1
3.155J/6.152J
Microelectronic Processing
Fall Term,2003
Bob O'Handley
Martin Schmidt
Problem set 1 Solutions Out Sept,8,2003 Due Sept.17,2002
1,a) Mean free path
l=
k
B
T
2pd
2
p
where k
B
= 1.38 ￥ 10
-23
,T = 293°K,p = 10 m Torr =133 Pa,d a 1A = 10
-10
m
l=
1.38￥10
-23
()
293°K
()
2p 10
-10
m
()
2
1.33Pa
()
=
6.9cm
1.7cm
ì
ó
b) Use the ideal gas law pV = N k
B
T
c) Flux on a surface:
J
x
=
p
2pk
B
T
m
where m = mass of particle = ( 28 amu) ( 1.67 ￥ 10
–27
kg/amu) = 4.76 ￥ 10
-26
kg
J
x
= 3.86￥10
22
molecules
m
2
s
= 3.9￥10
18
cm
-2
s
-1
d) We are trying to find the average molecular speed of N
2
molecules impinging a
surface in the chamber,Two ways to do this:
v
x
=
2J
n
=2.3
m
s
=2.3￥10
4
cm
s
or
v
x
=
2k
B
T
pm
=235
m
s
for d = 1 A
for d = 2 A
N
V
=
p
k
B
T
=
1.33Pa
(1.38￥10
-23
) 293°K
()
= 3.3￥10
20
m
-3
= 3.3￥10
14
cm
-3
2
some people calculated the average molecular speed in any direction,which is
v =
8kT
pm
=476
m
s
This is also accepted,(Note that the molecular speed is independent of pressure
because both J and n are proportional to p.)
2,SCCM stands for a flow of a,standard cubic centimeter per minute” (standard cc is
at temperature of 273 °K and pressure of 760 Torr).
100
cm
3
min
ê
á
á
ˉ
L
1000cm
3
ê
á
ˉ
min
60s
ê
á
ˉ
=0.001666
L
s
To change to 100 °K and 1 mTorr,
0.001666
L
s
ê
á
ˉ
100°K
273°K
ê
á
ˉ
760Torr
10
-3
Torr
ê
á
ˉ
=463
L
s
CVD
3,Assume chemical equilibrium is established in a CVD reactor according to the
equation:
SiH
4
(g) ′ SiH
2
(g) + H
2
(g)
The temperature is maintained at 650 C and the pressure at 0.1 atm,If the
equilibrium constant for the reaction is K(T)=2￥ 10
9
(Torr) exp[-1.8 eV/(k
B
T)],find
the partial pressure of each gas assuming p(H
2
) p(SiH
2
).
10 points.
K(T =923K)=2￥10
9
exp -2.9￥10
-19
J / k
B
￥923K
()
[]
=0.3 Torr
Also,K =
p
SiH
2
￥ p
H
2
p
SiH
4
=
p
SiH
2
2
p
SiH
4
=0.3 Torr (*) (assumption, p
H
2
= p
SiH
2
)
I
in addition,
p
SiH
4
+ p
SiH
2
+ p
H
2
= 76 Torr
p
SiH
4
+2p
SiH
2
= 76 Torr (**)
Substituting (**) into (*):
3
p
SiH
2
2
=0.3p
SiH
4
=0.3(76-2p
SiH
2
)
=22.8-0.6p
SiH
2
fi p
SiH
2
2
+0.6 p
SiH
2
-22.8=0
p
SiH
2
= 4.48 Torr
p
H
2
a 4.48 Torr
p
SiH
4
=67 Torr
Check that the sum of the partial pressures,p
i
= 76 Torr = 0.1 atm.
4,Assume a CVD process based on the reaction,2AB(g) ′ 2A(s) + B
2
(g).
a) Sketch and briefly describe the individual steps that control the reaction.
15 points.
1) Bulk transport governed by gas velocity outside boundary layer,u
.
2) Diffusion across boundary layer
J μ
D
d(x)
C
g
-C
s
()
,d(x) is the thickness of the boundary layer,a function
of x.
3) Adsorption involves sticking.
4) Dissociation of AB due to temperature and possibly catalyzed by interaction
of A with surface,B may remain adsorbed or desorb upon dissociation.
5) A(s) actually bonds with a site on substrate surface (film growth).
