6.152J/3.155J
1
1
Nov.26,2003
6.12J / 3.155J Microelectronic processing
CRYSTAL GROWTH
Si
Crystal
What do we need to know prior to crystal growth?
4,Critical nucleus size
5,Growth
6,Impurities,defects more stable at high T; how grow pure crystal?
7,Segregation solid vs,liquid
1,Reactants in molten form
2,Transport to S/L interface
3,Adsorbtion,entropy decreases
CRYSTAL GROWTH steps
and questions
DS
DH
-TDS increases DH decreases
(exo)
2
Nov.26,2003
6.12J / 3.155J Microelectronic processing
Defects and crystal growth
Defects
impurities,vacancies,dislocations…T dependence
Crystal growth techniques,
float zone,Bridgman,Czochralski
Segregation during growth
Segregation coefficients
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6.12J / 3.155J Microelectronic processing
Thermodynamics and phase diagrams
H
B
- H
A
= DH = heat of formation
of B from A
Do all reactions that give off heat proceed?
S
B
- S
A
= DS = Entropy (disorder) change
from A to B
Do all reactions that increase disorder proceed?
¨S
A
B
S
Configurations
H
A
B
¨H
G = H-TS
G
B
- G
A
= DG = DH - TDS
G must decrease if reaction is to proceed.
(From equilibrium,all changes increase G).
A
B
G
Answer in Gibbs free energy:
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Nov.26,2003
6.12J / 3.155J Microelectronic processing
DH =ofBfromA
DS =from A to B
¨S < 0
B more
ordered
A
B
S
Configurations
H
A
B
¨H
DG = DH - TDS
Will not go
above T = DH /DS
A
B
G
Exothermic
Do all
exothermal reactions
proceed?
¨S > 0
B more
disordered
A
B
S
Configurations
H
A
B
¨H
Will not go
below T=DH /DS
A
B
G
Endothermic
Does disorder always increase
in reactions?
Examples,freezing of water melting of copper
Thermodynamics and phase diagrams
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Nov.26,2003
6.12J / 3.155J Microelectronic processing
Under what conditions will Si melt crystallize?
T
T
m
Liquid Si
T = T
m
+
T = T
m
-
Crystal Si
high S
high H
low S
low H
For solidification:
DS = ST
-
()
final
- ST
+
()
initial
< 0
DH = HT
-
()- HT
+
()< 0
DG =DH - T
m
DS
< 0
>0
TDS must have smaller
magnitude than DH
for solidification;
this defines solidification temp.
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Nov.26,2003
6.12J / 3.155J Microelectronic processing
Note the relatively large solid solubility of As in Si
Whereas
this field
is As + SiAs
2
Arsenic solubility
in Si
increases with T
….then decreases
on approaching T
melt
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7
Nov.26,2003
6.12J / 3.155J Microelectronic processing
1-dimensional defects,We saw soluble impurities in Si.
