6.155/ 3.155J 9/29/03 1
P
+
Poly
AlAl
P
+
Poly
AlAl
Doping and diffusion I Motivation
Shorter channel but with
same source,drain depth =>
drain field dominates gate field
=>”drain-induced barrier lowering” DIBL
drainsource
Faster MOSFET requires
shorter channel
Requires shallower source,drain
Shallower source,drain depth
demands better control
in doping & diffusion.
CHANNEL ASPECT RATIO =>r
s
6.155/ 3.155J 9/29/03 2
Need
sharper diffusion profiles:
C
depth
How are shallow doped layers made?
1) Predeposition,controlled number of dopant species at surface
‘60s,film or gas phase of dopant at surface
Surface concentration is limited by equilib, solubility
Now,Ion implant (non-equilibrium),heat substrate to diffuse dopant
but ions damage target…requires anneal,changes doping,C (z )
Soon,film or gas phase of dopant at surface
C (x)
x
2) Drive-in process,heat substrate+predep,
diffusion determines junction depth,sharpness
C (x)
x
6.155/ 3.155J 9/29/03 4
Doping and diffusion
apparatus
Later…
Ion implantation
x
Later…
Ion implantation
x
6.155/ 3.155J 9/29/03 5
Doping,Diffusion I
a) Gas diffusion
F =U-TS.
If no chem’l interaction with air:
F = - T S
H
2
S
b)
I =
V
R
J =sE =s -
,f
,z
ê
ˉ
Electric potential gradient
fi charge flow
Initial state
Gas disperses,fills all
possible states randomly.
diffusion
Later time
order
C(z)
z
-
Electrons drift down potential gradient:
here f is imposed from outside
f
r
E =-
r
—f
what about solids…
6.155/ 3.155J 9/29/03 6
c) Mass (or heat) flow J,
due to concentration gradient
J
#
area u t
()
= D -
,C
,z
ê
á
ˉ
Fick I
t > 0
J =0
C ( z,t )
z
d)
z
Dz
J
in
= Jz()
J
out
= Jz+Dz()
z z+Dz
dC z()
dt
Dz =+Jz()- Jz+Dz()
#
area u t
dC
dt
=
Jz()- Jz+Dz()
Dz
Dz?0
-
dJ
dz
These Eqs => time evolution of some initial conditions,boundary conditions
dC(z)
dt
=-
d
dz
D -
,C
,z
ê
ˉ
…or if D is constant
dC z,t()
dt
= D—
2
Cz,t()
Fick II
Time dep,Schr?dinger Eq.
+ih
,y
,t
=-
h
2
2m
—
2
y + Vy
Diffusing
species
must be
soluble
6.155/ 3.155J 9/29/03 7
Atomistic picture of diffusion
See web site for movies:
http://www.tf.uni-kiel.de/matwis/amat/def_en/index.html
Most important is vacancy diffusion.
E
a
if
En
Initial and final states have same energy
Also possible is direct exchange (￥ = broken bond)
Higher energy barrier or break more bonds => lower value of D = D
0
exp -
E
kT
()
Atoms that bond with Si are
substitutional impurities P,B,As,Al,Ga,Sb,Ge
￥
￥
￥
￥
6.155/ 3.155J 9/29/03 8
2steps
for diffusion,1) create vacancy 2) achieve energy E
a
n
v
=
N
v
N
0
= exp -
2.6
kT
è
í
˙ n
v
=n
0
exp -
E
a
kT
ê
á
ˉ
D = D
0
exp -
E
VD
kT
è
D ~ a x v
= a
2
n
0
exp -
E
v
+ E
a
kT
è
í
˙
cm
2
s
ê
á
ˉ
Contains n
0
§ Debye frequency
a
3
2
k
B
T
h
= 9 ￥10
12
s
-1
@10
13
s
-1
e
-
DG
kT
e
-
E
VF
-TS
kT
= e
S
k
e
-
E
VF
kT
ê
á
ˉ
Atomistic picture of diffusion
E
VD
a
n
0
Vacancy diffusion
6.155/ 3.155J 9/29/03 9
Analytic Solution to Diffusion Equations,
Fick II:
,C
,t
= D
,
2
C
,z
2
Steady state,dC/dt = 0
C(z) = a + bz
z ?0
C(z)
We assumed this to be the case in oxidation:
O
2
diffusion through SiO
2
,where
flux is same everywhere
,J =-D
,C
,z
=-Db,
Conversely,
if C(z) is curved,
dC/dt ? 0.
?
6.155/ 3.155J 9/29/03 10
Diffusion couple,thin dopant layer on rod face,
press 2 identical pieces together,heat.
Then study diffusion profile in sections.
For other solutions,consider classical experiment:
symmetry
z ?0
t =0
dC 0,t()
dz
= 0
C?,t()= 0
Cz,0()= 0 z B0()
Cz,t()
-?
