1
3.155J/6.152J
Microelectronic Processing Technology
Fall Term,2003
Bob O'Handley
Martin Schmidt
Problem set 6 Out Nov,12,2003 Due Nov,26,2003
Sputter deposition,Read Plummer Chap,9,Sections 9.2.2.2 to 9.3.10,Consider
reading Ohring
1,You need to deposit a high quality (low electrical resistivity) Al film at a very high
rate (v > 1 micron/min) and achieve good step coveage using sputter deposition.
Referring to information in the class notes and text,answer the following three
questions,(Grade will depend more on how you justify your design,rather than on its
correctness – which is harder to determine.)
a) Design,and justify your design,for a sputtering system to deposit this Al with
attention paid to configuration of the anode(s),cathode(s) and substrate placement
(be creative here),as well as the use of DC or RF power source,and biasing,If
possible give some conditions on power requirements and critical dimensions in
the chamber.
b) What sputtering conditions would you use? (type of gas,gas pressure,sputtering
voltage,bias voltage,substrate temperature).
c) Assuming a sticking coefficient of unity for both Al and oxygen,what oxygen
partial pressure could you tolerate in the chamber to keep the oxygen content in
the film less than 1% for your chosen conditions?
Some possibly useful information is shown below:
SOLUTION
1,a) Text suggests DC (RF is acceptable),magnetron (for high deposition rate)
sputtering; biasing will need to be justified in terms of film stress and density.
Here is one way to configure magnets and electrodes for more grazing incidence
of ions on target (q just less than 90
0
) and greater sputter yield:
q
S
90°
2
Target(s)
Substrate(s)
Either planar or cylindrical
symmetry; the latter has advantage
of very large ratio of target area to
substrate area,Other geometries
acceptable but should be justified.
Magnets are arranged so their field lines run nearly parallel to target surface.
They could be oriented in opposition to each other (so overlapping fields cancel) to
minimize field at substrates,Target/substrate spacing,a,will be governed by radius of
curvature of electrons in B field,If the radius is comparable to a,then many Ar+ ions
may curve away from the target and strike the substrate,If r << a,then the Ar+ ions
will spiral about the B field with many possible strikes on the target,F = q(v x B) and
F = mv
2
/r indicate r = mv/(qB),The velocity may be governed by the power,P ≈
(1/2)mv
2
≈ eV > 1 keV (class notes slides 10,11).
b) Either Ne
20
or Ar
40
could be used to sputter Al
27
,Use large negative bias,
perhaps -200 V on substrate (slide 15),Chose pressure and temperature on basis of
ideas in slide 13,low pressure and high temperature for good step coverage,but to
avoid porous film,keep temperature quite high,text recommends 150 to 300C.
c) The given film growth rate of v
film
= 1 micron /min ≈ 17 nm/s = 1.7 x 10
-6
cm/s,
and assuming an Al film density of 10
23
cm
-3
,the Al flux on the substrate is J = nv
film
=
1.7 x 10
17
/s,So the oxygen flux should be less than about 10
14
/s,Using
J =
P
2pmkT
,
we get a pressure of 2.6 x 10
-9
Pa (1 atm = 10
5
Pa = 760 T),so P(O
2
) ≈ 10
-11
Torr.
2,It is very difficult to evaporate stoichiometric SiO
2; you often get SiO
x
with 1 < x < 2,
which implies a mixture of SiO and SiO
2
,But you still need to get as close as possible to
SiO
2
.
a) What partial pressure of O
2
must you have in your vacuum chamber so that the
flux of O
2
on the substrate is the same as that of Si? Your evaporation source has
a surface area of 1 cm
2
,the substrate is at a planetary radius of 20 cm,and your
crucible is heated to 1500
o
C.
b) What is the mean free path of O
2
in this partial pressure (assume the diameter of
an O2 molecule is 0.3 nm).
c) What does your measurement in b) mean for your process?
3
SOLUTION:
2,a) A
crucible
= 1 cm
2
,r = 20 cm,T
Si
=1500°C (1773K)
J
O
2
J
Si
=1 =
P
O
2
P
eq.vap
Si
￥
2pkT
Si
m
Si
2pkT
H
2
O
m
O
2
A
cru
4pr
2
,m
Si
= 28 amu,
m
O
2
= 32 amu
from table,
P
Si
1500°C
@ 2 ￥10
-3
Torr
P
O
2
=
2 ￥10
-3
￥ 133 Pa Torr
( )[ ]
1cm
2
293 ￥ 32
4p ￥20
2
cm
2
1773￥ 28
= 2.3 ￥10
-5
Pa =1.7 ￥10
-7
Torr
b)
l =
kT
2pd
2
P
Tot
= 23 cm (using T,P of Si)
c) There are virtually no collisions between O
2
molecules in the chamber,for that
matter between O
2
and Si vapor molecules,Both species do collide with walls of
chamber,Without collisions,the two gases (Si,O
2
) are not in equilibrium.
Also Si vapor follows a straight line path to substrate,It is only on the
substrate that Si has a chance to interact with O
2
.
