A MINI-COURSE ON THE PRINCIPLES OF LOW-PRESSURE DISCHARGES AND MATERIALS PROCESSING

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1 A MINI-COURSE ON THE PRINCIPLES OF LOW-PRESSURE DISCHARGES AND MATERIALS PROCESSING Michael A. Lieberman Department of Electrical Engineering and Computer Science, CA LiebermanMinicourse07 1

2 OUTLINE Introduction Summary of Plasma and Discharge Fundamentals Global Model of Discharge Equilibrium Break Inductive Discharges Free Radical Balance in Discharges Adsorption and Desorption Kinetics Plasma-Assisted Etch Kinetics LiebermanMinicourse07 2

3 INTRODUCTION TO DISCHARGES AND PROCESSING LiebermanMinicourse07 3

4 THE NANOELECTRONICS REVOLUTION Transistors/chip doubling every years since ,000,000-fold decrease in cost for the same performance In 2020: 6 nm gate length, transistors, 73 GHz on-chip clock, wiring levels 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 LiebermanMinicourse07 4

5 RANGE OF NANOELECTRONICS APPLICATIONS Etching Si, a-si, oxide, nitride, III-V s Ashing Photoresist removal Deposition (PECVD) Oxide, nitride, a-si Oxidation Si Sputtering Al, W, Au, Cu, YBaCuO Polymerization Various plastics Implantation H, He, B, P, O, As, Pd LiebermanMinicourse07 5

6 EVOLUTION OF ETCHING DISCHARGES FIRST GENERATION (1 source, multi-wafer, low density) SECOND GENERATION (2 sources, single wafer, high density) THIRD GENERATION (multi-source, single wafer, moderate density) LiebermanMinicourse07 6

7 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 LiebermanMinicourse07 7

8 DENSITY VERSUS TEMPERATURE LiebermanMinicourse07 8

9 RELATIVE DENSITIES AND ENERGIES Charged particle densities neutral particle densities LiebermanMinicourse07 9

10 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 LiebermanMinicourse07 10

11 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 LiebermanMinicourse07 11

12 SUMMARY OF FUNDAMENTALS LiebermanMinicourse07 12

13 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 Electron energy distribution function (EEDF) g e (E) E 1/2 exp( E/T e ) Pressure p = nkt For neutral gas at room temperature (300 K) n g [cm 3 ] p[torr] LiebermanMinicourse07 13

14 z y 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 Γ e 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 LiebermanMinicourse07 14

15 FORCES ON PARTICLES For a unit volume of electrons (or ions) mn e du e dt = qn e E p e mn e ν m u e 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 LiebermanMinicourse07 15

16 BOLTZMANN FACTOR FOR ELECTRONS 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 LiebermanMinicourse07 16

17 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 J c = en e ũ e is mainly due to electrons Newton s law (electric field and neutral drag) is jωmũ e = eẽ mν mũ e Solve for ũ e and evaluate J c to obtain ] ωpe ǫ p ǫ 0 κ p = ǫ 0 [1 2 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) For ν m ω, ǫ p is mainly imaginary (very lossy dielectric) LiebermanMinicourse07 17

18 CONDUCTIVITY σ p It is useful to introduce the plasma conductivity J c σ p Ẽ RF plasma conductivity σ p = DC plasma conductivity (ω ν m ) e 2 n e m(ν m + jω) σ dc = e2 n e mν m The plasma dielectric constant and conductivity are related by: jωǫ p = σ p + jωǫ 0 Due to Re(σ p ) [or Im(ǫ p )], rf current flowing through the plasma heats electrons (just like a resistor) LiebermanMinicourse07 18

19 SUMMARY OF DISCHARGE FUNDAMENTALS LiebermanMinicourse07 19

20 ELECTRON COLLISIONS WITH ARGON Maxwellian electrons collide with Ar atoms (density n g ) dn e = νn e = Kn g n e dt 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 + Ar e + Ar (elastic scattering) e e Ar Ar Rate coefficient K(T e ) is average over Maxwellian distribution of cross section σ [m 2 ] relative velocity v for the process K(T e ) = σv Maxwellian LiebermanMinicourse07 20

