6.152J/3.155J 1
PHYSICAL VAPOR DEPOSITION (PVD)
PVD II,Evaporation
We saw CVD
Gas phase reactants,p
g
≈ 1 mTorr to 1 atm.
Good step coverage,T > 350 K
?
We saw sputtering Noble (+ reactive gas) p ≈ 10 mTorr; ionized particles
Industrial process,high rate,reasonable step coverage
Extensively used in electrical,optical,magnetic devices.
?
Now see evaporation:
Source material heated,p
eq.vap.
=~ 10
-3
Torr,p
g
< 10
-6
Torr
Generally no chemical reaction (except in,reactive depos’n),
λ = 10’s of meters,Knudsen number N
K
>> 1
Poor step coverage,alloy fractionation,? p
vapor
Historical (optical,electrical)
?
Campbell,Ch,12 is more extensive than Plummer on evaporation
6.152J/3.155J 2
Standard vacuum chambers
Σp
i
≈ 10
-6
Torr (1.3 × 10
-4
N / m
2)
Mostly H
2
O,hydrocarbons,N
2
He by
residual gas analysis (RGA = mass spec.)
6.152J/3.155J 3
p < 10
-8
Torr demands:
Stainless steel chamber
Bakeable to 150
o
C
Turbo,ion,cryo pumps
Ultra-high vacuum chambers
6.152J/3.155J 4
-6 -4 -2 0 2 4
Log[P (N/m
2
)]
25
23
Log[n (#/m
3
)]
21
19
17
15
λ = 10 0.01 cm
p = 10
-10
10
-8
10
-6
10
-4
10
-2
10
0
Torr
1 Atm=
0.1 MPa
760 mm
≈14 lb/in
2
Generally λ/L < 1
Films less pure
Epitaxy is rare
Evaporation
Ballistic,molecular flow,λ/L >> 1
High purity films
Epitaxy
Sputtering
CVD
Knudson number ≈ 1
6.152J/3.155J 5
Atomic flux on surface due to residual gas
J
atoms
area?t
=
nv
x
2
=
p
2k
B
T
2k
B
T
πm
=
p
2πmk
B
T
= J
Given 10
-6
Torr of water vapor @ room temp,find flux
p =10
6
Torr ×
1atm
760 T
×
10
5
P
a
atm
,k
B
TRT
( )
= 0.025eV = 4 ×10
21
J
p =1.3×10
4
N
m
2
m
H
2
O
=
18
N
A
= 3×10
26
kg
What is atomic density in 1 monolayer (ML) of Si?
N = 5 x 10
22
cm
-3
=> 1.3 x 10
15
cm
-2
.
So at 10
-6
Torr,1 ML of residual gas hits surface every 3 seconds!
Epitaxy requires slow deposition,surface mobility,
So you must keep pressure low to maintain pure film
J = 4.8×10
14
atoms/molecules
cm
2
sec
6.152J/3.155J 6
Now add evaporation source
Equilibrium vapor pressure:
H = heat of vaporization
p
v
= p
0
exp?
H
k
B
T
Strong T dependence
e
-
+
_
e-beam
B field
Resistive heater
I
RF-induction
heater
Work
function
e
-
Heat of
vaporization
solid
V(x)
free
6.152J/3.155J 7
Vapor pressure of elements employed in
semiconductor materials,Dots
correspond to melting points
Rely on tables,attached,p
vapor
> > p
vac,
Elemental metals easy to evaporate,but…
alloys
compounds
Differential p
vapor
so use 2 crucibles
or deposit multilayers
and diffuse
Oxides,nitrides deposit in oxygen
(or other) partial p
6.152J/3.155J 8
(note W,Mo
can be used
as crucibles.)
6.152J/3.155J 9
The source also? flux
J =
p
vap
2πm
source
k
B
T
source
For Al at 1000 K,p
vap
= 10
-7
Torr (from figure)
m = species to be evaporated,= 27 amu for Al
J
Al
≈ 2 x 10
13
Al/cm
2
-s just above crucible
Mass flow out of crucible ~ J A
c
m ( mass / t)
We expressed flux of residual gas:
J ≈ 5×10
14
molecules
cm
2
s
( )
Chamber p = 10
-6
Torr
J =
p
2πmk
B
T
Net flow out of crucible ~ JA
c
(# / t)
A
c
Therefore heat Al to T > 800 C
but that’s not all…
Note,3 different temperatures,T
source
≈ T
evaporant
>> T
substrate
> T
chamber
= T
resid gas
≈ RT.
System not in thermal equilibrium; only thermal interaction among them is by radiation
and/or conduction through solid connects (weak contact),No convection when N
K
<< 1.
