Transcript Document

The Nernst-Einstein equation indicates that the ratio β /D for a
given material varies only with temperature. Calculate β/D for
oxygen ions in Zr0.8Y0.2O1.9 at 800°C.
1
The Nernst-Einstein equation indicates that the ratio β /D for a
given material varies only with temperature. Calculate β/D for
oxygen ions in Zr0.8Y0.2O1.9 at 800°C.
zq

 DT
k T
  m 2 V 1  s 1 

DT 
m 2  s 1
zq
2 1.6 1019
C


 23
1
k

T
1
.
38

10

1073
J

K
K

C
 21 .61
V  C  K 1  K
2
Simpson and Carter (J. Am. Ceram. Soc. 49 (1966) 139)
measured the self diffusion coefficient for oxygen in
Zr0.85Ca0.15O1.85 and found it to be DO = 2.0·10-7 cm2/s at 1100°C.
Calculate the electrical mobility and conductivity of oxygen ions
based on this. Assume density of Zr0.85Ca0.15O1.85 5.7g/cm3 and
molecular weight 113.15g/mole.
3
SI
O
Zr
O
Zr
O
Zr
Zr
O
Zr
O
Zr
O
O
Zr
O
Zr
O
Zr
Zr
O
Zr
O
Zr
O
O
Zr
O
Ca
O
Zr
Zr
O
Zr
O
Zr
O
zq

 DT
k T
  n z q
ZrO2
Zr0.85Ca0.15O1.85
 num ber
 m3 


4
Simpson and Carter (J. Am. Ceram. Soc. 49 (1966) 139)
measured the self diffusion coefficient for oxygen in
Zr0.85Ca0.15O1.85 and found it to be DO = 2.0·10-7 cm2/s at 1100°C.
Calculate the electrical mobility and conductivity of oxygen ions
based on this. Assume density of Zr0.85Ca0.15O1.85 5.7g/cm3.
zq
2 1.6 1019  2 1011 C  m 2  s 1

 DO 
k T
1.3810 23 1373 J  K 1  K
  3.381010
C  m 2  s 1
C  m 2  s 1
m2


1
1
J  K  K V C  K  K V  s
5
Electrical conductivity
Zahl der Zr0.85Ca0.15O1.85-Einheiten per m3
  n z q
5.7 103  6.0221023  0.15 kg  m3  m ole1
n

0.11315
kg  m ole1
n  4.54 10
27
1
m3
  n  z  q    4.5410  2 1.6 10  3.3810
27
3
19
10
1
1
 0.492
1
m  C  m V  s   m
2
6
1
For intrinsic silicon, the room-temperature electrical conductivity is
410-4 Ω-1m-1; the electron and hole mobilities are, respectively,
0.14 and 0.048 m2V-1s-1. Compute the electron and hole concentrations
at room temperature.
  n z q
  n  q  e  p  q   h
7
For intrinsic silicon, the room-temperature electrical conductivity is
410-4 Ω-1m-1; the electron and hole mobilities are, respectively,
0.14 and 0.048 m2V-1s-1. Compute the electron and hole concentrations
at room temperature.
Solution:
  n z q
  n  q  e  p  q   h

4 104
1m1
n p

q(e   h ) 1.6 1019  (0.14  0.048) Cm2V 1s 1
n  p  1.331016 m3
8
Calculate concentration of the charge carriers in intrinsic Si in a
function of temperature.
nil  e,  h
K g  Ne N h (mole fractions)
12
K g  0.9 10
Temperature dependence:
 T exp
3
 Eg
kT
Eg=1.14 eV energy gap,
k=8.63∙10-5 eV/K
Temperature (K)
300
Kg
Ne=Nh
700
1000
9
Intrinsic Silicon
nil  e,  h
K g  Ne N h (mole fractions)
12
K g  0.9 10
Temperature dependence:
 T exp
3
 Eg
kT
Eg=1.14 eV energy gap,
k=8.63∙10-5 eV/K
Temperature (K)
300
Kg
1.83∙10-24
Ne=Nh
1.35∙10-12
700
1000
10
What is the number of the oxygen vacancies in the unit cell of
Zr0.8Y0.2O1.9?
Assuming the lattice parameter of (cubic) YSZ is 0.54 nm, calculate
a concentration of the oxygen vacancies (number per m3).
Zr(Y)
O
Fluoritstruktur (CaF2-Typ)
11
In Zr0.8Y0.2O1.9, how many oxygen vacancies are there per unit
cell? If the lattice parameter of (cubic) YSZ is 0.54 nm, calculate
the density of vacancies (number per m3)
Zr(Y)
O
Formula
NVO
0.1