21 ELECTRON-ARGON RATE COEFFICIENTS LiebermanMinicourse07 21

22 ION COLLISIONS WITH ARGON Argon ions collide with Ar atoms Ar + + Ar Ar + + Ar Ar + + Ar Ar + Ar + (elastic scattering) (charge transfer) Total cross section for room temperature ions σ i cm 2 Ion-neutral mean free path λ i = 1 n g σ i Practical formula λ i [cm] = 1, [p in Torr] 330 p Ion-neutral collision frequency ν i = n g v i /λ i with v i = (8kT i /πm) 1/2 Ar + Ar + Ar Ar Ar + Ar Ar Ar + LiebermanMinicourse07 22

23 THREE ENERGY LOSS PROCESSES 1. Define 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 LiebermanMinicourse07 23

24 COLLISIONAL ENERGY LOSSES LiebermanMinicourse07 24

25 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 At the plasma-sheath edge (density n s ), electron-ion pairs are lost at the Bohm velocity u i = u B = ( kte M ) 1/2 LiebermanMinicourse07 25

26 AMBIPOLAR DIFFUSION AT HIGH PRESSURES Plasma bulk is quasi-neutral (n e n i = n) and the electron and ion loss fluxes are equal (Γ e Γ i Γ) Fick s law Γ = D a n with ambipolar diffusion coefficient D a = kt e /Mν i Density profile is sinusoidal n 0 Γ wall Γ wall l/2 0 l/2 n s x Loss flux to the wall is Γ wall = n s u B h l n 0 u B Edge-to-center density ratio is h l n s = π u B n 0 l ν i Applies for pressures > 100 mtorr in argon LiebermanMinicourse07 26

27 AMBIPOLAR DIFFUSION AT LOW PRESSURES The diffusion coefficient is not constant Density profile is relatively flat in the center and falls sharply near the sheath edge n 0 Γ wall Γ wall l/2 0 l/2 n s x The edge-to-center density ratio is h l n s n (3 + l/2λ i ) 1/2 where λ i = ion-neutral mean free path [p. 22] Applies for pressures < 100 mtorr in argon LiebermanMinicourse07 27

28 AMBIPOLAR DIFFUSION IN LOW PRESSURE CYLINDRICAL DISCHARGE nsr = hr n 0 R Plasma n e = n i = n 0 nsl = hl n 0 l For a cylindrical plasma of length l and radius R, loss fluxes to 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 ), h 0.8 1/2 R (4 + R/λ i ) 1/2 Applies for pressures < 100 mtorr in argon LiebermanMinicourse07 28

29 GLOBAL MODEL OF DISCHARGE EQUILIBRIUM LiebermanMinicourse07 29

30 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 Solve to obtain K iz n g n 0 πr 2 l = (2πR 2 h l n 0 + 2πRlh R n 0 )u B 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 LiebermanMinicourse07 30

31 ELECTRON TEMPERATURE IN ARGON DISCHARGE LiebermanMinicourse07 31

32 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 in the sheath Γ i Plasma Sheath Insulating wall Γ i Γ e Density n s + V s Γ i = n s u B, Γ e = 1 4 n s v e }{{} with v e = (8eT e /πm) 1/2 Random flux at sheath edge The ion and electron fluxes must balance V s = T ( ) e M 2 ln 2πm e V s /T e or V s 4.7 T e for argon Accounting for the initial ion energy, E i 5.2 T e LiebermanMinicourse07 32

33 ION ENERGY FOR HIGH VOLTAGE SHEATHS Large ion bombarding energies can be gained near rf- or dc-driven electrodes embedded in the plasma s Ṽ rf ~ C large Plasma V s 0.4 Ṽrf V s + + Vs Ṽ rf ~ V s + Plasma Low voltage sheath 5.2 T e V s 0.8 Ṽrf 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 LiebermanMinicourse07 33

