A MINI-COURSE ON THE PRINCIPLES OF LOW-PRESSURE DISCHARGES AND MATERIALS PROCESSING Michael A. Lieberman Department of Electrical Engineering and Computer Science, CA 94720 LiebermanMinicourse10 1
OUTLINE Introduction Summary of Plasma and Discharge Fundamentals Global Model of Discharge Equilibrium Break Inductive Discharges Reactive Neutral Balance in Discharges Adsorption and Desorption Kinetics Plasma-Assisted Etch Kinetics LiebermanMinicourse10 2
INTRODUCTION TO DISCHARGES AND PROCESSING LiebermanMinicourse10 3
THE NANOELECTRONICS REVOLUTION Transistors/chip doubling every 1 1 2 2 years since 1959 1,000,000-fold decrease in cost for the same performance EQUIVALENT AUTOMOTIVE ADVANCE 4 million km/hr 1 million km/liter Never break down Throw away rather than pay parking fees 3 cm long 1 cm wide Crash 3 a day LiebermanMinicourse10 4
EVOLUTION OF ETCHING DISCHARGES LiebermanMinicourse10 5
ISOTROPIC ETCHING 1. Start with inert molecular gas CF 4 2. Make discharge to create reactive species: CF 4 CF 3 + F 3. Species reacts with material, yielding volatile product: Si + 4F SiF 4 4. Pump away product ANISOTROPIC ETCHING 5. Energetic ions bombard trench bottom, but not sidewalls: (a) Increase etching reaction rate at trench bottom (b) Clear passivating films from trench bottom Mask Plasma Ions LiebermanMinicourse10 6
DENSITY VERSUS TEMPERATURE LiebermanMinicourse10 7
RELATIVE DENSITIES AND ENERGIES Charged particle densities neutral particle densities LiebermanMinicourse10 8
NON-EQUILIBRIUM Energy coupling between electrons and heavy particles is weak Input power Electrons weak Ions strong Walls Walls weak weak strong Neutrals strong Walls Electrons are not in thermal equilibrium with ions or neutrals T e T i Bombarding E i E e in plasma bulk at wafer surface High temperature processing at low temperatures 1. Wafer can be near room temperature 2. Electrons produce free radicals = chemistry 3. Electrons produce electron-ion pairs = ion bombardment LiebermanMinicourse10 9
ELEMENTARY DISCHARGE BEHAVIOR Uniform density of electrons and ions n e and n i at time t = 0 Low mass warm electrons quickly drain to the wall, forming sheaths Ion bombarding energy E i = plasma-wall potential V p Separation into bulk plasma and sheaths occurs for ALL discharges LiebermanMinicourse10 10
SUMMARY OF FUNDAMENTALS LiebermanMinicourse10 11
THERMAL EQUILIBRIUM PROPERTIES Electrons generally near thermal equilibrium Ions generally not in thermal equilibrium Maxwellian distribution of electrons ( ) 3/2 m f e (v) = n e exp( mv2 2πkT e 2kT e where v 2 = vx 2 + vy 2 + vz 2 f e (v x ) ) v Te = (kt e /m) 1/2 v x Pressure p = nkt For neutral gas at room temperature (300 K) n g [cm 3 ] 3.3 10 16 p[torr] LiebermanMinicourse10 12
AVERAGES OVER MAXWELLIAN DISTRIBUTION Average energy 1 2 mv2 = 1 n e d 3 v 1 2 mv2 f e (v) = 3 2 kt e Average speed v e = 1 ( ) 1/2 d 3 8kTe v vf e (v) = n e πm Average electron flux lost to a wall z Γ e [m 2 s 1 ] y Γ e = x dv x dv y dv z v z f e (v) = 1 4 n e v e [m 2 -s 1 ] 0 Average kinetic energy lost per electron lost to a wall E e = 2 T e LiebermanMinicourse10 13
