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Doping and diffusion I
Motivation
Faster MOSFET requires
shorter channel
P + Poly
Al
Requires shallower source, drain
Al
source
Al
P+
Poly
Al
drain
Shorter channel length; yes, but
same source & drain depth means
drain field dominates gate field
=>”drain-induced barrier lowering” DIBL
Shallower source, drain depth
demands better control
in doping & diffusion.
CHANNEL ASPECT RATIO =>ρs
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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 equilibrium solubility
c( x )
Now:
Ion implant (non-equilibrium), heat substrate to diffuse dopant
but ions damage target…requires anneal,
x
changes doping, c(z )
Soon: return to film or gas phase of dopant at surface
2) Drive-in process:
heat substrate after predeposition,
diffusion determines junction depth, sharpness
c ( x)
x
Need
sharper diffusion profiles:
c
depth
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1
Figure removed for copyright reasons.
Please see: Figures 4.26-4.28 in Ghandi, S. VLSI Fabrication Principles: Silicon and
Gallium Arsenide. 2nd ed. New York, NY: Wiley-Interscience, 1994. ISBN: 0471580058.
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c( z, t )
J=0
c) Mass (or heat) flow J,
due to concentration gradient
(
J #
d)
z
c
= D area t )
z J out = J ( z + z)
z
z
d C
dC(z)
= D dz z dt
…or if D is constant
t>0
Fick I
z+z
J in = J (z)
Diffusing
species
must be
soluble
In host
z
Inflow
out
dC(z)
#
z = + J (z) J (z + z)
dt
area t
dJ
dC J ( z) J ( z + z) z0
=
dz
z
dt
Fick II
dC(z, t)
2
= D C( z,t )
dt
Time dep. Schrödinger Eq.
+ih
h2 2
=
+ V
2m
t
These Eqs => time evolution of some initial conditions & boundary conditions
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Atomistic picture of diffusion in solids
S e e w e b s it e f o r m o v ie s :
http://www.tf.uni-kiel.de/matwis/amat/def_en/index.html
En
Most important is vacancy diffusion.
Ea
i
Initial and final states have same energy
f
Also possible is direct exchange
(
= b ro k en b o n d )
(
Higher energy barrier or break more bonds => lower value of D = D0 exp E
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September 21, 2005
kT
)
6
3
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Atomistic picture of vacancy diffusion
2 steps
for diffusion:
2) achieve energy Ea
1) create vacancy
" 2.6%
Nv
nv =
= exp$! '
# kT &
N0
#
&
! v = ! 0 exp% " E a kT (
$
'
# Ev + Ea &
2
D ~ a x v = a ! 0 exp%"
$
kT ('
! cm 2 $
#
&
" s %
Vacancy diffusion
ν0
" E $
D = D0 exp ! VD
# kT %
EVD
a
Contains ν0 ≈ Debye frequency
!
3 kBT
= 9 " 101 2s#1 $ 101 3s#1
2 h
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Solubility limits
Sample phase diagram
1.0
1.0
Liquid
T/Tm
From
phase
diagram:
T/Tm
Liq + Sol
B
Solubility dopant in Si
Solid
Si
100 at%
1022
1020
at% dopant
As
1021
Impurity content (cm-3)
10 at%
1021
P
1020
1019
1018
1017
1016
Impurity content (cm-3)
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4
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Solubility limits
Figure removed for copyright reasons.
Please see: Figure 2-4 in Campbell, S. The Science and Engineering of Microelectronic
Fabrication. 1st ed. New York, NY: Oxford University Press, 1996. ISBN: 0195105087.
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Analytic Solution to Diffusion Equations,
Fick II:
!C
! 2C
=D 2
!t
!z
There are many different solutions to this or any DE;
the correct solution is the one that satisfies the BC.
Steady state, dc/dt = 0
Implies either a) D = 0
c(z) may be curved
C(z)
C(z) = a + bz
or b) d2 c/dz2 = 0
c ( x) linear
0
z
∞
In oxidation we assumed steady state :
O2 diffusion through SiO2 , where
flux ,J = !D
"C
= !Db, is same everywhere.