34 POWER BALANCE AND n 0 Assume low voltage sheaths at all surfaces E T (T e ) = E c (T e ) }{{} + 2 T e }{{} 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 LiebermanMinicourse07 34

35 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) LiebermanMinicourse07 35

36 EXAMPLE 1 Let R = 0.15 m, l = 0.3 m, n g = 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. 22, 28, 30] From the T e versus n g d eff figure, T e 3.5 V [p. 31] From the E c versus T e figure, E c 42 V [p. 24]. Adding E e = 2T e 7 V and E i 5.2T e 18 V yields E T = 67 V [p. 23] Find u B m/s and find A eff 0.13 m 2 [pp. 25, 34] Power balance yields n m 3 [p. 34] Ion current density J il = eh l n 0 u B 2.9 ma/cm 2 Ion bombarding energy E i 18 V [p. 32] LiebermanMinicourse07 36

37 EXAMPLE 2 Apply a strong dc magnetic field along the cylinder axis = particle loss to radial wall is inhibited For no radial loss, d eff = l/2h l 0.5 m From the T e versus n g d eff figure, T e 3.3 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 m/s and find A eff = 2πR 2 h l m 2 Power balance yields n 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 = Small drop in T e and significant increase in n 0 EXPLAIN WHY! LiebermanMinicourse07 37

38 IMPROVEMENTS TO GLOBAL MODEL LiebermanMinicourse07 38

39 NON-MAXWELLIAN ELECTRON DISTRIBUTIONS Global model theory with EEDF [p. 13] g e E 1/2 exp( ce x ) x = 1 Maxwellian; x = 2 Druyvesteyn (J.T. Gudmundsson, Plasma Sources Sci. Technol. 10, 76, 2001) 1D PIC simulations with inductive source -type excitation, 5 cm gap, MHz Ar 10 9 x=2.5 x=1.25 x=1 EEPF EEPF p [mt] E [ev] (D. Monahan, Modelling the Electronegative Discharge, PhD Thesis submitted to Dublin City University, June 2007; Prof. Miles Turner, supervisor) LiebermanMinicourse07 39

40 NON-UNIFORM BULK Global model assumes a uniform bulk plasma (density n 0 ) dropping sharply to a sheath edge density n s The actual low-pressure profile is more like a piece of a circle Actual n n 0 n 0 n s x h l = n s n 0 = h l = n s n 0 LiebermanMinicourse07 40

41 PIC SIMULATION RESULTS 10 1 Max EEPF (Monahan, 2007) PIC (centre) PIC (average) PIC (homogeneous) Model (Max) Model (gen) Model (gen + mod h) n e [10 15 m 3 ] 10 0 Variable EEPF Center Variable EEPF + hl Volume-ave T e [ev] Max EEPF Variable EEPF pressure [mt] pressure [mt] LiebermanMinicourse07 41

42 ELECTRON HEATING 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) Achieving adequate electron heating is a central issue 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 LiebermanMinicourse07 42

43 INDUCTIVE DISCHARGES DESCRIPTION AND MODEL LiebermanMinicourse07 43

44 MOTIVATION Independent control of plasma density and ion energy Simplicity of concept RF rather than microwave powered No source magnetic fields LiebermanMinicourse07 44

45 CYLINDRICAL AND PLANAR CONFIGURATIONS Cylindrical coil Planar coil LiebermanMinicourse07 45

46 z HIGH DENSITY REGIME Inductive coil excites a decaying electromagnetic wave in plasma Ẽ H δ p Decaying wave Plasma Coil Window Wave decays exponentially into plasma Ẽ = Ẽ0 e z/δ p, δ p = c 1 ω Im(κ 1/2 p ) where κ p = plasma dielectric constant [p. 17] ωpe 2 κ p = 1 ω(ω jν m ) For typical high density, low pressure (ν m ω) discharge δ p c ( ) 1/2 m = ω pe e cm µ 0 n e LiebermanMinicourse07 46