ú
dz = Q = const,(# /area)
6.155/ 3.155J 9/29/03 11
Analytic Solution to Diffusion Equations J =-D
,C
,z
,
,C
,t
= D
,
2
C
,z
2
t =0
dC 0,t()
dz
= 0
C?,t()= 0
Cz,0()= 0 z B0()
z ?0
I., Drive in” of small,fixed amount of dopant,solution is Gaussian
Predeposition is
delta function,d(z).
Cz,t()=
Q
pDt
exp -
z
2
4Dt
è
í
˙
t >0
Units
Width of Gaussian = = diffusion length a
(a is large relative to width of predeposition)
1
2
a = 2 Dt
Dose,Q,amount of dopant in sample,is constant.
6.155/ 3.155J 9/29/03 12
erf
t =0
C
0
Solutions for other I.C./B.C,can be obtained by superposition:
II,Predeposition
(Limitless source
of dopant)
x
2
=
z
2
4Dt
(z - z
0
)
2
4Dt
C
imp
z,t()=
2
p
exp -x
2
[]
0
u=
z
2 Dt
ú
dx > erf u()= erf
z
2 Dt
ê
ˉ
C ( ?,t )
C 0,t()= C
0
Cz,0()= 0
z
Const C
0
t
C?,t()= 0
0
6.155/ 3.155J 9/29/03 13
Other I.C./B.C,(cont.):
Cz,t()= C
surf
erfc
z
2Dt
è
í
˙
,t >0
a = diffusion length
= 2 Dt
(D § 10
-15
) ￥ (t =10
3
) fi a § 0.2 mm
erfCu()=1-erf u()
z
C
z
erf z()
0 1 2
1
0.995
C ( ?,t )
C 0,t()= C
0
Cz,0()= 0
z
Const C
0
t
Dose,Q,
amount of dopant in sample,
increases as t
1/2
.
Dose > Q = C(z,t)dz =
2 Dt
p
0
ú
C
0
=
a
p
C
0
C
0
limited by solid solubility
6.155/ 3.155J 9/29/03 15
Heavily doped layer can generate its own field
due to displacement of mobile carriers from ionized dopants:
P-type Si
A
A
A
A A
z
C
A
+
+
+
+
+
+
+
++
z
C
C
A
Mobile hole
concentration
Net charge
-
+
E
E enhances diffusion of A
-
to right,(also down concentration gradient).
Internal E fields alter Fick’s Law
J
mass
=-D
,C
,z
+ Cm
r
E > D -
,C
,z
+
Cq
r
E
kT
è
í
˙
diffusion
A
-
drift
m =
Dq
kT
Einstein relation
from
Brownian motion
- -
-
- -
h
h
h
h
h
+
6.155/ 3.155J 9/29/03 16
Neutral and charged impurities,dopants
If impurity is Gp,IV (e.g,Ge) uncharged,no e or h
But if imp,= B,P As… it will be charged:
[ Higher activation
energy for charged
vacancy diffusion]
So vacancies can be charged
As
e
-
Si
electrons holes
For small dopant concentration,different diffusion processes are independent:
D
0
e
Ea
kT
= D fi D
0
+ D
1-
n
n
i
+ D
2-
n
n
i
ê
á
ˉ
2
+,.,D
+
p
p
i
ê
á
ˉ
+ D
2+
p
p
i
ê
á
ˉ
2
+,..
(these D
o
are NOT
same as single
activation energy
values)
6.155/ 3.155J 9/29/03 17
D
off
= D
0
+ D
-
n
n
i
ê
á
á
ˉ
+ D
2-
n
n
i
ê
á
á
ˉ
2
+,..+ D
+
p
n
i
ê
á
á
ˉ
+ D
2+
p
n
i
ê
á
á
ˉ
2
+,..
“What is n?”