FORCES ON PARTICLES For a unit volume of electrons (or ions) du e mn e = qn e E p e mn e ν m u e dt mass acceleration = electric field force + + pressure gradient force + friction (gas drag) force m = electron mass n e = electron density u e = electron flow velocity q = e for electrons (+e for ions) E = electric field p e = n e kt e = electron pressure ν m = collision frequency of electrons with neutrals p e p e p e (x) p e (x + dx) Drag force u e Neutrals x x + dx x LiebermanMinicourse10 14
BOLTZMANN FACTOR FOR ELECTRONS If electric field and pressure gradient forces almost balance 0 en e E p e Let E = Φ and p e = n e kt e Φ = kt e n e e n e Put kt e /e = T e (volts) and integrate to obtain Φ n e (r) = n e0 e Φ(r)/T e n e x n e0 x LiebermanMinicourse10 15
DIELECTRIC CONSTANT ǫ p RF discharges are driven at a frequency ω E(t) = Re(Ẽ ejωt ), etc. Define ǫ p from the total current in Maxwell s equations H = J c + jωǫ 0 Ẽ }{{} jωǫ pẽ Total current J Conduction current is J c = en e ũ e Newton s law is jωmũ e = eẽ mν mũ e Solve for ũ e and evaluate J c to obtain ǫ p ǫ 0 κ p = ǫ 0 [1 ω 2 pe ω(ω jν m ) with ω pe = (e 2 n e /ǫ 0 m) 1/2 the electron plasma frequency For ω ν m, ǫ p is mainly real (nearly lossless dielectric) ] LiebermanMinicourse10 16
CONDUCTIVITY σ p It is useful to introduce rf plasma conductivity J c σ p Ẽ Since J c is a linear function of Ẽ [p. 16] σ p = e 2 n e m(ν m + jω) DC plasma conductivity (ω ν m ) σ dc = e2 n e mν m RF current flowing through the plasma heats electrons (just like a resistor) LiebermanMinicourse10 17
SUMMARY OF DISCHARGE FUNDAMENTALS LiebermanMinicourse10 18
ELECTRON COLLISIONS WITH ARGON Maxwellian electrons collide with Ar atoms (density n g ) # collisions of a particular kind s-m 3 = νn e = Kn g n e ν = collision frequency [s 1 ], K(T e ) = rate coefficient [m 3 /s] Electron-Ar collision processes e + Ar Ar + + 2e (ionization) e + Ar e + Ar e + Ar + photon (excitation) e e + Ar e + Ar (elastic scattering) e Rate coefficient K(T e ) is average of cross section σ(v R ) [m 2 ] over Maxwellian distribution K(T e ) = σ v R Maxwellian v R = relative velocity of colliding particles Ar Ar LiebermanMinicourse10 19
ELECTRON-ARGON RATE COEFFICIENTS LiebermanMinicourse10 20
ION COLLISIONS WITH ARGON Argon ions collide with Ar atoms Ar + + Ar Ar + + Ar (elastic scattering) Ar + + Ar Ar + Ar + (charge transfer) Ar + Ar + Ar Ar Ar + Ar Ar Ar + Total cross section for room temperature ions σ i 10 14 cm 2 Ion-neutral mean free path (distance ion travels before colliding) λ i = 1 n g σ i Practical formula λ i (cm) = 1, p in Torr 330 p LiebermanMinicourse10 21
THREE ENERGY LOSS PROCESSES 1. Collisional energy E c lost per electron-ion pair created K iz E c = K iz E iz + K ex E ex + K el (2m/M)(3T e /2) = E c (T e ) (voltage units) E iz, E ex, and (3m/M)T e are energies lost by an electron due to an ionization, excitation, and elastic scattering collision 2. Electron kinetic energy lost to walls E e = 2 T e 3. Ion kinetic energy lost to walls is mainly due to the dc potential V s across the sheath E i V s Total energy lost per electron-ion pair lost to walls E T = E c + E e + E i LiebermanMinicourse10 22
