Upload
others
View
1
Download
0
Embed Size (px)
Citation preview
6.155/ 3.155J 9/29/03 1
P + Poly
AlAl
P +
Poly
AlAl
Doping and diffusion I Motivation
Shorter channel but withsame source, drain depth =>drain field dominates gate field
=>”drain-induced barrier lowering” DIBL
drainsource
Faster MOSFET requiresshorter channel
Requires shallower source, drain
Shallower source, drain depthdemands better controlin doping & diffusion.CHANNEL ASPECT RATIO =>rs
6.155/ 3.155J 9/29/03 2
Needsharper 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 . solubilityNow: 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 Ia) Gas diffusion
F =U - T S.
If no chem’l interaction with air:
F = - T S
H2S
b) I = VR
J =sE = s -fz
ÊË
ˆ¯
Electric potential gradientfi charge flow
Initial state
Gas disperses, fills allpossible states randomly.
diffusion
Later timeorderC(z)
z
-Electrons drift down potential gradient:
here f is imposed from outside
frE = -
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 t( ) = D -
Cz
Ê
ËÁ
ˆ
¯˜ Fick I
t > 0
J = 0C ( z, t )
z
d)
zDz
J in = J z( ) Jout = J z + Dz( )z z+Dz dC z( )
dtDz = +J z( ) - J z + Dz( ) #
area tdCdt
=J z( ) - J z + Dz( )
DzDzÆ0æ Ææææ -
dJdz
These Eqs => time evolution of some initial conditions, boundary conditions
dC(z)dt
= -ddz
D -Cz
ÊË
ˆ¯
…or if D is constantdC z, t( )
dt=D—2C z,t( )
Fick II
Time dep. Schrödinger Eq.
+ih yt= -
h2
2m— 2y + Vy
Diffusingspeciesmust besoluble
6.155/ 3.155J 9/29/03 7
Atomistic picture of diffusionSee web site for movies:http://www.tf.uni-kiel.de/matwis/amat/def_en/index.html
Most important is vvacancy diffusion.
Ea
i f
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 =D0 exp -EkT( )
Atoms that bond with Si aresubstitutional impurities P, B, As, Al, Ga, Sb, Ge
¥¥
¥¥
6.155/ 3.155J 9/29/03 8
2 stepsfor diffusion: 1) create vacancy 2) achieve energy Ea
nv =Nv
N0= exp - 2.6
kTÈ
ÎÍ
˘
˚˙ nv = n0 exp - Ea
kTÊËÁ
ˆ¯˜
D =D0 exp - EVD
kTÈÎ
˘˚
D ~ a x v = a2n 0 exp - Ev + Ea
kTÈ
Î͢
˚̇cm2
sÊ
ËÁ ˆ
¯˜
Contains n0 Debye frequency
ª32
kBTh
= 9 ¥1012s-1 @1013s-1
e-DGkT Æ e
-EVF -TS
kT = eSk e
-EVF
kTÊ
ËÁ
ˆ
¯˜
Atomistic picture of diffusion
EVD
a
n0Vacancy diffusion
6.155/ 3.155J 9/29/03 9
Analytic Solution to Diffusion Equations,Fick II: C
t= D
2Cz 2
Steady state, dC/dt = 0
C(z) = a + bz
z0
C(z)
We assumed this to be the case in oxidation:
O2 diffusion through SiO2 , where
flux is same everywhere,J = -D Cz= -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
z0
t = 0
dC 0,t( )dz
= 0
C •,t( ) = 0
C z,0( ) = 0 z 0( )
C z, t( )-•
•
Ú dz =Q = const. (# /area)
6.155/ 3.155J 9/29/03 11
Analytic Solution to Diffusion Equations J = -D Cz
, Ct= D
2Cz 2
t = 0
dC 0,t( )dz
= 0
C •,t( ) = 0
C z,0( ) = 0 z 0( )
z0
I. “ Drive in” of small, fixed amount of dopant, solution is Gaussian
Predeposition isdelta function, d(z).
