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PHYSICAL VAPOR DEPOSITION (PVD)PVD II: Evaporation

We saw CVD Gas phase reactants: pg ≈ 1 mTorr to 1 atm.

Good step coverage, T > 350 K

We saw sputtering Noble (+ reactive gas) p ≈ 10 mTorr; ionized particles

Industrial process, high rate, reasonable step coverage

Extensively used in electrical, optical, magnetic devices.

Now see evaporation: Source material heated, peq.vap. =~ 10-3 Torr, pg < 10-6 Torr

Generally no chemical reaction (except in “reactive depos’n),

λ = 10’s of meters, Knudsen number NK >> 1

Poor step coverage, alloy fractionation: ∆ pvapor

Historical (optical, electrical)Campbell, Ch. 12 is more extensive than Plummer on evaporation

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Standard vacuum chambers

Σpi ≈ 10-6 Torr (1.3 × 10-4 N / m2)

Mostly H2O, hydrocarbons, N2 He by residual gas analysis (RGA = mass spec.)

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p < 10-8 Torr demands:Stainless steel chamberBakeable to 150oCTurbo, ion, cryo pumps

Ultra-high vacuum chambers

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-6 -4 -2 0 2 4Log[P (N/m2)]

25

23Log[n (#/m3)]

21

19

17

15

λ = 10 0.01 cmp = 10-10 10-8 10-6 10-4 10-2 100 Torr

1 Atm=0.1 MPa760 mm≈14 lb/in2

Generally λ/L < 1Films less pureEpitaxy is rare

Evaporation

Ballistic, molecular flow, λ/L >> 1High purity films

Epitaxy

Sputtering

CVD

Knudson number ≈ 1

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Atomic flux on surface due to residual gas

J atomsarea ⋅ t

=

nv x2

=p

2kBT2kBTπm

=p

2πmkBT= J

Given 10-6 Torr of water vapor @ room temp, find flux

p =10−6 Torr ×1atm760 T

×105 Pa

atm, kBT RT( )= 0.025eV = 4 ×10−21J

p =1.3 ×10−4 Nm2 mH 2O =

18NA

= 3×10−26kg

What is atomic density in 1 monolayer (ML) of Si?

N = 5 x 1022 cm-3 => 1.3 x 1015 cm-2.So at 10-6 Torr, 1 ML of residual gas hits surface every 3 seconds!Epitaxy requires slow deposition, surface mobility,So you must keep pressure low to maintain pure film

J = 4.8 ×1014 atoms/moleculescm2sec

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Now add evaporation source

Equilibrium vapor pressure:

∆H = heat of vaporization

pv = p0 exp −∆HkBT

Strong T dependence

e-

+

_

e-beam

B field

Resistive heater

I

RF-induction heater

Workfunction

e-

Heat of vaporization

solid

V(x)

free

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Vapor pressure of elements employed in semiconductor materials. Dots correspond to melting points

Rely on tables, attached: pvapor > > pvac,

Elemental metals easy to evaporate, but…

alloys

compoundsDifferential pvapor

so use 2 crucibles or deposit multilayersand diffuse

Oxides, nitrides deposit in oxygen

(or other) partial p

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(note W, Mo can be used as crucibles.)

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The source also ⇒ flux

J =pvap

2πmsourcekBTsource

For Al at 1000 K, pvap = 10 -7 Torr (from figure)

m = species to be evaporated, = 27 amu for Al

JAl ≈ 2 x 1013 Al/cm2-s just above crucible

Mass flow out of crucible ~ J Ac m ( mass / t)

We expressed flux of residual gas:

J ≈ 5 ×1014 moleculescm2s( )Chamber p = 10-6 TorrJ =

p2πmkBT

Net flow out of crucible ~ J Ac (# / t)Ac

Therefore heat Al to T > 800 C

but that’s not all…

Note: 3 different temperatures: Tsource ≈ Tevaporant >> Tsubstrate > Tchamber = Tresid gas ≈ RT.System not in thermal equilibrium; only thermal interaction among them is by radiation and/or conduction through solid connects (weak contact). No convection when NK << 1.

