Transcript tv 82 cm

VACUUM PUMPS AND HARDWARE
OUTLINE
• Introduction
• Basic concepts of vacuum
• Vacuum Hardware (pumps, gauges)
• Mass Spectrometry
1
Research applications: impact on everyday life
GETTERS
NEED OF VACUUM
TV TUBES
LCD BACKLIGHT
GAS LIGHTS (NEON, HIGH POWER LAMPS)
DEWAR (FOR DRINKS)
Getters are stripes of material adsorbing the gas
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
2
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
Ultra High Vacuum Apparatus
4
Gas kinetics
velocity distribution 1D
m
e
2kBT
f v x  
kB = Boltzmann constant
 m
f v x ,v y ,v z  
 2kBT

velocity distribution 3D


mv x2
2kBT
3

 e


mv 2
2kBT
probability of finding a particle with speed in the element dv around v
dv x dv y dv z  dv vd v sin d 
2

 m
f v dv   f v x ,v y ,v z dV   d  d sin v 2 
 2kBT
0
0


Maxwell-Boltzmann distribution
3
v 2  v x2  v y2  v z2

 e

 m
f v   4v 2 
 2kBT

mv 2
2kBT
3

 e


 m
dv  4v 2 
 2kBT
3

 e


mv 2
2kBT
dv
mv 2
2kBT
probability density of finding a particle with speed in the element dv around v
5
Gas kinetics

f v   4v 2
v   f v vdv
0

3

mv 2
 m
m  2  2kBT
2





f
v
vdv

4

v
e
vdv

4

0
0
 2k T 
B


 2kBT
a   m 
2


mv


2
k
T
B
2k T
3


0 v e B dv 
x  v 2



 dx  2vdv 







3
 m

 2kBT
3
mv 2
 2  3  2kBT
  v e
dv
 0

3
2
 x e
0
ax
1
2x

1
dx   xe ax dx
1
20
2
n ax
 x e dx 
1 n ax n
x e   x n 1e ax dx
a
a


1
1
1  1 ax 
1
1
ax 
ax
ax
0 xe dx   a xe 0  a 0 e dx   a 0 e dx   a  a e 0  a 2
ax

 m



f
v
vdv

4

0
 2kBT
3
2
 1
2


1
2
 2a
2

 8k T 
v   f v vdv   B 
 m 
0

 m

 2kBT
3
2
mv 2
  2kBT
 e

2
1
2
  2kBT 
2  2k T 
 8k T 
 
  1  B   B 
m 
 m 
  m 
2 
1
2
1
2
6
Gas kinetics
3
mv 2
  2kBT
 e

f(v)
f v   4v 2
 m

 2kBT
Molecular speed
Most
probable
df v 

dv
2kBT
m
v   f v vdv 
8kBT
m
vp 

Mean
0
Quadratic
mean
v rms 
T (°C)

2


f
v
v
dv 

0
3kBT
m
Neon @ 300 K
mNe = 20 • 1.67 x 10-27 kg
v rms
3  1.38  10 23  300

 610 m / s
2  1.67  10 26
7
Gas kinetics
pV  NKBT
p  nKBT
for ideal gas
N = total number of molecules
n = N/V = number density (mol/cm3)
Consider n molecules with speed v moving towards a surface dS
dS
Arrival rate R:
number of particles landing at a surface per unit area and unit time
on a surface dS we take the molecules arriving with speed vx in a time dt
Nmolecule =
how many molecules in dV?
volume dV = vdt cosdS
total number of
molecules with speed vx
hitting the unit surface
in a time dt
Nmolecules  nf v x v x dtdS 
Nmolecules
dR 
 nv xf v x 
dtdS
8
Gas kinetics
N
dR  molecules  nv xf v x 
dtdS
f v x  

 m
R   nv xf v x  dv x  n 
 2kBT
0

v
x
e
mv x2

2kBT
dv x 
0
 m
R  n 
 2kBT
a   m



2kBT 

y v 2

x


 dy  2v x dv x 




1
2
 kBT


m


y
0
n  kBT 


2  m 
N
R 
4V
R 
p v
N v


V 4
kBT 4



1
2

1
2
v
x
e

mv x2
2kBT
dv x
0
1
2
e
ay
1
2y
kBT
1
1
ay
dy

e
dy



1
2 0
2a
m
2

1
2
kBT
n  8k T 
 kT 
 n B    B 
m
4  m 
 2m 
8kBT
m
p
2mkBT
mv x2

m
e 2kBT
2kBT
1
2
8kBT
m
pV  NKBT
v 
R 
p
2mkBT
9
Gas kinetics
molecules arrival rate R at a surface
(unit area, time)
if Mr=relative Molar mass
R 
p

2mkBT
1
2
1
0.399
m = Mr • amu
1
1.67 x 10 27 kB
2mkBT
mNe = 20 • 1.67 x 10-27 g
p