C z,t( ) = QpDt
exp - z 2
4DtÈ
Î͢
˚̇t > 0 Units
Width of Gaussian = = diffusion length a(a is large relative to width of predeposition)
12 a = 2 Dt
Dose, Q, amount of dopant in sample, is constant.
6.155/ 3.155J 9/29/03 12
erf
t = 0C
0
Solutions for other I.C./B.C. can be obtained by superposition:
II. Predeposition(Limitless source
of dopant)
x 2 =z 2
4DtÆ
(z - z0 )2
4Dt
Cimp z,t( ) = 2p
exp -x 2[ ]0
u=z
2 Dt
Ú dx erf u( ) = erf z2 DtÊË
ˆ¯
C ( , t )C 0,t( ) = C0
C z,0( ) = 0z
Const C0
t C •,t( ) = 0
0
6.155/ 3.155J 9/29/03 13
Other I.C./B.C. (cont.):
C z,t( ) = Csurferfc z2 DtÈÎÍ
˘˚̇, t > 0
a = diffusion length = 2 Dt
(D 10-15) ¥ (t = 103) fi a 0.2 mm
erfC u( ) =1-erf u( )
z
C
zerf z( )
0 1 2
10.995
C ( , t )C 0,t( ) = C0
C z,0( ) = 0z
Const C0
t Dose, Q,amount of dopant in sample,increases as t1/2.
Dose Q = C(z,t)dz = 2 Dtp0
•
Ú C0 =ap
C0
C0 limited by solid solubility
6.155/ 3.155J 9/29/03 15
Heavily doped layer can generate its own fielddue to displacement of mobile carriers from ionized dopants:
P-type SiA A
A
A A
z
CA
+ + + + + + + + +z
C CAMobile holeconcentration
Net charge- +EE enhances diffusion of A- to right, (also down concentration gradient).
Internal E fields alter Fick’s Law
Jmass = -D Cz+ Cm
rE D -
Cz+
CqrE
kTÈ
Î͢
˚̇diffusion A- drift
m =DqkT
Einstein relation from Brownian motion
- -
-
- -
h hh 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 activationenergy for chargedvacancy diffusion]
So vacancies can be chargedAs
e-
Si
electrons holes
For small dopant concentration, different diffusion processes are independent:
D0eEakT =Dfi D0 +D1- n
ni
+ D2- nni
Ê
ËÁ ˆ
¯˜
2
+ ... D+ ppi
Ê
ËÁ ˆ
¯˜+ D2+ p
pi
Ê
ËÁ ˆ
¯˜
2
+ ...
(these Do are NOTsame as singleactivation energyvalues)
6.155/ 3.155J 9/29/03 17
Doff =D0 +D- nni
Ê
ËÁÁ
ˆ
¯˜̃ +D2- n
ni
Ê
ËÁÁ
ˆ
¯˜̃
2
+ ...+ D+ pni
Ê
ËÁÁ
ˆ
¯˜̃+ D2+ p
ni
Ê
ËÁÁ
ˆ
¯˜̃
2
+ ...
“What is n?”