1.0
T/T
m
10
20
10
21
Impurity content (cm
-3
)
B As
P
S
mix
Si Impurity
-TS
mix
Si Impurity
From
Phase
diagram
8
Nov.26,2003
6.12J / 3.155J Microelectronic processing
1-dimensional defects,More point defects
Interstitial
impurity
Substitutional
impurity
Vacancy
Self
interstitial
V-I pair =
Frankel defect
Strain field of vacancy…
Strain and surface energy
can be reduced
by agglomeration:
Vacancy => void,
Interstitial => precipitate
…when concentration > equilibrium
interstitial
6.152J/3.155J
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9
Nov.26,2003
6.12J / 3.155J Microelectronic processing
Bonding-antibonding orbital energy separation
bonding/antibonding => energy gaps
in semiconductors,
insulators
E
Gp IV
atom
s
2
p
2
Gp IV
atom
s
2
p
2
s-p
3
anti-
bonding
s-p
3
bonding
Semicon
crystal
Gap
Conduction band
Valence band
10
Nov.26,2003
6.12J / 3.155J Microelectronic processing
Vacancy concentration,Vacancy requires breaking 4 bonds
n
vac
= n
0
exp -E
g
/k
B
T
[]
E
g
=1.12eV
Conduction
band
Valence
band
n
vac
= 5 ￥10
22
exp -2.6 eV /k
B
T
[]
Empirical:
ln
n
vac
n
0
ê
á
ˉ
=-E
a
/k
B
T ln
n
vac
n
0
ê
á
ˉ
1000 /k
B
T
E
a
Arrhenius
plot
At RT,n
vac
= 3.4 x 10
-23
/cm
3
( § 300 km between vacancies)
At 1273 K,n
vac
= 2.6 x 10
12
/cm
3
( § 700 nm between vacancies)
Vacancies abundant at high temperature
Equilibrium vacancy concentration:
6.152J/3.155J
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Nov.26,2003
6.12J / 3.155J Microelectronic processing
Oxygen impurities in Si:
Observed to follow Arrhenius
C
oxy
= 2 ￥10
22
exp -1.03eV /k
B
T[]
Want about 10 - 30 ppm (7 x 10
17
/cm
3
)… which occurs at T § 1250
0
C
Activation energies (eV),1.03 1.12 2.6 4.5
Oxygen E
g
Vacancy Interstitial
40
C
oxy
(ppm)
20
0
Agglomeration =>
warpage,stress,dislocations
No agglomeration
(many isolated O
2-
ions)
Optimal
1414
T (
0
C)
1200
High C
oxy
3 hr anneal
=> 15 ppm
Anneal => denuded zone
deeper than deepest feature
Agglomeration
Oxygen
§ 25
microns
12
Nov.26,2003
6.12J / 3.155J Microelectronic processing
Dopants,impurities (substitutional,interstitial)
At RT number of intrinsic carriers:
n
i
= (n
e
n
h
)
1/2
= n
0
exp(-E
g
/2k
B
T) => n
i
=2x10
10
/cm
3
5 x 10
22
J =sE = ne < v >
m =
v
E
=
s
ne
=
et
m *
s =
ne
2
t
m *
What do dopants do?
Very small doping concentration
=> large increase in carrier concentration
So doping at 1 ppm => 10
-6
= n
D,A
/(5 x 10
22
/cm
3
)
n
D,A
=5x10
16
/cm
3
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6.12J / 3.155J Microelectronic processing
CRYSTAL GROWTH
Confined1
Normal
freezing
(Bridgman)
Crystal
horizontal
vertical
Crystal
B
2
O
3
Si
CZ2
Meniscus controlled
LEC
Pull from
solution
Crystal
Crystal
Floating zone
feed rod
high-purity
crystal
3
Zone
melting
poly
poly
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16
Nov.26,2003
6.12J / 3.155J Microelectronic processing
1) Reactants,first need high-purity Si
1500°C
seed
tang
VAC
crystal
SiO
2
Si + CO(g) SiHCl
3
(l) Si + HCl
C
boat
HCl
+ H
2
distill H
2
1100°C
poly
Making
high-purity Si:
MGS -98% pure
(Metallurgical grade Si)
EGS 9 9s pure
(Electronic grade Si)
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6.12J / 3.155J Microelectronic processing
Growth rate μG
s
-G
L
>DG
DG = 0fi
DH = TDS,
DS =DH /T
eq.
