Transcript Document

Conductors and Dielectrics
• Conductors
–
–
–
–
–
Current, current density, drift velocity, continuity
Energy bands in materials
Mobility, micro/macro Ohm’s Law
Boundary conditions on conductors
Methods of Images
• Dielectrics
–
–
–
–
Polarization, displacement, electric field
Permittivity, susceptibility, relative permittivity
Dielectrics research
Boundary conditions on dielectrics
Conductors and Dielectrics
• Polarization
– Static alignment of charge in material
– Charge aligns when voltage applied, moves no further
– Charge proportional to voltage
• Conduction
– Continuous motion of charge through material
– Enters one side, exits another
– Current proportional to voltage
• Real-world materials
– Plastics, ceramics, glasses -> dielectrics (maybe some conductivity)
– Metals -> conductors, semiconductors, superconductors
– Cement, Biosystems -> Both (water high dielectric, salt conductivity)
Current and current density
• Basic definition of current C/s = Amps
• Basic current density (J perp. surface)
• Vector current density
n
Current density and charge velocity
• Basic definition of current
∆𝑄 𝜌𝑣 ∆𝑣 𝜌𝑣 𝑑𝑆 ∆𝑥
𝐼=
=
=
= 𝜌𝑣 𝑑𝑆 𝑣
∆𝑡
∆𝑡
∆𝑡
• Combining with earlier expression
• Gives current density
𝑱 = 𝜌𝑣 𝒗
Charge and current continuity
• Current leaving any closed surface is time rate of change of charge
within that surface
• Using divergence theorem on left
• Taking time derivative inside integral
Qi(t)
• Equating integrands
Example – charge and current continuity
•
Given spherically symmetric current density
•
Current increasing from r = 5m to r= 6m at t=1s
1
𝑱 = 𝑒 −𝑡 𝒂𝒓
𝑟
𝐼 = 𝐽𝑟 𝑆 =
•
1 −1
𝑒 4𝜋52 = 23.1𝐴 @ 5𝑚,
5
1 −1
𝑒 4𝜋62 = 27.7@ 6𝑚
6
^^ Why is current
increasing ?
Current density from continuity equation
𝜕𝜌
1 𝜕
1
1
= −𝛻 ∙ 𝐽 = − 2
𝑟 2 𝑒 −𝑡 = − 2 𝑒 −𝑡
𝜕𝑡
𝑟 𝜕𝑟
𝑟
𝑟
•
Charge density ρ integral w.r.t. time
𝜌=−
•
Drift velocity is thus
1
𝑟2
𝑒 −𝑡 𝑑𝑡 =
1 −𝑡
𝑒 + 𝑐𝑜𝑛𝑠𝑡(𝑟)
𝑟2
1 −𝑡
𝑒
𝐽𝑟
𝑣𝑟 =
= 𝑟
=𝑟 𝑚 𝑠
1 −𝑡
𝜌𝑣
𝑒
𝑟2
<<Some central
repulsive force!
Energy Band Structure in Three Material Types
Discrete quantum states broaden into energy bands in condensed materials with overlapping potentials
•
Valence band – outermost filled band
•
Conduction band – higher energy unfilled band
Band structure determines type of material
a)
Insulators show large energy gaps, requiring large amounts of energy to lift electrons into the conduction
band. When this occurs, the dielectric breaks down.
b)
Conductors exhibit no energy gap between valence and conduction bands so electrons move freely
c)
Semiconductors have a relatively small energy gap, so modest amounts of energy (applied through heat,
light,or an electric field) may lift electrons from valence to conduction bands.
Ohm’s Law (microscopic form)
Free electrons are accelerated by an electric field. The applied force on an electron of charge Q = -e is
𝐚=
−𝑒𝐄
𝑚
But in reality the electrons are constantly bumping into things (like a terminal velocity) so they attain an
equilibrium or drift velocity:
where e is the electron mobility, expressed in units of m2/V-s. The drift velocity is used in the current density
through:
