VACUUM PUMPS AND HARDWARE

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• Introduction

• Basic concepts of vacuum

• Vacuum Hardware (pumps, gauges)

• Mass Spectrometry

OUTLINE

VACUUM PUMPS AND HARDWARE

1

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GETTERS

Getters are stripes of material adsorbing the gas NEED OF VACUUM

TV TUBES

LCD BACKLIGHT

GAS LIGHTS (NEON, HIGH POWER LAMPS) DEWAR (FOR DRINKS)

Active material: alkali (Cs, Rb), rare earths (Yb, Lu), Hg Support: Al₂O₃, Zr

Interaction of gas (CO₂, O) with getter surface (passivation or oxidation) Role of the surface morphology: surface area/bulk

Research applications: impact on everyday life

2

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Basic concepts of vacuum

• UHV Apparatus

• Gas Kinetics

• Vacuum concepts

• Vacuum Pumps

• Vacuum Gauges

• Sample Preparation in UHV

• Cleaving

• Sputtering & Annealing

• Fracturing

• Exposure to gas/vapor

• Evaporation/Sublimation

3

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Ultra High Vacuum Apparatus

4

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5

velocity distribution 1D

k_B = Boltzmann constant

Gas kinetics

velocity distribution 3D

probability of finding a particle with speed in the element dv around v

probability density of finding a particle with speed in the element dv around v Maxwell-Boltzmann distribution

  ^mv^k^T

B

e x

T k v m

f _x ²

2

 



    ^e ^dv

T k v m

dv T e

k v m

d d dV

v v v f dv

v

f ^mv^k^T

B T

k mv

B z

y

x ^B 2 ^B

3 2

0

2 3 2 2

0

2 2

4 2 sin 2

,

, ^  ^



 



 



 



 

  ^ ^^ ^ ^ _ ^ _

  ^mv^k^T

z B y

x e ^B

T k v m

v v

f ²

3 ²

, 2

,  ^



 



 



vd^v ^d^

dv dv

dv

dv_x _y _z  sin _v²  _v_x² _v_y² _v_z²

  ^mv^k^T

B

e B

T k v m

v

f ²

3 2

2

4 2  ^



 



 

 

(6)

Gas kinetics

6

  ^mv^k^T

B

e B

T k v m

v

f ²

3 2

2

4 2  ^



 



 

 

  





0

vdv v

f v

   

 ^ ^ ^ ^





 



 



 



 

0

2 2 3

3

0

2 2 3 2

0

2 2

4 2

4 2 v e dv

T k vdv m

T e k v m

vdv v

f ^k^T

mv

B T

k mv

B

B  

 

 

 

0

2 3

2

dv e

v ^k^T

mv

B



^xⁿêâx^dx ^ _a¹ ^xⁿêâx ^ _aⁿ ^xⁿ^¹êâx^dx



















vdv dx

v x

T km a

B

2 2

2 ^ ^ ^ ^

0

0 23 21

2 1 2

1 dx xe dx

x e

x ^ax ^ax

0 2 0

0 0 0

1 1

e a a dx a

a e dx

a e a xe

dx

xeâx ^ _^ âx _^^ ^ ^ âx ^ ^ ^ âx ^ ^ _^ âx_^^ ^

  



  ²

1 2

1

21 2 2

3

21 2

2 3

0

8 2

2 2

2 2 2

1

4 2 



 



 



 



 



 



 



 



 



 



 



m T k m

T k m

T k T

k m a

T k vdv m

v

f ^B ^B ^B

B

B   

 

  ²

1

0

8 



 



 

 ^^f ^v ^vdv ^k_m^T

v ^B



(7)

Gas kinetics

Mean

T (°C)

Molecular speed

Quadratic mean

Mostprobable

Neon @ 300 K

m_Ne = 20 • 1.67 x 10^-27 kg

f(v)

7

s m

v_rms 610 /

10 67

. 1 2

300 10

38 . 1 3

26

23 



  ^ _

  ^mv^k^T

B

e B

T k v m

v

f ²

3 2

2

4 2  ^



 



 

 

 

m T k dvv

v_p ^ df ^ ² ^B

 v vdv kmT f

v ^B

 8

0



 ^

  m

T dv k

v v f

v_rms ³ ^B

0

2 

 ^

(8)

Gas kinetics

for ideal gas

n = N/V = number density (mol/cm³) N = total number of molecules

8

Arrival rate R:

number of particles landing at a surface per unit area and unit time

N_molecule =

how many molecules in dV?