6) B
2
(g) must diffuse across boundary layer,()
22
2
2
)(
B
s
B
g
B
B
CC
x
D
J -=
d
7) Bulk transport of B
2
(g) under carrier gas velocity u
.
C
s
4
3
A
B
2
C
g
1)
A
B A B
B
5
B
B
6
u
B
B
7
A
B
A
4
b) How would you distinguish between i) the reaction-limited and ii) a transport-
limited cases?
5 points.
In the reaction-limited regime,the temperature dependence of the film growth
rate exhibits Arrhenius behavior:
ˉ
á
á
ê D-
μ
ˉ
á
á
ê D-
==
Tk
G
v
Tk
G
kk
N
kC
v
B
B
o
g
exp
exp,
In the transport-limited regime,the deposition rate is given by:
v =
3lv
x
C
g
4N
f
x
ru
x
h
,
where
v
x
=
2k
B
T
pm
,
C
g
P
g
=
1
k
B
T
,and l=
k
B
T
2pd
2
P
total
so that
v μP
g
T u
Thus,temperature dependence is the biggest differentiator between the two
regimes.
c) Sketch the variation of the log of the CVD growth rate as functions of the square
root of the gas flow velocity and as a function of 1/T.
5 points.
ln(v)
u
Transport regime
Reaction regime
1/T
ln(v)
Transport regime
Reaction regime
5
d) If you wanted to increase the growth rate of a transport-limited CVD process,
what processing variables would be most effective? (List them in decreasing
order of efficacy,e.g,v μ exp(x) first,etc.)
5 points.
For the fixed length,in the transport-limited regime,in decreasing order of
efficacy,we have P
g
,u
,and T,
5,Phosphorus-doped polysilicon is produced by CVD based on the following reactions:
SiH
4
(g) ′ Si (s) + 2H
2
(g)
2PH
3
(g) ′ 2P (s) + 3H
2
(g).
Deposition occurs at 1000
o
C,N
Si
=5￥ 10
22
cm
-3
,the Si deposition is reaction limited
(c
g
=10
18
cm
-3
,k
o
=6￥ 10
5
(cm/s),DG = 1.0 eV),and the deposition of the P is
transport limited (c
g
= 10
18
cm
-3
,D
g
= 10
-5
￥ T
3/2
cm/s,and (ru/h)
0.5
= 0.01 cm
-0.5
).
a) What range of P concentration in the poly-silicon might you expect across a 30
cm diameter wafer? Explain the assumptions you must make to answer this question.
10 points.
The silicon deposition is reaction limited:
v
silicon
=
kC
g
N
k =6￥10
5
exp
-1eV
8.617￥10
-5
eV /K ￥1273K
ê
á
ˉ
=66 cm /s
C
g
=10
18
cm
-3
N
silicon
=5￥10
22
cm
-3
v
silicon
=1.32￥10
-3
cm /s
The phosphorus deposition is transport limited:
v
phosphorus
=
DC
g
LN
P
rLm
h
D=10
-5
￥(1273)
3
2
=0.45cm
2
/s
v
phosphorus
=
3￥10
-6
L
cm /s
6
We assume that a boundary layer develops at L = 0 cm:
v
P
(0)=?
But the phosphorus will not dope the silicon beyond its solid solubility limit,about 1.5 ￥
10
21
cm
-3
(Fig 2.4 in Campbell):
%3
105
105.1
22
21
=
￥
￥
At L = 30 cm:
v
silicon
=1.32￥10
-3
cm /s
v
phosphorus
=5.47￥10
-7
cm /s
C
phosphorus
C
silicon
=
5.47￥10
-7
cm /s
1.32￥10
-3
cm /s
= 4.15￥10
-4
= 41ppm
b) If the temperature at the far end of the wafer is constrained to be 10°C hotter than
that at the near end of the reactor,how does this affect your answer?
10 points.
With the hotter temperature,both deposition rates should increase:
v
si
=13.3
mm
s
v
si
0
12434
￥e
1eV
k
B
1
1273
-
1
1283
ê
á
ˉ
=1.4￥10
-3
cm /s=1.07￥v
si
0
v
p
=v
p
0
￥
1283
1273
ê
á
ˉ
3
2
=5.53￥10
-7
cm /sec=1.01v
p
0
Thus,the silicon deposition rate has increased by 7%,while the phosphorus
deposition rate increased by only 1%,The increase in temperature actually leads to a less
uniform doping.