n is free electron concentration,For strong donor doping n > n
i
So clearly,D
eff
= D
0
+ D
-
(n/n
i
) + … can be >> D = D
0
exp(-E
VD
/kT)
For intrinsic semiconductor or N
D
< n
i
,n=p=n
i
D
eff
= D
0
+ D
-
(1) + …
D
eff
As
= D
0
+ D
-
(n/n
i
)+
2.67 x 10
-16
+ 1.17 x 10
-15
(n/n
i
)
N
D
= 10
18
,D
eff
As
= 1.43 x 10
-15
N
D
= 10
20
,D
eff
As
= 16.6 x 10
-15
See example p,412 Plummer
n
i
= 7.14 x 10
18
(Single-activation-energy value,D = 1.5 x 10
-15
)
6.155/ 3.155J 9/29/03 18
Calculate diffusivity of P in Si at 1000° C for
a) C
P
< n
i
b) C
P
= 4 ￥ 10
19
cm
-3
c) compare diffusion length b) with uncharged estimate
a) C
P
< n
i
= 10
19
from F 1.16
Exercise
6.155/ 3.155J 9/29/03 19
Exercise
Calculate diffusivity of P in Si at 1000° C for
a) C
P
< n
i
b) C
P
= 4 ￥ 10
19
cm
-3
c) compare diffusion length b) with uncharged estimate
a) C
P
< n
i
= 10
19
D
P
0
= 3.85exp -
3.66
kT
è
í
˙
=1.3￥10
-14
(cm
2
s
-1
)
D
P
-
= 4.4exp -
4
kT
è
= 6.63￥10
-16
(cm
2
s
-1
)
D = D
P
0
+ D
P
-
n
n
i
ê
á
ˉ
=1.37 ￥10
-14
(cm
2
s
-1
)
Diffusion of P in Si
D
0
E
a
D
0
-
E
a
-
(cm
2
/s) (eV) (cm
2
/s) (eV)
3.85 3.66 4.4 4.0
Better than single-activation-energy:
D
P
= 4.7exp -
3.68
kT
è
í
˙
=1.3￥10
-14
(cm
2
s
-1
)
10
8
n
n
i
6
4
2
0
246810
N
D
/n
i
n a
N
D
2
+
N
D
2
ê
á
ˉ
2
+ n
i
2
Plummer,Eq,1.16
6.155/ 3.155J 9/29/03
Exercise
Calculate diffusivity of P in Si at 1000° C for
a) C
P
< n
i
b) C
P
= 4 ￥ 10
19
cm
-3
c) compare diffusion length b) with uncharged estimate
b) C
P
=N
D
= 4￥ 10
19
n a
N
D
2
+
N
D
2
ê
á
ˉ
2
+ n
i
2
=10
19
+ 4 ￥10
38
+10
38
= 3.24 ￥10
19
cm
-3
a = 2 Dt,
a
0
= 2 1.37 ￥10
-14
￥ 3600 = 0.14 mm
1 hr fi
c)
a
P
= 2 1.51￥10
-14
￥ 3600 = 0.147 mm
D = D
P
0
+ D
P
-
n
n
i
ê
á
ˉ
=1.3￥10
-14
+ 6.63￥10
-16
3.24
1
ê
á
ˉ
=1.5 ￥10
-14
(cm
2
s
-1
)
vs 1.37
Plummer,Eq,1.16
6.155/ 3.155J 9/29/03 21
Consequence:
Diffusion is enhanced
at high dopant
concentrations
fiSharper
diffusion profile
6.155/ 3.155J 9/29/03 22
Effect of oxidation of Si
on diffusion
B and P
observed to diffuse faster
when Si surface is oxidized,
Sb slower?
Different behavior of B and Sb under oxidation suggests
a different mechanism may dominate in these two dopants
So far
we have concentrated on
diffusion
by vacancy mechanism
6.155/ 3.155J 9/29/03 23
Effect of oxidation on diffusion in Si
Si + Oxygen? SiO
x
+?
Si interstitial
Because B and P can diffuse via vacancies as well as interstitial process
their diffusion is enhances by oxidation.
But Sb is large and diffuses only by vacancies.
Si interstitials created by oxidation,
recombine and reduce concentration of vacancies
suppressing diffusion of Sb atoms.
B
Si
Si
SiSi
Si
Si
Si
SiSi
Interstitial oxygen
=> lattice expansion,
induces stacking fault
formation
along (111) planes
6.155/ 3.155J 9/29/03 25
Review,Doping and diffusion,small dose
d fn
C
0
0 z
If, predeposition” is small dose,
followed by a higher T,
larger t ( larger )
,drive-in” process.
a ~ Dt
C(z,t) =
Q
pDt
exp -
z
a
ê
ˉ
2
è
í
˙
C(0,t) =
Q
pDt
decreases like t
-1/2
1) I.C,C ( z,0 ) = 0 z > 0
2) B.C,C ( ?,t )=0
3) Fixed dose
Q =
a
p
c
0
z
inc
Dt
C
0
(0,0)
C
0
(0,t)
dC(z,t)
dt
=
d
dz
D
dC
dz
ê
ˉ
Plus I.C,and B.C.s
6.155/ 3.155J 9/29/03 26
Diffusion preceded by, pre-deposition” to deliver a large amount of impurity.
If pre-dep is inexhaustible or equivalently,if is small,then
a ~ Dt
C(z,t) = C
0
erfc
z
a
è
a = 2 Dt
Dose > Q = C(z,t)dz =
2 Dt
p
0
ú
C
0
=
a
p
C
0
C
0
limited by solid solubility
Review,Doping and diffusion,large dose
1) I.C,C ( z,0 ) = 0 z > 0
2) B.C,C ( ?,t )=0
3) B.C,C ( 0,t ) = C
0
Might be
SiO
2
+ dopant
inc
Dt
C
0
0 z
6.155/ 3.155J 9/29/03 27
Junctions between different doped regions
Diffuse B at high concentration,into n-type Si,(uniformly doped,N
D
,with P).