MrT
p

MrT
1.67 x 10 27 x 1.38x 10 23
p
0.399x 0.659x 1025
 2.63x 1024
MrT
2.63x 1024
R 
7.5x 10 3 x 10 4
R 
p
p
 3.5x 1022
MrT
p
MrT
kB = Boltzmann’s constant (J/K)
p
MrT
T = Temperature (K)
p = Pressure (torr)
R  3.5  10
MOx =32
22
p
MrT
molecule s-1 cm 2
O2 at p = 760 torr, 293 K
R = 2.75 1023 molecules s-1cm2
O2 at p = 1 x 10-6 torr, 293 K
R = 3.61 1014 molecules s-1cm2
10
Gas kinetics
Mean free path
2r
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
  4r 2
N
N
v  4r 2
V
V
8kBT
m
11
Gas kinetics
For different molecules A and B
rA
rAB
rB
r r
 A B
2
N
  4r
V
2
AB

mA mB
mA  mB
8kBT

Mean free path


v


V
N 4 2r 2
5x 10 3

p
v

kBT
1

2r 2 p
p in torr
 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
12
13
Gas kinetics: why the UHV
Residual Gas
H2O
CO
O2
CO2
CH4
Solid Surface
N2
1 Monolayer ~
1014 – 1015 atoms/cm2
Bulk Solid
Adsorbed Atoms & Molecules
14
Why the UHV
O2 at p = 1 x 10-6 torr, 293 K
R = 3.61 1014
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
15
Plots of relevant vacuum features vs. pressure
16
Gas flow through a pipe
Pipe
d
dV
Throughput
p
p = pressure measured in the plane
dV = volume of matter crossing the plane
dV/dt = Volumetric flow rate (portata volumetrica)
d pV 
dV
Q 
p
dt
dt
[Q] = [p][L]3[t]-1
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. Qin = Qout
n 
Nat
; R  KB NAV ; M  mNAV
NAV
nRT  Nat KBT ;
pV  Nat KBT
d ( pV )
dNat
 KBT
Q
dt
dt
dNat
Q  KBT
dt
Particle flow rate: variation of number of molecules through an area
17
Gas flow through a pipe
Qp
dV
dt
Qm 
d
dt
Mass flow rate
Throughput
M = total mass
Variation of mass through an area
KB M dNat
dNat
d d (mNat ) mNAV dNat
M
M




KBT

Q  Qm
dt
dt
NAV
dt
KB NAV dt
RT
dt
RT
M
Qm 
Q
RT
M  mNAV
M=molar mass
Factors affecting the flow
•
•
•
•
Magnitude of flow rates
Pressure drop at the pipe ends
Surface and geometry of pipe
Nature of gases
18
Regimes of gas flow through a pipe
Throughput
dV
Qp
dt
For  < d
viscous
For   d
intermediate
For  > d
molecular
d
pipe
Viscous
S = layer contact area
dvx /dy = mol speed gradient
The mol-mol collisions are dominant
Ff  S
Friction force
dv x
dy
 = viscosity
laminar
turbulent
19
Regimes of gas flow through a pipe
Qp
Qm 
dV
dt
d
dt
d
pipe
mass flow
For a pipe with diameter d and section d2/4
Q '
Q’ mass flow per unit section
Reynolds number
Re  Q'
d
Qm
4Qm

area d 2
 = viscosity

Laminar: Re<1200
turbulent: Re>2200
20
Regimes of gas flow through a pipe
Reynolds number
Re  Q'
d

Re 
RT d
Q 
Re
M 4
4Qm d
M 4

Q
2
d  RT d
Laminar : Q < 8 103 (T/M)d [Pa m3/s]
: Q < 5.88 104 (T/M)d [Torr l3/s]
Turbulent: Q > 1.4 104 (T/M)d [Pa m3/s]
: Q > 1.08 105 (T/M)d [Torr l3/s]
21
Regimes of gas flow through a pipe
22
Regimes of gas flow through a pipe
23
Regimes of gas flow through a pipe
d
For  < d
viscous
For   d
intermediate
For  > d
molecular
Knudsen number = d/
intermediate
molecular
3  d/  80
d/  3
For air at RT
p in Torr, λ in cm
Only for intermediate and molecular flow

kBT
1
2r 2 p
intermediate: 10-2  p d  0.5
molecular: p d  10-2
24
Pipe conductance:
Q
C 
p  p0
gas flow across pipe
Pressures at pipe ends
SI: m3s-1
For small aperture
connecting two chambers
N2, V2
P2
R2 
[C] = [L]3[t]-1
cgs: lt s-1
N1, V1
P1
p v
N2 v
 2
V2 4
kBT 4
Q  KBT
R1 
p v
N1 v
 1
V1 4
kBT 4
Arrival rate R:
number of particles landing at a
surface per unit area and unit time
dNat d A R1  R2  v


A p1  p2 
dt
dt
4
C 
v
A
4
25
Pipe conductance
Viscous and intermediate regime (Poiseuille law)
d 4 p1  p2
C 
L
2
d3
C 
v
12
L

Laminar
d5
C  p  p 
L
2
1
2
2
Turbulent
Molecular regime
Long cylindrical pipe
d3
C 
4 

