3.155J/6.152J Homework set 3 solution,fall 2003
1
3.155J/6.152J
Microelectronic Processing
Fall Term,2003
Bob O'Handley
Martin Schmidt
Problem set 3 Solutions Out Sept,29,2003 Due Oct.08,2003
Problem 1
C z,t( ) =
Q
pDt
exp -
z
2
4Dt
ê
á
á
ˉ
dC
dt
= -
Q
2t pDt
+
Q
pDt
z
2
4Dt
2
ê
á
á
ˉ
exp -
z
2
4Dt
ê
á
á
ˉ
= C z,t( )
z
2
2Dt
-1
ê
á
á
ˉ
1
2t
dC
dz
=
-2z
4Dt
C z,t( )
d
2
C
dz
2
= -
2
4Dt
C z,t( ) +
4z
2
16D
2
t
2
C z,t( ) = C z,t( )
2z
2
4Dt
-1
ê
á
á
ˉ
2
4Dt
Thus,
dC z,t( )
dt
= D
d
2
C z,t( )
dz
2
Problem 2
a) From Fig,1.16 in Plummer or 3.4 in Campbell,
n
i
= 2 ￥10
19
b) i) From table,
D
0
.O = 0.011,D
0
.E = 3.44eV,D
-
.O = 31.0,D
-
.E = 4.15
Using relation
D = D
0
.0exp -
D.E
kT
ê
á
ˉ
D
eff
= D
0
= 2.66 ￥10
-15
cm
2
s
-1
ii) Use the equation
D
eff
= D
0
+ D
-
n
n
i
ê
á
ˉ
to get D
-
￥ (n/n
i
) = 9.35 ￥ 10
-15
cm
2
/s thus:
D
eff
= 1.2 ￥10
-14
cm
2
s
-1
c) Diffusion length
a = 2 Dt
for part i) a = 0.062 mm
for part ii) a = 0.13 mm
3.155J/6.152J Homework set 3 solution,fall 2003
2
Problem 3
a) Solubility limit for P at 1273 K is C
0
≈ 1 ￥ 10
21
cm
-3
,Intrinsic carrier
concentration is 10
19
cm
-3
,Thus,before including higher order terms,
D
O
=1.3￥10
-14
cm
2
/s,But with first order and second order correction terms
(n/
i
) for large P concentration,D = 237 ￥ 10
-14
cm
2
/s.
Using
Q =
C
0
a
p
=
2 DtC
0
p
we get Q = 1.9 ￥ 10
16
cm
-2
b) Find t
0
for z
jct
= 0.4 ￥ 10
-4
cm,Solubility limit for phosphorous at 1373 K is C
0
≈
6 ￥ 10
20
cm
-3
,Intrinsic carrier concentration at this temperature is 1.2 ￥ 10
19
cm
-3
.
With first order and second order correction terms (n/
i
) for large P concentration,
D = 110 ￥ 10
-13
cm
2
/s,Because the predep is at a higher concentration than the
solubility limit right inside the surface,the value of C(0,t) remains 6 ￥ 10
20
cm
-3
throughout this drive-in process,and the solution must be an erfc,The depth
dependence of the diffusion profile is:
Cz,t
()
=C
s
erfc-
z
2Dt
ê
á
ˉ
=N
A
=10
17
cm
-3
erfc-
z
2Dt
ê
á
ˉ
=
N
A
C
s
=
10
17
C
s
=1.67￥10
-4
Look up 1-1.67 ￥ 10
-4
= 0.99983 in the Plummer’s table for error functions and
you find the argument must be 2.66 = z/([2÷(Dt)],Solving for t gives t = 5.4 sec.
c) As stated above,the concentration at the surface is continually fed from the
predep and remains at the solubility limit of P,6 ￥ 10
20
cm
-3
.
Problem 4
Fig,8.3 gives the depth and standard deviation for boron implantation at 40 keV,For
ease of plotting I have expressed everything in nanometers,R
p
≈ 140 nm and DR
p
≈
57 nm,Given Q = 5 ￥ 10
15
cm
-2
= 50 nm
-2
,you can use Eq,8.4 to get C
p
= 0.35 nm
-3
(3.5 ￥ 10
20
cm
-3
),then plot Eq,8.1 as shown by the bold line below,Here the
concentration is plotted as number per cubic nm.
Following the arguments around Eqs,8.11 and 8.12 - the profile evolves like ÷(2Dt)
- you can plot Eq,8.12 using D at 950 C ≈ 1.94 ￥ 10
-16
cm
2
/s = 1.94 ￥ 10
-2
nm
2
/s to
plot the fine line in the figure below.