Consider n molecules with speed v moving towards a surface dS

on a surface dS we take the molecules arriving with speed v_x in a time dt

volume dV = vdt cosdS

total number of

molecules with speed v_x hitting the unit surface in a time dt

dS

T nK p

T NK pV

B B



nf  v  v dtdS 

N_molecules ^ _x _x

 _x

molecules nvxf v dtdS

dR ^ N ^

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9

Gas kinetics

T mk

p v

T k

p v

VN R

B 4 ₂ B

4  



m T k V

R N ^B

 8

 4

T mk R p

 B

 2

  

 ^ ^





 



 



0 2 2

1

0

2

2 ^k^T ^x

mv x

B x

x

x v e dv

T k n m

dv v

f nv

R ^B^x



 _x

molecules nvxf v dtdS

dR ^ N ^ ^{ } ^mv^k^T

B

e x

T k v m

f _x ²

2

 



 

 

0

2

T x k mv

xe dv

v ^B^x



















x x x

B

dv v dy

v y

T km a

2 2

2 m

T k dy a

e dy

y e

y ^ay   ^ay    ^B

 ^



2 1 2

1 2

1

0

0 2 21

1

2 1 2

1 2

2 1

1 8

4 2 2

2 



 



 



 



 



 



 



 



 



m T n k

m T k m

T n k

m T k T

k n m

R ^B ^B ^B ^B ^B

B   



T NK

pV m

T v k

B B



  8

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10

p = Pressure (torr) T = Temperature (K)

m = M_r • amu

O₂ at p = 760 torr, 293 K R = 2.75 10²³molecules s^-1cm² O₂ at p = 1 x 10^-6 torr, 293 K R = 3.61 10¹⁴molecules s^-1cm²

k_B = Boltzmann’s constant (J/K)

molecules arrival rate R at a surface (unit area, time)

Gas kinetics

M_Ox =32

if M_r=relative Molar mass

m_Ne = 20 • 1.67 x 10^-27 g

2 -1

22 s

10 5

.

3 molecule cm

T M R p

r





T mk R p

 B

 2

T M x p

T M

p x

x x R

T M x p

x

T M

p x

x x

T M

p k

T x mk R p

r r

r B r B

22 4

3 24

24 25

23 27

27

10 5 . 10 3

10 5 . 7

10 63 . 2

10 63 . 2 10

659 . 0 399 . 0

10 38 . 1 10

67 . 1 399 1 . 0

10 67 . 1

1 2





 



(11)

11

Mean free path

Gas kinetics

2r

The sphere with 2r is the hard volume

The surface of the sphere is the effective section or cross section for impact

The number of impacts per unit time is

m T k VN

r VN v

r ^B

 



  ₄ ²  ₄ ² ⁸

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12

Mean free path

Gas kinetics

For different molecules A and B

r_B r_A

l is so large that the collisions with walls are dominant with respect to molecular collisions

2 depends on the fact that we did not consider the presence of other molecules

p in torr

 

 kT

V

r_AB N ⁸ ^B 4 ²



2 ^B

AB A

r r ^ r ^

B A

B

A m

m m m

 



l _ v

2 2

4 r

N V v

  l  

r p T

k_B 1 2 ² l 

p x10 ³

5 ^

l 

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13

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14

Gas kinetics: why the UHV

1 Monolayer ~

10¹⁴ – 10¹⁵ atoms/cm²

Residual Gas H₂O

CO₂ CO

CH₄ O₂

N₂ Solid Surface

Bulk Solid

Adsorbed Atoms & Molecules

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15

Sticking probability = 1

1 monolayer of atoms or molecules from

the residual gas is adsorbed at the surface in:

1 sec @ p = 1 x 10^-6 torr 10 sec @ p = 1 x 10^-7 torr 100 sec @ p = 1 x 10^-8 torr 1,000 sec @ p = 1 x 10^-9 torr 10,000 sec @ p = 1 x 10^-10 torr 100,000 sec @ p = 1 x 10^-11 torr

Utra High Vacuum (UHV): p < 10^-10-10^-11 torr

Why the UHV

O₂ at p = 1 x 10^-6 torr, 293 K R = 3.61 10¹⁴

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16

Plots of relevant vacuum features vs. pressure

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17

Gas flow through a pipe

[Q] = [p][L]³[t]^-1

Throughput

Pipe

p = pressure measured in the plane

dV = volume of matter crossing the plane

dV/dt = Volumetric flow rate (portata volumetrica)

d

dV p

Particle flow rate: variation of number of molecules through an area

Quantity of gas (the V of gas at a known p) that passes a plane in a known time at constant temperature For steady flow, Q is continuous, i.e., it has the same value at every position along

the pipe, reflecting the conservation of mass. Q_in = Q_out

 

dt p dV dt

pV

Q ^ d ^

dt Q T dN dtpV K

d

T K N pV T

K N nRT

mN M

N K NN R

n

B at

at B at B

AV AV

AV B at



) (

;

dNdt T

K

Q ^ _B ^at

(18)

18

Mass flow rate

Variation of mass through an area

Throughput

• Magnitude of flow rates

• Pressure drop at the pipe ends

• Surface and geometry of pipe

• Nature of gases

Gas flow through a pipe

M=molar mass M = total mass

Factors affecting the flow

dt p dV Q ^

at m at B

AV B

B at

AV AV

at Q Q

RTM dt

T dN RTM K

dt dN N

K

M K dt

dN N

mN dt

mN d

dt

d^ _ ( ) _ _ _ _ _

RTM Q Q_m ^

dt Q_m ^ d^

mNAV

M 

(19)

19

Regimes of gas flow through a pipe

For l < d viscous

For l  d intermediate For l > d molecular

Viscous

laminar

turbulent

pipe

The mol-mol collisions are dominant

Friction force  = viscosity

S = layer contact area

dv_x /dy = mol speed gradient

Throughput dt d

p dV Q ^

dy S dv

F_f   ^x

(20)

20

Laminar: R_e<1200

turbulent: Re>2200

mass flow

For a pipe with diameter d and section d²/4 Q’ mass flow per unit section

Reynolds number ^ = viscosity

pipe

dt d p dV Q ^

dt Q_m ^ d^

2

' 4

d Q area

Q Q^m ^m

 



 Q d R_e  '

(21)

21

Laminar : Q < 8 10³ (T/M)d [Pa m³/s]

: Q < 5.88 10⁴ (T/M)d [Torr l³/s]

Reynolds number

Turbulent: Q > 1.4 10⁴ (T/M)d [Pa m³/s]

: Q > 1.08 10⁵ (T/M)d [Torr l³/s]

 Q d R_e  '

d Q RTM

d d R_e Q^m





 4 4

2 



Re

d M

Q RT

4

 

(22)

22

(23)

23

(24)

24

For l < d For l  d For l > d

viscous

intermediate molecular

Knudsen number = d/l Only for intermediate and molecular flow

intermediate molecular

3  d/l  80 d/l  3

intermediate: 10^-2  p d  0.5 molecular: p d  10^-2

For air at RT

p in Torr, λ in cm

d

r p T k_B 1

2 ² l 

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25

Pipe conductance: gas flow across pipe Pressures at pipe ends

[C] = [L]³[t]^-1

SI: m³s^-1 cgs: lt s^-1 N₁, V₁

P₁ N₂, V₂

P₂

Arrival rate R:

number of particles landing at a surface per unit area and unit time

For small aperture

connecting two chambers

p0

p C Q

 

4

4 ²

2

2 2 v

T kp v

V R N

B



 

 ¹ ²   ₁ ₂

4 A p p v

dt R R

A d dt

T dN K

Q ^ _B ^at ^ ^ ^ ^

4

4 ¹

1

1 1 v

T kp v

V R N

B



v A C  4

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26

Pipe conductance

Viscous and intermediate regime (Poiseuille law)