47 z TRANSFORMER MODEL For simplicity consider 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 LiebermanMinicourse07 47

48 RESISTANCE AND INDUCTANCE Plasma resistance R p R p = 1 circumference of plasma loop σ dc average cross sectional area of loop where σ dc = e2 n es mν m [p. 18] = 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 LiebermanMinicourse07 48

49 Model the source as a transformer COUPLING OF COIL TO Coil Plasma 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 Plasma resistance V p = I p R p LiebermanMinicourse07 49

50 SOURCE CURRENT AND VOLTAGE Solve for 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 l Power balance = Ĩrf ( ) b 2 R 2 1 From source impedance = V rf P abs = 1 2Ĩ2 rf R s Ṽ rf = ĨrfZ s LiebermanMinicourse07 50

51 EXAMPLE Assume plasma radius R = 10 cm, coil radius b = 15 cm, length l = 20 cm, N = 3 turns, gas density n g = cm 3 (20 mtorr argon at 300 K), ω = 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, and d eff 34 cm [pp. 22, 28, 30] Particle balance (T e versus n g d eff figure) yields T e 2.1 V [p. 31] Collisional energy losses (E c versus T e figure) are E c 110 V [p. 24]. Adding E e + E i = 7.2T e yields total energy losses E T 126 V [p. 23] u B cm/s and A eff 185 cm 2 [pp. 25, 34] Power balance yields n e cm 3 and n se cm 3 [p. 34] Use n se to find skin depth δ p 2.0 cm [p. 46]; estimate ν m = K el n g (K el versus T e figure) to find ν m s 1 [p. 21] Use ν m and n se to find σ dc 61 Ω 1 -m 1 [p. 18] Evaluate impedance elements R s 23.5 Ω and L s 2.2 µh; Z s ωl s 190 Ω [p. 50] Power balance yields Ĩrf 7.1A; from impedance Ṽrf 1360 V [p. 50] LiebermanMinicourse07 51

52 MATCHING DISCHARGE TO POWER SOURCE Consider an rf power source connected to a load + Z T Ĩ + Ṽ T ~ Ṽ Z L Source Load Source impedance Z T = R T + jx T is given Load impedance Z L = R L + jx L Time-average power delivered to load P abs = 1 2 Re (Ṽ Ĩ ) For fixed source ṼT and Z T, to maximize power delivered to load, X L = X T R L = R T LiebermanMinicourse07 52

53 MATCHING NETWORK Insert lossless matching network between power source and coil Power source Matching network Discharge coil Continue EXAMPLE with R s = 23.5 Ω and ωl s = 190 Ω; assume R T = 50 Ω (corresponds to conductance of 1/50 Ω 1 ) Choose C 1 such that the conductance seen looking to the right at terminals AA is 1/R T = 1/50 Ω 1 = 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 ) = I rf R T = 245 V LiebermanMinicourse07 53

54 z PLANAR COIL DISCHARGE Magnetic field produced by planar coil RF power is deposited in 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 LiebermanMinicourse07 54

55 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) LiebermanMinicourse07 55

56 INDUCTIVE DISCHARGES POWER BALANCE LiebermanMinicourse07 56

57 RESISTANCE AT HIGH AND LOW DENSITIES Plasma resistance seen by the coil [p. 50] ω 2 L 2 12 R s = R p Rp 2 + ω2 L 2 p High density (normal inductive operation) 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 1 n e n e δ p plasmasize n e LiebermanMinicourse07 57

58 POWER BALANCE Drive discharge with rf current Power absorbed by discharge is P abs = 1 2 Ĩrf 2 R s (n e ) Power lost by discharge P loss n e [p. 34] Intersection 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, the critical current LiebermanMinicourse07 58 n e

59 z CAPACITIVE COUPLING OF COIL TO For Ĩrf below the critical 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 n e LiebermanMinicourse07 59