6.152J/3.155J 10
How much evaporant strikes substrate? At 10
-6
Torr,trajectories are uninterrupted.
While a point source deposits uniformly on a sphere about it,a planar source does not,
Geometric factor =
A
c
2πR
2
cosθ
1
cosθ
2
cosθ
1
= cosθ
2
=
R
2r
Deposition rate =
Jm
A
c
4πr
2
m
area? t
or
J
A
c
4πr
2
#
area?t
substrate
θ
1
θ
2
R
R
θ
1
θ
2
r
r
substrate
Convenient
geometry
Geometric factor =
A
c
4πr
2
Film growth rate
= Jm
A
c
4πr
2
1
ρ
f
thick
t
∝cosθ
1
J
6.152J/3.155J 11
v
ox
=
H p
g
N
1
h
+
t
ox
D
+
1
k
s
oxide
In PVD growth,strike balance
R =
deposition rate
Surface diffusion rate
R > 1 stochastic growth,rough
R < 1 layer by layer,smooth (can heat substrate)
Film growth rate
for evaporation
= Jm
A
c
4πr
2
1
ρ
f
thick
t
v =
p
vap
2πm
source
k
B
T
source
m
ρ
m
A
c
4πr
2
=
p
vap
ρ
m
A
c
4πr
2
m
source
2πk
B
T
source
Cf,CVD
v
f
=
C
g
N
1
n
g
+
1
k
6.152J/3.155J 12
Exercise
Deposit Al (2.7 g/cm
3
) at r = 40 cm from 5 cm diam,
crucible heated to 1100°C (cf T
melt
) p
Al vap
≈ 10
-3
Torr,
p
H
2
O
=10
6
Torr
Compare arrival rate of Al and H
2
O at substrate…and calculate film growth
rate
J
H
2
O
=
10
6
760
()
×10
5
2π × 0.025eV× e × 18 N
A
()
= 4.8×10
18
molecules
m
2
s
J
Al
=
10
3
×10
5
760
()
2π × 1373k
B
()
× 27 N
A
()
A
c
4πr
2
=1.76×10
18
atoms
m
2
s
A
c
=π
5
2
2
Leave shutter closed so initial Al deposition can getter O
2
and H
2
O,
Also,better done at lower or higher deposition rate.
p
H
2
O
(this is not good)
v =
p
vap
ρ
m
A
c
4πr
2
m
source
2πk
B
T
source
≈10
11
m /s
slow!
6.152J/3.155J 13
Step coverage is poor in evaporation (ballistic) - Shadow effects
Heat substrate to increase
surface diffusion,
D
S
= D
0
S
exp?
E
a
Surf
kT
E
a
S
<< E
a
bulk
e.g,WF
6
+ 3H
2
→ W + 6HF
←
G ≈ 70
kJ
mole
(0.73eV)
Can do below 400°C
By contrast,metal CVD and sputtering => better step coverage
W
6.152J/3.155J 14
Other methods of heating charge.
Resistive heater
I
RF-induction
heater
e
-
+
_
e-beam
B field
Can you suggest other methods?
Laser,Pulsed Laser Deposition (PLD),laser ablation ?
Ion beam deposition (IB)D,
keep substrate chamber at low P,bring in ion beam
through differentially pumped path.
Source,
target
Substrate,
film
Ion beam
?
Ion beam
Can also use ion beam on film to add energy
(ion beam assisted deposition,IBAD)
6.152J/3.155J 15
Surface energy in a growing film depends on the number of bonds
the adsorbed atom forms with the substrate (or number unsatisfied).
This depends on the crystallography of the surface face
and on the type of site occupied (face,edge,corner,crevice),
Macroscopically,a curved surface has higher surface energy
(more dangling bonds) than a flat surface.
6.152J/3.155J 16
Microstructure types observed in sputtered films
with increasing substrate temperature
normalized to melting temperature of deposited species.
Quenched growth Thermally activated growth
λ < a λ < aT
s
/T
m
> 0.3 T
s
/T
m
> 0.5
6.152J/3.155J 17
Interfaces between dissimilar materials
Four characteristic equilibrium,binary phase diagrams,above,
and the types of interface structures they may lead to,below (non-equilibrium),
The first-column figures would apply to Ga-As,the second to Si-Ge,Third case,
B diffuses into A causing swelling; A is forced by swelling into B as a second phase.
6.152J/3.155J 18
Schematic stress strain curve
showing plastic deformation
beyond the yield point.