So Ohm’s Law in point form (material property)
With the conductivity given as:
S/m (electrons)
S/m (electrons/holes)
Ohm’s Law (macroscopic form)
•
For constant electric field
𝑎
𝑉𝑎𝑏 =
𝑬 ∙ 𝒅𝑳 = 𝐸𝐿
𝑏
•
Ohm’s Law becomes
•
Rearranging gives
•
•
•
𝐼
𝑉
= 𝐽 = 𝜎𝐸 = 𝜎
𝑆
𝐿
𝐼=
𝜎𝑆
𝑉
𝐿
𝐺=
𝜎𝑆
𝐿
𝑉=
𝐿
𝐼
𝜎𝑆
𝑅=
𝐿
𝜌𝐿
=
𝜎𝑆
𝑆
𝑠𝑖𝑒𝑚𝑒𝑛𝑠
(𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒)
Or
Variation with geometry
Conductance vs. Resistance
𝑜ℎ𝑚𝑠
(𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒)
Ohm’s Law example 1
•
Checking ohms law microscopic form
7.65 × 106
•
𝐴
= 𝐽 = 𝜎𝐸 = 5.8 × 107 𝑆 𝑚 (0.132 𝑉 𝑚)
𝑚2
Mobility of copper is 0.0032 m2/V-s
𝑣𝑑 = −𝑢𝑒 𝐸 = .0032 𝑚2 𝑉𝑠 0.132 𝑉 = 0.000422 𝑚 𝑠
•
Charge density
𝜎 = −𝜌𝑒 𝜇𝑒
𝜌𝑒 = −1.81 × 1010 𝐶 𝑚3
Ohm’s Law example 2
𝑣
a)
𝐸 = 𝑢𝑑
b)
𝐽 = 𝜎𝐸
c)
𝐸=𝑑
d)
𝐽 = 𝐴𝑟𝑒𝑎
𝑒
𝑉
𝐼
𝐽 = 𝜎𝐸
𝐽 = 𝜎𝐸
Boundary conditions for conductors
• No electric field in interior
– Otherwise charges repel to the surface
• No tangential electric field at surface
– 𝐸𝑡 = 0
– Otherwise charges redistribute along surface
• Normal electric field at surface
– 𝜀𝑜 𝐸𝑛 = 𝐷𝑛 = 𝜌𝑠
– Displacement Normal equals Charge Density
(Gauss’s Law)
Boundary Condition for Tangential Electric Field E
Over the rectangular integration path, we use
or
To find:
These become negligible as h approaches zero.
dielectric
Therefore
n
More formally:
conductor
Boundary Condition for the Normal Displacement D
Gauss’ Law is applied to the cylindrical surface shown below:
This reduces to:
as h approaches zero
dielectric
Therefore
n
More formally:
s
conductor
Summary
Tangential E is zero
At the surface:
Normal D is equal to the surface charge density
Example - Boundary Conditions for Conductors
•
Potential given by
𝑉 = 100(𝑥 2 − 𝑦 2 )
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Potential at (2,-1,3) is 300 V. Also 300 V
along entire surface where
300 = 100(𝑥 2 − 𝑦 2 )
•
Thus we can “insert” conductor in region
provided the conductor follow hyperbola
3 = 𝑥2 − 𝑦2
•
The Electric Field is at all times normal
to conducting surface
𝐸 = −𝛻𝑉 = −200𝑥𝒂𝒙 + 200𝑦 𝒂𝒚
•
Electric field at point 2,-1,3)
– Ex = -400 V/m, Ey = -200 V/m
– Down and to left
Example – Streamlines of Electric Field
•
Slope of line equals electric field ratio
𝑑𝑦
𝑑𝑥
•
=
𝐸𝑦
=
𝐸𝑥
200 𝑦
−200 𝑥
Rearranging
𝑑𝑥
𝑥
+
𝑑𝑦
𝑦
=0
ln 𝑥 + ln 𝑦 = 𝐶1
𝑥𝑦 = 𝐶2
•
Evaluate at P(2,-1,3)
𝑥𝑦 =-2
=−
𝑦
𝑥
Boundary condition example (from my phone)*
* www.mathstudio.net
Method of Images
The Theorem of Uniqueness states that if we are given a configuration of charges and
boundary conditions, there will exist only one potential and electric field solution.
In the electric dipole, the surface along the plane of symmetry is an equipotential with V = 0.
The same is true if a grounded conducting plane is located there.
So the boundary conditions and charges are identical in the upper half spaces of both configurations
(not in the lower half).
In effect, the positive point charge images across the conducting plane, allowing the conductor to be
replaced by the image. The field and potential distribution in the upper half space is now found much
more easily!
Forms of Image Charges
Each charge in a given configuration will have its own image