COLLISIONAL ENERGY LOSSES LiebermanMinicourse10 23
BOHM (ION LOSS) VELOCITY u B u B Plasma Sheath Wall Density n s Due to formation of a presheath, ions arrive at the plasma-sheath edge with directed energy kt e /2 1 2 Mu2 i = kt e 2 Electron-ion pairs are lost at the Bohm velocity at the plasma-sheath edge (density n s ) u i = u B = ( kte M ) 1/2 LiebermanMinicourse10 24
x STEADY STATE DIFFUSION Particle balance: losses to walls = creation in volume Γ e,i = ν iz n e Ambipolar: equal fluxes (and densities) of electrons and ions Γ = D a n D a = kt e /Mν i = ambipolar diffusion coefficient Boundary condition: Γ wall = n s u B at plasma-sheath edge n 0 Γ wall Γ wall l/2 0 l/2 n s = h l n s /n 0 = edge-to-center density ratio LiebermanMinicourse10 25
x DIFFUSION AT LOW PRESSURES Plasma density profile is relatively flat in the center and falls sharply near the sheath edge n 0 Γ wall Γ wall l/2 Ion and electron loss flux to the wall is Γ wall = n s u B h l n 0 u B The edge-to-center density ratio is 0 h l n s n 0 l/2 0.86 n s (3 + l/2λ i ) 1/2 where λ i = ion-neutral mean free path [p. 21] Applies for pressures < 100 mtorr in argon LiebermanMinicourse10 26
DIFFUSION IN LOW PRESSURE CYLINDRICAL DISCHARGE n sr = h R n 0 R Plasma n e = n i = n 0 n sl = h l n 0 Loss fluxes to the axial and radial walls are Γ axial = h l n 0 u B, Γ radial = h R n 0 u B where the edge-to-center density ratios are 0.86 h l (3 + l/2λ i ) 1/2, h 0.8 R (4 + R/λ i ) 1/2 Applies for pressures < 100 mtorr in argon l LiebermanMinicourse10 27
GLOBAL MODEL OF DISCHARGE EQUILIBRIUM LiebermanMinicourse10 28
PARTICLE BALANCE AND T e Assume uniform cylindrical plasma absorbing power P abs R P abs Plasma n e = n i = n 0 Particle balance l Production due to ionization = loss to the walls K iz n g n/ 0 πr 2 l = (2πR 2 h l n/ 0 + 2πRlh R n/ 0 )u B Solve to obtain K iz (T e ) u B (T e ) = 1 n g d eff where d eff = 1 Rl 2 Rh l + lh R is an effective plasma size Given n g and d eff = electron temperature T e T e varies over a narrow range of 2 5 volts LiebermanMinicourse10 29
ELECTRON TEMPERATURE IN ARGON DISCHARGE LiebermanMinicourse10 30
ION ENERGY FOR LOW VOLTAGE SHEATHS E i = energy entering sheath + energy gained traversing sheath Ion energy entering sheath = T e /2 (voltage units) Sheath voltage determined from particle conservation Γ i Density n s Plasma Sheath ~ 0.2 mm + V s Insulating wall Γ i Γ e Γ i = n s u B, Γ e = 1 4 n s v e }{{} e V s/t e with v e = (8eT e /πm) 1/2 Random flux at sheath edge The ion and electron fluxes at the wall must balance V s = T ( ) e M 2 ln 2πm or V s 4.7 T e for argon Accounting for the initial ion energy, E i 5.2 T e LiebermanMinicourse10 31
ION ENERGY FOR HIGH VOLTAGE SHEATHS Large ion bombarding energies can be gained near rf-driven electrodes s V rf ~ C large Plasma V s ~ 0.4 V rf + + V s V s Low voltage sheath ~ 5.2 T e V rf ~ Plasma V s ~ 0.8 V rf V s + The sheath thickness s is given by the Child law J i = en s u B = 4 ( ) 1/2 3/2 2e 9 ǫ V s 0 M s 2 Estimating ion energy is not simple as it depends on the type of discharge and the application of rf or dc bias voltages LiebermanMinicourse10 32