Upon thermal cycling,
a film deposited under conditions
that leave it in tensile stress
may evolve through compression
then even greater tension,
PHYSICAL VAPOR DEPOSITION (PVD)
PVD II,Evaporation
We saw CVD
Gas phase reactants,p
g
≈ 1 mTorr to 1 atm.
Good step coverage,T > 350 K
?
We saw sputtering Noble (+ reactive gas) p ≈ 10 mTorr; ionized particles
Industrial process,high rate,reasonable step coverage
Extensively used in electrical,optical,magnetic devices.
?
Now see evaporation:
Source material heated,p
eq.vap.
=~ 10
-3
Torr,p
g
< 10
-6
Torr
Generally no chemical reaction (except in,reactive depos’n),
λ = 10’s of meters,Knudsen number N
K
>> 1
Poor step coverage,alloy fractionation,? p
vapor
Historical (optical,electrical)
?
Campbell,Ch,12 is more extensive than Plummer on evaporation
6.152J/3.155J 2
Standard vacuum chambers
Σp
i
≈ 10
-6
Torr (1.3 × 10
-4
N / m
2)
Mostly H
2
O,hydrocarbons,N
2
He by
residual gas analysis (RGA = mass spec.)
6.152J/3.155J 3
p < 10
-8
Torr demands:
Stainless steel chamber
Bakeable to 150
o
C
Turbo,ion,cryo pumps
Ultra-high vacuum chambers
6.152J/3.155J 4
-6 -4 -2 0 2 4
Log[P (N/m
2
)]
25
23
Log[n (#/m
3
)]
21
19
17
15
λ = 10 0.01 cm
p = 10
-10
10
-8
10
-6
10
-4
10
-2
10
0
Torr
1 Atm=
0.1 MPa
760 mm
≈14 lb/in
2
Generally λ/L < 1
Films less pure
Epitaxy is rare
Evaporation
Ballistic,molecular flow,λ/L >> 1
High purity films
Epitaxy
Sputtering
CVD
Knudson number ≈ 1
6.152J/3.155J 5
Atomic flux on surface due to residual gas
J
atoms
area?t
=
nv
x
2
=
p
2k
B
T
2k
B
T
πm
=
p
2πmk
B
T
= J
Given 10
-6
Torr of water vapor @ room temp,find flux
p =10
6
Torr ×
1atm
760 T
×
10
5
P
a
atm
,k
B
TRT
( )
= 0.025eV = 4 ×10
21
J
p =1.3×10
4
N
m
2
m
H
2
O
=
18
N
A
= 3×10
26
kg
What is atomic density in 1 monolayer (ML) of Si?
N = 5 x 10
22
cm
-3
=> 1.3 x 10
15
cm
-2
.
So at 10
-6
Torr,1 ML of residual gas hits surface every 3 seconds!
Epitaxy requires slow deposition,surface mobility,
So you must keep pressure low to maintain pure film
J = 4.8×10
14
atoms/molecules
cm
2
sec
6.152J/3.155J 6
Now add evaporation source
Equilibrium vapor pressure:
H = heat of vaporization
p
v
= p
0
exp?
H
k
B
T
Strong T dependence
e
-
+
_
e-beam
B field
Resistive heater
I
RF-induction
heater
Work
function
e
-
Heat of
vaporization
solid
V(x)
free
6.152J/3.155J 7
Vapor pressure of elements employed in
semiconductor materials,Dots
correspond to melting points
Rely on tables,attached,p
vapor
> > p
vac,
Elemental metals easy to evaporate,but…
alloys
compounds
Differential p
vapor
so use 2 crucibles
or deposit multilayers
and diffuse
Oxides,nitrides deposit in oxygen
(or other) partial p
6.152J/3.155J 8
(note W,Mo
can be used
as crucibles.)
6.152J/3.155J 9
The source also? flux
J =
p
vap
2πm
source
k
B
T
source
For Al at 1000 K,p
vap
= 10
-7
Torr (from figure)
m = species to be evaporated,= 27 amu for Al
J
Al
≈ 2 x 10
13
Al/cm
2
-s just above crucible
Mass flow out of crucible ~ J A
c
m ( mass / t)
We expressed flux of residual gas:
J ≈ 5×10
14
molecules
cm
2
s
( )
Chamber p = 10
-6
Torr
J =
p
2πmk
B
T
Net flow out of crucible ~ JA
c
(# / t)
A
c
Therefore heat Al to T > 800 C
but that’s not all…
Note,3 different temperatures,T
source
≈ T
evaporant
>> T
substrate
> T
chamber
= T
resid gas
≈ RT.
System not in thermal equilibrium; only thermal interaction among them is by radiation
and/or conduction through solid connects (weak contact),No convection when N
K
<< 1.
