PHYSICAL VAPOR DEPOSITION (PVD PVD II: Evaporation ◆ We saw oⅴD Gas phase reactants: po l mTorr to l atm Good step coverage, T> 350 K ◆ We saw sputtering Noble (+ reactive gas)p 10 mTorr; ionized particles Industrial process high rate reasonable step coverage Extensively used in electrical, optical, magnetic devices Now see evaporation: Source material heated, Peg, vap. 10- Torr, Pg 10- Torr Generally no chemical reaction(except in "reactive depos'n n=10s of meters. Knudsen number n >>1 Poor step coverage, alloy fractionation: 4 pvapor Historical(optical, electrical Campbell, Ch 12 is more extensive than Plummer on evaporation 6.152J3.155J
6.152J/3.155J 1 PHYSICAL VAPOR DEPOSITION (PVD) PVD II: Evaporation We saw CVD Gas phase reactants: pg ≈ 1 mTorr to 1 atm. Good step coverage, T > 350 K We saw sputtering Noble (+ reactive gas) p ≈ 10 mTorr; ionized particles Industrial process, high rate, reasonable step coverage Extensively used in electrical, optical, magnetic devices. Now see evaporation: Source material heated, peq.vap. =~ 10-3 Torr, pg > 1 Poor step coverage, alloy fractionation: ∆ pvapor Historical (optical, electrical) Campbell, Ch. 12 is more extensive than Plummer on evaporation
Standard vacuum chambers 1≈106Tor(1.3×104N Mostly H,O, hydrocarbons, N, He by residual gas analySis(RGa=mass spec 6.152J3.155J
6.152J/3.155J 2 Standard vacuum chambers Σpi ≈ 10-6 Torr (1.3 × 10-4 N / m2) Mostly H2O, hydrocarbons, N2 He by residual gas analysis (RGA = mass spec.)
Ultra-high vacuum chambers p<10-8 Torr demands Stainless steel chamber Bakeable to150°C Turbo, ion, cryo pumps 6.152J3.155J
6.152J/3.155J 3 p < 10-8 Torr demands: Stainless steel chamber Bakeable to 150oC Turbo, ion, cryo pumps Ultra-high vacuum chambers
Knudson number≈1 0.01cm P=10-10 106 10 10-2 100Torr 23 Log[n(#/m5)] 21 I At 0.1MP 760mm 14 lb/in2 Sputtering CVD 15 vaporation Log[P(N/m2) Ballistic. molecular flOw AL>> Generally /L<I High purity films Films less pure Epita Epitaxy is rare 6.152J3.155J
6.152J/3.155J 4 -6 -4 -2 0 2 4 Log[P (N/m2)] 25 23 Log[n (#/m3)] 21 19 17 15 λ = 10 0.01 cm p = 10-10 10-8 10-6 10-4 10-2 100 Torr 1 Atm= 0.1 MPa 760 mm ≈14 lb/in2 Generally λ/L > 1 High purity films Epitaxy Sputtering CVD Knudson number ≈ 1
Atomic flux on surface due to residual gas atoms nv 2k T area. t 2k V zm, t Given 10-b Torr of water vapor room temp, find flux latm 10'P p=10°Tor 760t atm kgT(R7)=0.025eV=4×102J P=1.3×10-4 18 3×10kg J=4.8x104atoms/molecules cm sec What is atomic density in I monolayer(MD) of Si? N=5x1022cm3=1.3x1015cm So at 10-6 Torr, 1 ML of residual gas hits surface every 3 seconds Epitaxy requires slow deposition, surface mobility So you must keep pressure low to maintain pure film 6.152J3.155J
6.152J/3.155J 5 Atomic flux on surface due to residual gas J atoms area ⋅ t = nv x 2 = p 2kBT 2kBT πm = p 2πmkBT = J Given 10-6 Torr of water vapor @ room temp, find flux p =10−6Torr × 1atm 760 T × 105Pa atm , kBT RT ( )= 0.025eV = 4 ×10−21 J p =1.3×10−4 N m 2 mH2O = 18 NA = 3×10−26 kg What is atomic density in 1 monolayer (ML) of Si? N = 5 x 1022 cm-3 => 1.3 x 1015 cm-2. So at 10-6 Torr, 1 ML of residual gas hits surface every 3 seconds! Epitaxy requires slow deposition, surface mobility, So you must keep pressure low to maintain pure film J = 4.8 ×1014 atoms/molecules cm2sec