Molecular regime Long cylindrical pipe

Elbow pipe

Laminar Turbulent

The molecules must collide with walls at least once before exiting

Equivalent to a longer pipe

For air at 0 C:

11,6 d³/L [lt/s]

2 ²

4 p1 p

L

C ^ d ^



 

 





d L

C d

L v d C

3 1 4

12

3

 3

 

L p d

p

C ^ ₁² ^ ₂² ⁵

(27)

27

Pipe conductance:

In parallel

Pipe impedance:

p0

p C Q

 



 ^N

i Ci

C

1

P C

Q

P C

Q









2 2

1 1

C C  P

Q Q

Q ^ ₁ ^ ₂ ^ ₁ ^ ₂ ^

2

1 C

C C_T ^ ^

Z ^ C¹

(28)

28

In series

Q₁ = Q₂ = Q_T or gas would accumulate



 ^N

i Ci

C 1

1 1

2 2

2

1 1 1

P C

Q

P C

Q









T

T CT

Q C

P Q P

P ^ ^ ^ ^ ^ ^ ^



2 2 1

2 1 1

T T T

T T C

Q C

P ^ Q ^ ^



2 1

1 1

1

C C

C_T ^ ^

(29)

29

The Concept of Transmission Probability

R₁ R₂

R₁A = total number of molecules /s crossing the plane EN to enter the pipe

They approach it from all directions within a solid angle 2π in the left-hand volume Few molecules (1) will pass right through the pipe without touching the sides

The majority (2) collide with the wall at a place such as X and return to the vacuum in a random direction

After collision the molecule may:

(a) return to the left-hand volume

(b) go across the pipe to Y, and then another “three-outcome” event (c) leave the pipe through the exit plane EX into the right-hand volume.

These three outcomes occur with different probabilities

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30

[S] = [L]³[t]^-1 Pumping speed S = Q/p₀

Q= flow through aspiration aperture p = Vessel Pressure

V = Vessel Volume

p₀

Relevant physical parameters of a pumping system

SI: m³s^-1 cgs: lt s^-1 In the presence of a pipe

Effective pumping speed in the vessel

Q at the pump inlet is the same as Q in pipe C

p₀ = pressure at pump inlet

Volumetric flow rate

S S

S SS S

pp Sp

p p Q

p p

C ^e _e

1 1 1

1 0 1

0 0

0   



 



 

 



 



 

 



p S Sp

Q ^ ₀ ^ _e

p0

p S

S

e



C S S_e

1 1

1  

(31)

31

[S] = [L]³[t]^-1 Pumping speed S = Q/p₀

V = Vessel Volume p₀

if C = S

Effective pumping speed

the S is halved C

C S S_e

1 1

1  

SC S

C CS

S S

CS SS

e e



 







1 1 1

1 1

(32)

32

V = Vessel Volume

p₀

Q₁ = True leak rate (leaks from air,

wall permeability)

Q₂ = Virtual leak rate (outgas from materials, walls)

Outgas rate for stainless steel after 2 hours pumping: 10^-8 mbar Ls^-1 cm^-2 Sources of flow (molecules)

1

0 Q

Q

Q ^ ^

(33)

33

Pump-down equation for a constant volume system

True leak rate Only the gas initially present contributes

Virtual leak rate Other outgassing sources contribute

Short time limit Long time limit

Q = Q₀ +Q₁

S = Pumping speed p = Vessel Pressure V = Vessel Volume

Q dt pS

V dp ^ ^



0

1 Q

Q

Q ^ ^

(34)

34

True leak rate

Short time limit

Q = Q₀ +Q₁

Suppose:

Constant S Q = 0

Time needed to reduce p by 50 %

V= 1000 L P₀ = 133 Pa S= 20 L/s

t = 331,6 s 7.5 L/s = 27 m³/h

Vol of 1 m³ = 10³ L to be pumped down from 1000 mbar to 10 mbar in 10 min = 600 s