L1  d 
3 

For air at 0 C:
11,6 d3/L [lt/s]
Elbow pipe
The molecules must collide with
walls at least once before exiting
Equivalent to a longer pipe
26
Pipe conductance:
Pipe impedance:
C 
Q
p  p0
1
Z 
C
In parallel
Q1  C 1 P
Q2  C 2 P
Q  Q1  Q2  C1  C 2 P
CT  C1  C 2
N
C  Ci
i 1
27
In series
Q1  C 1 P1
Q2  C 2 P2
Q1 Q2 QT
PT  P1  P2 


C 1 C 2 CT
Q1 = Q2 = QT or gas would accumulate
QT QT QT
PT 


C 1 C 2 CT
1
1
1


CT C 1 C 2
N
1
1

C
i 1 C i
28
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
29
Relevant physical parameters of a pumping system
Q= flow through aspiration aperture
p = Vessel Pressure
V = Vessel Volume
p0 = pressure at pump inlet
p0
C
Pumping speed S = Q/p0
[S] = [L]3[t]-1
Volumetric flow rate
SI: m3s-1
In the presence of a pipe
cgs: lt s-1
Q at the pump inlet is the same as Q in pipe
Q  Sp0  Se p
p
S

Se
p0
 p


 1 
p  p0 p  p0  p0
1




C
Q
Sp0
S
Effective pumping speed
in the vessel
S


 1 
 Se
 1  1
S
Se S
1
1
1
 
Se S C
30
Relevant physical parameters of a pumping system
p0
Q= flow through aspiration aperture
p = Vessel Pressure
V = Vessel Volume
C
Pumping speed S = Q/p0
1
1
1
 
Se S C
Effective pumping speed
[S] = [L]3[t]-1
S
S
1
Se
C
Se
1
C
1


S
S
S 1C
1
C
S
if C = S
the Se is halved
31
Relevant physical parameters of a pumping system
Q= flow through aspiration aperture
p = Vessel Pressure
V = Vessel Volume
p0
Sources of flow (molecules)
Q  Q0  Q1
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
32
Pump-down equation for a constant volume system
dp
V
 pS  Q
dt
Q = Q0 +Q1
S = Pumping speed
p = Vessel Pressure
V = Vessel Volume
Q  Q1  Q0
Short time limit
True leak rate
Only the gas
initially present
contributes
Long time limit
Virtual leak rate
Other outgassing
sources contribute
33
Pump-down equation for a constant volume system
dp
V
 pS  Q
dt
Q = Q0 +Q1
S = Pumping speed
p = Vessel Pressure
V = Vessel Volume
Short time limit
True leak rate
Suppose:
Constant S
Q=0
dp
V
 pS
dt
V  p0 
t  ln 
S p 
dp
S
  dt
p
V
Time needed to reduce p by 50 %
V
S
  0,69
V= 1000 L
P0 = 133 Pa
S= 20 L/s
t = 331,6 s
p  p0e
S
 t
V
Vol of 1 m3 = 103 L to be pumped down from
1000 mbar to 10 mbar in 10 min = 600 s
V  p  1000
S  ln 0  
ln102   7.5L / s
t  p  600
7.5 L/s = 27 m3/h
34
Pump-down equation for a constant volume system
dp
V
 pS  Q
dt
Q = Q0 +Q1
S = Pumping speed
p = Vessel Pressure
V = Vessel Volume
Long time limit
Virtual leak rate
Other outgassing
sources contribute
dp/dt = 0
0  pS  Q
pu  Q S
Ultimate pressure
35
Pressure versus distance
p  x2
36
Differential pumping
operate adjacent parts of a vacuum system at distinctly different pressures
A, B to be maintained at pressures P1 and P2, P1 >> P2
A: gas in with flow QL
gas to B with flow q
Q1 = flow pumped
S1 = Q1/p1  QL/p1
B: gas in with flow q
To keep pressure p2
S2 = Q/p2
q = C(p1 − p2)  C p1
S2 = Cp1/p2
The size of the aperture depends by its function  conductance C is determined.
37
Modern Vacuum Physics, Ch. 5.8
Example
CVD coatings on panels
Antireflective coatings, p-n junction growth for solar panels
P1
P0
C
P2
C
S1
S1 = Cp0/p1
P1
P0
C
S2
S2 = Cp1/p2
S3
S3 = Cp2/p1
38
Gas-solid interaction
H2O
CO
CO2
N2
H2
inelastic
trapped
physical adsorption (shortened to Physisorption):
bonding with structure of the molecule unchanged
CH4
He
elastic
Chemisorption:
bonding involves electron transfer
or sharing between the molecule
and atoms of the surface
Can be thought of as a chemical reaction
O2
39
Gas-solid interaction
Physisorption
CO
Origin:
Van der Waals forces
H2O
CH4
CO2
O2
U z  
b
c

r 12 r 6
N2
H2
He
Typical q:
6 - 40 kJ/mol =
0,062 - 0,52 eV /molecule
The well depth is the energy of adsorption
E to be supplied to desorb the molecule
40
Gas-solid interaction
Chemisorption
CO
Origin:
Electron sharing or transfer
between molecules and surface atoms
H2O
CH4
CO2
O2
U z   Q z  
b
c

r 12 r 6
N2
H2
He
Typical q:
40 - 1000 kJ/mol =
0,52 - 10 eV /molecule
The well depth is the energy of adsorption
P is a precursor state the molecules have to overcome
41
Gas-solid interaction
How does this affect vacuum?
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?
O2
Surface atoms have Evib = h = KBT
 = KBT/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.
probability that fluctuations in the energy
will result in an energy q
  e
q