Want to know depth of p.n,junction ( N
A
= N
D
)
boron phosphorus
N
D
= C
0
erfc
z
jct
a
è
í
˙
|
junction
z
jct
= a erfc
-1
N
D
C
0
è
í
˙
a = 2 Dt
Limitless
“predep”
Small
Dt
C
B
z,t()= C
0
erfc
z
2 Dt
è
í
˙
Boron
C
0
z
N
D
(n-type Si)
Log C
6.155/ 3.155J 9/29/03 28
Thin predep
“Drive in”
Cz,t()=
Q
pDt
exp -
z
2
a
2
è
í
˙
Q
n
N
D
p
N
D
=
Q
pDt
exp -
z
a
ê
ˉ
2
è
í
˙
at jct
Jct
z
Net carrier
concentration
z
exp +
z
a
ê
ˉ
2
è
í
˙
=
Q
N
D
pDt
ê
á
ˉ
z
jct
= a ln
Q
N
D
pDt
è
í
˙
Junctions between different doped regions
6.155/ 3.155J 9/29/03 29
C
B
z,t()= C
B
0,t()erfc
z
2 Dt
è
í
˙
+
z
jct
C
P
-
Cz()
z
Cz()
C
B
+ C
P
-
z
Log[Cz()]
z
C
B
C
P
-
2 Dt erfc
-1
C
P
C
B
0
()
ì
ó
= z
jct
6.155/ 3.155J 9/29/03 30
N-type Si,N
D
=10
16
cm
-3
is doped with boron by a,predep”
from a const source with C
0
(boron) = 10
18
cm
-3
Question:
If predep is done at 1000°C for 1 hr,
what is junction depth?
Let erfc
-1
[10
-2
] = x,erfc[x] = 0.01 = 1-erf[x],erf [x] = 0.99.
From appendix,x = 1.82
D boron()= 0.037exp -
3.46
kT
è
í
˙
7.6 ￥ 10
-16
cm
2
/s
1273 K
a = 2 Dt = 3.31￥10
-6
cm = 0.033mm
z
jct
= 0.033￥1.82 = 0.06mm
Exercise
N
D
= 10
16
cm
-3
Cz,t()= C
0
erfc
z
a
è
,
C
0
(boron)
10
18
/cm
3
z
jct
= a erfc
-1
N
D
C
0
è
í
˙
= 2 Dt erfc
-1
10
-2
[]
6.155/ 3.155J 9/29/03 31
Q = Cz,1hr()
ú
dz =
aC
0
p
=1.87 ￥10
12
cm
-2
7.57 ￥10
-15
cm
2
s
D (boron)
1373 K
Dt = 5.22 ￥10
-6
cm
z
jct
= 0.033￥1.82 = 0.06mm
From z
jct
(t) and z
jct
(0)
you can calculate
junction depth
at different time:
z
jct
1 hr t
z
jct
= 2 Dt erfc
-1
10
-2
[]
Question,Now surface film (const,source) is removed
and this dose,C(z,1hr),is, driven in” for 1 hr at 1100°C.
Now where is junction?
z
jct
= a ln
Q
N
D
pDt
ê
á
ˉ
=1.8 ￥10
-5
cm = 0.18mm
6.155/ 3.155J 9/29/03 32
Measuring diffusion profiles
R =r
L
A
(W),
s =
1
r
= nqm
r=
RA
L
(Wm)
But n,m are functions of position due to doping
s =
q
t
nz()
0
t
ú
m n()dz
r =
1
s
=
t
qnmdz
ú
Define
sheet resistance
r
t
> R
s
=
RA
Lt
= R
W
L
(
W
sq
)
an average measurement of n
A
t
I
L
Resistance resistivity conductivity
+
_
f
Spreading resistance probe:
(Developed at Bell Labs in ‘40s)
6.155/ 3.155J 9/29/03 33
4-point probe
I
V
Equipotentials
These fi average n if done from surface.
These are most useful if done on beveled wafer:
sample
Polish off fi depth profile
Measuring diffusion profiles
Also square array
(Van der Pauw method)
I
V
f
25 micron
spacing
6.155/ 3.155J 9/29/03 34
Hall effect,electrical transport in magnetic field.
F = qv ￥ B
J = nq v
B
I
V
+
+
+
+
_
_
_
_
_ e
-
e
-
v
e
-
e
-
I
Again an average measurement
I
I
+
+
+
–
–
–
h
+
h
+
v
V
B
F
q
= E
H
=
J
nq
B
E
H
= R
H
J ￥ B()
R
H
=
1
nq
Hall coefficient => charge sign
and concentration
R
H
is slope of V vs B data
B
V
H
Large hole
Concentration p
Small electron
concentration n
6.155/ 3.155J 9/29/03 35
C =
eA
d
Depletion width
Capacitance
Useful for lightly doped regions
Use MOS structure,
gate & substrate
are electrodes
SIMS
Depth profile
Ion source
1 - 5 k eV
Sputters
surface
Secondary ions
to mass spectrometer
(Secondary ion
mass spectroscopy)
RBS
Ions penetrate
Backscatter intensity μ (mass impurities )
2
Backscatter energy
depends on depth and impurities
He
+
1MeV
(Rutherford back scattering)
P
+
Poly
AlAl
P
+
Poly
AlAl
Doping and diffusion I Motivation
Shorter channel but with
same source,drain depth =>
drain field dominates gate field
=>”drain-induced barrier lowering” DIBL
drainsource
Faster MOSFET requires
shorter channel
Requires shallower source,drain
Shallower source,drain depth
demands better control
in doping & diffusion.
