EE143 F2010 Lecture 12 Vacuum Basicsee143/fa10/lectures/Lec_12.pdf · Professor N Cheung, U.C....
Transcript of EE143 F2010 Lecture 12 Vacuum Basicsee143/fa10/lectures/Lec_12.pdf · Professor N Cheung, U.C....
Professor N Cheung, U.C. Berkeley
Lecture 12EE143 F2010
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Vacuum Basics
1. Units– 1 atmosphere = 760 torr = 1.013x105 Pa– 1 bar = 105 Pa = 750 torr– 1 torr = 1 mm Hg– 1 mtorr = 1 micron Hg– 1Pa = 7.5 mtorr = 1 newton/m2
– 1 torr = 133.3 Pa
2. Ideal Gas Law: PV = NkT– k = 1.38E-23 Joules/molecule –K
= 1.37E-22 atm cm3/K– N = # of molecules– T = absolute temperature in K– [Note] At T = 300 K ; kT = 3.1E-20 torr-liter
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Professor N Cheung, U.C. Berkeley
Lecture 12EE143 F2010
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For mixture of non-reactive gases in a common vessel,each gas exerts its pressure independent of others.
Ptotal = P1 + P2 + … + PN (Total P = Sum of partial pressure)Ntotal = N1 + N2 + … + NNP1V = N1kTP2V = N2kT...................PNV = NNkT
3. Dalton’s Law of Partial Pressure
Professor N Cheung, U.C. Berkeley
Lecture 12EE143 F2010
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4. Average Molecular Velocity
v = (8kT/ m)1/2
where m = molecular weight of gas molecule
5. Mean Free Path ofmolecular collision
=1
2 d2o n
where n = molecular density = N/V,do = molecular diameter
[Note] For air at 300 °K, =6.6
P( in Pa) =0.05
P( in torr)with in mm
Assumes Maxwell-Boltzman Velocity Distribution
Professor N Cheung, U.C. Berkeley
Lecture 12EE143 F2010
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6. Impingement Rate,
=n v
4 = # of molecules striking unit surface /unit time.
= 3.51022 × PmT
in #/cm2-secwith P in torr, m in amu
[Note] For air at 300 °K ; (in #/cm2 -sec) =3.81020 P
Example Calculation : Contamination from Residual Vacuum
For a residual vacuum of 10-6 torr, = 4 1014/cm2-secIf each striking molecule sticks to the surface, the equivalentdeposition rate of the residual gas is ~ 1/3 of a monolayer of solid persecond.
Professor N Cheung, U.C. Berkeley
Lecture 12EE143 F2010
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Pressure (Torr)
Tim
e to
form
a m
onol
ayer
(sec
)
Impi
ngm
ent R
ate
(Mol
ecul
es/c
m2 s)
Mea
n fr
ee P
ath
(mm
)
At 25oC
M
I
P
1 meter!
m/min
Vacuum Basics (Cont.)
ResidualVacuum
PlasmaProcessing
CVD
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Lecture 12EE143 F2010
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Thin Film Deposition
substratefilm
Applications:Metalization (e.g. Al, TiN, W, silicide)Poly-Sidielectric layers; surface passivation.
EvaporationSputtering
Reactive Sputtering
Chemical Vapor DepositionLow Pressure CVD
Plasma Enhanced CVD
Physical Methods Chemical Methods
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Lecture 12EE143 F2010
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(1) Evaporation Deposition
Al film
wafer
Al vapor
Al
hot heatingboat (e.g. W)
I
electronsource
crucible iswater cooled
Al vapor
PmkT2
P P eHkT
0
Vapor PressureEvaporation Rate (max)
m=molecular weight of vapor molecule
Log P
1/T
e
Vacuum Torr 10 -5
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Lecture 12EE143 F2010
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Vapor Pressure versus Temperature
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Lecture 12EE143 F2010
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Basic Properties of Plasma
• The bulk of plasma contains equal concentrations of ionsand electrons.
• Electric potential is constant inside bulk of plasma. Thevoltage drop is mostly across the sheath regions.
• Plasma used in IC processing is a “weak” plasma,containing mostly neutral atoms/molecules. Degree ofionization is 10-3 to 10-6.
Professor N Cheung, U.C. Berkeley
Lecture 12EE143 F2010
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Professor N Cheung, U.C. Berkeley
Lecture 12EE143 F2010
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DiffusionFlux
DriftFlux
E 0
E ~ 0
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Lecture 12EE143 F2010
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Al
(2) Sputtering Deposition
Al AlAr+
Deposited Al film
Al target
Ar plasma
waferheatsubstrate to~ 300oC (optiona12l)
NegativeBias ( kV)
I
Gas Pressure 1-10 m TorrDeposition rate =
sputtering yield
ion current
constant I S
Ar+
Example:DC plasma
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Professor N Cheung, U.C. Berkeley
Lecture 12EE143 F2010
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Professor N Cheung, U.C. Berkeley
Lecture 12EE143 F2010
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Professor N Cheung, U.C. Berkeley
Lecture 12EE143 F2010
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S # of ejected target atoms
incoming ion.