n is free electron concentration. For strong donor doping n > ni
So clearly, Deff = D0 + D-(n/ni) + … can be >> D = D0exp(-EVD/kT)
For intrinsic semiconductor or ND < ni, n = p = ni
Deff = D0 + D-(1) + …
DeffAs = D0 + D-(n/ni) +
2.67 x 10-16 + 1.17 x 10-15(n/ni)
ND = 1018: DeffAs = 1.43 x 10-15
ND = 1020: DeffAs = 16.6 x 10-15
See example p. 412 Plummer
ni = 7.14 x 1018
(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 fora) CP < ni b) CP = 4 ¥ 1019 cm-3
c) compare diffusion length b) with uncharged estimate
a) CP < ni = 1019
from F 1.16
Exercise
6.155/ 3.155J 9/29/03 19
ExerciseCalculate 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 DP0 = 3.85exp - 3.66
kTÈ
Î͢
˚̇=1.3¥10-14 (cm2s-1)
DP- = 4.4exp - 4
kTÈÎ
˘˚= 6.63¥10-16(cm2s-1)
D =DP0 + DP
- nni
Ê
ËÁ ˆ
¯˜ =1.37 ¥10-14 (cm2s-1)
Diffusion of P in SiD0 Ea D0
- Ea-
(cm2/s) (eV) (cm2/s) (eV)
3.85 3.66 4.4 4.0
Better than single-activation-energy:
DP = 4.7exp - 3.68kT
È
Î͢
˚̇=1.3¥10-14 (cm2s-1)
10
8nni 6
4
2
0 2 4 6 8 10ND/ni
n ª ND
2+
ND
2Ê
ËÁ
ˆ
¯˜
2
+ ni2
Plummer, Eq. 1.16
6.155/ 3.155J 9/29/03
ExerciseCalculate 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 ª ND
2+
ND
2Ê
ËÁ
ˆ
¯˜
2
+ ni2 =1019 + 4 ¥10 38 +1038 = 3.24 ¥1019cm-3
a = 2 Dt, a0 = 2 1.37 ¥10-14 ¥ 3600 = 0.14 mm1 hr fic)
aP = 2 1.51¥10-14 ¥ 3600 = 0.147 mm
D = DP0 + DP
- nni
Ê
ËÁ
ˆ
¯˜ =1.3¥10-14 + 6.63¥10-16 3.24
1Ê
ËÁ
ˆ
¯˜ =1.5 ¥10-14 (cm2s-1) vs 1.37
Plummer, Eq. 1.16
6.155/ 3.155J 9/29/03 21
Consequence:
Diffusion is enhancedat high dopant concentrationsfiSharper diffusion profile
6.155/ 3.155J 9/29/03 22
Effect of oxidation of Si
on diffusion
B and Pobserved 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 ondiffusion
by vacancy mechanism
6.155/ 3.155J 9/29/03 23
Effect of oxidation on diffusion in SiSi + Oxygen Æ SiOx
+ ÆSi interstitial
Because B and P can diffuse via vacancies as well as interstitial processtheir diffusion is enhances by oxidation.
But Sb is large and diffuses only by vacancies.Si interstitials created by oxidation,recombine and reduce concentration of vacanciessuppressing diffusion of Sb atoms.
B
SiSi
SiSi
Si
Si
Si
SiSi
Interstitial oxygen=> lattice expansion,induces stacking faultformationalong (111) planes
6.155/ 3.155J 9/29/03 25
Review: Doping and diffusion, small dose
d fnC0
0 z
If “ predeposition” is small dose,
followed by a higher T,
larger t ( larger )
“ drive-in” process.