Equilibrium at
\T
interface
< T
eq
DH < 0
H
s
- H
L
larger
(latent) heat
content
Crystal
DG =DH
T
eq
-T
T
eq
è
í
í
˙
˙
Must decrease
for
solidification
to occur
DG = DH - TDS
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Nov.26,2003
6.12J / 3.155J Microelectronic processing
1) Pulling crystal from melt
At S-L interface:
k
S
A
,T
,z
ˉ
S
= k
L
A
,T
,z
ˉ
L
+ L
,m
,t
,m
,t
= vAr
m
Heat of fusion Liquid?Solid
L a 340 cal mole
Thermal
conductivities
k
s
a1.5 W /(cm-°C)
T
z
Q
If growth velocity too large,
solid cannot dissipate heat
(Typical v = 1 mm / min.)
v
max
=
k
s
Lr
m
,T
,z
solid
If too large
fi thermal stress
Tz()
￠
S
0
1500°C
seed
tang
VAC
crystal
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Nov.26,2003
6.12J / 3.155J Microelectronic processing
dT/dx § 100
0
C/cm
Seed
Melt
Dead boundary layer
Czochralski growth of single crystals,stress,dislocations
For large temperature gradients,
e.g,dT/dx § 100
0
C/cm.,and given a = 2.6 x 10
-6
/
0
C,
then Dl/l = aDT => strains of 0.6%,
which exceeds the yield stress of Si,=> dislocations
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6.12J / 3.155J Microelectronic processing
Dislocations originate in shear strains,
mostly induced by
thermal gradients
during growth.
Line defects,dislocations
A couple of dislocations/wafer is typical.
Why so few?
1),Tang” (neck at beginning of xtl)
allows dislocations to move to surface
2) Large number of atoms are involved in a dislocation,
=> high energy,U
Dislocation has low entropy (most atoms are in unique place)
G = H - TS is very positive
large small
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Impurities
must diffuse across
still boundary layer.
B field suppresses convection of ions (reactive)
deflecting them from interface
Boundary layer keeps impurities
away from solid-liquid interface
Czochralski growth of single crystals,impurities
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6.12J / 3.155J Microelectronic processing
Bridgman growth of GaAs
dT/dx < 10
0
C/cm =>
dislocation density < 10
3
/cm
2
but melt-ampule contact =>
lower resistivity.
GaAs in quartz ampule,GaAs As
T § 620
0
C
Convection barrier
T
m
§
v § mm/hr.
r § 10 MWcm
2 - 5 x 10
3
dislocations /cm
2
GaAs in
BN ampule:
T
m
Vertical Bridgman
To correct contact problems,
use vertical Bridgman
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6.12J / 3.155J Microelectronic processing
Liquid encapsulated CZ or Bridgman growth
…because Ga As
vapor pressure at T
m
= 1238
0
C,0.001 10 atmospheres.
Seed
BN
Crucible
(keep Si out)
B
2
O
3
liquid plug
GaAs + As
r § 100 MWcm
10
4
defects/cm
2
100+ dislocations /cm
2
B
2
O
3
minimizes loss of As;
v § 1 cm/hr.
Pressure may
build to 20 atm:
HPLEC T
melt
k Crit,resolved
shear stress
Si 1414 0.21 1.85
Ge 960 0.17 0.70
GaAs 1238 0.07 0.4
GaAs cannot dissipate heat;
grow slowly
GaAs cannot take stress;
grow slowly
24
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6.12J / 3.155J Microelectronic processing
Floating zone For very high purity Si (not used for GaAs)
high-purity
crystal
Polycrystal
feed rod
seed
Top seed
(molten part
supported
by upper
crystal)
seed
high-purity
crystal
Bottom
seed
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6.12J / 3.155J Microelectronic processing
CRYSTAL GROWTH reviewed