Example of the Image Method
Want to find surface charge density on conducting plane at the point (2,5,0). A 30-nC line of charge
lies parallel to the y axis at x=0, z = 3.
First step is to replace conducting plane with image line of charge -30 nC at z = -3.
Example of the Image Method (continued)
Vectors from each line charge to observation point:
Electric Fields from each line charge
-
Add both fields to get: (x component cancels)
Example of the Image Method (continued)
Electric Field at P is thus:
Displacement is thus
Charge density is
n
𝜌𝑠 = 𝑫 ∙ 𝒏
𝑆
= −2.20 𝒂𝒛 ∙ 𝒂𝒛
= −𝟐. 𝟐𝟎 nC/𝑚2
D
Image Method using Potentials
•
Conducting plane at x = 4 with vertical wire in front.
•
Potential for wire in front at x = 6, y=3:
𝑉 𝜌 =−
•
+𝑘
Boundary condition for wire in front at x = 6, y=3:
0=−
•
𝜌𝑙
ln 𝜌
2𝜋𝜀𝑜
40 𝑛𝐶
ln 2
2𝜋𝜀𝑜
+𝑘
𝑘 = 498.6
Boundary condition for image wire in back at x=2, y=3:
0=−
−40 𝑛𝐶
ln 2
2𝜋𝜀𝑜
+𝑘
𝑘 = −498.6
Image Method using Potentials (cont)
•
Total potential becomes
𝑉 𝜌 =−
•
40 𝑛𝐶
−40 𝑛𝐶
2𝜋𝜀𝑜
ln 𝜌𝑖𝑚𝑎𝑔𝑒
At point (7,-1,5) 𝜌 = 17 𝜌𝑖𝑚𝑎𝑔𝑒 = 41 gives
𝑉 𝜌 =−
•
ln 𝜌 −
2𝜋𝜀𝑜
40 𝑛𝐶
2𝜋𝜀𝑜
ln 17 +
40 𝑛𝐶
2𝜋𝜀𝑜
41 = −1019 + 1335 = 316
To get electric field must write V(ρ) as V(x,y) and take gradient
𝑉 𝑥, 𝑦 = −
𝐸 = −𝛻𝑉 𝑥, 𝑦
=
40 𝑛𝐶
2𝜋𝜀𝑜 2
40 𝑛𝐶
ln
2𝜋𝜀𝑜
𝐸𝑥 = −
𝑥−6
2
+ 𝑦−3
2
+ 𝑦−3
= 719.34
+
40 𝑛𝐶
ln
2𝜋𝜀𝑜
𝑥−2
2
+ 𝑦−3
2
𝜕𝑉
=
𝜕𝑥
2(𝑥 − 6)
𝑥−6
2
2
𝑥−6
2
+ 𝑦−3
7−6
7 − 6 2 + −1 − 3
−
2
2
−
2(𝑥 − 2)
2
𝑥−2
2
+ 𝑦−3
7−2
7 − 2 2 + −1 − 3
2
2
𝑥−2
= −45.3 𝑉/𝑚
2
+ 𝑦−3
2
Dielectrics
•
•
•
•
•
Material has random oriented dipoles
Applied field aligns dipoles (negative at (+) terminal, positive at (-)
terminal
Effect is to cancel applied field, lower voltage
OR, increase charge to maintain voltage
Either increases capacitance C= Q/V
Review Dipole Moment
• Define dipole moment
𝒑 = 𝑄𝒅
• Potential for dipole
• Written in terms of dipole
moment and position
• Dipole moment determines “strength” of polar molecule
amount of charge (Q) and offset (d) of charge
Polarization as sum of dipole moments
(per volume)
Introducing an electric field may increase the charge separation in each dipole, and possibly re-orient dipoles so that
there is some aggregate alignment, as shown here. The effect is small, and is greatly exaggerated here!
E
The effect is to increase P.
= 𝑛𝒑 = 𝑛𝑄𝒅
n = charge/volume
p = polarization of individual dipole
P = polarization/volume
Polarization near electrodes
•
From diagram
–
---------------positive
–
E
–
•
Excess positive bound charge near top negative
electrode
Excess negative bound charge near bottom
positive electrode
Rest of material neutral
Excess charge in bound (red) volumes
∆𝑄 = 𝑛𝑄𝑑 ∆𝑆
neutral
•
Writing in terms of polarization
∆𝑄 = 𝑷 ∆𝑆
•
Writing similar to Gauss’s law
negative
+++++++++++++
𝑄𝑏 = −
𝑃 ∙ 𝑑𝑠
(Note dot product sign, outward normal
leaves opposite charge enclosed)
Combining total, free, and bound charge
•
Total, free, and bound charge
𝑄𝑡 = 𝑄𝑓 + 𝑄𝑏
---------------positive
E
•
Total
𝑄𝑡 =
•
Free
neutral
𝑄𝑓 =
•
+++++++++++++
•
𝐷 ∙ 𝑑𝑠
Bound
𝑄𝑏 = −
negative
𝜀𝑜 𝐸 ∙ 𝑑𝑠
𝑃 ∙ 𝑑𝑠
Combining
𝐷 = 𝜀𝑜 𝐸 + 𝑃
D, P, and E in Dielectric
• D continuous
• Polarization increases