POWER BALANCE AND n 0 Assume low voltage sheaths at all surfaces E T (T e ) = E c (T e ) }{{} + 2 T e }{{} + 5.2 T e }{{} Collisional Electron Ion Power balance Power in = power out Solve to obtain P abs = (h l n 0 2πR 2 + h R n 0 2πRl) u B ee T n 0 = P abs A eff u B ee T where A eff = 2πR 2 h l + 2πRlh R is an effective area for particle loss Density n 0 is proportional to the absorbed power P abs Density n 0 depends on pressure p through h l, h R, and T e LiebermanMinicourse10 33
PARTICLE AND POWER BALANCE Particle balance = electron temperature T e (independent of plasma density) Power balance = plasma density n 0 (once electron temperature T e is known) LiebermanMinicourse10 34
EXAMPLE 1 Let R = 0.15 m, l = 0.3 m, n g = 3.3 10 19 m 3 (p = 1 mtorr at 300 K), and P abs = 800 W Assume low voltage sheaths at all surfaces Find λ i = 0.03 m. Then h l h R 0.3 and d eff 0.17 m [pp. 21, 27, 29] From the T e versus n g d eff figure, T e 3.5 V [p. 30] From the E c versus T e figure, E c 42 V [p. 23]. Adding E e = 2T e 7 V and E i 5.2T e 18 V yields E T = 67 V [p. 22] Find u B 2.9 10 3 m/s and find A eff 0.13 m 2 [pp. 24, 33] Power balance yields n 0 2.0 10 17 m 3 [p. 33] Ion current density J il = eh l n 0 u B 2.9 ma/cm 2 Ion bombarding energy E i 18 V [p. 31] LiebermanMinicourse10 35
EXAMPLE 2 Apply a strong dc magnetic field along the cylinder axis = particle loss to radial wall is inhibited Assume no radial losses, then d eff = l/2h l 0.5 m From the T e versus n g d eff figure, T e 3.3 V (was 3.5 V) From the E c versus T e figure, E c 46 V. Adding E e = 2T e 6.6 V and E i 5.2T e 17 V yields E T = 70 V Find u B 2.8 10 3 m/s and find A eff = 2πR 2 h l 0.043 m 2 Power balance yields n 0 5.8 10 17 m 3 (was 2 10 17 m 3 ) Ion current density J il = eh l n 0 u B 7.8 ma/cm 2 Ion bombarding energy E i 17 V = Slight decrease in electron temperature T e = Significant increase in plasma density n 0 EXPLAIN WHY! What happens to T e and n 0 if there is a sheath voltage V s = 500 V at each end plate? LiebermanMinicourse10 36
ELECTRON HEATING MECHANISMS Discharges can be distinguished by electron heating mechanisms (a) Ohmic (collisional) heating (capacitive, inductive discharges) (b) Stochastic (collisionless) heating (capacitive, inductive discharges) (c) Resonant wave-particle interaction heating (Electron cyclotron resonance and helicon discharges) Although the heated electrons provide the ionization required to sustain the discharge, the electrons tend to short out the applied heating fields within the bulk plasma Achieving adequate electron heating is a key issue LiebermanMinicourse10 37
INDUCTIVE DISCHARGES DESCRIPTION AND MODEL LiebermanMinicourse10 38
MOTIVATION High density (compared to capacitive discharge) Independent control of plasma density and ion energy Simplicity of concept RF rather than microwave powered No source magnetic fields LiebermanMinicourse10 39
CYLINDRICAL AND PLANAR CONFIGURATIONS Cylindrical coil Planar coil LiebermanMinicourse10 40
