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6.152J/3.155J 1
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
6.152J/3.155J 2
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.)
6.152J/3.155J 3
p < 10-8 Torr demands:Stainless steel chamberBakeable to 150oCTurbo, ion, cryo pumps
Ultra-high vacuum chambers
6.152J/3.155J 4
-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
6.152J/3.155J 5
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
6.152J/3.155J 6
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
6.152J/3.155J 7
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
6.152J/3.155J 8
(note W, Mo can be used as crucibles.)
6.152J/3.155J 9
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.
6.152J/3.155J 10
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
6.152J/3.155J 11
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
6.152J/3.155J 12
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!
6.152J/3.155J 13
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
6.152J/3.155J 14
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)
6.152J/3.155J 15
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.
6.152J/3.155J 16
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
6.152J/3.155J 17
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.
6.152J/3.155J 18
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 .
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