• E decreases
• 𝑫 = 𝜀𝑜 𝑬 + 𝑷
• C/m2
Charge Densities
Taking the previous results and using the divergence theorem, we find the point form expressions:
Bound Charge:
Total Charge:
Free Charge:
Electric Susceptibility and the Dielectric Constant
A stronger electric field results in a larger polarization in the medium. In a linear medium, the relation
between P and E is linear, and is given by:
where e is the electric susceptibility of the medium.
We may now write:
where the dielectric constant, or relative permittivity is defined as:
Leading to the overall permittivity of the medium:
where
Isotropic vs. Anisotropic Media
In an isotropic medium, the dielectric constant is invariant with direction of the applied electric field.
This is not the case in an anisotropic medium (usually a crystal) in which the dielectric constant will vary
as the electric field is rotated in certain directions. In this case, the electric flux density vector components
must be evaluated separately through the dielectric tensor. The relation can be expressed in the form:
Permittivity of Materials
• Typical permittivity for various solids and liquids.
–
–
–
–
–
–
Teflon – 2
Plastics - 3-6
Ceramics 8-10
Titanates>100
Acetone 2 1
Water 78
• Actual dielectric “constant” varies with:
– Temperature
– Direction
– Field Strength
– Frequency
– Real & Imaginary components
Variation with frequency
•
Charge polarization due to:
–
–
–
–
•
Dielectric relaxation
–
–
•
Ionic (low frequency)
Orientation (medium, microwave)
Atomic (IR)
Electronic (Visible, UV)
As frequency is raised, molecule can
no longer “track”.
Real permittivity decreases and
imaginary permittivity peaks
In medium and microwave range
– Rotation, reorientation, etc >>
•
Modeling:
–
Permittivity & impedance diagrams.
–
Statistical relaxation functions
(Debye, Cole Davidson).
IEEE – March 2005
Application to Polymer Composites
• Dielectric Permittivity in Epoxy Resin 10Hz -10 MHz
•
•
•
•
Polar-group rotation in epoxy resin.
Low-frequency range 10 Hz – 10 MHz.
Permittivity-loss transition at 1 MHz, at –4°C.
Transition frequency increases with temperature.
www.msi-sensing.com
IEEE – March 2005
Dielectric Permittivity in Epoxy Resin 1 MHz -1 GHz
•
•
•
•
Aerospace resin Hexcel 8552.
High frequency range 1 MHz – 1 GHz.
Temperature constant 125°C, transition decreases with cure.
TDR measurement method.
www.msi-sensing.com
IEEE – March 2005
Permittivity in Epoxy Resin during Complete Cure Cycle
www.msi-sensing.com
IEEE – March 2005
Application to cement hydration
•
Cement Conductivity - Variation with Cure
•
Imaginary counterpart of real permittivity (’’).
•
Multiply by  to remove power law (o’’).
•
Decrease in ion conductivity, growth of intermediate feature with cure
•
Frequency of intermediate feature does not match permittivity
www.msi-sensing.com
IEEE – March 2005
Cement Cure -Dielectric Relaxation Model
Requirements:
•
•
Provide free-relaxation, two intermediate-frequency relaxations
Provide conductivity and electrode polarization
Debye for free & medium. Cole-Davidson for low. (literature, biosystems)
 ( )    
Cm
1  i  m
 ( )   
C
l
(1 i  ) 
l
Combined