z HIGH DENSITY REGIME Inductive coil launches decaying wave into plasma Decaying wave Ẽ H δ p Plasma Coil Window Wave decays exponentially into plasma Ẽ = Ẽ0 e z/δ p, δ p = c ω 1 Im(κ 1/2 p ) where κ p = plasma dielectric constant [p. 16] ωpe 2 κ p = 1 ω(ω jν m ) For typical high density, low pressure (ν m ω) discharge δ p c ω pe 1 2 cm LiebermanMinicourse10 41
z TRANSFORMER MODEL For simplicity consider a long cylindrical discharge N turn coil Ĩ rf b R δ p Ĩ p Plasma l Current Ĩrf in N turn coil induces current Ĩp in 1-turn plasma skin = A transformer LiebermanMinicourse10 42
RESISTANCE AND INDUCTANCE Plasma resistance R p R p = 1 circumference of plasma loop σ dc average cross sectional area of loop where [p. 17] σ dc = e2 n es mν m with n es = density at plasma-sheath edge = R p = πr σ dc lδ p Plasma inductance L p magnetic flux produced by plasma current L p = plasma current Using magnetic flux = πr 2 µ 0 Ĩ p /l = L p = µ 0πR 2 l LiebermanMinicourse10 43
Model the source as a transformer COUPLING OF COIL TO Ṽ rf = jωl 11 Ĩ rf + jωl 12 Ĩ p Ṽ p = jωl 21 Ĩ rf + jωl 22 Ĩ p Transformer inductances magnetic flux linking coil L 11 = = µ 0πb 2 N 2 coil current l magnetic flux linking plasma L 12 = L 21 = = µ 0πR 2 N coil current l L 22 = L p = µ 0πR 2 l LiebermanMinicourse10 44
SOURCE CURRENT AND VOLTAGE Put Ṽp = ĨpR p in transformer equations and solve for the impedance Z s = Ṽrf/Ĩrf seen at coil terminals Z s = jωl 11 + ω2 L 2 12 R p + jωl p R s + jωl s Equivalent circuit at coil terminals R s = N 2 πr σ dc lδ p L s = µ 0πR 2 N 2 ( ) b 2 l R 2 1 Power balance = Ĩrf P abs = 1 2Ĩ2 rf R s From source impedance = V rf Ṽ rf = ĨrfZ s LiebermanMinicourse10 45
EXAMPLE Assume plasma radius R = 10 cm, coil radius b = 15 cm, length l = 20 cm, N = 3 turns, gas density n g = 6.6 10 14 cm 3 (20 mtorr argon at 300 K), ω = 85 10 6 s 1 (13.56 MHz), absorbed power P abs = 600 W, and low voltage sheaths At 20 mtorr, λ i 0.15 cm, h l h R 0.1, d eff 34 cm [pp. 21, 27, 29] Particle balance (T e versus n g d eff figure [p. 30]) yields T e 2.1 V Collisional energy losses (E c versus T e figure [p. 23]) are E c 110 V. Adding E e + E i = 7.2T e yields total energy losses E T 126 V [p. 22] u B 2.3 10 5 cm/s [p. 24] and A eff 185 cm 2 [p. 33] Power balance yields n e 7.1 10 11 cm 3 and n se 7.4 10 10 cm 3 [p. 33] Use n se to find skin depth δ p 2.0 cm [p. 41]; estimate ν m = K el n g (K el versus T e figure [p. 20]) to find ν m 3.4 10 7 s 1 Use ν m and n se to find σ dc 61 Ω 1 -m 1 [p. 17] Evaluate impedance elements R s 23.5 Ω and L s 2.2 µh; Z s ωl s 190 Ω [p. 45] Power balance yields Ĩrf 7.1A; from source impedance Z s = 190 Ω, Ṽ rf 1360 V [p. 45] LiebermanMinicourse10 46
MATCHING DISCHARGE TO POWER SOURCE Consider an rf power source connected to a discharge + Z T Ĩ + Ṽ T ~ Ṽ Z L Source Discharge Source impedance Z T = R T + jx T is given Discharge impedance Z L = R L + jx L Time-average power delivered to discharge P abs = 1 2 Re (Ṽ Ĩ ) For fixed source ṼT and Z T, maximize power delivered to discharge X L = X T R L = R T LiebermanMinicourse10 47