3.155J/6.152J
Microelectronic Processing Technology
Fall Term,2003
Bob O'Handley
Martin Schmidt
Problem set 6 Out Nov,12,2003 Due Nov,26,2003
Sputter deposition,Read Plummer Chap,9,Sections 9.2.2.2 to 9.3.10,Consider
reading Ohring
1,You need to deposit a high quality (low electrical resistivity) Al film at a very high
rate (v > 1 micron/min) and achieve good step coveage using sputter deposition.
Referring to information in the class notes and text,answer the following three
questions,(Grade will depend more on how you justify your design,rather than on its
correctness – which is harder to determine.)
a) Design,and justify your design,for a sputtering system to deposit this Al with
attention paid to configuration of the anode(s),cathode(s) and substrate placement
(be creative here),as well as the use of DC or RF power source,and biasing,If
possible give some conditions on power requirements and critical dimensions in
the chamber.
b) What sputtering conditions would you use? (type of gas,gas pressure,sputtering
voltage,bias voltage,substrate temperature).
c) Assuming a sticking coefficient of unity for both Al and oxygen,what oxygen
partial pressure could you tolerate in the chamber to keep the oxygen content in
the film less than 1% for your chosen conditions?
Some possibly useful information is shown below:
SOLUTION
1,a) Text suggests DC (RF is acceptable),magnetron (for high deposition rate)
sputtering; biasing will need to be justified in terms of film stress and density.
Here is one way to configure magnets and electrodes for more grazing incidence
of ions on target (q just less than 90
0
) and greater sputter yield:
q
S
90°
2
Target(s)
Substrate(s)
Either planar or cylindrical
symmetry; the latter has advantage
of very large ratio of target area to
substrate area,Other geometries
acceptable but should be justified.
Magnets are arranged so their field lines run nearly parallel to target surface.
They could be oriented in opposition to each other (so overlapping fields cancel) to
minimize field at substrates,Target/substrate spacing,a,will be governed by radius of
curvature of electrons in B field,If the radius is comparable to a,then many Ar+ ions
may curve away from the target and strike the substrate,If r << a,then the Ar+ ions
will spiral about the B field with many possible strikes on the target,F = q(v x B) and
F = mv
2
/r indicate r = mv/(qB),The velocity may be governed by the power,P ≈
(1/2)mv
2
≈ eV > 1 keV (class notes slides 10,11).
b) Either Ne
20
or Ar
40
could be used to sputter Al
27
,Use large negative bias,
perhaps -200 V on substrate (slide 15),Chose pressure and temperature on basis of
ideas in slide 13,low pressure and high temperature for good step coverage,but to
avoid porous film,keep temperature quite high,text recommends 150 to 300C.
c) The given film growth rate of v
film
= 1 micron /min ≈ 17 nm/s = 1.7 x 10
-6
cm/s,
and assuming an Al film density of 10
23
cm
-3
,the Al flux on the substrate is J = nv
film
=
1.7 x 10
17
/s,So the oxygen flux should be less than about 10
14
/s,Using
J =
P
2pmkT
,
we get a pressure of 2.6 x 10
-9
Pa (1 atm = 10
5
Pa = 760 T),so P(O
2
) ≈ 10
-11
Torr.
2,It is very difficult to evaporate stoichiometric SiO
2; you often get SiO
x
with 1 < x < 2,
which implies a mixture of SiO and SiO
2
,But you still need to get as close as possible to
SiO
2
.
a) What partial pressure of O
2
must you have in your vacuum chamber so that the
flux of O
2
on the substrate is the same as that of Si? Your evaporation source has
a surface area of 1 cm
2
,the substrate is at a planetary radius of 20 cm,and your
crucible is heated to 1500
o
C.
b) What is the mean free path of O
2
in this partial pressure (assume the diameter of
an O2 molecule is 0.3 nm).
c) What does your measurement in b) mean for your process?
3
SOLUTION:
2,a) A
crucible
= 1 cm
2
,r = 20 cm,T
Si
=1500°C (1773K)
J
O
2
J
Si
=1 =
P
O
2
P
eq.vap
Si
￥
2pkT
Si
m
Si
2pkT
H
2
O
m
O
2
A
cru
4pr
2
,m
Si
= 28 amu,
m
O
2
= 32 amu
from table,
P
Si
1500°C
@ 2 ￥10
-3
Torr
P
O
2
=
2 ￥10
-3
￥ 133 Pa Torr
( )[ ]
1cm
2
293 ￥ 32
4p ￥20
2
cm
2
1773￥ 28
= 2.3 ￥10
-5
Pa =1.7 ￥10
-7
Torr
b)
l =
kT
2pd
2
P
Tot
= 23 cm (using T,P of Si)
c) There are virtually no collisions between O
2
molecules in the chamber,for that
matter between O
2
and Si vapor molecules,Both species do collide with walls of
chamber,Without collisions,the two gases (Si,O
2
) are not in equilibrium.
Also Si vapor follows a straight line path to substrate,It is only on the
substrate that Si has a chance to interact with O
2
.