Confined1
Normal
freezing
(Bridgman)
Crystal
horizontal
vertical
Crystal
B
2
O
3
Si
CZ2
Meniscus controlled
LEC
Pull from
solution
Crystal
Crystal
Floating zone
feed rod
highest-
purity
crystal;
Larger radial
variations
than CZ
3
Zone
melting
poly
poly
Largest crystals:
300 mm diameter
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Nov.26,2003
6.12J / 3.155J Microelectronic processing
Introduction of dopants via melt
Different dopants have different solubilities in solid
Define segregation coefficient,k,
as ratio of dopant concentrations solid/melt,k =
C
s
C
l
k = 0.002 0.3 0.8 0.25 0.35 0.023
Al As B O P Sb
k < 1 implies only a small fraction of dopant moves into solid;
concentration builds up in melt
as S/L interface advances…
This in turn drives up amount transferred to solid…
We can calculate the dopant/impurity concentration
as a function of position in crystal
(let x = fraction solidified)…
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Dopant or impurity concentration vs,Position
assumptions,no solid state diffusion,perfect liquid mixing
CZ growth
C
x=0 x=L
S L
C
liq
= C
0
C
L
(x) - C
S
(x)[]dx = dC
L
(x) L - x[]
C
L
(x)1- k[]dx = dC
L
(x) L - x[]
1- k[]
dx
L - x
0
x
ú =
dC
L
(x)
C
L
(x)
C
0
C
L
(x )
ú
C
L
(x) = C
0
L
L - x
è
1-k
C
S
(x) = kC
0
L
L - x
è
1-k
C
0
- kC
L
(x)[]dx = dC
L
(x)l
0
dx
l
0
0
x
ú =
dC
L
(x)
C
0
- kC
L
(x)
C
0
C
L
(x )
ú
x
l
0
=-
1
k
C
0
- kC
L
(x)[]
C
0
CL (x )
-
kx
l
0
= ln
C
0
- kC
L
(x)
C
0
(1 - k)
è
í
˙
C
0
C
L
(x )
C
S
(x) = C
0
1-(1 - k)exp(-kx / l
0
)[]
Float zone growth
C
x = 0 l
o
x=L
C
0
S L S
C
0
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6.12J / 3.155J Microelectronic processing
CZ growth
C
S
(x) = kC
0
L
L - x
è
1-k
x=0 x=L
10
1.0
0.1
k = 0.1,L = 1
Float zone growth
C
S
(x) = C
0
1-(1 - k)exp(-kx / l
0
)[]
k = 0.1,l = 0.1L
C
s
/C
0
Dopant/inpurity concentration vs,Position
1
1
Nov.26,2003
6.12J / 3.155J Microelectronic processing
CRYSTAL GROWTH
Si
Crystal
What do we need to know prior to crystal growth?
4,Critical nucleus size
5,Growth
6,Impurities,defects more stable at high T; how grow pure crystal?
7,Segregation solid vs,liquid
1,Reactants in molten form
2,Transport to S/L interface
3,Adsorbtion,entropy decreases
CRYSTAL GROWTH steps
and questions
DS
DH
-TDS increases DH decreases
(exo)
2
Nov.26,2003
6.12J / 3.155J Microelectronic processing
Defects and crystal growth
Defects
impurities,vacancies,dislocations…T dependence
Crystal growth techniques,
float zone,Bridgman,Czochralski
Segregation during growth
Segregation coefficients
6.152J/3.155J
2
3
Nov.26,2003
6.12J / 3.155J Microelectronic processing
Thermodynamics and phase diagrams
H
B
- H
A
= DH = heat of formation
of B from A
Do all reactions that give off heat proceed?
S
B
- S
A
= DS = Entropy (disorder) change
from A to B
Do all reactions that increase disorder proceed?
¨S
A
B
S
Configurations
H
A
B
¨H
G = H-TS
G
B
- G
A
= DG = DH - TDS
G must decrease if reaction is to proceed.
(From equilibrium,all changes increase G).
A
B
G
Answer in Gibbs free energy:
4
Nov.26,2003
6.12J / 3.155J Microelectronic processing
DH =ofBfromA
DS =from A to B
¨S < 0
B more
ordered
A
B
S
Configurations
H
A
B
¨H
DG = DH - TDS
Will not go
above T = DH /DS
A
B
G
Exothermic
Do all
exothermal reactions
proceed?
