EE143 S06 Lecture 13 Vacuum Basics - EECS Instructional …ee143/sp06/lectures/Lec_13.pdf ·...

Professor N Cheung, U.C. Berkeley

Lecture 13EE143 S06

1

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

1


Lecture 13EE143 S06

2

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


Lecture 13EE143 S06

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4. Average Molecular Velocity

v = (8kT/πm)1/2

where m = molecular weight of gas molecule

5. Mean Free Path of molecular collision

λ = 1

2 πd2o n

where n = molecular density = N/V,do = molecular diameter

[Note] For air at 300 °K, λ = 6.6

P( in Pa) = 0.05

P( in torr)with λ in mm

Assumes Maxwell-Boltzman Velocity Distribution


Lecture 13EE143 S06

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6. Impingement Rate, Φ

Φ = n v

4 = # of molecules striking unit surface /unit time.

= 3.5×1022 × PmT

in #/cm2-secwith P in torr, m in amu

[Note] For air at 300 °K ; Φ(in #/cm2 -sec) = 3.8×1020 ⋅P

Example Calculation : Contamination from Residual Vacuum

For a residual vacuum of 10-6 torr, Φ = 4 × 1014/cm2-secIf each striking molecule sticks to the surface, the equivalent deposition rate of the residual gas is ~ 1/3 of a monolayer of solid per second.


Lecture 13EE143 S06

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Pressure (Torr)

Tim

e to

form

a m

onol

ayer

(sec

)

Impi

ngm

entR

ate

(Mol

ecul

es/c

m2 s

)

Mea

n fr

ee P

ath

(mm

)

At 25oC

M

I

P

1 meter!

1 µm/min

Vacuum Basics (Cont.)

ResidualVacuum

Plasma Processing

CVD


Lecture 13EE143 S06

6

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


Lecture 13EE143 S06

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(1) Evaporation Deposition

Al film

wafer

Al vapor

Al

hot heatingboat (e.g. W)

I

electronsource

crucible is water cooled

Al vapor

=PmkT2π

P P eHkT=

−

0∆

Vapor PressureEvaporation Rate (max)

m=molecular weight of vapor molecule

Log P

1/T

e

Vacuum Torr< 10 -5


Lecture 13EE143 S06

8

Vapor Pressure versus Temperature


Lecture 13EE143 S06

9

Basic Properties of Plasma

• The bulk of plasma contains equal concentrations of ions and electrons.

• Electric potential is ≈ constant inside bulk of plasma. The voltage drop is mostly across the sheath regions.

• Plasma used in IC processing is a “weak” plasma, containing mostly neutral atoms/molecules. Degree of ionization is ≈ 10-3 to 10-6.


Lecture 13EE143 S06

10Source: www.icknowledge.com


Lecture 13EE143 S06

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DiffusionFlux

DriftFlux

E ≠ 0

E ~ 0


Lecture 13EE143 S06

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Al

(2) Sputtering Deposition

Al AlAr+

Deposited Al film

Al target

Ar plasma

waferheat substrate to ~ 300oC (optiona12l)

Negative Bias ( kV)

⋅

I

Gas Pressure 1-10 m TorrDeposition rate =

≅sputtering yield

ion current

constant I S• •

Ar+

Example:DC plasma

12


Lecture 13EE143 S06

13Source: www.icknowledge.com


Lecture 13EE143 S06

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Lecture 13EE143 S06

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S ≡# of ejected target atoms

incoming ion.

0.1 < S < 30

Al

Al

Al

Ar

Sputtering Yield S


Lecture 13EE143 S06

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Sputtering Yield of bombarding ion atomic number

For reference only


Lecture 13EE143 S06

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The Sputtering Yield with incidence angle


Lecture 13EE143 S06

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Ar+

Aflux

Bflux

AxBy

targetFilm has same composition of target at steady state.

Because SA ≠ SB, Target surface will acquire a composition Ax’By’ at steady state.

Sputtering of Compound Targets


Lecture 13EE143 S06

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Target

AxBy

Deposited Film on substrate

AxBy

AxBy

Ax’By

’

Ax’By

’

t=t1

t=t2Difference betweent2 and t1 must becomposition depositedon substrate

Bulk target composition

Proof


Lecture 13EE143 S06

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Reactive Sputtering

• Sputter a Ti target with a nitrogen plasma

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. When sandwiched between two thin films, it can also be used as a “diffusion barrier” to block thin-film reactions


Lecture 13EE143 S06

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Radio Frequency Plasma Generation

For reference only

•More efficient plasma generation•Needed if substrate is electricallyinsulating


Lecture 13EE143 S06

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Lecture 13EE143 S06

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Thickness Uniformity with various PVD sources

(i) Point-like Source

(ii) Plane-like small-area SourceF

θsurface

θ

wafer

isotropic

•Ideal situation•Flux F leaving source surface isindependent of θ (isotropic source)

F

•Flux F leaving surface ∝ cos θExample: E-beam evaporation


Lecture 13EE143 S06

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Let F = emission flux from sourceTherefore receiving flux F’ at a distance r from source = F/ r2

If the receiving surface is making an angle φ with respect to the F’ vectorThickness deposited ∝ F’ cos φ = F cos φ / r2

Film Thickness Deposition on Wafer

F

F’

source

waferφr


Lecture 13EE143 S06

25

wafer-l +lx =0

R0R

point-source(emission flux has no θ dependence)

φ

Thickness t at x=0

∝ cos φ / Ro2

= 1 / Ro2

Note : φ =0 at position x=0

For this special geometry, θ = φ

θ

Example: Flat Wafer directly on top of Point Source


Lecture 13EE143 S06

26

( ) ( )

( )( )

23

2

01

0

3

coscos

3

22

0

0

0

220

220

−

+=

+=

=+=

=

+=

+=

Rl

lR

Rxt

lxt

R

RlRlR

θφ

x=0 x=+l

Thickness t at position x= +l ∝ cos φ / R2


Lecture 13EE143 S06

27

Example: Plane-like Source & Spherical Receiving Surface

θ

φθ

φ

θ

θφθ

2cos21

coscos21

cos'waferonThickness

cos21'fluxReceiving

cos flux F Emission

r

r

Fr

F

∝

∝

⋅∝

⋅∝

∝=

Since cos θ = (r/2)/R

Thickness∴ ∝ = constant(1/r2) (r2/4R2)

waferR

R

r

plane-source

θ

φ F ’


Lecture 13EE143 S06

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Step Coverage Problem with PVD

• Both evaporation and sputtering have directional fluxes.

wafer

step

filmFlux

film

“geometricalshadowing”


Lecture 13EE143 S06

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stepfilm

“self-shadowing”

t=t1

t=0 t=t11 244444 344444

shadowing distance

t=0


Lecture 13EE143 S06

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Lecture 13EE143 S06

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Film Thinningat stepsstep step

wafer

“key hole”

Key Hole Formation

Thinner film depositionat steps


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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


Lecture 13EE143 S06

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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


Lecture 13EE143 S06

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Sputtering Target

Profile due to one small-area source

Superposition of all small-area sources

Sputtering Target

Profile due to one small-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 film deposition uniformity on wafer.

•For multi-component thin films, sputtering gives better composition control using compound targets. Evaporation depends on vapor pressure of various vapor components and is difficult to control.

Advantages of Sputtering over Evaporation

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EE143 S06 Lecture 13 Vacuum Basics - EECS Instructional …ee143/sp06/lectures/Lec_13.pdf ·...

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