 0.05
N0
2
VO per unit cell
NVO  0.05 8  0.4
Vc =0.543∙10-27m3
Fluoritstruktur (CaF2-Typ)
nVO  0.05 8 
1
1
27  1 
 0.05 8 

2
.
54

10
 m3 
Vc
0.543 1027
12
Defektkonzentration n/N0 bei verschiedenen
Temperaturen
n
  EV 
 exp

N0
 kT 
Temperatur
[oC]
Aktivierungsenergie eV
1
2
8
100
3·10-14
1·10-27
1·10-108
500
3·10-7
1·10-13
8·10-53
1000
1·10-4
1·10-8
2·10-32
1500
1·10-3
2·10-6
2·10-23
2000
6·10-3
4·10-5
2·10-18
13
Write the Kröger-Vink notation for the following fully charged
species in MgO:


O
MgMg
• Cation and anion on their normal sites
O

• Oxygen vacancy
VO
//
VMg
• Magnesium vacancy

Mg
• Interstitial magnesium ion
i
14
Write the Kröger-Vink notation for the following species in
ZrO2:
•Cation and anion on their normal sites
•Oxygen vacancy
•Zirkonium vacancy
•Yttrium dopant substituting Zr
•Nitrogen ion (N3-) substituting for oxygen ion
Write the Kröger-Vink notation for the following fully charged
species in CaTiO3:
•Calcium vacancies
•Titanium vacances
•Oxygen vacances
•Ti ions on Ca sites and vice versa
•Ti interstitials
15
Write the Kröger-Vink notation for the following species in
ZrO2:
•Cation and anion on their normal sites
ZrZr OO
•Oxygen vacancy
VO
•Zirkonium vacancy
VZr////
•Yttrium dopant substituting Zr
YZr/
•Nitrogen ion (N3-) sobstituting for oxygen ion
N O/
Write the Kröger-Vink notation for the following fully charged
species in CaTiO3:
VCa//
•Calcium vacancies
////
V
Ti
•Titanium vacances

V
•Oxygen vacances
O

•Ti ions on Ca sites and vice versa TiCa
CaTi//
•Ti interstitials
Ti 
i
16
Write the electroneutrality condition for defects in silicon :
•pure
/

•boron-doped
nil  e  h
•phosphorous-doped
Si
B

BSi/  h
Si
P

PSi  e/

[h ]  [e/ ]
/
Si

[ B ]  [h ]
[ BSi/ ]  p

Si
/
[ PSi ]  [e/ ]
[ PSi ]  n
nil  e  h
/
B
 B  h
Si
P
 P  e
Si
/
Si

n p
17
Write the electroneutrality condition for MO1-x

O
O  2M

M
1
 V  2M  O2
2

O
/
M
M M/  M M  e/
Write the electroneutrality condition for MO1+x (oxygen
interstitial sites)
Write the electroneutrality condition for M1-xO
1
//  2h  O
O2  VM
O
2
1
/  h  O
O2  VM
O
2
Write the electroneutrality condition for M1+xO (metal
interstitial sites)
18
Write the electroneutrality condition for MO1-x

O
O  2M

M
1
 V  2M  O2
2

O
/
M
M M/   2VO 
M M/  M M  e/
M M/  e /   2VO 
M M/  n  2VO 
19
Write the electroneutrality condition for MO1+x (oxygen
interstitial sites)
1
2M  O2  2M M  Oi//
2