KBT
e

q
KBT
Boltzmann factor
probability per second that a molecule will desorb
42
Gas-solid interaction
  e

q
KBT
probability per second
that a molecule will desorb
p(t) = probability that it is still adsorbed after elapsed t
O2
p(t+dt) = p(t) x (1-dt)
probability of not being
desorbed after dt
dp = p(t+dt) - p(t) = - dt p(t)
dp
 p
dt
p t   e t
average time of stay
a 
1


1

e
q
KBT
43
Gas-solid interaction
average time of stay
O2
a 
 a  10 13e
q
KBT
1


1

e
q
KBT
At RT   1013 s−1
Molecular dependance
97 kJ / mol = 1 eV / molecule
Temperature dependance
Note: Simple model
Neglects all other interactions, surface diffusion, adsorption sites so a can change
44
Gas-solid interaction
monolayer (ML): monomolecular layer adsorbed on a surface
number of molecules in a monolayer: N0  1015
na = number of adsorbed molecules per unit area

na
N0
fractional coverage
s() = sticking coefficient
Probability that, on striking the surface already
having coverage θ, a molecule becomes adsorbed.
45
Gas-solid interaction
For dilute adsorbed layers (<<1) simple model for the equilibrium state
dna
 s ( )R
dt
n
 dna 

  na    a
a
 dt des
rate of adsorption
rate of desorption
at equilibrium
dna  dna 


dt
dt

des
na ,eq  seq ( )R a
N2 at p = 1 x 10-5 mbar, 293 K
R = 2.9 1015 cm-2 s-1
na ,eq  seq ( )R a
seq= 0.2 a = 5 x 10-3 s
N0  seq
na,eq = 1 x 1012 cm-2
p
1
e
2mkBT 
 = 0.003
q
KBT
equation of state for the adsorbed phase
Coverage is proportional to pressure
46
Desorption
P = 1000 mbar
P = 10-7 mbar
pumping
Equilibrium
Experimental
relation
q1  qG
1h
qG 
q1
t
t 
t
1h
Far from equilibrium till….
Gas flow /area
 = 0.5
 = 1 for metals
 = 0.5 for elastomers
=1
q1metals  1x10−7 mbar L s−1cm-2
For 1 mbar L at RT
Nat  2.6x1019
pV  Nat KBT
with q1metals, outgassing rate  1012 molec s−1cm-2

10-3
ML /second of released gas, principally water vapor
47
Desorption
How important is the molecule/surface interaction energy?
H2O
Rate of desorption
n
 dna 

  na    a
a
 dt des
 a  10 13e
q
KBT
N2
q

 dna 
K BT
13

n
10
e


a
 dt des
n
 dna 

  a
a
 dt des
na  e

integrate
t
a
decays exponentially from the initial state
with a time constant equal to the stay time
Simple model calculation
idealized UHV system
RT, V= 1 L, A = 100 cm2
S = 1 L/s
only gas source: initially complete ML
of specified binding energy adsorbed
at the wall
fall of pressure at RT
H2 O
q
48
Outgassing
Gas is continuously released, (at relatively small rates) from walls
Principally water vapor
Limit to attainable vacuum achievable in reasonable times (hours) ∼10−6 mbar
Origin of flowes:
Permeation
Adsorption
Solubility
Desorption
49
Gas-solid permeation
p1 = 1000 mbar
p2 = 1x10-8 mbar
H2O
CO
CO2
CH4
N2
H2
He
O2
Residual Gas
50
Gas-solid permeation
p1 = 1000 mbar
Permeation is a
complex process
Adsorption
p2 = 1x10-9 mbar
Residual Gas
Dissociation
Solution into the solid
Diffusion
Recombination
Desorption
51
Gas-solid permeation
p1 = 1000 mbar
Permeation process
can be quantified trough
phenomenological
quantities
p2 = 1x10-9 mbar
Residual Gas
permeability
=Q/(p1-p2)A
Q=flow trough wall
A= unit area
[Q] = [p][L]3[t]-1
=[L]3[t]-1[L]-2
m3s-1m-2
ls-1cm-2
52
Gas-solid permeation
cm3s-1cm-2 Pa-1
For a given gas
A = wall area
d = wall thickness
Qp  K p f p1   f p2 
He
A
d
f p   p, p
depending on diffusion
mechanisms
Kp = Permeability coefficient
p = 13 mbar
d = 1 mm
m3s-1m-1Pa-1
53
Gas-solid permeation
Metal – gas Kp
Table of gas permeability
Glass
He,
Ne,
O2
p
Metals
H2, No rare gas
Ar,
  p
Polymers
All gases
p
54
Solubility
Is the quantity of substance A that can be dissolved in B at given T and p
For a gas
Gas quantity dissolved in solid volume unit at standard conditions
For undissociated molecular gas (interstitial)
c = gas concentration
c s p
Henry’s law
Valid for low concentrations and for glass and plastic materials
No formation of alloys
55
Solubility
H2 on metals
For dissociated gas
Interstitial or substitutional
c s
p
Sievert’s law
Valid for low concentrations
and for metals
Note the high solubility of H2 in Ti,Zr
56
Vacuum Pumps
Throughput pumps
• Pistons
• Gears
• Turbines
• Jet stream
Capture pumps
• Cold traps
• Ionization
• Getters
Differences: pressure range, speed, gas selectivity
57
Pressure Ranges Spanned by Different Vacuum Pumps
More than one pump to HV and UHV
58
What pump to use?
Pumping speed S = Q/p
p = inlet pressure
• Depends on the gas type
• S varies with p
S 
V
V dp
p  pu dt
dp
 pS  Q
dt
S = [L]3[t]-1
Q=Q0 cost
puS = Q0
For a pressure range where S does not depend on p, i.e. the pumping speed is constant
Sdt  
V
dp
p  pu
 p  pu 
S
 t
log