6.152J/3.155J 10
How much evaporant strikes substrate? At 10
-6
Torr,trajectories are uninterrupted.
While a point source deposits uniformly on a sphere about it,a planar source does not,
Geometric factor =
A
c
2πR
2
cosθ
1
cosθ
2
cosθ
1
= cosθ
2
=
R
2r
Deposition rate =
Jm
A
c
4πr
2
m
area? t
or
J
A
c
4πr
2
#
area?t
substrate
θ
1
θ
2
R
R
θ
1
θ
2
r
r
substrate
Convenient
geometry
Geometric factor =
A
c
4πr
2
Film growth rate
= Jm
A
c
4πr
2
1
ρ
f
thick
t
∝cosθ
1
J
6.152J/3.155J 11
v
ox
=
H p
g
N
1
h
+
t
ox
D
+
1
k
s
oxide
In PVD growth,strike balance
R =
deposition rate
Surface diffusion rate
R > 1 stochastic growth,rough
R < 1 layer by layer,smooth (can heat substrate)
Film growth rate
for evaporation
= Jm
A
c
4πr
2
1
ρ
f
thick
t
v =
p
vap
2πm
source
k
B
T
source
m
ρ
m
A
c
4πr
2
=
p
vap
ρ
m
A
c
4πr
2
m
source
2πk
B
T
source
Cf,CVD
v
f
=
C
g
N
1
n
g
+
1
k
6.152J/3.155J 12
Exercise
Deposit Al (2.7 g/cm
3
) at r = 40 cm from 5 cm diam,
crucible heated to 1100°C (cf T
melt
) p
Al vap
≈ 10
-3
Torr,
p
H
2
O
=10
6
Torr
Compare arrival rate of Al and H
2
O at substrate…and calculate film growth
rate
J
H
2
O
=
10
6
760
()
×10
5
2π × 0.025eV× e × 18 N
A
()
= 4.8×10
18
molecules
m
2
s
J
Al
=
10
3
×10
5
760
()
2π × 1373k
B
()
× 27 N
A
()
A
c
4πr
2
=1.76×10
18
atoms
m
2
s
A
c
=π
5
2
2
Leave shutter closed so initial Al deposition can getter O
2
and H
2
O,
Also,better done at lower or higher deposition rate.
p
H
2
O
(this is not good)
v =
p
vap
ρ
m
A
c
4πr
2
m
source
2πk
B
T
source
≈10
11
m /s
slow!
6.152J/3.155J 13
Step coverage is poor in evaporation (ballistic) - Shadow effects
Heat substrate to increase
surface diffusion,
D
S
= D
0
S
exp?
E
a
Surf
kT
E
a
S
<< E
a
bulk
e.g,WF
6
+ 3H
2
→ W + 6HF
←
G ≈ 70
kJ
mole
(0.73eV)
Can do below 400°C
By contrast,metal CVD and sputtering => better step coverage
W
6.152J/3.155J 14
Other methods of heating charge.
Resistive heater
I
RF-induction
heater
e
-
+
_
e-beam
B field
Can you suggest other methods?
Laser,Pulsed Laser Deposition (PLD),laser ablation ?
Ion beam deposition (IB)D,
keep substrate chamber at low P,bring in ion beam
through differentially pumped path.
Source,
target
Substrate,
film
Ion beam
?
Ion beam
Can also use ion beam on film to add energy
(ion beam assisted deposition,IBAD)
6.152J/3.155J 15
Surface energy in a growing film depends on the number of bonds
the adsorbed atom forms with the substrate (or number unsatisfied).
This depends on the crystallography of the surface face
and on the type of site occupied (face,edge,corner,crevice),
Macroscopically,a curved surface has higher surface energy
(more dangling bonds) than a flat surface.
6.152J/3.155J 16
Microstructure types observed in sputtered films
with increasing substrate temperature
normalized to melting temperature of deposited species.
Quenched growth Thermally activated growth
λ < a λ < aT
s
/T
m
> 0.3 T
s
/T
m
> 0.5
6.152J/3.155J 17
Interfaces between dissimilar materials
Four characteristic equilibrium,binary phase diagrams,above,
and the types of interface structures they may lead to,below (non-equilibrium),
The first-column figures would apply to Ga-As,the second to Si-Ge,Third case,
B diffuses into A causing swelling; A is forced by swelling into B as a second phase.
6.152J/3.155J 18
Schematic stress strain curve
showing plastic deformation
beyond the yield point.
Upon thermal cycling,
a film deposited under conditions
that leave it in tensile stress
may evolve through compression
then even greater tension,