M
M M  2Oi// 
M M  M M  h
M M  h  2Oi// 
M M  p  2Oi// 
20
Write the electroneutrality condition for M1-xO
1
//  2h  O
O2  VM
O
2

V  h V
//
M
/
M
h   p  V  2V 

/
M
//
M
1
2 M  O2  2 M M  VM//  OO
2
1

2M M  O2  2M M  2h  VM//  OO
2

M
21
Write the electroneutrality condition for M1+xO (metal
interstitial sites)

O
O M

M
M

i
V
//
M

O
V
1
 2e  O2
2
/
    
n  2 VM//  2 VO  2 M i

VM//  VM/  e /
      
n  2 VM//  VM/  2 VO  2 M i

22
Mf   10
Metal oxide MeO2 is doped with Mf2O3 at the doping level Me 
/
Me

Me
3
At a certain temperature T and oxygen partial pressure 10-9 atm, concentration
of oxygen vacancies is 10-3. Make a plot showing dependence of point defects
/
concentration ( VO , MfMe
 and e / ) on oxygen partial pressure at T.
Identify the charge carriers and regions of intrinsic and extrinsic conductivity.
 
 
/
Mf2O3  2MfMe
 VO  3OO
1

/
Me Me
 OO  2Me Me
 VO  O2
2
e   Me   2K 
10
-1
10
-2
10
-3
10
-4
10
-5
10
-6
-15
10
e   Me  
1/ 6
O2
1/ 3
'
Me
Konzentration [mol/mol]
/
'
p
-14
10
-13
10
-12
10
-11
10
 2K
 
/
 Mf Me
-10
10
-9
10
-8
10
-7
10
-6
10
-5
10
-4
10
Sauerstoffpartialdruck [atm]
'
Me
-3
10
-2
10
-1
10




0.5
 pO12 / 4
0
10
23
Me

Me

O
 O  2Me
1
 V  O2
2

O
/
Me
/
Mf2O3  2MfMe
 VO  3OO
e   Me   2  V   2K 
/

O
'
Me
1/ 3
Konzentration [mol/mol]
Brouwer (Patterson)-Diagramm
-1
10
10
-2
10
-3
2K 

e   Me    Mf


'
1/ 6
O2
p
'
Me
0.5
/
Me
 pO12 / 4
T=const
-1/6
-1/4
Mf 
0 V 
/
Me

O
10
-4
10
-5
10
-6
e 
/
extrinsic
intrinsic
-15
10
-14
10
-13
10
-12
10
-11
10
-10
10
-9
10
-8
10
-7
10
-6
10
-5
10
-4
10
Sauerstoffpartialdruck [atm]
-3
10
-2
10
-1
10
0
10
24
Cobalt oxide: The electronic conductivity of Co1-yO at 1350°C and pO2 = 0.1
atm is 25 S/cm. Thermogravimetric measurements show that y = 0.008 under
the same conditions. It is assumed that singly charged cobalt vacancies are the
dominating point defects. Identify the charge carriers responsible for the
conductivity and calculate their charge mobility. (Assume that the density of
CoO at 1350°C equals that at room temperature, 6.4 g/cm3. Atomic weights MCo
= 58.93; MO = 16.00; q=1.6∙10-19 C)


0.5O2  2CoCo
CoO

 VCo//  2CoCo
 OO

Co

Co

Co

O
O2  2Co 
 2V  2Co  2O
Co CoO

 Co  h
CoO
/
Co

Co
Platzverhältnis
Die Anzahl an
Kationenplätzen (K) einer
Verbindung KxAy muss
immer im richtigen
Verhältnis zur Anzahl der
Anionenplätze (A) stehen