V
 pi  pu 
St  V log( p  pu )
This can be used to measure S
or to estimate the time to reach pu
pout
Compression ratio: CR 
pinlet
59
What pump to use?
• Ultimate pressure
• Time to reach the u.p.
• Residual gas composition
• Other (absence of
magnetic fields)
60
Rotary Roughing Pump
inlet
Exhaust valve
Oil
Rotor blade
Eccentric rotor
Spring
Cylindric
body
S: 2,5 ÷ 102 m3/h
0.7 ÷ 28 l/s
CR: 105
Starting operating pressure: 103 mbar
1 m3/h = 0.28 l/s
Pu: 10-2 mbar
61
Dual stage Rotary Roughing Pump
inlet
Exhaust valve
Rotor blade
Eccentric rotor
Spring
Pu: 10-3 ÷ 10-4 mbar
Advantages
Disadvantages
• No saturation
• Heavy duty
• Low cost (2500 €)
• Oil backstreaming
• Need traps for oil vapor
• Noisy
62
Rotary Roughing Pump: gas ballast
CR=105
Op. temp
T 70 °C
The gas can liquefy
inside the rotation chamber
Vapor pressure
Pump water vapor at 70 °C
when P reaches 3.3 104 Pa
The vapor liquefies and does not reach P > 1 105 Pa
So the exhaust valve does not open
The vapor remains inside the pump and is mixed with oil
Decrease pump speed, and can damage the rotor by increasing the friction
63
Rotary Roughing Pump: gas ballast
Solution: gas ballast
NO gas ballast
Gas ballast
The valve is set to
decrease the CR to 10
liquid
Ballast valve
The vapor does not
liquefy
64
Diaphragm Pump
Housing
Valves
Head cover
Diaph. clamping disc
Diaphragm
Diaphragm supp. disc
Connecting rod
Eccentric bushing
CR: 102  103
Starting operating pressure: 103 mbar
Pu: ~ 1 mbar
65
Diaphragm Pump
Advantages
Oil-free
No saturation
Low cost
Disadvantages
High ult.pressure (4 mbar)
Low pump speed
Noisy
66
Root Pump
Eight-shaped rotor turning
in opposite direction
•Clearance between rotors ~ 0.3 mm
•No lubricants
•CR depends on clearance
Advantages
Oil-free
No saturation
High throughput
Disadvantages
Need prevacuum
Medium cost
delicate
67
Root Pump
S and CR of a root pump depend
on the preliminary pump
installed ahead
patm
pp
pr
root
palette
Sr
Sp
The gas flow is the same for both pumps
Sr Pr  Sp Pp
CRr 
Pp
Pr
Sr  SpCRr
Palette: 60 m3/h = 16,8 l/s
Sr = 16,8 x 40 = 672 l/s
68
Turbomolecular Pump
S: 50 ÷ 5000 l/s
CR: 105  109
Starting operating
pressure: 10-2 mbar
Pu: 10-10 mbar
69
Turbomolecular Pump
2kBT
df v 