"z
Not necessarily so in diffusion where
non-linear c ( z) can exist and be frozen in at D = 0
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5
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For solutions, boundary condition, consider classical experiment:
Diffusion couple: thin dopant layer on rod face,
press 2 identical pieces together, heat.
Then study diffusion profile in sections.
symmetry
BC
dC(0,t )
=0
dz
IC
C( z,0) = 0
( z 0)
C(,t ) = 0
C(z, t)dz = Q = const. (# /area)
BC
t=0
0
z
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11
c
c
2c
,
=D 2
z
t
z
I. “ Drive in” of fixed amount, Q, of dopant; solution is Gaussian
J = D
Analytic Solution to Diffusion Equations
z2 c ( z,t ) = const exp
4Dt c ( z,0) = 0 ( z 0)
dc (0,t )
=0
dz
c (,t ) = 0
P r e d e p o s it io n i s
d e lt a f u n c t io n , ( z ) .
t=0
0
z
D o s e , Q , a m o u n t o f d o p a n t i n s a m p le , is c o n s t a n t .
C(z, t)dz = Q = const. (# /area)
C( z,t ) =
z2 Q
exp 4Dt
Dt
U n it s
t>0
#/vol = Q/(length2)1/2
S o Q h a s u n it s # / a r e a
1
Width of Gaussian = a = 2 Dt = diffusion length a
2
(a is large relative to width of predeposition)
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6
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Solutions for other I.C./B.C. can be obtained by superposition:
II. Limitless source
BC
Const C0
C(0, t) = C0
of dopant
(e.g. growth in
presence of vapor)
C(C!,t
) =t 0)
(∞,
BC
t
IC
C( z,0) = 0
0
u=
2
Cimp (z, t) =
!
z
z
$ exp["# ]d# % erf (u) = erf &'2
2 Dt
2
0
!2 =
z (
Dt )
t=0
C
erf
z2
(z # z0 )2
"
4Dt
4Dt
0
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Other I.C./B.C. (cont.):
1
erf (z )
c
0.995
erfc( u) = 1! erf (u)
z
0
1
z
2
! z $
C( z,t ) = Csurf erfc#
,
" 2 Dt &%
a = diffusion length = 2 Dt
t>0
(D ≈ 10-15) × (t = 103) ⇒ a ≈ 0.2 µm
Dose ! Q =
#
$ C(z,t)dz = 2
0
C(0, t) = C0
Const c0
=
a
C
! 0
c0 limited by solid solubility
t
c(∞, t) = 0
C( z,0) = 0
z
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September 21, 2005
Dt
C
" 0
Dose, Q,
amount of dopant
in sample,
increases as t 1/2.
14
7
Figure removed for copyright reasons.
Please see: Figure 7-28 in Plummer, J., M. Deal, and P. Griffin. Silicon VLSI Technology:
Fundamentals, Practice, and Modeling. Upper Saddle River, NJ: Prentice Hall, 2000.
ISBN: 0130850373.
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Internal E fields alter Fick’s Law
Heavily doped layer can generate its own field
due to displacement of mobile carriers from ionized dopants:
A
−
A
p-type Si
A
−
A
A
h
−
C
CA
h+
h
h
−
Mobile hole
concentration
CA
++
z
−
h
(dopant
ions)
+++
-
+ + + +
Net charge
z
+
E
E enhances diffusion of A- to right, (also down concentration gradient).
J mass = !D
!"
!"
$ " C CqE '
"C
+ Cµ E # D &!