MATCHING NETWORK Insert lossless matching network between power source and coil Power source Matching network Discharge coil Continue EXAMPLE [p. 46] with R s = 23.5 Ω and ωl s = 190 Ω; assume R T = 50 Ω (corresponds to a conductance 1/R T = 1/50 S) Choose C 1 such that the conductance seen looking to the right at terminals AA is 1/50 S = C 1 = 71 pf Choose C 2 to cancel the reactive part of the impedance seen at AA = C 2 = 249 pf For P abs = 600 W, the 50 Ω source supplies Ĩrf = 4.9 A Voltage at source terminals (AA ) = ĨrfR T = 245 V LiebermanMinicourse10 48
z PLANAR COIL DISCHARGE Magnetic field produced by planar coil RF power is deposited in a ring-shaped plasma volume δ p Ĩ rf N turn coil Primary inductance Coupling inductance Ĩ p Plasma Plasma inductance As for a cylindrical discharge, there is a primary (L 11 ), coupling (L 12 = L 21 ) and secondary (L p = L 22 ) inductance LiebermanMinicourse10 49
PLANAR COIL FIELDS A ring-shaped plasma forms because { 0, on axis Induced electric field = max, at r 1 2 R wall 0, at r = R wall Measured radial variation of B r (and E θ ) at three distances below the window (5 mtorr argon, 500 W, Hopwood et al, 1993) LiebermanMinicourse10 50
INDUCTIVE DISCHARGES POWER BALANCE LiebermanMinicourse10 51
RESISTANCE AT HIGH AND LOW DENSITIES Plasma resistance seen by the coil [p. 45] ω 2 L 2 12 R s = R p Rp 2 + ω2 L 2 p High density (normal inductive operation) [p. 45] R s R p 1 1 σ dc δ p ne Low density (skin depth > plasma size) R s number of electrons in the heating volume n e R s n e Low density High density δ p ~ plasma size n e 1 n e LiebermanMinicourse10 52
POWER BALANCE Drive discharge with rf current Power absorbed by discharge is P abs = 1 2 Ĩrf 2 R s (n e ) [p. 45] Power lost by discharge P loss n e [p. 33] Intersection (red dot) gives operating point; let Ĩ1 < Ĩ2 < Ĩ3 P loss Power P abs = 1 2Ĩ2 3R s P abs = 1 2Ĩ2 2R s P abs = 1 2Ĩ2 1R s Inductive operation impossible for Ĩrf Ĩ2 LiebermanMinicourse10 53 n e
z CAPACITIVE COUPLING OF COIL TO For Ĩrf below the minimum current Ĩ2, there is only a weak capacitive coupling of the coil to the plasma + Ṽ rf Capacitive coupling Ĩ p Plasma A small capacitive power is absorbed = low density capacitive discharge P loss Power Cap Ind P abs = 1 2Ĩ2 3R s P abs = 1 2Ĩ2 1R s Cap Mode Ind Mode LiebermanMinicourse10 54 n e
MEASURMENTS OF ARGON ION DENSITY Above 100 W, discharge is inductive and n e P abs Below 100 W, a weak capacitive discharge is present LiebermanMinicourse10 55
REACTIVE NEUTRAL BALANCE IN DISCHARGES LiebermanMinicourse10 56
PLANE-PARALLEL DISCHARGE Example of N 2 discharge with low fractional ionization (n g n N2 ) and planar 1D geometry (l R) Low voltage sheaths P abs A=πR 2 nis N 2 gas ni nis l Determine T e Ion particle balance is [p. 29] K iz n g n i la 2n is u B A where n is = h l n i with h l = 0.86/(3 + l/2λ i ) 1/2 [p. 26] K iz (T e ) u B (T e ) 2h l n g l = T e LiebermanMinicourse10 57
PLANE-PARALLEL DISCHARGE (CONT D) Determine edge plasma density n is Overall discharge power balance [p. 33] gives the plasma density at the sheath edge n is P abs 2eE T u B A Determine central plasma density [p. 26] n i = n is h l Determine ion flux to the surface [p. 26] Γ is n is u B Determine ion bombarding energy [p. 31] E i = 5.2 T e LiebermanMinicourse10 58