C


l
Re 



 (1  i ) 

l 


C


l
 Im
 o 
 (1  i )  

l 
 C
m 
Re 
1 i m 


 C
m   
Im
o
1  i m 
 C
f
Re 
1 i

f




 C
f
 Im
1 i

 C p 
f


 o 

 Ci
( permittivity)
(conductivity)
9 variables fit over entire range, real & imaginary, 2-stage fit, f = 8.2 ps
www.msi-sensing.com
IEEE – March 2005
Cement Cure - Model Fitting
•
Fits permittivity – both low and free relaxation.
•
Fits conductivity – both medium and free relaxation.
•
Fits permittivity polarization.
•
Fits conductivity baseline.
www.msi-sensing.com
IEEE – March 2005
Other applications
• Other Applications
–
–
–
–
–
–
–
–
Bio
Liquid Crystal
Composite polymers
Titanates
Wireless characterization
MRI dyes
Ground water monitoring
Oil Drilling fluid characterization (GPR)
Boundary Condition for Tangential Electric Field E
Since E is conservative, we setup line integral straddling both dielectrics:
Left and right sides cancel, so
Leading to Continuity for tangential E
And Discontinuity for tangential D
E same, D higher in high permittivity material
Boundary Condition for Normal Displacement D
Apply Gauss’ Law to the cylindrical volume straddling both dielectrics
n
s
Flux enters and exits only through top and bottom
surfaces, zero on sides
Leading to Continuity for normal D (for ρS = 0)
And Discontinuity for normal E
𝜀1 𝐸𝑁1 = 𝜀2 𝐸𝑁2
D same. E lower in high permittivity material
Bending of D at boundary
• Boundary conditions
– DN continuous
–
𝐷𝑇1
𝐷𝑇2
=
𝜀1
𝜀2
• Trigonometry
– tan 𝜃1 =
𝐷𝑇1
𝐷𝑁
– tan 𝜃2 =
𝐷𝑇2
𝐷𝑁
high
low
• Eliminating DN
tan
tan
𝜃1
𝜃2
=
𝐷𝑇1
𝐷𝑇2
=
𝜀1
𝜀2
Example
•
Teflon εr = 2.1
• Displacement and Polarization outside
𝑫 = 𝜀𝑜 𝑬𝒐 + 𝑷
𝑃 = χ 𝜀𝑜 𝐸𝑜 = 0
𝑫 = 𝜀𝑜 𝐸𝑜
• Displacement and Polarization inside
𝑫 = 𝜀𝑜 𝑬𝒐 + 𝑷
𝑷 = χ 𝜀𝑜 𝐸𝑜 = (𝜀𝑟 − 1) 𝜀𝑜 𝐸𝑜
𝑫 = 𝜀𝑟 𝜀𝑜 𝑬𝒐
• At boundary D is continuous, so inside
• 𝐸𝑖𝑛 =
𝐸𝑜𝑢𝑡
𝜀𝑟
= 0.476 𝐸𝑜
• 𝑷 = (𝜀𝑟 −1) 𝜀𝑜 𝐸𝑖𝑛
=
(𝜀𝑟 −1) 𝜀𝑜 (0.476) 𝐸𝑜
=
0.524 𝜀𝑜 𝐸𝑜
Example (continued)
• Polarization up, E field down, D maintains continuity
Example
Quiz 2 – Problem 4.21