dv
m
N2  418m / s
vp 
Molecular speed distribution
without blades (only v)
Molecular speed distribution
plus blade speed
70
Turbomolecular Pump
Principle of operation
Molecular regime
Low pressure side
High pressure side
The speed distribution (ellipse) depends on the angle between
V and blade
The pumping action is provided by the collisions
between blades and molecules
71
Turbomolecular Pump
Pumping speed: depends on gas type
After bake out
Residual gas: H2
72
Compression Ratio of a Turbomolecular Pump
73
Turbomolecular Pump
Rotor suspension
Ball bearings (lubricant required)
Magnetic (lubricant absent)
Advantages
Disadvantages
No saturation
Clean (magnetic)
UHV
Any orientation
Cost
Delicate
Quite noisy
70 l/s ~ 4000 €
250 l/s ~ 9000 €
2000 l/s ~ 20000 €
74
Molecular drag pump
Turbo disk
Threaded stator
Safety ball bearing
Magnetic bearing
Cylindrical Rotor
Operating principle:
Same as turbo but
different geometry
Threaded stator
No blades but threads
Forevacuum flange
(outlet)
Gas entry
Electrical socket
Lubricant reservoir
75
Molecular drag pump
S: 40 ÷ 100 l/s
CR:
H2: 102  109
He: 103  104
Starting operating
pressure: 1-20 mbar
Pu: 10-7 mbar
N2: 107  109
They are use in combination with turbo
in a single mounting so
Higher backing vacuum pressure
Use a low CR backing pump
(i.e. membrane for clean
operation)
76
vapor diffusion pump
Fluid is heated and ejected from nozzles at high speed
baffle
due to the nozzle shape and pressure difference between
inside and pump cylinder.
Fluid speed up to Mach 3-5
The gas molecules are compressed to the pump base through collisions with oil vapor
77
vapor diffusion pump
The pumping speed and the pressure strongly depends on oil type
S: 20 ÷ 600 l/s
Advantages
No saturation
Heavy duty
Low cost
Starting operating
pressure: 10-2 mbar
Pu: 10-9 mbar
Disadvantages
gas reaction
Liquid vapor tension
Contamination
Needs water cooling
78
Getter pumps
Pumping mechanism
- Gas-surface chemical interaction
- Chemisorption
- Solution of gas inside material
Sublimation getters
The active material is sublimated
by thermal heating
Non evaporable getters
The active material is
constituted by porous medium
79
Sublimation getter pumps
Pumping mechanism
- Gas-surface chemical interaction
- Chemisorption
- Solution of gas inside material
Sublimation getters
Ti or Ti – Mo filaments
The material form a thin film
on the pump walls that becomes
the active layer
The molecules are chemisorbed on the film
80
Non evaporable getter pumps
Pumping operation
Cartridge of porous material (Zr-16%Al)
Activated by heating (750 °C) and kept
at operating T 300 °C to increase
molecule diffusion
Problem: saturation of getter material requires cartridge change
81
Pumping speed (l/s)
Getter pumps
T
S 
A
m
Adsorption
probability
area
Molecular weight (g)
S strongly depends on gas
sublimation
> 103 l/s
Non evaporable
800- 2x103 l/s
A’= sublimation, A=non evaporable
Zr-Al
S depends on active surface saturation
82
Getter pumps
With a number of panels
one can obtain S > 1x104 l/s
Stripes of active material
Plus: Wall cooling
Gas-surface weak interaction
Physisorption and
diffusion into the bulk
But if warmed it releases the gas
Advantages
Pump H2
Heavy duty
Low cost
No contamination
Disadvantages
Saturation
Metal vapours
No rare gas pumping
Pressure limit:
10-10 ÷ 10-12 mbar
83
Ion-getter pump
Pumping mechanism
- Gas-surface chemical interaction
- Chemisorption
- Solution of gas inside material
- Ionization of gas molecules
- Burying inside the active material
Ion-getter with cathodic grinding
Ti
~1 Tesla
7 KV
84
Basic processes occurring within a single cell
• e- ionize molecules
• Secondary e- ionize molecules
Ions are accelerated to cathodes
• produce secondary e• grind up cathode material
• make craters
Ions buried into
cathode material
H2: accumulates into the cathodes
Produce cathode vapors
Depositing also on anodes
to work as getters
Need regeneration by annealing
85
Ion-getter pump
S: 4 ÷ 1000 l/s
Advantages
Heavy duty
No traps
No contamination
Any mounting position
Silent
Pressure limit:
10-11 ÷ 10-12 mbar
Starting operating
pressure: 10-3 10-4 mbar
Disadvantages
High magnetic fields
Low pump S for H2
Medium - high cost
86
Cryogenic pumps
Pumping mechanism
Liquid He cooled
Cold walls
- Gas – cold surface interaction
- Physisorption, condensation
Liquid N2 cooled
Adsorbing material
Adsorbing pumps
Pumping mechanism
- Gas – cold surface interaction
- Physisorption
Adsorbing porous material
High surface/volume ratio
Zeolites
Al2O3, SiO2
H2O and N2 pumping
87
Cryopump
Pumping mechanism
- Gas – cold surface interaction
- Physisorption and condensation
Metal wall
88
vapor pressure
Cryopump
The gas condensation
if gas pressure > vapor pressure
at wall T
S: 4 ÷ 100 l/s
Starting operating
pressure: 10-9 mbar
Advantages
Heavy duty
No contamination
Low cost
Disadvantages
Pressure limit:
10-10 ÷ 10-11 mbar
Saturation
Noisy
Needs other UHV pumps
89
Ionization in gases
Type of collisions:
- neutral Molecule – electron
- neutral Molecule – ions
- neutral molecule – neutral molecule (Penning)
- radiation absorption
- neutral Molecule – hot metal surface
Ionization of a molecule (atom) from collisions with e-
-
+
-
Ion -
Ion +
90
Ionization in gases
Ionization energy
Electron affinity
Ion -
Ion +
-
+
Less probable
eV
More probable
91
Atom or neutral molecule – electron collision
Collision type:
- elastic
- atom excitation
- molecule dissociation
- Ionising ( e)
Elastic collision
Considering the relative speed and energy conservation
ve  
Eke  Ekmol
mmol  me
relative energy loss
for electrons
mmol
vmol
me
In the collision the kinetic energy of electrons (and of the molecules)
remains almost unchanged
Ke 
Eki  Ekf
Eki
Very small
92
Elastic collision
e- suffers very small energy loss for each elastic collision
e- mean free path e = average space between two elastic collisions
e- collision rate e = collisions number per unit time
L
L   ete
number of collisions
L
e
  et
L / e
Total energy loss
K E
i
e
ki
93
Elastic collision
Apply external electric field E
If e- has vin~ 0
eEe
Ek 
2Ke
v max
eE