3.155J/6.152J
Microelectronic Processing
Fall Term,2003
Bob O'Handley
Martin Schmidt
Problem set 1 Solutions Out Sept,8,2003 Due Sept.17,2002
1,a) Mean free path
l=
k
B
T
2pd
2
p
where k
B
= 1.38 ￥ 10
-23
,T = 293°K,p = 10 m Torr =133 Pa,d a 1A = 10
-10
m
l=
1.38￥10
-23
()
293°K
()
2p 10
-10
m
()
2
1.33Pa
()
=
6.9cm
1.7cm
ì
ó
b) Use the ideal gas law pV = N k
B
T
c) Flux on a surface:
J
x
=
p
2pk
B
T
m
where m = mass of particle = ( 28 amu) ( 1.67 ￥ 10
–27
kg/amu) = 4.76 ￥ 10
-26
kg
J
x
= 3.86￥10
22
molecules
m
2
s
= 3.9￥10
18
cm
-2
s
-1
d) We are trying to find the average molecular speed of N
2
molecules impinging a
surface in the chamber,Two ways to do this:
v
x
=
2J
n
=2.3
m
s
=2.3￥10
4
cm
s
or
v
x
=
2k
B
T
pm
=235
m
s
for d = 1 A
for d = 2 A
N
V
=
p
k
B
T
=
1.33Pa
(1.38￥10
-23
) 293°K
()
= 3.3￥10
20
m
-3
= 3.3￥10
14
cm
-3
2
some people calculated the average molecular speed in any direction,which is
v =
8kT
pm
=476
m
s
This is also accepted,(Note that the molecular speed is independent of pressure
because both J and n are proportional to p.)
2,SCCM stands for a flow of a,standard cubic centimeter per minute” (standard cc is
at temperature of 273 °K and pressure of 760 Torr).
100
cm
3
min
ê
á
á
ˉ
L
1000cm
3
ê
á
ˉ
min
60s
ê
á
ˉ
=0.001666
L
s
To change to 100 °K and 1 mTorr,
0.001666
L
s
ê
á
ˉ
100°K
273°K
ê
á
ˉ
760Torr
10
-3
Torr
ê
á
ˉ
=463
L
s
CVD
3,Assume chemical equilibrium is established in a CVD reactor according to the
equation:
SiH
4
(g) ′ SiH
2
(g) + H
2
(g)
The temperature is maintained at 650 C and the pressure at 0.1 atm,If the
equilibrium constant for the reaction is K(T)=2￥ 10
9
(Torr) exp[-1.8 eV/(k
B
T)],find
the partial pressure of each gas assuming p(H
2
) p(SiH
2
).
10 points.
K(T =923K)=2￥10
9
exp -2.9￥10
-19
J / k
B
￥923K
()
[]
=0.3 Torr
Also,K =
p
SiH
2
￥ p
H
2
p
SiH
4
=
p
SiH
2
2
p
SiH
4
=0.3 Torr (*) (assumption, p
H
2
= p
SiH
2
)
I
in addition,
p
SiH
4
+ p
SiH
2
+ p
H
2
= 76 Torr
p
SiH
4
+2p
SiH
2
= 76 Torr (**)
Substituting (**) into (*):
3
p
SiH
2
2
=0.3p
SiH
4
=0.3(76-2p
SiH
2
)
=22.8-0.6p
SiH
2
fi p
SiH
2
2
+0.6 p
SiH
2
-22.8=0
p
SiH
2
= 4.48 Torr
p
H
2
a 4.48 Torr
p
SiH
4
=67 Torr
Check that the sum of the partial pressures,p
i
= 76 Torr = 0.1 atm.
4,Assume a CVD process based on the reaction,2AB(g) ′ 2A(s) + B
2
(g).
a) Sketch and briefly describe the individual steps that control the reaction.
15 points.
1) Bulk transport governed by gas velocity outside boundary layer,u
.
2) Diffusion across boundary layer
J μ
D
d(x)
C
g
-C
s
()
,d(x) is the thickness of the boundary layer,a function
of x.
3) Adsorption involves sticking.
4) Dissociation of AB due to temperature and possibly catalyzed by interaction
of A with surface,B may remain adsorbed or desorb upon dissociation.
5) A(s) actually bonds with a site on substrate surface (film growth).