CHANNEL ASPECT RATIO =>r
s
6.155/ 3.155J 9/29/03 2
Need
sharper diffusion profiles:
C
depth
How are shallow doped layers made?
1) Predeposition,controlled number of dopant species at surface
‘60s,film or gas phase of dopant at surface
Surface concentration is limited by equilib, solubility
Now,Ion implant (non-equilibrium),heat substrate to diffuse dopant
but ions damage target…requires anneal,changes doping,C (z )
Soon,film or gas phase of dopant at surface
C (x)
x
2) Drive-in process,heat substrate+predep,
diffusion determines junction depth,sharpness
C (x)
x
6.155/ 3.155J 9/29/03 4
Doping and diffusion
apparatus
Later…
Ion implantation
x
Later…
Ion implantation
x
6.155/ 3.155J 9/29/03 5
Doping,Diffusion I
a) Gas diffusion
F =U-TS.
If no chem’l interaction with air:
F = - T S
H
2
S
b)
I =
V
R
J =sE =s -
,f
,z
ê
ˉ
Electric potential gradient
fi charge flow
Initial state
Gas disperses,fills all
possible states randomly.
diffusion
Later time
order
C(z)
z
-
Electrons drift down potential gradient:
here f is imposed from outside
f
r
E =-
r
—f
what about solids…
6.155/ 3.155J 9/29/03 6
c) Mass (or heat) flow J,
due to concentration gradient
J
#
area u t
()
= D -
,C
,z
ê
á
ˉ
Fick I
t > 0
J =0
C ( z,t )
z
d)
z
Dz
J
in
= Jz()
J
out
= Jz+Dz()
z z+Dz
dC z()
dt
Dz =+Jz()- Jz+Dz()
#
area u t
dC
dt
=
Jz()- Jz+Dz()
Dz
Dz?0
-
dJ
dz
These Eqs => time evolution of some initial conditions,boundary conditions
dC(z)
dt
=-
d
dz
D -
,C
,z
ê
ˉ
…or if D is constant
dC z,t()
dt
= D—
2
Cz,t()
Fick II
Time dep,Schr?dinger Eq.
+ih
,y
,t
=-
h
2
2m
—
2
y + Vy
Diffusing
species
must be
soluble
6.155/ 3.155J 9/29/03 7
Atomistic picture of diffusion
See web site for movies:
http://www.tf.uni-kiel.de/matwis/amat/def_en/index.html
Most important is vacancy diffusion.
E
a
if
En
Initial and final states have same energy
Also possible is direct exchange (￥ = broken bond)
Higher energy barrier or break more bonds => lower value of D = D
0
exp -
E
kT
()
Atoms that bond with Si are
substitutional impurities P,B,As,Al,Ga,Sb,Ge
￥
￥
￥
￥
6.155/ 3.155J 9/29/03 8
2steps
for diffusion,1) create vacancy 2) achieve energy E
a
n
v
=
N
v
N
0
= exp -
2.6
kT
è
í
˙ n
v
=n
0
exp -
E
a
kT
ê
á
ˉ
D = D
0
exp -
E
VD
kT
è
D ~ a x v
= a
2
n
0
exp -
E
v
+ E
a
kT
è
í
˙
cm
2
s
ê
á
ˉ
Contains n
0
§ Debye frequency
a
3
2
k
B
T
h
= 9 ￥10
12
s
-1
@10
13
s
-1
e
-
DG
kT
e
-
E
VF
-TS
kT
= e
S
k
e
-
E
VF
kT
ê
á
ˉ
Atomistic picture of diffusion
E
VD
a
n
0
Vacancy diffusion
6.155/ 3.155J 9/29/03 9
Analytic Solution to Diffusion Equations,
Fick II:
,C
,t
= D
,
2
C
,z
2
Steady state,dC/dt = 0
C(z) = a + bz
z ?0
C(z)
We assumed this to be the case in oxidation:
O
2
diffusion through SiO
2
,where
flux is same everywhere
,J =-D
,C
,z
=-Db,
Conversely,
if C(z) is curved,
dC/dt ? 0.
?
6.155/ 3.155J 9/29/03 10
Diffusion couple,thin dopant layer on rod face,
press 2 identical pieces together,heat.
Then study diffusion profile in sections.
For other solutions,consider classical experiment:
symmetry
z ?0
t =0
dC 0,t()
dz
= 0
C?,t()= 0
Cz,0()= 0 z B0()
Cz,t()
-?