me e
Maximum kinetic energy of an emoving in a gas
kBT 1
e  2
r p
e kBT E
Ek 
2Ke r 2 p
Depends on electric field
and pressure
94
Ionization
if
Ek 
eEe
e kBT E

 Wi
2
2Ke r p
2Ke
Ionization energy
e- can ionize an atom
-
But it can also
- Increase the atom kinetic energy
- Excite an e- to unoccupied bound states
Ionization probability i = ionizing collisions/total collisions
+
-
95
Ionization
e- can ionize an atom
But it can also
- e- trapped inside atom
with formation of negative ions
-
Due to practical measurements
e-
with Ek
unit pressure
unit lenght
Specific ionization coefficient
-
number of (e-  ion)
i 
incident electron
i
i 
e p
Long path to produce more ions
96
Vacuum measurement
Different types of vacuometers depending on pressure range
Mechanical, thermal, ionization
97
Vacuum measurement
Bourdon
Mechanical
Membrane
tube
Pin wheel
index
To vacuum
105  102 Pa
(103  1 mbar)
The tube curvature changes
with pressure
Needs calibration
Precision: 1-2% fsr
105  102 Pa
(103  1 mbar)
R2
x0 
p
4T
The membrane or bellow bends
with pressure
Needs calibration
Precision: 1-2% fsr
98
Thermal conductivity vacuometers
Pirani
heated filament
The filament temperature, and hence the resistance
depends on heat dissipation in the gas,
i.e. on the gas pressure
Pressure variation means T variation i.e.
resistance variation.
This is measured through the W. bridge V variation
V
W 
Rf
2
2
 R3 


 R  R   pK gasT  T
3 
 2
4
 KfT
99
Thermal conductivity vacuometers
unbalanced
V
W 
Rf
2
2
Thermal
dissipation
contact
dissipation
 R3 

  pK gasT  T 4  KfT
R  R 
3 
 2
radiative
dissipation
= cost Stephan-Boltzmann
=wire emissivity
Kgas= gas thermal conductivity
Kf= wire thermal conductivity
=coefficient
For small p, the reference bridge is
2
V  R3 

  T
W0 

Rf  R2  R3 
2
0
Hence
V V
W W0 
Rf
2
2
0
4
2
 KfT
 R3 


 R  R   pK gasT
3 
 2
p  cost (V 2 V02 )
The pressure is obtained by measuring
the Wheatstone voltage
In general it depends
on the gas type
100
Ionization vacuum gauges
Hot cathode
Cold cathode
Based on gas ionization and current measurements
101
Ionization vacuum gauge
I+ = ion current
i = specific ionization coefficient
I- = electron current
from filament
Sensitivity K = σi · λe
e = electron mean free path
I+ = I- i e p
Directly proportional to pressure
The gauge measure the total pressure
Range: 10-4 – 10-12 mbar
Sensitivity K = i e
K depends on gas, gauge geometry,
gauge potential
102
Usually one increases  by designing the gometry
Ionization vacuum gauge
1 tesla
Cold cathode
electrons from gas or field emission
similar to the behavior inside the
ion getter pumps
Less precise due to problem of
discharge current at low pressure
No filament so less subject to
Filament faults
Range: 10-4 – 5 x10-10 mbar
Note: discharge starts only by mag field
to avoid high E field - induced currents
103
Mass Spectrometry
Need to distinguish the intensity of specific gas molecules
Collect molecules
• Molecule ionization
• Separation of different molecules
• Current measurement
Specific mass = ion mass (a.u.)/ion charge =
m
ne
n = ion ionization multiplicity
Specific mass of Ar+ = 40
Specific mass of Ar++ = 20
For a single molecule there are
many peaks, depending on n
104
Mass Spectrometry
Specific mass table
105
Mass Spectrometry
detector
To remove
secondary electrons
Faraday cup
All ions measured
No filaments
Low sensitivity
sturdy
Amplifier time constant large
Channeltron - electron multiplyer
High sensitivity
Delicate
Fast response
106
Quadrupole Mass Spectrometry (QMS)
Storing
system
Detector
(Channeltron)
Analyser
(Quadrupole field)
Ion source
(filament)
Vacuum
Chamber
Quadrupole field between the rods
Ions of varying mass are shot axially into the rod
The applied quadrupole field deflects the ions in
the X and Y directions, causing them to describe
helical trajectories through the mass filter.
107
Quadrupole Mass Spectrometry (QMS)
The forces are uncoupled along x,y,z axis
Quadrupole potential
U (x 2  y 2 )