+
)
"z
kT (
% "z
µ=
diffusion A- drift
Dq
kT
Einstein relation
from
Brownian motion
(units)
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Neutral and charged impurities, dopants
If impurity is Gp. IV (e.g. Ge): uncharged, no e or h
Si
But if impurity = B, P As… it will be charged:
As
e-
So vacancies can be charged
For small dopant concentration, different diffusion processes are independent,
but generally:
D0e
Ea
kT
2
2
# n&
# p&
# p&
n
=D! D +D
+ D2" % ( + ... D+ % ( + D2+% ( + ...
n
$
n
'
$
p
'
$ pi '
i
i
electrons i
0
1"
Power series representation;
holes
Intrinsic Diffusivities and Activation Energies of Substitutional Diffusers in Silicon*
(these Do ar e NOT
same as single
activation ener gy
values)
P
Dio
Di+
Di-
Di2
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Al
Ga
0.037
1.385
0.374
3.46
3.41
3.39
Do
0.76
2480
28.5
Eo
3.46
4.20
3.92
Do
Eo
As
Sb
3.85
0.066
0.214
3.66
3.44
3.65
Do
4.44
22.9
13
Eo
4.0
4.1
4.0
Do
44.2
Eo
4.37
B
* D0 in cm2/s; E0 in eV.
higher orders in n describe
dopant-dopant interactions
[ Higher activation
energy for charged
vacancy diffusion;
prefactor is greater]
18
Figure by MIT OCW.
9
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“What is n?”
"n%
" n %2
" p%
" p %2
Doff = D0 + D!$$ '' + D2!$$ '' + ...+ D+$$ '' + D2+$$ '' + ...
# ni &
# ni &
# ni &
# ni &
n is local free electron concentration in host. n > ni always
n!
" N %2
ND
+ $ D ' + ni2
# 2 &
2
So clearly, Deff = D0 + D-(n/ni) + … can be >> D = D0exp(-EVD/kT)
(provided D1- etc not too small)
For intrinsic semiconductor or ND << ni, n = p = ni
Deff = D0 + D-(1) + …
See example Plummer, p. 412, As
ni = 7.14 x 1018
DeffAs
= D0
+
at 10000C
D-(n/ni)
+…
2.67 x 10-16 + 1.17 x 10-15(n/ni )
ND = 1018:
ND = 1020:
DeffAs = 1.43 x 10-15
DeffAs = 16.6 x 10-15
Caution: these
numbers not from
table on prior slide
(Single-activation-ener gy value: D = 1.5 x 10-15)
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Internal E field
Is related to
concentration dependent diffusion
−
h
−
−
C
−
h
h
+++
-
2
"n%
"n%
Doff = D0 + D!$ ' + D2!$ ' + ...
# ni &
# ni &
h
−
Mobile hole
concentration
CA
++
h+
+ + + +
Net charge
E
+
z
E enhances diffusion of A- to right,
Higher dopant concentration means
greater charge separation, Δq,
thus larger internal electric fields.
(also down concentration gradient).
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10
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Exercise
Calculate diffusivity of P in Si at 1000° C for
a) cP < ni
b) cP = 4 × 1019 cm-3
c) compare diffusion length b) with uncharged estimate
a) cP < ni = 1019
from Fig 1.16
Figure removed for copyright reasons.
Please see: Figure 1.16 in Plummer et al., 2000.
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Exercise
Calculate diffusivity of P in Si at 1000° C for
a) cP < ni
b) cP = 4 × 1019 cm-3
c) compare diffusion length b) with uncharged estimate
a) cP < ni = 1019
Diffusion of P in Si
D0
D0-
Ea
(cm2/s)
(eV)
(cm2/s)
3.85 3.66 4.4
10
8
n
ni
6
(eV)
4.0
"n%
D = DP0 + D!P $ ' = 1.37 (10!1 4(cm 2 s!1)
# ni &
Linear approximation
" N %2
ND
+ $ D ' + ni2
# 2 &
2
Plummer, Eq. 1.16
n!
Single-valued D
Better than single-activation-energy :
4
" 3.68 %
DP = 4.7exp$!
= 1.3 (10!14 (cm2 s!1 )
# kT &'
2
0
Ea-
" 3.66 %
DP0 = 3.85exp$!
= 1.3 (10!14 (cm2 s!1 )
# kT '&
" 4$
!
!1 6
2 !1
DP = 4.4exp !