ú
dz = Q = const,(# /area)
6.155/ 3.155J 9/29/03 11
Analytic Solution to Diffusion Equations J =-D
,C
,z
,
,C
,t
= D
,
2
C
,z
2
t =0
dC 0,t()
dz
= 0
C?,t()= 0
Cz,0()= 0 z B0()
z ?0
I., Drive in” of small,fixed amount of dopant,solution is Gaussian
Predeposition is
delta function,d(z).
Cz,t()=
Q
pDt
exp -
z
2
4Dt
è
í
˙
t >0
Units
Width of Gaussian = = diffusion length a
(a is large relative to width of predeposition)
1
2
a = 2 Dt
Dose,Q,amount of dopant in sample,is constant.
6.155/ 3.155J 9/29/03 12
erf
t =0
C
0
Solutions for other I.C./B.C,can be obtained by superposition:
II,Predeposition
(Limitless source
of dopant)
x
2
=
z
2
4Dt
(z - z
0
)
2
4Dt
C
imp
z,t()=
2
p
exp -x
2
[]
0
u=
z
2 Dt
ú
dx > erf u()= erf
z
2 Dt
ê
ˉ
C ( ?,t )
C 0,t()= C
0
Cz,0()= 0
z
Const C
0
t
C?,t()= 0
0
6.155/ 3.155J 9/29/03 13
Other I.C./B.C,(cont.):
Cz,t()= C
surf
erfc
z
2Dt
è
í
˙
,t >0
a = diffusion length
= 2 Dt
(D § 10
-15
) ￥ (t =10
3
) fi a § 0.2 mm
erfCu()=1-erf u()
z
C
z
erf z()
0 1 2
1
0.995
C ( ?,t )
C 0,t()= C
0
Cz,0()= 0
z
Const C
0
t
Dose,Q,
amount of dopant in sample,
increases as t
1/2
.
Dose > Q = C(z,t)dz =
2 Dt
p
0
ú
C
0
=
a
p
C
0
C
0
limited by solid solubility
6.155/ 3.155J 9/29/03 15
Heavily doped layer can generate its own field
due to displacement of mobile carriers from ionized dopants:
P-type Si
A
A
A
A A
z
C
A
+
+
+
+
+
+
+
++
z
C
C
A
Mobile hole
concentration
Net charge
-
+
E
E enhances diffusion of A
-
to right,(also down concentration gradient).
Internal E fields alter Fick’s Law
J
mass
=-D
,C
,z
+ Cm
r
E > D -
,C
,z
+
Cq
r
E
kT
è
í
˙
diffusion
A
-
drift
m =
Dq
kT
Einstein relation
from
Brownian motion
- -
-
- -
h
h
h
h
h
+
6.155/ 3.155J 9/29/03 16
Neutral and charged impurities,dopants
If impurity is Gp,IV (e.g,Ge) uncharged,no e or h
But if imp,= B,P As… it will be charged:
[ Higher activation
energy for charged
vacancy diffusion]
So vacancies can be charged
As
e
-
Si
electrons holes
For small dopant concentration,different diffusion processes are independent:
D
0
e
Ea
kT
= D fi D
0
+ D
1-
n
n
i
+ D
2-
n
n
i
ê
á
ˉ
2
+,.,D
+
p
p
i
ê
á
ˉ
+ D
2+
p
p
i
ê
á
ˉ
2
+,..
(these D
o
are NOT
same as single
activation energy
values)
6.155/ 3.155J 9/29/03 17
D
off
= D
0
+ D
-
n
n
i
ê
á
á
ˉ
+ D
2-
n
n
i
ê
á
á
ˉ
2
+,..+ D
+
p
n
i
ê
á
á
ˉ
+ D
2+
p
n
i
ê
á
á
ˉ
2
+,..
“What is n?”