¨S > 0
B more
disordered
A
B
S
Configurations
H
A
B
¨H
Will not go
below T=DH /DS
A
B
G
Endothermic
Does disorder always increase
in reactions?
Examples,freezing of water melting of copper
Thermodynamics and phase diagrams
6.152J/3.155J
3
5
Nov.26,2003
6.12J / 3.155J Microelectronic processing
Under what conditions will Si melt crystallize?
T
T
m
Liquid Si
T = T
m
+
T = T
m
-
Crystal Si
high S
high H
low S
low H
For solidification:
DS = ST
-
()
final
- ST
+
()
initial
< 0
DH = HT
-
()- HT
+
()< 0
DG =DH - T
m
DS
< 0
>0
TDS must have smaller
magnitude than DH
for solidification;
this defines solidification temp.
6
Nov.26,2003
6.12J / 3.155J Microelectronic processing
Note the relatively large solid solubility of As in Si
Whereas
this field
is As + SiAs
2
Arsenic solubility
in Si
increases with T
….then decreases
on approaching T
melt
6.152J/3.155J
4
7
Nov.26,2003
6.12J / 3.155J Microelectronic processing
1-dimensional defects,We saw soluble impurities in Si.
1.0
T/T
m
10
20
10
21
Impurity content (cm
-3
)
B As
P
S
mix
Si Impurity
-TS
mix
Si Impurity
From
Phase
diagram
8
Nov.26,2003
6.12J / 3.155J Microelectronic processing
1-dimensional defects,More point defects
Interstitial
impurity
Substitutional
impurity
Vacancy
Self
interstitial
V-I pair =
Frankel defect
Strain field of vacancy…
Strain and surface energy
can be reduced
by agglomeration:
Vacancy => void,
Interstitial => precipitate
…when concentration > equilibrium
interstitial
6.152J/3.155J
5
9
Nov.26,2003
6.12J / 3.155J Microelectronic processing
Bonding-antibonding orbital energy separation
bonding/antibonding => energy gaps
in semiconductors,
insulators
E
Gp IV
atom
s
2
p
2
Gp IV
atom
s
2
p
2
s-p
3
anti-
bonding
s-p
3
bonding
Semicon
crystal
Gap
Conduction band
Valence band
10
Nov.26,2003
6.12J / 3.155J Microelectronic processing
Vacancy concentration,Vacancy requires breaking 4 bonds
n
vac
= n
0
exp -E
g
/k
B
T
[]
E
g
=1.12eV
Conduction
band
Valence
band
n
vac
= 5 ￥10
22
exp -2.6 eV /k
B
T
[]
Empirical:
ln
n
vac
n
0
ê
á
ˉ
=-E
a
/k
B
T ln
n
vac
n
0
ê
á
ˉ
1000 /k
B
T
E
a
Arrhenius
plot
At RT,n
vac
= 3.4 x 10
-23
/cm
3
( § 300 km between vacancies)
At 1273 K,n
vac
= 2.6 x 10
12
/cm
3
( § 700 nm between vacancies)
Vacancies abundant at high temperature
Equilibrium vacancy concentration:
6.152J/3.155J
6
11
Nov.26,2003
6.12J / 3.155J Microelectronic processing
Oxygen impurities in Si:
Observed to follow Arrhenius
C
oxy
= 2 ￥10
22
exp -1.03eV /k
B
T[]
Want about 10 - 30 ppm (7 x 10
17
/cm
3
)… which occurs at T § 1250
0
C
Activation energies (eV),1.03 1.12 2.6 4.5
Oxygen E
g
Vacancy Interstitial
40
C
oxy
(ppm)
20
0
Agglomeration =>
warpage,stress,dislocations
No agglomeration
(many isolated O
2-
ions)
Optimal
1414
T (
0
C)
1200
High C
oxy
3 hr anneal
=> 15 ppm
Anneal => denuded zone
deeper than deepest feature
Agglomeration
Oxygen
§ 25
microns
12
Nov.26,2003
6.12J / 3.155J Microelectronic processing
Dopants,impurities (substitutional,interstitial)
At RT number of intrinsic carriers:
n
i
= (n
e
n
h
)
1/2
= n
0
exp(-E
g
/2k
B
T) => n
i
=2x10
10
/cm
3
5 x 10
22
J =sE = ne < v >
m =
v
E
=
s
ne
=
et
m *
s =
ne
2
t
m *
What do dopants do?
