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Effusion Cells

Effusion Cells. What is the flux from an effusion cell ?. z-direction. temperature, T. Effusion cell. Kinetic Theory of Gases. Treat the effusion cell as an isothermal container

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Effusion Cells

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  1. Effusion Cells • What is the flux from an effusion cell ? z-direction temperature, T Effusion cell

  2. Kinetic Theory of Gases • Treat the effusion cell as an isothermal container • Energy of particles (atoms or molecules) in a gas at thermodynamic equilibrium will have a Maxwell-Boltzmann probability distribution of energies: f(E) ~ exp(-E/kT) E = ½mv2 = ½ m (vx2 + vy2 + vz2)

  3. Kinetic Theory of Gases • Probability for a particle having velocity (vx, vy, vz) is: exp[-½m (vx2+vy2+vz2)/kT]dvxdvydvz ∫ ∫ ∫exp [-½m (vx2+vy2+vz2)/kT]dvxdvydvz = exp [-½m (vx2+vy2+vz2)/kT]dvxdvydvz / (2pkT/m)3/2 + ∞ - ∞ from Manos, Fig. 13, p. 206

  4. Kinetic Theory of Gases • Probability distribution function for velocity (vx, vy, vz) is: f(vx, vy, vz) = exp [-½m (vx2+vy2+vz2)/kT] /(2pkT/m)3/2 • Probability for velocity vz is: fz(vz) =   f(vx, vy, vz)dvxdvy = (m / 2pkT)½ exp [-mvz2/2kT ]

  5. Impingment Rate • z = impingement rate in z-direction = flux = # particles per unit area per unit time striking the interior surface of a gas container z-direction n = # particles per unit volume vz = average velocity in z direction Effusion cell z = n vz

  6. Impingment Rate z = n vz = n ∫ fzvzdvz = n (kT / 2pm)½ +∞ 0 z-direction Effusion cell

  7. Impingment Rate • z = n (kT / 2pm)½ • Using ideal gas law, n = P / kT z = P / ( 2pmkT)½ impingement rate, z Effusion cell

  8. Impingment Rate Langmuir equation : z = P / ( 2pmkT)½ • P is called the beam equivalent pressure (BEP)

  9. Impingment Rate z = P / ( 2pmkT)½ = 3.513 x 1022 P / (MT)½ where [T] = K, [P] = Torr, [M] = g (molar weight) • e.g., N2 at P = 10-6 Torr and T = 300 K gives z = 3.8 x 1014 cm-2s-1

  10. Monolayer Coverage • How much time is required to cover a surface with one atomic layer (monolayer) ? impingement rate, z Effusion cell

  11. Monolayer Coverage • Time for monolayer adsorption, t: t = Ns / z Ns= surface atom density ~ 1015 cm-2 for Si surface z = 3.8 x 1014 cm-2s-1 → t = 2.6 s

  12. Monolayer Coverage • Need BEP ~ 10-6 Torr to achieve 1 ml/s growth rate • 1 Langmuir (1 L) = coverage achieved in 1 sec at a pressure of 10-6 Torr • 1 L = 10-6 Ts • 1 L/s = flux at a pressure of 10-6 Torr

  13. Film Purity UHV required (< 10-9 Torr) leak Effusion cell • The deposition rate must be much greater than the impingement rate of contaminants • Need Pimpurities << 10-6 Torr to achieve desired purity of thin film (typically UHV)

  14. Film Purity • · Affected by: • Impurities initially present in the source material • Source contamination from the crucible or surrounding materials • Residual gases in the vacuum chamber • → use cryopanels as a fast pump

  15. Beam Equivalent Pressure z = P / ( 2pmkT)½ • P = BEP = equilibrium vapor pressure • P = Po exp (–DGvap/kT) • Po = standard pressure • = 1 atm = 105 Pa = 760 Torr • DGvap = Gibbs free energy of vaporization • DGvap= DHvap – TDSvap • DHvap = heat of vaporization • DSvap = entropy of vaporization • Assumes thermodynamic equilibrium between the condensed phase in the cell and the vapor

  16. Equilibrium VP • P = Po exp(DSvap/k) exp (-DHvap / kT) • Plot of the log of equilibrium VP versus 1/T (Arrhenius plot) will be linear with slope = -DHvap from Ohring, Fig. 3-1, p. 83

  17. Equilibrium VP from Mahan, Table III.2, p. 78

  18. Evaporation of Alloys • What is the VP of an element evaporating from an alloy? • Raoult's law for an alloy: • Poi, alloy = Xi Pio • i = component of the alloy • Poi, alloy = equilibrium vapor pressure of component i from alloy source • Pio = equilibrium vapor pressure of pure component i • Xi = mole fraction of pure component i in the alloy

  19. Evaporation of Alloys z = P / ( 2pmkT)½ For an alloy AB, the ratio of impingement rates is: zA / zB = (XAPAo/ XBPBo) (mB / mA)½ = (PAo/PBo)(mB/mA)½ (XA/XB)