n is free electron concentration,For strong donor doping n > n
i
So clearly,D
eff
= D
0
+ D
-
(n/n
i
) + … can be >> D = D
0
exp(-E
VD
/kT)
For intrinsic semiconductor or N
D
< n
i
,n=p=n
i
D
eff
= D
0
+ D
-
(1) + …
D
eff
As
= D
0
+ D
-
(n/n
i
)+
2.67 x 10
-16
+ 1.17 x 10
-15
(n/n
i
)
N
D
= 10
18
,D
eff
As
= 1.43 x 10
-15
N
D
= 10
20
,D
eff
As
= 16.6 x 10
-15
See example p,412 Plummer
n
i
= 7.14 x 10
18
(Single-activation-energy value,D = 1.5 x 10
-15
)
6.155/ 3.155J 9/29/03 18
Calculate diffusivity of P in Si at 1000° C for
a) C
P
< n
i
b) C
P
= 4 ￥ 10
19
cm
-3
c) compare diffusion length b) with uncharged estimate
a) C
P
< n
i
= 10
19
from F 1.16
Exercise
6.155/ 3.155J 9/29/03 19
Exercise
Calculate diffusivity of P in Si at 1000° C for
a) C
P
< n
i
b) C
P
= 4 ￥ 10
19
cm
-3
c) compare diffusion length b) with uncharged estimate
a) C
P
< n
i
= 10
19
D
P
0
= 3.85exp -
3.66
kT
è
í
˙
=1.3￥10
-14
(cm
2
s
-1
)
D
P
-
= 4.4exp -
4
kT
è
= 6.63￥10
-16
(cm
2
s
-1
)
D = D
P
0
+ D
P
-
n
n
i
ê
á
ˉ
=1.37 ￥10
-14
(cm
2
s
-1
)
Diffusion of P in Si
D
0
E
a
D
0
-
E
a
-
(cm
2
/s) (eV) (cm
2
/s) (eV)
3.85 3.66 4.4 4.0
Better than single-activation-energy:
D
P
= 4.7exp -
3.68
kT
è
í
˙
=1.3￥10
-14
(cm
2
s
-1
)
10
8
n
n
i
6
4
2
0
246810
N
D
/n
i
n a
N
D
2
+
N
D
2
ê
á
ˉ
2
+ n
i
2
Plummer,Eq,1.16
6.155/ 3.155J 9/29/03
Exercise
Calculate diffusivity of P in Si at 1000° C for
a) C
P
< n
i
b) C
P
= 4 ￥ 10
19
cm
-3
c) compare diffusion length b) with uncharged estimate
b) C
P
=N
D
= 4￥ 10
19
n a
N
D
2
+
N
D
2
ê
á
ˉ
2
+ n
i
2
=10
19
+ 4 ￥10
38
+10
38
= 3.24 ￥10
19
cm
-3
a = 2 Dt,
a
0
= 2 1.37 ￥10
-14
￥ 3600 = 0.14 mm
1 hr fi
c)
a
P
= 2 1.51￥10
-14
￥ 3600 = 0.147 mm
D = D
P
0
+ D
P
-
n
n
i
ê
á
ˉ
=1.3￥10
-14
+ 6.63￥10
-16
3.24
1
ê
á
ˉ
=1.5 ￥10
-14
(cm
2
s
-1
)
vs 1.37
Plummer,Eq,1.16
6.155/ 3.155J 9/29/03 21
Consequence:
Diffusion is enhanced
at high dopant
concentrations
fiSharper
diffusion profile
6.155/ 3.155J 9/29/03 22
Effect of oxidation of Si
on diffusion
B and P
observed to diffuse faster
when Si surface is oxidized,
Sb slower?
Different behavior of B and Sb under oxidation suggests
a different mechanism may dominate in these two dopants
So far
we have concentrated on
diffusion
by vacancy mechanism
6.155/ 3.155J 9/29/03 23
Effect of oxidation on diffusion in Si
Si + Oxygen? SiO
x
+?
Si interstitial
Because B and P can diffuse via vacancies as well as interstitial process
their diffusion is enhances by oxidation.
But Sb is large and diffuses only by vacancies.
Si interstitials created by oxidation,
recombine and reduce concentration of vacancies
suppressing diffusion of Sb atoms.
B
Si
Si
SiSi
Si
Si
Si
SiSi
Interstitial oxygen
=> lattice expansion,
induces stacking fault
formation
along (111) planes
6.155/ 3.155J 9/29/03 25
Review,Doping and diffusion,small dose
d fn
C
0
0 z
If, predeposition” is small dose,
followed by a higher T,
larger t ( larger )
,drive-in” process.
a ~ Dt
C(z,t) =
Q
pDt
exp -
z
a
ê
ˉ
2
è
í
˙
C(0,t) =
Q
pDt
decreases like t
-1/2
1) I.C,C ( z,0 ) = 0 z > 0
2) B.C,C ( ?,t )=0
3) Fixed dose
Q =
a
p
c
0
z
inc
Dt
C
0
(0,0)
C
0
(0,t)
dC(z,t)
dt
=
d
dz
D
dC
dz
ê
ˉ
Plus I.C,and B.C.s
6.155/ 3.155J 9/29/03 26
Diffusion preceded by, pre-deposition” to deliver a large amount of impurity.
If pre-dep is inexhaustible or equivalently,if is small,then
a ~ Dt
C(z,t) = C
0
erfc
z
a
è
a = 2 Dt
Dose > Q = C(z,t)dz =
2 Dt
p
0
ú
C
0
=
a
p
C
0
C
0
limited by solid solubility
Review,Doping and diffusion,large dose
1) I.C,C ( z,0 ) = 0 z > 0
2) B.C,C ( ?,t )=0
3) B.C,C ( 0,t ) = C
0
Might be
SiO
2
+ dopant
inc
Dt
C
0
0 z
6.155/ 3.155J 9/29/03 27
Junctions between different doped regions
Diffuse B at high concentration,into n-type Si,(uniformly doped,N
D
,with P).
