of x, and a constant D field is directed along x as well.
dielectric
Inhomogeneous dielectrics
We can consider an inhomogeneous dielectric as being made up of a stack of thin sheets of thickness dx and permittivity (x).
x
D In each sheet, positive charges will
accumulate on the right and negative ones on the left, according to the permittivity of the sheet.
Inhomogeneous dielectrics
We can consider an inhomogeneous dielectric as being made up of a stack of thin sheets of thickness dx and permittivity (x).
x
D The charges will mostly cancel by
adjacent sheets, but any difference in permittivity between adjacent sheets d will leave some net charge density.
Inhomogeneous dielectrics
We can consider an inhomogeneous dielectric as being made up of a stack of thin sheets of thickness dx and permittivity (x).
x
D We can express this net bound charge
easily as the difference in polarizations, so that we have:
dx
dP
dx
xPdxxPbound
Inhomogeneous dielectricsIn the more general case when the permittivity is varies in all directions, i. e. (x,y,x).
x
D
We can express this net bound charge easily as the difference in polarizations, so that we have:
Pbound
y
z
dx
dP
dx
xPdxxPbound
PED o
Inhomogeneous dielectricsIn the more general case when the permittivity is varies in all directions, i. e. (x,y,x).
x
D
Take divergence on each side:
y
z
PED o
PED o
freeboundtotal