  20. Evaporation of Alloys • zA/zB = (PAo/PBo) (mB/mA)½ (XA/XB) • Film will have different composition than the source • The source gradually becomes depleted in the more volatile component; the film composition may change over time (although mixing may reduce this effect) • Usually need separate sources for components A and B

  21. Evaporation of Alloys From Ohring, Table 3-1, p. 85

  22. Example: Beer Can Evaporator(from Mahan, pp. 8-12) • Top of beer can: Aluminum alloy containing 1% Mg and 1.3% Mn placed in effusion cell • DSvap (J/K) DHvap (kJ) X • Mg 99 134 0.010 • Mn 106 247 0.013 • Al 118 314 0.977 • Pi = XiPo exp(DSvap/k) exp (-DHvap / kT) • PMg (900 K) = 0.0185 Torr • PMn (900 K) = 1.5 x 10-8 Torr • PAl (900 K) = 6.3 x 10-10 Torr • PMg >> PMn > PAl

  23. Beer Can Evaporator z = P / ( 2pmkT)½ zMg (900 K) = 440 Å-2s-1 zMn (900 K) = 2.4 x 10-4 Å-2s-1 zAl (900 K) = 1.42 x 10-5 Å-2s-1 Beer can is a good source of Mg and a poor source of Al A more practical example is the use of GaAs as a source of As

  24. Temporal Variation of Flux

  25. Spatial Variation of Flux ·JW =# particles per unit time per unit solid angle emitted in direction q by orifice of area A A from Mahan, Fig. V.1, p. 116

  26. Spatial Variation of Flux   JW = z Acosq / p (cosine law of emission) Beer can evaporator (A = 1 cm2 orifice, q = 0°): JW, Mg = 1.4 x 1018 s-1 JW, Mn = 7.7 x 1011 s-1 JW, Al = 4.5 x 1010 s-1

  27. Spatial Variation of Flux • Ji = # particles per unit time per unit area dA emitted in direction b b dA dW S

  28. Spatial Variation of Flux b dA R dW S • Flux incident on a substrate: • Ji = JWcosb / R2 • R = source-to-substrate distance • = deposition angle = angle between R and substrate normal

  29. Spatial Variation of Flux Ji = JWcosb / R2 = zAcosqcosb / pR2 Beer can evaporator (R = 10 cm, b = 0°): Ji, Mg = 1.4 Å-2s-1 Ji, Mn = 7.7 x 10-7 Å-2s-1 Ji, Al = 4.5 x 10-8 Å-2s-1

  30. Spatial Variation of Flux · 3 types of receiving surface: Hemisphere Planar  Spherical from Mahan, Fig. VIII.5, p. 277 Ji = zAcosqcosb / pR2

  31. Spatial Variation of Flux from Mahan, Fig. VIII.5, p. 277 hemisphere: Ji = zAcosq / ph2        planar: Ji = zA(cosq)4 / ph2        spherical: Ji = zA / ph2

  32. Spatial Variation of Flux · Spherical geometry provides most uniform deposition from Mahan, Fig. VIII.6, p. 279

  33. Spatial Variation of Flux · Planetary deposition systems from Mahan, Fig. VIII.7, p. 279

  34. Spatial Variation of Flux • · Deviations from cosine law occurs due to: • Shape of effusion cell • 2. Finite thickness & diameter of orifice • a. Particles adsorbed • b. Particles reemitted • (specular + diffusive)

  35. Spatial Variation of Flux from Mahan, Fig. V.7, p. 125 from Mahan, Fig. V.8, p. 128

  36. Spatial Variation of Flux • Design of effusion cell affects the uniformity of film thickness from Panish & Temkin, Fig. 3.3, p. 59

  37. Spatial Variation of Flux • Determines the compositional variations across the film from Ohring, Fig. 7-21(a), p. 338

  38. Spatial Variation of Flux ·  Rotate substrate: 1. Improve uniformity of flux distribution 2. Prevent shadowing 3. Achieve conformal coverage from Jaeger, Fig. 6.5, p. 114

  39. PVD Process from Mahan, Fig. I.6, p. 7

  40. Growth Rate • · Condensation flux, • Jc ~ a [ Ji – zeq(Tsub) ] • = accommodation coefficient From Mahan, Fig. I.7, p. 11

  41. Growth Rate • · For a thin film to form • Ji (Tsource) > zeq (Tsub) • (supersaturation condition) • · Condensation flux, • Jc ~ a (Ji – zeq) • = accommodation coefficient • ·  Growth rate, • GR = Jc / nf • nf = film density

  42. Growth Rate • · Beer can evaporator: • Suppose Tsub = 580 K • Tsource = 900 K • Then Ji,Mg (900 K)= 1.40 Å-2s-1 • zeq, Mg (580 K) = 2.79 Å-2s-1 • No growth occurs even though Tsub < Tsource • Need Tsub < 570 K

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