Want to know depth of p.n,junction ( N
A
= N
D
)
boron phosphorus
N
D
= C
0
erfc
z
jct
a
è
í
˙
|
junction
z
jct
= a erfc
-1
N
D
C
0
è
í
˙
a = 2 Dt
Limitless
“predep”
Small
Dt
C
B
z,t()= C
0
erfc
z
2 Dt
è
í
˙
Boron
C
0
z
N
D
(n-type Si)
Log C
6.155/ 3.155J 9/29/03 28
Thin predep
“Drive in”
Cz,t()=
Q
pDt
exp -
z
2
a
2
è
í
˙
Q
n
N
D
p
N
D
=
Q
pDt
exp -
z
a
ê
ˉ
2
è
í
˙
at jct
Jct
z
Net carrier
concentration
z
exp +
z
a
ê
ˉ
2
è
í
˙
=
Q
N
D
pDt
ê
á
ˉ
z
jct
= a ln
Q
N
D
pDt
è
í
˙
Junctions between different doped regions
6.155/ 3.155J 9/29/03 29
C
B
z,t()= C
B
0,t()erfc
z
2 Dt
è
í
˙
+
z
jct
C
P
-
Cz()
z
Cz()
C
B
+ C
P
-
z
Log[Cz()]
z
C
B
C
P
-
2 Dt erfc
-1
C
P
C
B
0
()
ì
ó
= z
jct
6.155/ 3.155J 9/29/03 30
N-type Si,N
D
=10
16
cm
-3
is doped with boron by a,predep”
from a const source with C
0
(boron) = 10
18
cm
-3
Question:
If predep is done at 1000°C for 1 hr,
what is junction depth?
Let erfc
-1
[10
-2
] = x,erfc[x] = 0.01 = 1-erf[x],erf [x] = 0.99.
From appendix,x = 1.82
D boron()= 0.037exp -
3.46
kT
è
í
˙
7.6 ￥ 10
-16
cm
2
/s
1273 K
a = 2 Dt = 3.31￥10
-6
cm = 0.033mm
z
jct
= 0.033￥1.82 = 0.06mm
Exercise
N
D
= 10
16
cm
-3
Cz,t()= C
0
erfc
z
a
è
,
C
0
(boron)
10
18
/cm
3
z
jct
= a erfc
-1
N
D
C
0
è
í
˙
= 2 Dt erfc
-1
10
-2
[]
6.155/ 3.155J 9/29/03 31
Q = Cz,1hr()
ú
dz =
aC
0
p
=1.87 ￥10
12
cm
-2
7.57 ￥10
-15
cm
2
s
D (boron)
1373 K
Dt = 5.22 ￥10
-6
cm
z
jct
= 0.033￥1.82 = 0.06mm
From z
jct
(t) and z
jct
(0)
you can calculate
junction depth
at different time:
z
jct
1 hr t
z
jct
= 2 Dt erfc
-1
10
-2
[]
Question,Now surface film (const,source) is removed
and this dose,C(z,1hr),is, driven in” for 1 hr at 1100°C.
Now where is junction?
z
jct
= a ln
Q
N
D
pDt
ê
á
ˉ
=1.8 ￥10
-5
cm = 0.18mm
6.155/ 3.155J 9/29/03 32
Measuring diffusion profiles
R =r
L
A
(W),
s =
1
r
= nqm
r=
RA
L
(Wm)
But n,m are functions of position due to doping
s =
q
t
nz()
0
t
ú
m n()dz
r =
1
s
=
t
qnmdz
ú
Define
sheet resistance
r
t
> R
s
=
RA
Lt
= R
W
L
(
W
sq
)
an average measurement of n
A
t
I
L
Resistance resistivity conductivity
+
_
f
Spreading resistance probe:
(Developed at Bell Labs in ‘40s)
6.155/ 3.155J 9/29/03 33
4-point probe
I
V
Equipotentials
These fi average n if done from surface.
These are most useful if done on beveled wafer:
sample
Polish off fi depth profile
Measuring diffusion profiles
Also square array
(Van der Pauw method)
I
V
f
25 micron
spacing
6.155/ 3.155J 9/29/03 34
Hall effect,electrical transport in magnetic field.
F = qv ￥ B
J = nq v
B
I
V
+
+
+
+
_
_
_
_
_ e
-
e
-
v
e
-
e
-
I
Again an average measurement
I
I
+
+
+
–
–
–
h
+
h
+
v
V
B
F
q
= E
H
=
J
nq
B
E
H
= R
H
J ￥ B()
R
H
=
1
nq
Hall coefficient => charge sign
and concentration
R
H
is slope of V vs B data
B
V
H
Large hole
Concentration p
Small electron
concentration n
6.155/ 3.155J 9/29/03 35
C =
eA
d
Depletion width
Capacitance
Useful for lightly doped regions
Use MOS structure,
gate & substrate
are electrodes
SIMS
Depth profile
Ion source
1 - 5 k eV
Sputters
surface
Secondary ions
to mass spectrometer
(Secondary ion
mass spectroscopy)
RBS
Ions penetrate
Backscatter intensity μ (mass impurities )
2
Backscatter energy
depends on depth and impurities
He
+
1MeV
(Rutherford back scattering)