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How much evaporant strikes substrate? At 10-6 Torr, trajectories are uninterrupted.While a point source deposits uniformly on a sphere about it, a planar source does not:

Geometric factor = Ac

2πR 2 cos θ1 cos θ 2cosθ1 = cosθ2 =

R2r

Deposition rate = JmAc

4πr 2m

area ⋅ t

or J

Ac

4πr2#

area ⋅ t

substrate

θ1

θ2R R

θ1

θ2

r

r

substrate

Convenient geometry

Geometric factor = Ac

4πr2

Film growth rate = JmAc

4πr 21ρ f

thickt

∝cosθ1J

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vox =H pg N

1h

+tox

D+

1ks

oxide

In PVD growth, strike balance

R = deposition rate

Surface diffusion rate

R > 1 stochastic growth, rough

R < 1 layer by layer, smooth (can heat substrate)

Film growth ratefor evaporation

= JmAc

4πr 21ρ f

thickt

v =pvap

2πmsourcekBTsource

mρm

Ac

4πr2 =pvap

ρm

Ac

4πr2msource

2πkBTsource

Cf. CVD v f =Cg

N1ng

+ 1k

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ExerciseDeposit Al (2.7 g/cm3) at r = 40 cm from 5 cm diam. crucible heated to 1100°C (cf Tmelt) pAl vap ≈ 10-3 Torr,

pH2O =10−6 Torr

Compare arrival rate of Al and H2O at substrate…and calculate film growth rate

JH2O =10−6 760( )×105

2π × 0.025eV × e( )× 18 NA( )= 4.8 ×1018 molecules

m2s

JAl =10−3 ×105 760( )

2π × 1373kB( )× 27 NA( )Ac

4πr2

=1.76 ×1018 atoms

m2s

Ac = π52

2

Leave shutter closed so initial Al deposition can getter O2 and H2O.

Also, better done at lower or higher deposition rate.pH2O

(this is not good)

v =pvap

ρm

Ac

4πr2msource

2πkBTsource

≈10−11 m /s slow!

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Step coverage is poor in evaporation (ballistic) - Shadow effects

Heat substrate to increase surface diffusion.

DS = D0S exp −

EaSurf

kT

Ea

S << Eabulk

e.g. WF6 + 3H2 → W + 6HF←∆G ≈ 70kJ

mole(0.73eV)

Can do below 400°C

By contrast, metal CVD and sputtering => better step coverage

W

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Other methods of heating charge.

Resistive heater

I

RF-induction heater e-

+

_

e-beam

B field

Can you suggest other methods?

Laser: Pulsed Laser Deposition (PLD), laser ablation

Ion beam deposition (IB)D: keep substrate chamber at low P, bring in ion beam

through differentially pumped path.

Source, target

Substrate, film

Ion beam

Ion beamCan also use ion beam on film to add energy (ion beam assisted deposition, IBAD)

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Surface energy in a growing film depends on the number of bondsthe adsorbed atom forms with the substrate (or number unsatisfied).

This depends on the crystallography of the surface face and on the type of site occupied (face, edge, corner, crevice). Macroscopically, a curved surface has higher surface energy(more dangling bonds) than a flat surface.

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Microstructure types observed in sputtered films with increasing substrate temperature normalized to melting temperature of deposited species.

Quenched growth Thermally activated growth

λ < a λ < a Ts/Tm > 0.3 Ts/Tm > 0.5

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Interfaces between dissimilar materials

Four characteristic equilibrium, binary phase diagrams, above, and the types of interface structures they may lead to, below (non-equilibrium). The first-column figures would apply to Ga-As, the second to Si-Ge. Third case, B diffuses into A causing swelling; A is forced by swelling into B as a second phase.

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Schematic stress strain curve showing plastic deformation beyond the yield point.

Upon thermal cycling,a film deposited under conditionsthat leave it in tensile stress may evolve through compression then even greater tension .