Doping and diffusion I Motivation - DSpace@MIT Homedspace.mit.edu/.../0/lecture7a.pdf · 2...

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

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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

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Doping and diffusion

apparatus

Later…Ion implantation

x

Later…Ion implantation

x

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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…

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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

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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

¥¥

¥¥

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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

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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.

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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)

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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.

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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

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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

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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+

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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)

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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)

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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

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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

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Consequence:

Diffusion is enhancedat high dopant concentrationsfiSharper diffusion profile

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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

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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

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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

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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

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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)

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

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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

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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

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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[ ]

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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

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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)

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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

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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

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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)