Download EGN xxx Electronic MAterials - Help-A-Bull

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
CHAPTER 7: Dielectrics …
What is a capacitor ?
a passive two-terminal electrical component used to store energy in an
electric field
Al metallization
Metal termination
Ceramic
(a)
Epoxy
Polymer film
Leads
(b)
Metal electrode
(a) Single layer ceramic capacitor
(e.g. disk capacitors)
(b) Multilayer ceramic capacitor
(stacked ceramic layers)
Single and multilayer dielectric capacitors
Two polymer tapes in (a) each with a metallized film electrode on the
surface (offset from each other) can be rolled together (like a Swiss rollcake) to obtain a polymer film capacitor as in (b). As the two separate
metal films are lined at oppose edges, electroding is done over the whole
side surface.
Capacitors
What is a dielectric material ?
an electrical insulator that can be polarized by an applied electric field
ε oε r A
dQ
C
... C 
dV
d
What are the ways of increasing capacitance?
Dielectric
+Qo
Co
Qo
+Q
C
i (t)
E
E
V
V
V
Q
Polarization
Polarization
… polar molecules vs. non-polar molecules
Can non-polar molecules become polar ?
E
C x O
Electron cloud
Atomic
nucleus
Electric Dipole Moment,

p  Qâ
Center of negative
charge
pinduced
(a) A neutral atom in E = 0.
(b) Induced dipole moment in a field
The origin of electronic polarization.
The definition of electric dipole mome
Polarization
How do dielectrics increase the capacitance of a capacitor ?
Bound polarization
chargeson thesurfaces
QP
+ QP
Area =A
p
+Q
E
Q
total
(c)
QP
P
+QP
d
Dielectric
+Qo
Co
Qo
V
+Q
C
i (t)
E
E
V
V
V
Q
Polarization,
QP
P
A
Polarization
QP
P
A
P  χeεoE
χe  1  εr
p induced  αE
Susceptibility, Χe: A proportionality constant that indicates the degree of
polarization of a dielectric material in response to an applied electric
field. (dimesnionless)
Permittivity: A measure of how an electric field affects, and is affected by, a
dielectric medium. (F/m)
polarizability … dielectric constant … are material properties.
In a way they are equivalent to what mobility is for conductors and
semiconductors.
Polarization Mechanism
Dependence of ε’r & ε”r frequency depends on
polarization mechanism:
a.
b.
c.
d.
Electronic
Ionic
Interfacial
Orientational
POLARIZATION MECHANISMS
Ionic:
polarization caused by relative displacements between positive and
negative ions in ionic crystals
p+
p-
(a)
x
Ð
+
Cl
Na
p'+
p'-
(b)
E
(a) A NaCl chain in the NaCl crystal without an applied field. Average or net dipole moment per ion
is zero. (b) In the presence of an applied field the ions become slightly displaced which leads to a
net average dipole moment per ion.
POLARIZATION MECHANISMS
Electronic:
the stretching of atoms/electronic clouds under an applied E-field (in
covalent solids)
(a) Valence electrons in covalent bonds in the absence of an applied field. (b) When
an electric field is applied to a covalent solid, the valence electrons in the covalent
bonds are shifted very easily with respect to the positive ionic cores. The whole solid
becomes polarized due to the collective shift in the negative charge distribution of
the valence electrons.
POLARIZATION MECHANISMS
Interfacial:
charge accumulation at defective interfaces (2 material or 2 regions of same
material) leads to the formation of a net polarization vector
Electrode
Electrode
Dielectric
Fixed charge
Mobile charge
(a)
E
Accumulated charge
(b)
E
Grain boundary or interface
(c)
(a) A crystal with equal number of mobile positive ions and fixed negative
ions. In the basence of a field there is no net separation between all the
POLARIZATION MECHANISMS
Orientational (Dipolar):
in “rigid polarized molecule” materials, an applied field aligns the
permanent dipoles to yield a net polarization vector
+Q
t
po = aQ
H+
Cl
po
F
Q
(a)
pav = 0
(c)
pav ¹ 0
(b)
E
(d)
q
F =QE
E
POLARIZATION MECHANISMS
DIELECTRIC LOSS
The dielectric constant is frequency dependent … why ?
The polarization process is not instantaneous … i.e. it takes a finite amount of
time for the molecules to align themselves.
If the applied field is changing so fast that the molecules cannot respond to it
at all … then the polarization is … zero!
Therefore ε is frequency dependent …
εr  εr'  jεr''
DIELECTRIC LOSS
P = P osin(t - )
p
d(0)Eo
E = Eosint
r' and r''
pÐd(0)E
d(0)E
t
r'
E
r (0)
Eo
E
t
0
r''
The dc field is suddenly changed from Eo to E at time t = 0. The
induced dipole moment p has to decrease from d(0)Eo to a final value
of d(0)E. The decrease is achieved by random collisions of molecules
in the gas.
1
v = Vosint
(a)
0.01/


0.1/

1/

(b)
10/

100/

εr  εr'  jεr''
The imaginary part represents dielectric “losses” due to “slow” polarization.
DIELECTRIC LOSS
ε r  ε 'r  jε 'r'
ε o (ε 'r  jε 'r' )A jω  Aε o ε 'r ωAε o ε 'r'
Y  jω  C  jω


d
d
d
P = Posin(t -)
Conductance = Gp = 1/Rp
ε 'r'
loss tangent ... tanδ  '
εr
C
v = V osint
Dielectric Loss
per unit volume,
v = V osint
2
Wvol
V
1
V2