To get the junction depth,set Eq Eq,8.12 equal to 2 ￥ 10
15
cm
-3
= 2 ￥ 10
-6
nm
3
,and
solve it for x,the distance at what that concentration now occurs from the peak,If
you’re prone to math errors,you could plot it as shown in the second figure,to find
that the junction occurs at about 464 nm.
3.155J/6.152J Homework set 3 solution,fall 2003
3
Problem 5
At 30 keV,from Fig,8-3,R
p
≈ 120 nm and DR
p
≈ 50 nm,Given Q= 10
12
cm
-2
= 0.01
nm
-2
.
a) Peak occurs at depth where x = R
p
≈ 120 nm.
b) From Eq,8.4,C
p
= 8 ￥ 10
-5
nm
-3
= 8 ￥ 10
16
cm
-3
.
c) At x = 300 nm,Eq,8.1 gives C
p
= 1.23 ￥ 10
-7
nm
-3
= 1.23 ￥ 10
14
cm
-3
.
d) If the dose at 0.3 microns is 10 times larger than this even though the
values of R
p
and C
p
are correct,then the wafer was not pure,but rather
doped with boron to 1.1 ￥ 10
15
cm
-3
(1.11 ￥ 10
15
+ 1.23 ￥ 10
14
= 10 ￥
3.155J/6.152J Homework set 3 solution,fall 2003
4
1.23 ￥ 10
14
),A later exposure to elevated temperature could not cause
the increased concentration at 0.3 microns without shifting the standard
deviation and peak value,Also,while the actual distribution may not be
exactly Gaussian,for boron,the skew is toward increased concentration
near the surface (due to enhanced backscatter of this light species) not
what is asked here.
Problem 6
R
p
= 0.3 μm (300 nm) demands an implant of boron at about 100 keV (Fig,8-3),At
this energy,DR
p
≈ 85 nm,so the dose giving a peak concentration of 10
17
cm
-3
is
easily calculated to be 2.13 ￥ 10
12
cm
-2
,For a background doping of 1015 cm-3,the
junction depth before diffusion is given by inverting Eq,8.1,x = 558 nm or 0.558
μm.
1
3.155J/6.152J
Microelectronic Processing
Fall Term,2003
Bob O'Handley
Martin Schmidt
Problem set 3 Solutions Out Sept,29,2003 Due Oct.08,2003
Problem 1
C z,t( ) =
Q
pDt
exp -
z
2
4Dt
ê
á
á
ˉ
dC
dt
= -
Q
2t pDt
+
Q
pDt
z
2
4Dt
2
ê
á
á
ˉ
exp -
z
2
4Dt
ê
á
á
ˉ
= C z,t( )
z
2
2Dt
-1
ê
á
á
ˉ
1
2t
dC
dz
=
-2z
4Dt
C z,t( )
d
2
C
dz
2
= -
2
4Dt
C z,t( ) +
4z
2
16D
2
t
2
C z,t( ) = C z,t( )
2z
2
4Dt
-1
ê
á
á
ˉ
2
4Dt
Thus,
dC z,t( )
dt
= D
d
2
C z,t( )
dz
2
Problem 2
a) From Fig,1.16 in Plummer or 3.4 in Campbell,
n
i
= 2 ￥10
19
b) i) From table,
D
0
.O = 0.011,D
0
.E = 3.44eV,D
-
.O = 31.0,D
-
.E = 4.15
Using relation
D = D
0
.0exp -
D.E
kT
ê
á
ˉ
D
eff
= D
0
= 2.66 ￥10
-15
cm
2
s
-1
ii) Use the equation
D
eff
= D
0
+ D
-
n
n
i
ê
á
ˉ
to get D
-
￥ (n/n
i
) = 9.35 ￥ 10
-15
cm
2
/s thus:
D
eff
= 1.2 ￥10
-14
cm
2
s
-1
c) Diffusion length
a = 2 Dt
for part i) a = 0.062 mm
for part ii) a = 0.13 mm
3.155J/6.152J Homework set 3 solution,fall 2003
2
Problem 3
a) Solubility limit for P at 1273 K is C
0
≈ 1 ￥ 10
21
cm
-3
,Intrinsic carrier
concentration is 10
19
cm
-3
,Thus,before including higher order terms,
D
O
=1.3￥10
-14
cm
2
/s,But with first order and second order correction terms
(n/
i
) for large P concentration,D = 237 ￥ 10
-14
cm
2
/s.