0.1 < S < 30
Al
Al
Al
Ar
Sputtering Yield S
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Sputtering Yield of bombarding ion atomic number
For reference only
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Lecture 12EE143 F2010
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The Sputtering Yield with incidence angle
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Ar+
Aflux
Bflux
AxBy
targetFilm has same composition oftarget at steady state.
Because SA SB, Target surface will acquire a compositionAx’By’ at steady state.
Sputtering of Compound Targets
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Lecture 12EE143 F2010
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Target
AxBy
Deposited Filmon substrate
AxBy
AxBy
Ax’By
’
Ax’By
’
t=t1
t=t2Difference betweent2 and t1 must becomposition depositedon substrate
Bulk targetcomposition
Proof
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Reactive Sputtering
• Sputter a Ti targetwith a nitrogenplasma
Ti Target
N2 plasma
Ti, N2+
TiN
Example: Formation of TiN
Substrate
Note: TiN is a metallic compoundwith a golden color. It is used frequentlyin ICs as a metallic conductor. Whensandwiched between two thin films, it canalso be used as a “diffusion barrier” toblock thin-film reactions
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Radio Frequency Plasma Generation
For reference only
•More efficient plasma generation•Needed if substrate is electricallyinsulating
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Thickness Uniformity with various PVD sources
(i) Point-like Source
(ii) Plane-like Surface SourceF
surface
•Flux F leaving source surface isindependent of (isotropic source)Example: Laser vaporization of a wire tip
F
•Flux F leaving surface cosExample: E-beam evaporation
Emission source
surface
Emission source
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Let F = emission flux from source
Receiving flux F’ at a distance r from source = F/ r2
If the receiving surface is making an angle with respectto the F’ vector
Thickness deposited F’ cos = F cos / r2
Film Thickness Deposition on Wafer
F
F’
source
waferr
Professor N Cheung, U.C. Berkeley
Lecture 12EE143 F2010
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wafer-l +lx =0
R0R
point-source(emission flux has no dependence)
Thickness t at x=0
cos / Ro2
= 1 / Ro2
Note : =0at position x=0
For this special geometry, =
Example 1 : Flat Wafer directly on top of Point Source
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23
2
01
0
3
coscos
3
22
0
0
0
220
220
Rl
lR
Rxt
lxt
R
RlRlR
x=0 x=+l
Thickness t at position x= +l cos / R2
Professor N Cheung, U.C. Berkeley
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Thickness Variation of PVD
(n=0)
(n=1)
*For this particular geometry, =
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Lecture 12EE143 F2010
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Example 2 : Surafce Source & Spherical Receiving Surface
2cos21
coscos21
cos'waferonThickness
cos21'fluxReceiving
cosflux FEmission
r
r
Fr
F
Since cos = (r/2)/R
Thickness = constant !(1/r2) (r2/4R2)
waferR
R
r
plane-source
F ’
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Step Coverage Issues
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Lecture 12EE143 F2010
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Step Coverage Problem with PVD
• Both evaporation and sputtering have directional fluxes.
wafer
step
filmFlux
film
“geometricalshadowing”
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Lecture 12EE143 F2010
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stepfilm
“self-shadowing”
t=t1
t=0 t=t1
shadowingdistance
t=0
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Lecture 12EE143 F2010
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Sputtering Target
Methods to Minimize Step Coverage Problems
• Rotate + Tilt substrate during deposition• Elevate substrate temperature (enhance
surface diffusion)• Use large-area deposition source
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Lecture 12EE143 F2010
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Dip Photoresist in Chlorobenzene to slow down developingrate of surface layer .
Surface layer treated by chlorobenzene
Photoresist
Directional Deposition Flux
film
Patterning of deposited layer using directional deposition.
Lift-off Technique
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Lecture 12EE143 F2010
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Sputtering Target
Profile due to onesmall-area source
Superposition of all small-area sources
Sputtering Target
Profile due to onesmall-area source
Superposition of all small-area sources
•Better lateral thickness uniformityArea of sputtering target can be made much larger than that of an evaporating source.A larger area can be considered as a superposition of many small-area sources.By adding the flux from all the sources, a large area source will provide better lateral filmdeposition uniformity on wafer.
•For multi-component thin films, sputtering gives better compositioncontrol using compound targets. Evaporation depends on vapor pressure ofvarious vapor components and is difficult to control.
Advantages of Sputtering over Evaporation