a ~ Dt
C(z,t) = QpDt
exp - zaÊˈ¯
2È
Î͢
˚̇
C(0,t) = QpDt
decreases like t -1/2
1) I.C. C ( z, 0 ) = 0 z > 0
2) B.C. C ( , t ) = 0
3) Fixed dose Q =ap
c0
z
inc Dt
C0(0,0)
C0(0,t)
dC(z, t)dt
=d
dzD dC
dzÊË
ˆ¯ 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, thena ~ Dt
C(z,t) = C0erfc zaÈ΢˚
a = 2 Dt
Dose Q = C(z,t)dz = 2 Dtp0
•
Ú C0 =ap
C0
C0 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 ) = C0Might beSiO2+ dopant
inc Dt
C0
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, ND, with P).Want to know depth of p.n. junction ( NA = ND)
boron phosphorus
ND = C0erfczjct
aÈ
Î͢
˚̇junction
zjct = a erfc-1 ND
C0
È
Î͢
˚̇Æ a = 2 Dt
Limitless“predep”
SmallDt
CB z, t( ) = C0erfc z2 DtÈÎÍ
˘˚̇
¿
Boron
C0
z
ND (n-type Si)
Log C
6.155/ 3.155J 9/29/03 28
Thin predep
“Drive in”
C z,t( ) = QpDt
exp - z 2
a2È
Î͢
˚̇¿
Q
n ND
p
ND =QpDt
exp - zaÊˈ¯
2È
Î͢
˚̇¡ at jct
Jct z
Net carrierconcentration
z
exp + zaÊˈ¯
2È
Î͢
˚̇=
QND pDt
Ê
ËÁ
ˆ
¯˜
zjct = a ln QND pDtÈ
Î͢
˚̇¬
Junctions between different doped regions
6.155/ 3.155J 9/29/03 29
CB z, t( ) = CB 0,t( )erfc z2 DtÈÎÍ
˘˚̇+
zjct CP-
C z( )
z
C z( )
CB + CP-z
Log[C z( )]
z
CB CP-
2 Dt erfc -1 CP
CB 0( )ÏÌÓ
¸˝˛= zjct
6.155/ 3.155J 9/29/03 30
N-type Si, ND = 1016 cm-3 is doped with boron by a “predep”
from a const source with C0 (boron) = 1018 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 - 3.46kT
È
ÎÍ
˘
˚˙ 7.6 ¥ 10-16 cm2/s1273 K
a = 2 Dt = 3.31¥10-6cm = 0.033mm zjct = 0.033¥1.82 = 0.06mm
Exercise
ND = 1016 cm-3
C z,t( ) = C0erfc zaÈ΢˚,C0(boron)
1018/cm3
zjct = a erfc-1 ND
C0
È
Î͢
˚̇= 2 Dt erfc -1 10-2[ ]
6.155/ 3.155J 9/29/03 31
Q = C z,1hr( )Ú dz = aC0
p=1.87 ¥1012cm-2
7.57 ¥10-15 cm2
sD (boron)1373 K
Dt = 5.22 ¥10-6cm
zjct = 0.033¥1.82 = 0.06mm
From zjct(t) and zjct(0)you can calculate junction depth at different time:
z jct
1 hr t
zjct = 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?
zjct = a ln QND pDtÊ
ËÁ ˆ
¯˜ =1.8 ¥10-5 cm = 0.18mm
6.155/ 3.155J 9/29/03 32
Measuring diffusion profiles
R = rLA
(W), s =1r= nqmr =
RAL
(Wm)
But n, m are functions of position due to doping
s =qt
n z( )0
t
Ú m n( )dz
r =1s
=t
q nmdzÚDefine
sheet resistancert
Rs =RALt
= R WL
(Wsq
) an average measurement of n
At
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 probeI
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
f25 micronspacing
6.155/ 3.155J 9/29/03 34
Hall effect: electrical transport in magnetic field.
F = q v ¥ B
J = nq v
B I
V
++
++
___
__ e-
e-v
e-
e-
I
Again an average measurement
I
I
+
++ –
––
h+
h+v
V
B
Fq= EH =
Jnq
B E H = RH J ¥ B( )
RH =1
nqHall coefficient => charge signand concentrationRH is slope of V vs B data
B
VHLarge holeConcentration p
Small electronconcentration n
6.155/ 3.155J 9/29/03 35
C =eAd Depletion width
CapacitanceUseful for lightly doped regionsUse MOS structure,
gate & substrateare electrodes
SIMS
Depth profile
Ion source1 - 5 k eVSputterssurface
Secondary ionsto mass spectrometer
(Secondary ionmass spectroscopy)
RBS
Ions penetrate
Backscatter intensity µ (mass impurities ) 2
Backscatter energy
depends on depth and impurities
He+1 MeV
(Rutherford back scattering)