6) B
2
(g) must diffuse across boundary layer,()
22
2
2
)(
B
s
B
g
B
B
CC
x
D
J -=
d
7) Bulk transport of B
2
(g) under carrier gas velocity u
.
C
s
4
3
A
B
2
C
g
1)
A
B A B
B
5
B
B
6
u
B
B
7
A
B
A
4
b) How would you distinguish between i) the reaction-limited and ii) a transport-
limited cases?
5 points.
In the reaction-limited regime,the temperature dependence of the film growth
rate exhibits Arrhenius behavior:
ˉ
á
á
ê D-
μ
ˉ
á
á
ê D-
==
Tk
G
v
Tk
G
kk
N
kC
v
B
B
o
g
exp
exp,
In the transport-limited regime,the deposition rate is given by:
v =
3lv
x
C
g
4N
f
x
ru
x
h
,
where
v
x
=
2k
B
T
pm
,
C
g
P
g
=
1
k
B
T
,and l=
k
B
T
2pd
2
P
total
so that
v μP
g
T u
Thus,temperature dependence is the biggest differentiator between the two
regimes.
c) Sketch the variation of the log of the CVD growth rate as functions of the square
root of the gas flow velocity and as a function of 1/T.
5 points.
ln(v)
u
Transport regime
Reaction regime
1/T
ln(v)
Transport regime
Reaction regime
5
d) If you wanted to increase the growth rate of a transport-limited CVD process,
what processing variables would be most effective? (List them in decreasing
order of efficacy,e.g,v μ exp(x) first,etc.)
5 points.
For the fixed length,in the transport-limited regime,in decreasing order of
efficacy,we have P
g
,u
,and T,
5,Phosphorus-doped polysilicon is produced by CVD based on the following reactions:
SiH
4
(g) ′ Si (s) + 2H
2
(g)
2PH
3
(g) ′ 2P (s) + 3H
2
(g).
Deposition occurs at 1000
o
C,N
Si
=5￥ 10
22
cm
-3
,the Si deposition is reaction limited
(c
g
=10
18
cm
-3
,k
o
=6￥ 10
5
(cm/s),DG = 1.0 eV),and the deposition of the P is
transport limited (c
g
= 10
18
cm
-3
,D
g
= 10
-5
￥ T
3/2
cm/s,and (ru/h)
0.5
= 0.01 cm
-0.5
).
a) What range of P concentration in the poly-silicon might you expect across a 30
cm diameter wafer? Explain the assumptions you must make to answer this question.
10 points.
The silicon deposition is reaction limited:
v
silicon
=
kC
g
N
k =6￥10
5
exp
-1eV
8.617￥10
-5
eV /K ￥1273K
ê
á
ˉ
=66 cm /s
C
g
=10
18
cm
-3
N
silicon
=5￥10
22
cm
-3
v
silicon
=1.32￥10
-3
cm /s
The phosphorus deposition is transport limited:
v
phosphorus
=
DC
g
LN
P
rLm
h
D=10
-5
￥(1273)
3
2
=0.45cm
2
/s
v
phosphorus
=
3￥10
-6
L
cm /s
6
We assume that a boundary layer develops at L = 0 cm:
v
P
(0)=?
But the phosphorus will not dope the silicon beyond its solid solubility limit,about 1.5 ￥
10
21
cm
-3
(Fig 2.4 in Campbell):
%3
105
105.1
22
21
=
￥
￥
At L = 30 cm:
v
silicon
=1.32￥10
-3
cm /s
v
phosphorus
=5.47￥10
-7
cm /s
C
phosphorus
C
silicon
=
5.47￥10
-7
cm /s
1.32￥10
-3
cm /s
= 4.15￥10
-4
= 41ppm
b) If the temperature at the far end of the wafer is constrained to be 10°C hotter than
that at the near end of the reactor,how does this affect your answer?
10 points.
With the hotter temperature,both deposition rates should increase:
v
si
=13.3
mm
s
v
si
0
12434
￥e
1eV
k
B
1
1273
-
1
1283
ê
á
ˉ
=1.4￥10
-3
cm /s=1.07￥v
si
0
v
p
=v
p
0
￥
1283
1273
ê
á
ˉ
3
2
=5.53￥10
-7
cm /sec=1.01v
p
0
Thus,the silicon deposition rate has increased by 7%,while the phosphorus
deposition rate increased by only 1%,The increase in temperature actually leads to a less
uniform doping.