Q dt pS

V dp ^ ^



V t

e S

p

p ^ ₀ ^

dt pS V dp ^

 V dt

S p

dp _ _



 



 

p p S

t V ln ⁰

S 69V , 0

 

  L s p

p t

S V ln 10 7.5 /

600

ln ⁰   1000 ² 

 



 

(35)

35

Q = Q₀ +Q₁

Ultimate pressure

dp/dt = 0

Virtual leak rate Other outgassing sources contribute

Long time limit

Q dt pS

V dp ^ ^



S Q p_u ^

Q pS ^ 0 

(36)

36

Pressure versus distance x 2

p ^

(37)

37

Differential pumping

operate adjacent parts of a vacuum system at distinctly different pressures

The size of the aperture depends by its function  conductance C is determined.

A, B to be maintained at pressures P₁ and P₂, P₁ >> P₂ A: gas in with flow Q_L

gas to B with flow q Q₁ = flow pumped S1 = Q₁/p₁  Q_L/p₁

B: gas in with flow q To keep pressure p₂ S₂ = Q/p₂

q = C(p₁ − p₂)  C p₁

S₂ = Cp₁/p₂

Modern Vacuum Physics, Ch. 5.8

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38

Example

CVD coatings on panels

Antireflective coatings, p-n junction growth for solar panels

P₀ P₁ P₂ P₁ P₀

S₁ S₂ S₃

S₁ = Cp₀/p₁

C C

C

S₂ = Cp₁/p₂ S₃ = Cp₂/p₁

(39)

39

Gas-solid interaction

H₂O

CO₂ CO

CH₄

O₂ N₂

H₂ He

elastic inelastic trapped

physical adsorption (shortened to Physisorption):

bonding with structure of the molecule unchanged Chemisorption:

bonding involves electron transfer or sharing between the molecule

and atoms of the surface

Can be thought of as a chemical reaction

(40)

40

H₂O

CO₂ CO

CH₄

O₂

N₂

H₂ He

Origin:

Van der Waals forces

The well depth is the energy of adsorption

E to be supplied to desorb the molecule Typical q:

6 - 40 kJ/mol =

0,062 - 0,52 eV /molecule

Physisorption

  ₁₂ ₆

rc rb

z

U ^ ^

(41)

41

H₂O

CO₂ CO

CH₄

O₂

N₂

H₂ He

Origin:

Electron sharing or transfer

between molecules and surface atoms

The well depth is the energy of adsorption Typical q:

40 - 1000 kJ/mol = 0,52 - 10 eV /molecule

Chemisorption

P is a precursor state the molecules have to overcome

    ₁₂ ₆

rc rb

z Q z

U ^ ^ ^

(42)

42

How does this affect vacuum?

probability per second that a molecule will desorb

O₂

Molecule trapped in the adsorbed state at temp. T potential well of depth q

Dilute layer (no interactions with other mol.)

How long does it stays?

Surface atoms have E_vib = h = K_BT  = K_BT/h

At RT  = 0.025/(6.63 × 10⁻³⁴ ÷ 1.6 × 10⁻¹⁹) = 6 × 10¹² s⁻¹  10¹³ s⁻¹

 = number of attempts per second to overcome the potential barrier and break free of the surface.

Boltzmann factor probability that fluctuations in the energy

will result in an energy q ^K ^T

q

e ^ B

T K

q

e ^ B





(43)

43

probability per second that a molecule will desorb

O₂

p(t) = probability that it is still adsorbed after elapsed t

p(t+dt) = p(t) x (1-dt)

probability of not being desorbed after dt

dp = p(t+dt) - p(t) = - dt p(t)

average time of stay T K

q

e ^ B





dt p

dp  

 _t _e ^t

p ^ ^^

T K

q

a e ^B



  ¹  ¹

(44)

44

O₂

average time of stay

At RT   10¹³ s−1

97 kJ / mol = 1 eV / molecule

Temperature dependance

Molecular dependance

Note: Simple model

Neglects all other interactions, surface diffusion, adsorption sites so _a can change T

K q

a e ^B



  ¹  ¹

T K

q a  10^¹³e ^B