 2 ωε o ε "r  ωE 2 ε o ε 'r tanδ
R P dA d
DIELECTRIC LOSS
1. For the dielectric material as in the figure, calculate the loss tangent
at 10 MHz frequency.
If the parallel plate capacitor is formed with the material, where the
separation is 1mm, plate area is 1mm2 and 1V is applied across…
2. Calculate the value of resistance and capacitance.
3. Calculate the power loss per unit volume.
ε 'r'
1. loss tangent  tanδ  '
εr
d
2. R P 
ωAε o ε 'r'
=1.8x106 ohm
3. Wvol
=0.5
Aε o ε 'r
CP 
d
=1.7x10-14 F
V2
 2 ωε o ε "r  ωE 2 ε o ε 'r tanδ
d
=556
Wcm-3
ε
2
ε’r
1
ε”r
MHz
0
10
f
Matter Polarization & Permittivity

p  Qaˆ
Electric Dipole
Moment,
The definition of electric dipole moment.
Q
C
εr 

Q0 C0
pinduced  αE
Spring action
C x O
Electron cloud
Fr   β x
Atomic
nucleus
Fe   Ze E
Induced Electric Dipole Moment,
Center of negative
charge
pinduced
(a) A neutral atom in E = 0.
E
(b) Induced dipole moment in a field
The origin of electronic polarization.
Z 2e 2
pe  (Ze)x  (
)E
β
Matter Polarization & Permittivity
E
C x O
Electron cloud
Induced Electric Dipole Moment,
Atomic
nucleus
pe  (Ze)x  (
Center of negative
charge
Z 2e 2

)E
pinduced
(b) Induced dipole moment in a field
(a) A neutral atom in E = 0.
The origin of electronic polarization.
Removal of applied E field cause the vibration with a resonant frequency…
F=ma
d 2x
 x  Zme 2
dt
 β
ω0  
 Zme



2
Electronic polarization
resonance frequency
x(t)  xo cos (ω0t)
Electronic Polarizability
 Ze
 e   2
 ω0 me
2



2
Permittivity: Electronic Polarization
Bound polarization
charges on the surfaces
-QP
+Q
E
+QP
Total Polarization
-Q
(b)
Area = A
 p  P   e o E
ptotal
(c)
V
(a)
-QP
Qp d Qp
ptotal
P


volume
Ad
A
P
+QP
Total Polarization
=surface polarization charge density!
d
(a) When a dilectric is placed in an
electric
field,polarization
bound polarization
Also,
total
charges appear on the opposite surfaces. (b) The origin of these
polarization charges is the polarization of the molecules of the
medium. (c) We can represent the whole dielectric in terms of its
surface polarization charges +QP and -QP.
P  Npinduced  N e E
e 
1
o
N e
Permittivity: Electronic Polarization
before insertion of dielectric medium
Bound polarization
charges on the surfaces
-QP
+Q
E
+QP
-Q
(b)
Qo
Qo
o
V
E 


d Co d  0 A  0
o 
Area = A
(a)
-QP
free surface
charge density
ptotal
(c)
V
Qo
A
P
+QP
after insertion of dielectric medium
Q  Qo  Q p
d
(a) When a dilectric is placed in an electric field, bound polarization
charges appear on the opposite surfaces. (b) The origin of these
polarization charges is the polarization of the molecules of the dividing by the area, A
medium. (c) We can represent the whole dielectric in terms of its
0
surface polarization charges +QP and -QP.
   E  p
Permittivity: Electronic Polarization
Bound polarization
charges on the surfaces
-QP
+Q
E
+QP
Substitution for surface charge density
-Q
(b)
Area = A
ptotal
(c)
V
(a)
-QP
  (1   e ) 0 E
P
+QP
r 
Q
 Relative

Qo  o permittivity
d
(a) When a dilectric is placed in an electric field, bound polarization
charges appear on the opposite surfaces. (b) The origin of these
polarization charges is the polarization of the molecules of the
medium. (c) We can represent the whole dielectric in terms of its
surface polarization charges +QP and -QP.
r  1
N e
0
 r  1  e
Relation between polarization
mechanism to relative permittivity
Clausius-Mossotti Equation
The bulk electric field assumption is
not valid in the atomic level
Eloc
Elocal  E 
1
P
3ε0
Lorentz
Field
Electric field at
atomic scale
Eloc
E
E = V/d
x
  1 N

 1

The electric field inside a polarized dielectric at the atomic scale is
not uniform. The local field is the actual field that acts on a
e
r
molecules. It can be calculated by removing that molecules and
evaluating the field at that point from the charges on the plates and
the dipoles surrounding the point.
r
0
Relation between polarization
mechanism to relative permittivity
Clausius-Mossotti Equation
For electronic polarization
Total Polarization
 r  1 N i i

 r  1 3 0
Total Polarization
Clausius-Mossotti Equation
For ionic polarization
pav   e Eloc   i Eloc   d Eloc
Relative permittivity due to ionic and electronic polarization
 r 1
1
N e e  N i i 

 r  1 3 0
Total Polarization
7.9 Electronic and ionic polarization in KCl
KCl has the FCC crystal structure. Lattice parameter is 0.629 nm.
The ionic polarizability per ion pair (per K+-Cl- ion) is 4.58 10-40 F m2.
The electronic polarizability of K+ is 1.264  10-40 F m2 and Cl- is 3.408  10-40 F m2.
Calculate the dielectric constant under dc operation and at optical frequencies.
Experimental values are 4.84 and 2.19.
FCC… 4 KCl ion pairs per unit cell. The number of ion pairs, or individual ions, per unit volume (N) is:
N
4
4

a 3 0.629 10 9 m 3


= 1.607  1028 m-3
 r 1
1


N i e ( K  )  N i e (Cl  )  N i i 
 r  1 3 0
 rop  1
1


N i e ( K  )  N i e (Cl  ) 
 rop  1 3 0
r(op) = 2.18
under dc operation
under optical frequency
Example 7.2, 7.3