O2  2CoCo
CoO

 2VCo/  2CoCo
 2h  2OO
O2 CoO

 2VCo/  2h  2OO
25
O2 CoO

 2VCo/  2h  2OO
h   p  V 

/
Co
  n z q
6.4 106  6.0221023
26 1
n
 0.008  4.1110 [ 3 ]
(58.93  16)
m

2500
5




3
.
8

10
Holes mobility
26
19
n  z  q 4.1110 11.6 10
  1  m 1 A  m 2 C  m 2
m2 



 3

V  C s V  C s V 26
 m C
Nickel oxide: Assume that doubly charged nickel vacancies and holes are the
dominating defects in Ni1-yO under oxidising conditions. At 1245°C and pO2 = 1 atm
we know the following for the compound:
The self diffusion coefficient for nickel: DNi = 9∙10-11 cm2/s
Electrical conductivity: σ = 1.4 S/cm
(Data from M.L. Volpe and J. Reddy, J. Chem. Phys., 53 (1970) 1117)
Nickel vacancy concentration, in site or mole fraction: [VNi’’] = 2.510-4
(Data from W.C. Tripp and N.M. Tallan, J. Am. Ceram. Soc., 53 (1970) 531)
(Atomic weights MNi = 58.71, MO = 16.00, density of NiO = 6.67 g/cm3)
a) Calculate the charge mobility of the nickel vacancies and the ionic conductivity
under the conditions referred to above (Nernst-Einstein Gleichung)
b) Calculate the concentration of electron holes under the given conditions, given as
site fraction and as volume concentration ( number/m3).
c) Calculate the charge mobility of the holes.
q=1.6∙10-19 C
k=1.38∙10-23 J/K
27


0.5O2  2NiNi
NiO

 VNi//  2NiNi
 OO


NiNi
NiO

 NiNi
 h
V
//
Ni
zq

 DNi
k T
  n  z  q  V
Nernst-Einstein
Point „a“
//
Ni
28
a)
V
//
Ni
nickel vacancies
zq
2 1.6 1019  9 1015
13

 DNi 

1
.
37

10
k T
1.381023  (1245 273)
C  m 2  s 1
C  m 2  s 1
m2


1
1
J  K  K V C  K  K V  s
  n  z  q  V
//
Ni
6.67 106  6.0221023
4
25 1
n
 2.5 10  1.3410 [ 3 ]
(58.71 16)
m
  1.341025  2 1.6 1019 1.37 1013  5.87 107
m3  C  m2 V 1  s 1  1m1
Compare the obtained value with σ = 1.4 S/cm =140 S/m
VNi// are not dominating carriers
29
b)
holes


0.5O2  2NiNi
NiO

 VNi//  2NiNi
 OO


NiNi
NiO

 NiNi
 h


0.5O2  2NiNi
NiO

 VNi//  2NiNi
 2h  OO
0.5O2 NiO

 VNi//  2h  OO
site fraction
 
p  h
 
 2 VNi//  2  2.5 10 4
Volume concentration
 
6
23
6
.
67

10

6
.
022

10

4
25 1
h  p  2  2.5 10
 2.6910 [ 3 ]
(58.71 16)
m
30
c)
holes
σ for nickel vacances

140 5.87 107
5
h 


3
.
25

10
n  z  q 2.691025 11.6 1019

2
  1  m 1 A  m 2 C  m 2
m 



 3

V  C s V  C s V 
 m C
V  1.37 10 13
//
Ni
31
1. Calculate EMF (EMK) at 500 and 1100K for fuel cells in which Methane
(CH4) or Hydrogen is used as a fuel. Assume that the partial pressures of all
the gaseous reactants are equal 1 bar (pure oxygen at the cathode!).
2. Calculate what will be change of EMF at 1100K in the case of CH4 fuel,
assuming total pressure of the gases at both the electrodes 1 bar (pure oxygen
at the cathode!) and composition at anode 50%H2O, 25%CO2 and 25%CH4.
I.
Barin,
O.
Knacke,
„Thermochemical properties of
inorganic substances“, SpringerVerlag, 1973
32
½ O2 + H2  H2O
Kathode: ½ O2 + 2e-  O2-
Gi  Gio  RT ln ai
Anode: O2- + H2  H2O + 2e-
ai  1
G  GHo 2O  GHo 2  0.5GOo2
500K G=-80.965-(-15.996)-0.5·(-24.910)=-52.514 kcal/mol
-52.514·1000·4.184=-219719 J/mol
 G
E
zF
1100K
E=-(-219719)/(2·96486)=1.139V
-44.889 kcal/mol
-187816 J/mol
0.973V
33
2O2 + CH4  2H2O+CO2
Kathode: 2O2 + 8e-  4O2-
Anode: 4O2- + CH4  2H2O + CO2+8e-
o
o
o
Go  2GHo 2O  GCO