Very small doping concentration
=> large increase in carrier concentration
So doping at 1 ppm => 10
-6
= n
D,A
/(5 x 10
22
/cm
3
)
n
D,A
=5x10
16
/cm
3
6.152J/3.155J
7
13
Nov.26,2003
6.12J / 3.155J Microelectronic processing
CRYSTAL GROWTH
Confined1
Normal
freezing
(Bridgman)
Crystal
horizontal
vertical
Crystal
B
2
O
3
Si
CZ2
Meniscus controlled
LEC
Pull from
solution
Crystal
Crystal
Floating zone
feed rod
high-purity
crystal
3
Zone
melting
poly
poly
6.152J/3.155J
8
16
Nov.26,2003
6.12J / 3.155J Microelectronic processing
1) Reactants,first need high-purity Si
1500°C
seed
tang
VAC
crystal
SiO
2
Si + CO(g) SiHCl
3
(l) Si + HCl
C
boat
HCl
+ H
2
distill H
2
1100°C
poly
Making
high-purity Si:
MGS -98% pure
(Metallurgical grade Si)
EGS 9 9s pure
(Electronic grade Si)
6.152J/3.155J
9
17
Nov.26,2003
6.12J / 3.155J Microelectronic processing
Growth rate μG
s
-G
L
>DG
DG = 0fi
DH = TDS,
DS =DH /T
eq.
Equilibrium at
\T
interface
< T
eq
DH < 0
H
s
- H
L
larger
(latent) heat
content
Crystal
DG =DH
T
eq
-T
T
eq
è
í
í
˙
˙
Must decrease
for
solidification
to occur
DG = DH - TDS
18
Nov.26,2003
6.12J / 3.155J Microelectronic processing
1) Pulling crystal from melt
At S-L interface:
k
S
A
,T
,z
ˉ
S
= k
L
A
,T
,z
ˉ
L
+ L
,m
,t
,m
,t
= vAr
m
Heat of fusion Liquid?Solid
L a 340 cal mole
Thermal
conductivities
k
s
a1.5 W /(cm-°C)
T
z
Q
If growth velocity too large,
solid cannot dissipate heat
(Typical v = 1 mm / min.)
v
max
=
k
s
Lr
m
,T
,z
solid
If too large
fi thermal stress
Tz()
￠
S
0
1500°C
seed
tang
VAC
crystal
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dT/dx § 100
0
C/cm
Seed
Melt
Dead boundary layer
Czochralski growth of single crystals,stress,dislocations
For large temperature gradients,
e.g,dT/dx § 100
0
C/cm.,and given a = 2.6 x 10
-6
/
0
C,
then Dl/l = aDT => strains of 0.6%,
which exceeds the yield stress of Si,=> dislocations
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Dislocations originate in shear strains,
mostly induced by
thermal gradients
during growth.
Line defects,dislocations
A couple of dislocations/wafer is typical.
Why so few?
1),Tang” (neck at beginning of xtl)
allows dislocations to move to surface
2) Large number of atoms are involved in a dislocation,
=> high energy,U
Dislocation has low entropy (most atoms are in unique place)
G = H - TS is very positive
large small
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Impurities
must diffuse across
still boundary layer.
B field suppresses convection of ions (reactive)
deflecting them from interface
Boundary layer keeps impurities
away from solid-liquid interface
Czochralski growth of single crystals,impurities
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Bridgman growth of GaAs
dT/dx < 10
0
C/cm =>
dislocation density < 10
3
/cm
2
but melt-ampule contact =>
lower resistivity.