r02
-(U+Vcos(t))
U+Vcos(t)
r0 = rod separation (~3mm)
Superimpose an oscillating field Vcos(t)
108
Quadrupole Mass Spectrometry (QMS)
(U V cos t )(x 2  y 2 )

r02

x

my  e
y
mz  0
mx  e
ion equation of motion
x
0
r02
y
my  2e (U V cos t ) 2  0
r0
mz  0
mx  2e (U V cos t )
Constant speed along z
a 
q 
8eU
m 2r02
4eV
m 2r02
Stability parameters

t
2
109
Quadrupole Mass Spectrometry (QMS)
d 2x
 (a  2q cos 2 )x  0
2
d
Solved numerically for different a and q
d 2y
 (a  2q cos 2 )y  0
d 2
Ions oscillate in the xy plane
Only some e/m values reach detector
Solutions inside are real (stable trajectory)
All solutions outside are imaginary
and give increasing oscillation amplitudes
Neutralization of the ions on the rods
110
Quadrupole Mass Spectrometry (QMS)
d 2x
 (a  2q cos 2 )x  0
d 2
Zoom to region I
d 2y
 (a  2q cos 2 )y  0
d 2
a
q
fixed U, V and  the overall ion motion
can (depending on the values of a and q)
result in a stable trajectory
causing ions of a certain m/z value
to pass the quadrupole
8eU
a 
m 2r02
for
4eV
q 
m 2r02
Stable solutions
U
a

 0,336
V
2q
The line shrink to one point
Only one ion with m/e ratio can reach detector
m
4V

e q 2r02
111
Quadrupole Mass Spectrometry (QMS)
Reducing U relative toV, an increasingly wider m/z
range can be transmitted simultaneously.
Zoom to region I
Work line
a 
for
8eU
m 2r02
q 
4eV
m 2r02
U
a

 0,336
V
2q
The line enter the stable solutions region
All the ions with a/q on the line will reach detector
the width q of the stable region determines the resolution.
By varying the magnitude of U and V at constant U/V ratio
an U/V = constant scan is obtained
ions of increasingly higher m/e values to travel through the quadrupole
q
V=V0cos(t)
112
Quadrupole Mass Spectrometry (QMS)
How to select different m/e ratios
Change U, V keeping the ratio constant
This results in different m/e values
allowed to pass the quadrupole
In the U,V space a changing in m/e ratio means
moving along the straight line
113
Quadrupole Mass Spectrometry (QMS)
VAC > UDC
Ion traveling in the z direction
-(U+Vcos(t)) (y)
X Motion will tend to be stable
for (t) >0 and unstable (t) <0
U+Vcos(t) (x)
Greatest effect on the lighter ions (smaller m/e)  ejected to the electrodes
Heavier ions will be less perturbed and stay in a stable transmitting trajectory.
The pair of X rods acts as a high pass filter.
Y Motion will tend to be stable
for (t) <0 and unstable (t) >0
Greatest effect on the heavier ions (larger m/e)  ejected to the electrodes
Lighter ions will be less perturbed and stay in a stable transmitting trajectory.
The pair of Y rods acts as a low pass filter.
the two pairs act as a band pass filter, low and high mass ions outside the band
being rejected, with limits determined by the values of U, V, and ω.
114
Quadrupole Mass Spectroscopy (QMS) profiles of the residual gas
H2O
p ≈ 3x10-7 mbar
Before bake-out
H2
H2O
CO+N2
CO2
p ≈ 5x10-11 mbar
After bake-out
115
VACUUM SEALING
Low Vacuum
Clamps
Viton rings
No bake at high temperatures
Reusable
116
VACUUM SEALING
UHV
HV
Plastic deformation
and shear
Bake at high temperatures
Reusable (maybe once)
117
VALVES
Diaphragm
Butterfly
118
VALVES
Stem
All metal
Dynamometric sealing
119
VALVES
Leak
Gate
High conductance
UHV to air compatible
Large clearance for instruments
Bakeable
120
FEEDTHROUGH
Multi-pin for signal
or
Low currents
Multi-pin for high
currents
121
MANIPULATION
Rotation
122