= 6.63& 10 (cm s )
# kT %
2
4
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2005
D/ni
6
8
10
22
11
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Exercise
Calculate diffusivity of P in Si at 1000° C for
a) cP < ni
b) cP = 4 × 1019 cm-3
c) compare diffusion length b) with uncharged estimate
b)
cP = ND = 4× 1019
n!
" N %2
ND
+ $ D ' + ni2= 2 ! 1019 +
# 2 &
2
Plummer, Eq. 1.16
4 ! 10 38 + 10 38 = 4.24 ! 1019 cm "3
" n%
" 4.24 %
D = DP0 + DP! $ ' = 1.3 ( 10 !14 + 6.63 ( 10 !16 $
= 1.57 ( 10 !14 (cm 2 s !1 )
# 1 '&
# ni &
vs 1.37
c) a = 2 Dt, 1 hr ⇒a0 = 2 1.30 ! 10 "14 ! 3600 = 0.137 µ m
aP = 2 1.57 ! 10 "14 ! 3600 = 0.151 µ m
(This is a measure of the depth of doping.)
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Consequence:
Diffusion is enhanced
at high dopant
concentrations,
giving sharper
diffusion profile
Deff(T)
D1
Deff
4.15 eV
Graph removed for copyright reasons.
3.99 eV
D0
3.44 eV
103/T
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12
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Plummer
Fig. 7.36
Effect of oxidation of Si
on diffusion
B and P
observed to diffuse faster
when Si surface is oxidized,
Sb slower. Why?
So far
we have concentrated on
diffusion
by vacancy mechanism
Figure removed for copyright reasons.
Please see: Figure 7-36 in Plummer et al., 2000.
Different behavior of B and Sb under oxidation suggests
a different mechanism may dominate in these two dopants…
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Effect of surface oxidation on diffusion in Si
Si
+ Oxygen
+
→
SiOx
→
Si interstitial
Interstitial oxygen
=> lattice expansion,
stacking faults form
along (111) planes
Because B and P diffuse mainly by an interstitial process;
their diffusion is enhanced by oxidation.
Si
Si
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.
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Si
Si
Si
Sb
Si
Si
Si
Si
26
13
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Diffusion process is
different for different species.
Figure removed for copyright reasons.
Please see: Figure 7-15 in Plummer et al., 2000.
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Review: Doping and diffusion,
small dose
dC(z, t) d ! dC #
=
D
dt
dz " dz $
Plus I.C. and B.C.s
If “ predeposition” is small dose,
C0
followed by a higher T,
δ fn
longer t ( larger a ~ Dt )
“ drive-in” process.
0
z
1) I.C.
C ( z, 0 ) = 0
z>0
2) B.C. C ( ∞, t ) = 0, dC(0,t)/dz = 0
3) Fixed dose Q =
C0(0,0)
a
c
! 0
C0(0,t)
inc Dt
C(z,t) =
z
C(0, t) =
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September 21, 2005
Q
!Dt
' # z 2*
Q
exp)" % ,
!Dt
( $ a& +
decreases like t -1/2
28
14
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Review: Doping and diffusion,
large dose
Diffusion preceded by “ pre-deposition” to deliver a large amount of impurity.
If pre-dep is inexhaustible or equivalently, if a ~ Dt is small, then
1) I.C.
C0
C ( z, 0 ) = 0
2) B.C.
z>0
C ( ∞, t ) = 0
3) B.C. C ( 0, t ) = C0
Might be
SiO2+ dopant
inc
Dt
!z#
C(z,t) = C0 erfc
" a$
z
0
Dose ! Q =
#
a = 2 Dt
$ C(z,t)dz = 2
0
Dt
C
" 0
=
a
C
! 0
C0 limited by solid solubility
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Junctions between different doped regions
Diffuse B at high concentration, into n-type Si, (uniformly doped, ND, with P).