GaAs in quartz ampule,GaAs As
T § 620
0
C
Convection barrier
T
m
§
v § mm/hr.
r § 10 MWcm
2 - 5 x 10
3
dislocations /cm
2
GaAs in
BN ampule:
T
m
Vertical Bridgman
To correct contact problems,
use vertical Bridgman
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Liquid encapsulated CZ or Bridgman growth
…because Ga As
vapor pressure at T
m
= 1238
0
C,0.001 10 atmospheres.
Seed
BN
Crucible
(keep Si out)
B
2
O
3
liquid plug
GaAs + As
r § 100 MWcm
10
4
defects/cm
2
100+ dislocations /cm
2
B
2
O
3
minimizes loss of As;
v § 1 cm/hr.
Pressure may
build to 20 atm:
HPLEC T
melt
k Crit,resolved
shear stress
Si 1414 0.21 1.85
Ge 960 0.17 0.70
GaAs 1238 0.07 0.4
GaAs cannot dissipate heat;
grow slowly
GaAs cannot take stress;
grow slowly
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Floating zone For very high purity Si (not used for GaAs)
high-purity
crystal
Polycrystal
feed rod
seed
Top seed
(molten part
supported
by upper
crystal)
seed
high-purity
crystal
Bottom
seed
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CRYSTAL GROWTH reviewed
Confined1
Normal
freezing
(Bridgman)
Crystal
horizontal
vertical
Crystal
B
2
O
3
Si
CZ2
Meniscus controlled
LEC
Pull from
solution
Crystal
Crystal
Floating zone
feed rod
highest-
purity
crystal;
Larger radial
variations
than CZ
3
Zone
melting
poly
poly
Largest crystals:
300 mm diameter
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Introduction of dopants via melt
Different dopants have different solubilities in solid
Define segregation coefficient,k,
as ratio of dopant concentrations solid/melt,k =
C
s
C
l
k = 0.002 0.3 0.8 0.25 0.35 0.023
Al As B O P Sb
k < 1 implies only a small fraction of dopant moves into solid;
concentration builds up in melt
as S/L interface advances…
This in turn drives up amount transferred to solid…
We can calculate the dopant/impurity concentration
as a function of position in crystal
(let x = fraction solidified)…
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Dopant or impurity concentration vs,Position
assumptions,no solid state diffusion,perfect liquid mixing
CZ growth
C
x=0 x=L
S L
C
liq
= C
0
C
L
(x) - C
S
(x)[]dx = dC
L
(x) L - x[]
C
L
(x)1- k[]dx = dC
L
(x) L - x[]
1- k[]
dx
L - x
0
x
ú =
dC
L
(x)
C
L
(x)
C
0
C
L
(x )
ú
C
L
(x) = C
0
L
L - x
è
1-k
C
S
(x) = kC
0
L
L - x
è
1-k
C
0
- kC
L
(x)[]dx = dC
L
(x)l
0
dx
l
0
0
x
ú =
dC
L
(x)
C
0
- kC
L
(x)
C
0
C
L
(x )
ú
x
l
0
=-
1
k
C
0
- kC
L
(x)[]
C
0
CL (x )
-
kx
l
0
= ln
C
0
- kC
L
(x)
C
0
(1 - k)
è
í
˙
C
0
C
L
(x )
C
S
(x) = C
0
1-(1 - k)exp(-kx / l
0
)[]
Float zone growth
C
x = 0 l
o
x=L
C
0
S L S
C
0
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CZ growth
C
S
(x) = kC
0
L
L - x
è
1-k
x=0 x=L
10
1.0
0.1
k = 0.1,L = 1
Float zone growth
C
S
(x) = C
0
1-(1 - k)exp(-kx / l
0
)[]
k = 0.1,l = 0.1L
C
s
/C
0
Dopant/inpurity concentration vs,Position