G

2

G
CH 4
O2
2
 G o
E 
zF
o
1100K G =-801864 J/mol
Eo=-(-801864)/(8·96486)=1.039V
2
p
RT
H 2 O  pCO2
o
EE 
ln
zF
pCH 4  pO2 2
8.311100 0.52  0.25
E  1.039
ln
 1.039 (0.016)  1.055V
2
8  96486
0.251
34
On the diagram show the doping regions for intrinsic and doped silicon at room temperature.
K g  Ne N h
(mole fractions)
K g  0.9 10 12  298 3 exp
 1.14
 1.34 10  24
5
8.63 10  298
n
Concentration
-4
10
-5
10
-6
10
-7
10
-8
10
-9
10
-10
10
-11
10
-12
10
-13
10
-14
10
-15
10
-16
10
-17
10
-18
10
-19
10
-20
10
-16
-15
-14
-13
-12
-11
-10
10
10
10
10
10
10
10
n-type
K g  1.161012
-9
10
-8
10
Dopand concentration
-7
10
-6
10
-5
10
-4
10
p
35
Doped silicon
1. Phosphorus is added to high-purity silicon to give a concentration of 1023m-3 of
charge carriers at room temperature.
a) Is the material n-type or p-type?
b) Calculate the room-temperature conductivity of this material, assuming that electron
and hole mobilities (respectively, 0.14 and 0.048 m2V-1s-1) are the same as for the
intrinsic material
Density of Si 2.33 g/cm3; molecular weight 28.09 g/mol
K g  0.9 10
12
 1.14
 24
 298 exp

1
.
34

10
(mole fractions)
8.63 10 5  298
3
q=1.6∙10-19 C
36
1.
a) Phosphorus- V group, will act as a donor in silicon
b) 1023 m-3 electron concentration is greater than that
for the intrinsic case
K g  ne  n p  1.161012
(mole fractions)
2.33106  6.0231023
28 1
nSi 
 5 10 [ 3 ]
28.09
m
ne  np  5 1028  (1.341024 )1/ 2
  n  z  q    1023 11.6 1019  0.14  2240
3
1
1
1
m  C  m V  s   m
2
1
37
Doped silicon
2. The room-temperature conductivity of intrinsic silicon is 410-4 Ω-1m-1. An extrinsic n-type
silicon material is desired having a room-temperature conductivity of 150 Ω-1m-1.
a) Specify a donor element type that may be used and its concentration in atom percent.
b) Calculate the equilibrium hole concentration
Assume that electron and hole mobilities (respectively, 0.14 and 0.048 m2V-1s-1) are the same
as for the intrinsic material, and that at room temperature the donor atoms are already ionized.
Density of Si 2.33 g/cm3, molecular weight 28.09 g/mol.
nil  e ,  h 
K g  Ne N h
12
K g  0.9 10
 T exp
3
(mole fractions)
 Eg
kT
Eg=1.14 eV, k=8.63∙10-5 eV/K
38
2. a)
P, As, Sb

150
21
n  nd 


6
.
7

10
q   1.6 1019  0.14
 1m 1
3

m
Cm 2V 1s 1
2.33106  6.0231023
28 1
nSi 
 5 10 [ 3 ]
28.09
m
nd
6.7 1021
5
100% 

100
%

1
.
34

10
%
28
nSi  nd
5 10
39
2. b)
12
K g  0.9 10
 T exp
3
K g  0.9 10 12  298 3 exp
 Eg
kT
 1.14
 24

1
.
34

10
8.63 10 5  298
1.341024
17
Nh 


1

10
Ne 1.34107
Kg
40