For nitrogen atoms REACTIVE NEUTRAL BALANCE K e + N diss 2 2N + e Assume low fractional dissociation and loss of N atoms only due to a vacuum pump S p (m 3 /s) Al dn N = Al 2K diss n g n i S p n NS = 0 dt Solve for reactive neutral density at the surface Flux of N atoms to the surface n NS = K diss 2Aln g S p n i Γ NS = 1 4 n NS v N where v N = (8kT N /πm N ) 1/2 LiebermanMinicourse10 59
LOADING EFFECT Consider recombination and/or reaction of N atoms on surfaces N + wall γ rec 1 2 N 2 N + substrate γ reac product Pumping speed S p in the expression for n NS [p. 59] is replaced by 1 S p S p + γ rec 4 v 1 N(2A A subs ) + γ reac 4 v NA subs A subs is the part of the substrate area reacting with N atoms n NS is reduced due to recombination and reaction losses n NS, and therefore etch and deposition rates, now depend on the part of the substrate area A subs exposed to the reactive neutrals, a loading effect LiebermanMinicourse10 60
ADSORPTION AND DESORPTION KINETICS LiebermanMinicourse10 61
ADSORPTION Reaction of a molecule with the surface A + S K a A: S Physisorption (due to weak van der Waals forces) U 1 3 Å Kd 0.01 0.25 V x Chemisorption (due to formation of chemical bonds) U 1 1.5 Å 0.4 4 V x LiebermanMinicourse10 62
Adsorbed flux [p. 13] STICKING COEFFICIENT Γ ads = sγ A = s 1 4 n AS v A s(θ, T) = sticking coefficient θ = fraction of surface sites covered with absorbate n AS = gas phase density of A near the surface v A = (8kT A /πm A ) 1/2 = mean thermal speed of A Langmuir kinetics s(θ, T) = s 0 (1 θ) s 0 = zero-coverage sticking coefficient (s 0 10 6 1) s 0 1 s Langmuir s 0 0 0 1 θ 0 0 T LiebermanMinicourse10 63
DESORPTION A: S K d A + S Rate coefficient has Arrhenius form K d = K d0 e E desor/t where E desor = E chemi or E physi Pre-exponential factors are typically K d0 10 14 10 16 s 1 10 13 10 15 s 1 physisorption chemisorption LiebermanMinicourse10 64
ADSORPTION-DESORPTION KINETICS Consider the reactions A + S K a Kd A: S Adsorbed flux is [p. 63] Γ ads = K a n AS n 0 (1 θ) n 0 = area density (m 2 ) of adsorption sites n AS = the gas phase density at the surface 1 K a = s 0 4 v A/n 0 [m 3 /s] (adsorption rate coef) Desorbed flux area density n 0θ of covered sites [p. 64] Γ desor = K d n 0 θ LiebermanMinicourse10 65
ADSORPTION-DESORPTION KINETICS (CONT D) Equate adsorption and desorption fluxes (Γ ads = Γ desor ) where K = K a /K d = θ = Kn AS 1 + Kn AS 1 Note that T K θ θ 0 0 Langmuir isotherm (T=const) 5 10 Kn AS LiebermanMinicourse10 66
-ASSISTED ETCH KINETICS LiebermanMinicourse10 67
ION-ENHANCED ETCHING 1. Low chemical etch rate of silicon substrate in XeF 2 etchant gas 2. Tenfold increase in etch rate with XeF 2 + 500 V argon ions, simulating ion-enhanced plasma etching 3. Very low etch rate due to the physical sputtering of silicon by ion bombardment alone LiebermanMinicourse10 68
STANDARD MODEL OF ETCH KINETICS O atom etching of a carbon substrate O + CO K a K i K d Y i K i 1 θ θ C(s) CO(s) Let n 0 = active surface sites/m 2 Let θ = fraction of surface sites covered with C : O bonds O(g) + C(s) K a C : O C : O K d CO(g) ion + C : O Y ik i CO(g) (O atom adsorption) (CO thermal desorption) (CO ion-assisted desorption) LiebermanMinicourse10 69