Now add evaporation source e-beam ooooo ooooo B field Resistive heater RF-induction heater △H Equilibrium vapor pressure py=po exp 4H= heat of vaporization Strong t dependence Work function ee Heat of vaporization solid 6.152J3.155J
6.152J/3.155J 6 Now add evaporation source Equilibrium vapor pressure: ∆H = heat of vaporization pv = p0 exp − ∆H kBT Strong T dependence e- + _ e-beam B field Resistive heater I RF-induction heater Work function e- Heat of vaporization solid V(x) free
Vapor pressure of elements employed in semiconductor materials. Dots correspond to melting points Rely on tables, attached: Pvapor>>Pvac Elemental metals easy to evaporate, but.y -2 alloys Differential Vapor compounds so use 2 crucibles or deposit multilayers and diffuse Oxides, nitrides deposit in oxygen (or other) partial p 00 400600800100015062000 6.152J3.155J TEMPERATURE (K
6.152J/3.155J 7 Vapor pressure of elements employed in semiconductor materials. Dots correspond to melting points Rely on tables, attached: pvapor > > pvac, Elemental metals easy to evaporate, but… alloys compounds Differential pvapor so use 2 crucibles or deposit multilayers and diffuse Oxides, nitrides deposit in oxygen (or other) partial p
9001000 1500 2000 10 10 b9-5o5000 IIIA c Ni」 10 10 Ge 10 te W. mo w as crucibles. 10 9001000 1500 004000 Temperature in Degrees Centigrade
6.152J/3.155J 8 (note W, Mo can be used as crucibles.)
We expressed flux of residual gas molecule 27Imk t Chamber p=106Tor(J≈5×10 cm The source also→flux P For Al at 1000 K, p vap/10-7 Torr (from tigure vap 2rmsourcekgT. m=species to be evaporated, =27 amu for Al Al 800 C but that's not all Net flow out of crucible -A #/t) Mass flow out of crucible -J A m(mass /t) Note: 3 different temperatures: source=Evaporant>>Substrate chamber=Tresid gas=Rt. System not in thermal equilibrium; only thermal interaction among them is by radiation and/or conduction through solid connects(weak contact). No convection when NK <<1 6.152J3.155J
6.152J/3.155J 9 The source also ⇒ flux J = pvap 2πmsourcekBTsource For Al at 1000 K, pvap = 10 -7 Torr (from figure) m = species to be evaporated, = 27 amu for Al JAl ≈ 2 x 1013 Al/cm2-s just above crucible Mass flow out of crucible ~ J Ac m ( mass / t) We expressed flux of residual gas: J ≈ 5 ×1014 molecules cm2 Chamber p = 10 ( s) -6 Torr J = p 2πmkBT Net flow out of crucible ~ J Ac (# / t) Ac Therefore heat Al to T > 800 C but that’s not all… Note: 3 different temperatures: Tsource ≈ Tevaporant >> Tsubstrate > Tchamber = Tresid gas ≈ RT. System not in thermal equilibrium; only thermal interaction among them is by radiation and/or conduction through solid connects (weak contact). No convection when NK << 1
How much evaporant strikes substrate? At 10-6 Torr, trajectories are uninterrupted While a point source deposits uniformly on a sphere about it, a planar source does not J∝ccos, substrate ubstrate R R Convenient geometry R Geometric factor cos 6, cos 6. cos0,=cos0 2R 2r Geometric factor eposition rate 4丌-area·borJ 4mr2\area·t A 1 thick Film growth rate =Jm 4m2(t 6.152J3.155J
6.152J/3.155J 10 How much evaporant strikes substrate? At 10-6 Torr, trajectories are uninterrupted. While a point source deposits uniformly on a sphere about it, a planar source does not: Geometric factor = Ac 2πR2 cosθ1 cosθ 2 cosθ1 = cosθ 2 = R2r Deposition rate = Jm Ac 4πr 2 m area⋅ t or J Ac 4πr2 # area ⋅ t substrate θ1 θ2 R R θ1 θ2 r r substrate Convenient geometry Geometric factor = Ac 4πr2 Film growth rate = Jm Ac 4πr 2 1 ρ f thick t ∝cosθ1 J