Using
Q =
C
0
a
p
=
2 DtC
0
p
we get Q = 1.9 ￥ 10
16
cm
-2
b) Find t
0
for z
jct
= 0.4 ￥ 10
-4
cm,Solubility limit for phosphorous at 1373 K is C
0
≈
6 ￥ 10
20
cm
-3
,Intrinsic carrier concentration at this temperature is 1.2 ￥ 10
19
cm
-3
.
With first order and second order correction terms (n/
i
) for large P concentration,
D = 110 ￥ 10
-13
cm
2
/s,Because the predep is at a higher concentration than the
solubility limit right inside the surface,the value of C(0,t) remains 6 ￥ 10
20
cm
-3
throughout this drive-in process,and the solution must be an erfc,The depth
dependence of the diffusion profile is:
Cz,t
()
=C
s
erfc-
z
2Dt
ê
á
ˉ
=N
A
=10
17
cm
-3
erfc-
z
2Dt
ê
á
ˉ
=
N
A
C
s
=
10
17
C
s
=1.67￥10
-4
Look up 1-1.67 ￥ 10
-4
= 0.99983 in the Plummer’s table for error functions and
you find the argument must be 2.66 = z/([2÷(Dt)],Solving for t gives t = 5.4 sec.
c) As stated above,the concentration at the surface is continually fed from the
predep and remains at the solubility limit of P,6 ￥ 10
20
cm
-3
.
Problem 4
Fig,8.3 gives the depth and standard deviation for boron implantation at 40 keV,For
ease of plotting I have expressed everything in nanometers,R
p
≈ 140 nm and DR
p
≈
57 nm,Given Q = 5 ￥ 10
15
cm
-2
= 50 nm
-2
,you can use Eq,8.4 to get C
p
= 0.35 nm
-3
(3.5 ￥ 10
20
cm
-3
),then plot Eq,8.1 as shown by the bold line below,Here the
concentration is plotted as number per cubic nm.
Following the arguments around Eqs,8.11 and 8.12 - the profile evolves like ÷(2Dt)
- you can plot Eq,8.12 using D at 950 C ≈ 1.94 ￥ 10
-16
cm
2
/s = 1.94 ￥ 10
-2
nm
2
/s to
plot the fine line in the figure below.
To get the junction depth,set Eq Eq,8.12 equal to 2 ￥ 10
15
cm
-3
= 2 ￥ 10
-6
nm
3
,and
solve it for x,the distance at what that concentration now occurs from the peak,If
you’re prone to math errors,you could plot it as shown in the second figure,to find
that the junction occurs at about 464 nm.
3.155J/6.152J Homework set 3 solution,fall 2003
3
Problem 5
At 30 keV,from Fig,8-3,R
p
≈ 120 nm and DR
p
≈ 50 nm,Given Q= 10
12
cm
-2
= 0.01
nm
-2
.
a) Peak occurs at depth where x = R
p
≈ 120 nm.
b) From Eq,8.4,C
p
= 8 ￥ 10
-5
nm
-3
= 8 ￥ 10
16
cm
-3
.
c) At x = 300 nm,Eq,8.1 gives C
p
= 1.23 ￥ 10
-7
nm
-3
= 1.23 ￥ 10
14
cm
-3
.
d) If the dose at 0.3 microns is 10 times larger than this even though the
values of R
p
and C
p
are correct,then the wafer was not pure,but rather
doped with boron to 1.1 ￥ 10
15
cm
-3
(1.11 ￥ 10
15
+ 1.23 ￥ 10
14
= 10 ￥
3.155J/6.152J Homework set 3 solution,fall 2003
4
1.23 ￥ 10
14
),A later exposure to elevated temperature could not cause
the increased concentration at 0.3 microns without shifting the standard
deviation and peak value,Also,while the actual distribution may not be
exactly Gaussian,for boron,the skew is toward increased concentration
near the surface (due to enhanced backscatter of this light species) not
what is asked here.
Problem 6
R
p
= 0.3 μm (300 nm) demands an implant of boron at about 100 keV (Fig,8-3),At
this energy,DR
p
≈ 85 nm,so the dose giving a peak concentration of 10
17
cm
-3
is
easily calculated to be 2.13 ￥ 10
12
cm
-2
,For a background doping of 1015 cm-3,the
junction depth before diffusion is given by inverting Eq,8.1,x = 558 nm or 0.558
μm.