SURFACE COVERAGE The steady-state surface coverage is found from [pp. 65 66] dθ dt = K an OS (1 θ) K d θ Y i K i n is θ = 0 n OS is the O-atom density near the surface n is is the ion density at the plasma-sheath edge K a is the rate coefficient for O-atom adsorption K d is the rate coefficient for thermal desorption of CO K i = u B /n 0 is the rate coefficient for ions incident on the surface Y i is the yield of CO molecules desorbed per ion incident on a fully covered surface Typically Y i 1 and Y i Y i0 Ei E thr (as for sputtering) = θ = K a n OS K a n OS + K d + Y i K i n is LiebermanMinicourse10 70
ETCH RATES The flux of CO molecules leaving the surface is Γ CO = (K d + Y i K i n is ) θ n 0 [m 2 -s 1 ] with n 0 = number of surface sites/m2 The vertical etch rate is E v = Γ CO [m/s] n C where n C is the carbon atom density of the substrate The vertical (ion-enhanced) etch rate is E v = n 0 1 n C 1 + K d + Y i K i n is 1 K a n OS The horizontal (non ion-enhanced) etch rate is E h = n 0 1 n C 1 1 + K d K a n OS LiebermanMinicourse10 71
NORMALIZED ETCH RATES 10 Y i K i n is = 5 K d 8 En C K d n 0 6 4 Vertical (E v ) 2 0 0 Horizontal (E h ) 4 8 12 16 20 24 K a n OS K d High O-atom flux highest anisotropy E v /E h = 1 + Y i K i n is /K d Low O-atom flux low etch rates with E v /E h 1 LiebermanMinicourse10 72
SIMPLEST MODEL OF ION-ENHANCED ETCHING In the usual ion-enhanced regime Y i K i n is K d 1 = n C 1 1 E v Y i K i n is n + 0 K a n OS n 0 }{{}}{{} Γ is Γ OS The ion and neutral fluxes and the yield (a function of ion energy) determine the ion-assisted etch rate The discharge parameters set the ion and neutral fluxes and the ion bombarding energy LiebermanMinicourse10 73
ADDITIONAL CHEMISTRY AND PHYSICS Sputtering of carbon Γ C = γ sput K i n is n 0 Associative and normal desorption of O atoms, C : O C + O(g) 2C : O 2C + O 2 (g) Ion energy driven desorption of O atoms ions + C : O C + O(g) Formation and desorption of CO 2 as an etch product Non-zero ion angular bombardment of sidewall surfaces Deposition kinetics (C-atoms, etc) LiebermanMinicourse10 74
SUMMARY Plasma discharges are widely used for materials processing and are indispensible for microelectronics fabrication The charged particle balance determines the electron temperature and ion bombarding energy to the substrate = Y i (E i ) The energy balance determines the plasma density and the ion flux to the substrate = Γ is A transformer model determines the relation among voltage, current, and power for inductive discharges The reactive neutral balance determines the flux of reactive neutrals to the surface = Γ OS Hence the discharge parameters (power, pressure, geometry, etc) set the ion and neutral fluxes and the ion bombarding energy The ion and neutral fluxes and the yield (a function of ion energy) determine the ion-assisted etch rate LiebermanMinicourse10 75
THANK YOU FOR ATTENDING THIS COURSE MIKE LIEBERMAN LiebermanMinicourse10 76