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Introduction Basic concepts of vacuum Vacuum Hardware ( pumps , gauges ) Mass Spectrometry

VACUUM PUMPS AND HARDWARE. OUTLINE. Introduction Basic concepts of vacuum Vacuum Hardware ( pumps , gauges ) Mass Spectrometry. Research applications: impact on everyday life. GETTERS. NEED OF VACUUM. TV TUBES LCD BACKLIGHT GAS LIGHTS (NEON, HIGH POWER LAMPS) DEWAR (FOR DRINKS).

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Introduction Basic concepts of vacuum Vacuum Hardware ( pumps , gauges ) Mass Spectrometry

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  1. VACUUM PUMPS AND HARDWARE OUTLINE • Introduction • Basic concepts of vacuum • Vacuum Hardware (pumps, gauges) • Mass Spectrometry

  2. 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

  3. 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

  4. Ultra High Vacuum Apparatus

  5. Gas kinetics velocity distribution 1D kB = Boltzmann constant velocity distribution 3D probability of finding a particle with speed in the element dv aroundv Maxwell-Boltzmann distribution probability density of finding a particle with speed in the element dv aroundv

  6. Gas kinetics

  7. Gas kinetics T (°C) f(v) Molecular speed Most probable Mean Quadratic mean Neon @ 300 K mNe = 20 • 1.67 x 10-27 kg

  8. Gas kinetics 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 areaand unit time on a surface dS we take the molecules arriving with speed vx in a time dt total number of molecules with speed vx hitting the unit surface in a time dt Nmolecule = how many molecules in dV? volume dV = vdtcosdS

  9. Gas kinetics

  10. Gas kinetics molecules arrival rate R at a surface (unit area, time) if Mr=relative Molar mass m = Mr•amu mNe = 20 • 1.67 x 10-27 g kB = Boltzmann’s constant (J/K) T = Temperature (K) p = Pressure (torr) MOx =32 O2at 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

  11. 2r Gas kinetics Mean free path 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

  12. rA rB Gas kinetics For different molecules A and B Mean free path 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

  13. Gas kinetics: why the UHV H2O Residual Gas CO O2 CO2 CH4 N2 Solid Surface 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 10-6torr 10 sec @ p = 1 x 10-7torr 100 sec @ p = 1 x 10-8torr 1,000 sec @ p = 1 x 10-9torr 10,000 sec @ p = 1 x 10-10torr 100,000 sec @ p = 1 x 10-11torr Utra High Vacuum (UHV): p < 10-10-10-11torr

  15. Plots of relevant vacuum features vs. pressure

  16. Gas flow through a pipe Pipe d p = pressure measured in the plane dV = volume of matter crossing the plane dV p dV/dt = Volumetric flow rate (portatavolumetrica) [Q] = [p][L]3[t]-1 Throughput 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 Particle flow rate: variation of number of molecules through an area

  17. Gas flow through a pipe M = total mass Mass flow rate Throughput Variation of mass through an area 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 d pipe For  < d viscous For   d intermediate For  > d molecular Viscous S = layer contact area dvx /dy = mol speed gradient The mol-mol collisions are dominant Friction force  = viscosity laminar turbulent

  19. Regimes of gas flow through a pipe pipe d mass flow For a pipe with diameter d and section d2/4 Q’ mass flow per unit section  = viscosity Reynolds number Laminar: Re<1200 turbulent: Re>2200

  20. Regimes of gas flow through a pipe Reynolds number 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 viscous For  < d For   d For  > d intermediate molecular Only for intermediate and molecular flow Knudsen number = d/ intermediate 3  d/  80 d/  3 molecular intermediate: 10-2 p d  0.5 molecular: p d  10-2 For air at RT p in Torr, λ in cm

  24. gas flow across pipe Pipe conductance: [C] = [L]3[t]-1 Pressures at pipe ends SI: m3s-1 For small aperture connecting two chambers cgs: lt s-1 N1, V1 P1 N2, V2 P2 Arrival rate R: number of particles landing at a surface per unit areaand unit time

  25. Pipe conductance Viscous and intermediate regime (Poiseuille law) Laminar Turbulent Molecular regime For air at 0 C: 11,6 d3/L [lt/s] Long cylindrical pipe Elbow pipe The molecules must collide with walls at least once before exiting Equivalent to a longer pipe

  26. Pipe conductance: Pipe impedance: In parallel

  27. In series Q1 = Q2 = QT or gas would accumulate

  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 Pumping speed S = Q/p0 [S] = [L]3[t]-1 C Volumetric flow rate SI: m3s-1 cgs: lt s-1 In the presence of a pipe Q at the pump inlet is the same as Q in pipe Effective pumping speed in the vessel