Want to know depth of p-n junction ( NA = ND)
Consider limitless dopant source (i.e. erfc)
C0
Small
Dt
Log |C|
boron
phosphorus
! z $
" 2 Dt %&
 CB (z, t) = C0erfc#
Boron
ND (n-type Si)
z
junction
!z $
N D = C0erfc# jct &
" a%
!N $
zjct = a erfc -1# D &
" C0 %
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September 21, 2005


a = 2 Dt
30
15
Figure removed for copyright reasons.
Please see: Figure 4.14 in Ghandi, 1994.
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Internal E fields alter Fick’s Law
D is not a constant, but depends on c(x)
Figure removed for copyright reasons.
Please see: Figure 7-26 in Plummer et al., 2000.
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Exercise
n-type Si, ND = 1016 cm-3 is doped with boron
from a const source with C0 (boron) = 1018 cm -3
C0(boron)
1018/cm3
!z#
C( z,t ) = C0 erfc ,
"a $
ND = 1016 cm-3
Question:
If exposed to c0 at 1000°C for 1 hr,
what is junction depth?
!N $
zjct = a erfc -1 D = 2 Dt erfc -1[10'2]
#" C &%
0
Let erfc-1 [10-2] = x , erfc[x] = 0.01 = 1-erf[x],
erf [x] = 0.99
From appendix, x = 1.82 = z/[2(Dt)1/2 ]
" 3.46%
D(boron) = 0.037exp$!
' 1273 K
# kT &
a = 2 Dt = 3.31!10"6cm = 0.033µm
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7.6 × 10-16 cm2/s
zjct = 0.033! 1.82 = 0.06µm
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17
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September 21, 2005, Wednesday
zjct = 0.033! 1.82 = 0.06µm
z jct
From zjct(t) and zjct(0) = 0,
you can calculate
junction depth
at different time:
zjct = 2 Dt erfc -1[10!2]
1 hr
t
Question: Now constant source is removed
and this dose,C(z, 1hr) , is “ driven in”farther for 1 hr at 1100°C.
aC
Now where is junction?
Q = ! C(z,1hr)dz = 0 = 1.87 # 101 2cm -2
"
D (boron)
1373 K
2
7.57 !10"1 5 cm s
Dt = 5.22 ! 10"6cm
"
Q
%'
zjct = a ln$
= 1.8 (10)5 cm = 0.18µm
# N D !Dt &
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Measuring diffusion profiles
Resistance
L
R = ! ("),
A
resistivity
!=
RA
("m)
L
t
L
Define
sheet resistance
!=
1
= nqµ
"
But n, µ are functions of position due to doping
A
I
conductivity
! =
q
t
t
" n(z)µ( n) dz
0
1
t
! =
=
"
q # nµdz
!
RA
W #
" Rs =
=R ( )
t
Lt
L sq
an average measurement of n
+
_
Spreading resistance probe:
(Developed at Bell Labs in ‘40s)
φ
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18
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Measuring diffusion profiles
4-point probe
I
Also square array
(Van der Pauw method)
I
Equipotentials
25 micron
spacing
V
V
These average n if done from surface.
These are most useful if done on beveled wafer:
sample
Polish off depth profile
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Hall effect: electrical transport in magnetic field.
_
I
I
B
_
_
_
_ e-
e-
v
e-
e-
B
+
+
+
+
+
+
h+
+
v h
+
–
–
–
I
I
V
F =qv B
J = nq v
J
F
= EH =
B
nq
q
Hall coefficient => charge sign
and concentration
RH is slope of V vs B data
RH =
V
E H = RH (J B)
1
nq
VH
Large hole
Concentration p
B
Again an average measurement
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Small electron
concentration n
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19
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September 21, 2005, Wednesday
Use MOS structure,
gate & substrate
are electrodes
Capacitance
(Secondary ion
mass spectroscopy)
!A
d
Useful for lightly doped regions
Depletion width
Secondary ions
to mass spectrometer
Ion source
SIMS
C=
1 - 5 k eV
Sputters
surface
Depth profile
RBS
(Rutherford back scattering)
He+
Backscatter energy
1 MeV
Ions penetrate
depends on depth and impurities
Backscatter intensity ∝ (mass impurities ) 2
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20