  30. p0 Relevant physical parameters of a pumping system Q= flow through aspiration aperture p = Vessel Pressure V = Vessel Volume C Pumping speed S = Q/p0 Effective pumping speed [S] = [L]3[t]-1 if C = S the Se is halved

  31. p0 Relevant physical parameters of a pumping system Q= flow through aspiration aperture p = Vessel Pressure V = Vessel Volume Sources of flow (molecules) 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 Q = Q0 +Q1 S = Pumping speed p = Vessel Pressure V = Vessel Volume Long time limit Short time limit True leak rate Only the gas initially present contributes Virtual leak rate Other outgassing sources contribute

  33. Pump-down equation for a constant volume system Q = Q0 +Q1 S = Pumping speed p = Vessel Pressure V = Vessel Volume Short time limit True leak rate Suppose: Constant S Q= 0 Time needed to reduce p by 50 % Vol of 1 m3 = 103 L to be pumped down from 1000 mbar to 10 mbar in 10 min = 600 s V= 1000 L P0 = 133 Pa S= 20 L/s t = 331,6 s 7.5 L/s = 27 m3/h

  34. Pump-down equation for a constant volume system 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 Ultimate pressure

  35. Pressure versus distance

  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. ModernVacuumPhysics, Ch. 5.8

  37. Example CVD coatings on panels Antireflective coatings, p-n junction growth for solar panels P1 P2 P1 P0 P0 C C C S2 S3 S1 S1 = Cp0/p1 S2 = Cp1/p2 S3 = Cp2/p1

  38. Gas-solid interaction H2O CO inelastic trapped elastic CO2 physicaladsorption (shortened to Physisorption): bonding with structure of the moleculeunchanged Chemisorption: bondinginvolves electron transfer or sharingbetween the molecule and atoms of the surface Can be thought of as a chemicalreaction CH4 N2 O2 He H2

  39. Gas-solid interaction Physisorption CO Origin: Van der Waals forces H2O CH4 CO2 O2 N2 Typical q: 6 - 40 kJ/mol = 0,062 - 0,52 eV /molecule He H2 The welldepthis 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 N2 Typical q: 40 - 1000 kJ/mol = 0,52 - 10 eV /molecule He H2 The well depth is the energy of adsorption P is a precursor state the moleculeshave 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 potentialbarrier and break free of the surface. probabilitythatfluctuations in the energy willresult in an energyq Boltzmann factor probability per second that a molecule will desorb

  42. Gas-solid interaction probability per second that a molecule will desorb p(t) = probability that it is still adsorbed after elapsed t p(t+dt) = p(t) x (1-dt) O2 probability of not being desorbed after dt dp = p(t+dt) - p(t) = - dt p(t) average time of stay

  43. Gas-solid interaction average time of stay At RT  1013 s−1 Molecular dependance O2 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): monomolecularlayeradsorbed on a surface number of molecules in a monolayer: N0  1015 na = number of adsorbed molecules per unit area fractionalcoverage Probability that, on striking the surface already having coverage θ, a molecule becomes adsorbed. s() = sticking coefficient

  45. Gas-solid interaction For dilute adsorbed layers (<<1) simple model for the equilibrium state rate of adsorption rate of desorption at equilibrium N2at p = 1 x 10-5 mbar, 293 K R = 2.9 1015 cm-2 s-1 na,eq = 1 x 1012 cm-2 seq= 0.2 a = 5 x 10-3 s  = 0.003 equation of state for the adsorbedphase Coverageisproportional to pressure

  46. Desorption P = 1000 mbar P = 10-7 mbar pumping Equilibrium Far from equilibrium till…. Experimental relation Gas flow /area  = 0.5  = 1 for metals  = 1  = 0.5 for elastomers q1metals1x10−7mbar L s−1cm-2 For 1 mbarL at RT with q1metals, outgassingrate  1012molec s−1cm-2 Nat2.6x1019  10-3ML /second of released gas, principally water vapor

  47. Desorption How important is the molecule/surface interaction energy? H2O Rate of desorption N2 integrate fall of pressure at RT decays exponentially from the initial state with a time constant equal to the stay time H2O Simple model calculation idealized UHV system RT, V= 1 L, A = 100 cm2 S = 1 L/s only gas source: initially complete ML of specifiedbindingenergyadsorbed at the wall 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 p2 = 1x10-8 mbar p1 = 1000 mbar H2O CO CO2 CH4 N2 O2 He H2 Residual Gas

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