60 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 LiebermanMinicourse07 60

61 FREE RADICAL BALANCE IN DISCHARGES LiebermanMinicourse07 61

62 ONE-DIMENSIONAL SLAB MODEL N 2 discharge with low fractional ionization (n g n N2 ) Determine T e : Assume R l = plate separation. Then ion particle balance is K iz n g n i la 2n is u B A where n is = h l n i. This yields [p. 30] K iz (T e ) u B (T e ) 2h l n g l = T e Determine n i and Γ is : The overall discharge power balance is [p. 34] P abs 2AeE T n is u B This yields n is P abs 2eE T u B A and Γ is n is u B Determine ion bombarding energy [p. 32] E i = 5.2 T e LiebermanMinicourse07 62

63 For nitrogen: FREE RADICAL BALANCE K e + N diss 2 2N + e Assume low fractional dissociation and that the only loss of N atoms is 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 Solving for the free radical density at the surface The flux of N atoms is n NS = 2Aln g S p K diss n i Γ NS = 1 4 n NS v N where v N = (8kT N /πm N ) 1/2 But how does K diss (T e ) depend on discharge pressure? LiebermanMinicourse07 63

64 DISSOCIATION RATE COEFFICIENT Assume K diss and K iz have Arrhenius forms K diss = K diss0 e E diss/t e K iz = K iz0 e E iz/t e Eliminating T e from these equations yields K diss = C 0 K E diss/e iz iz where C 0 = K diss0 /K E diss/e iz iz0 Using ion particle balance [p. 62] to eliminate K iz yields K diss = C 0 ( 2hl u B n g l ) Ediss /E iz In this form, K diss depends only weakly on T e, and the dependance of K diss on n g l is made explicit Inserting n i [p. 62] and K diss [above] into n NS [p. 63] yields n NS = 2C 0 P abs ee T S p Typically, E diss /E iz ( ng l 2h l u B ) 1 Ediss /E iz LiebermanMinicourse07 64

65 RECOMBINATION, REACTION AND LOADING EFFECT Recall that [p. 64]: n NS = 2C 0 P abs ee T S p ( ng l 2h l u B ) 1 Ediss /E iz Consider a recombination coefficient γ rec for N atoms on the walls and a reaction coefficient γ reac for N atoms on a substrate. The pumping speed S p is replaced by: S p S p + γ rec 1 4 v N(2A A subs ) + γ reac 1 4 v NA subs Note that n NS is reduced due to recombination and reaction losses n NS now depends on the substrate area A subs, a loading effect LiebermanMinicourse07 65

66 ADSORPTION AND DESORPTION KINETICS LiebermanMinicourse07 66

67 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 V x Chemisorption (due to formation of chemical bonds) U Å V x LiebermanMinicourse07 67

68 Adsorbed flux STICKING COEFFICIENT Γ ads = sγ A = 1 4 s v An AS s(θ, T) = sticking coefficient θ = fraction of surface sites covered with absorbate Langmuir kinetics s(θ, T) = s 0 (T)(1 θ) s 0 s Langmuir θ Zero-coverage sticking coefficient s s T LiebermanMinicourse07 68

69 DESORPTION Rate constant A: S A + S K desor = K 0 e E desor/t [s 1 ] where E desor = E chemi or E physi K s s 1 physisorption chemisorption LiebermanMinicourse07 69

70 ADSORPTION-DESORPTION KINETICS Consider the reactions The adsorption flux is A + S K a Kd A: S 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] The desorption flux is Γ desor = K d n 0 θ K d = K d0 e E desor/t [s 1 ] LiebermanMinicourse07 70

71 LANGMUIR ISOTHERM Equate adsorption and desorption fluxes, Γ ads = Γ desor θ = where K = K a /K d Kn AS 1 + Kn AS 1 Note that T K θ θ Example: 2XeF 2 (g) + Si(s) SiF 4 (g) + 2Xe log ER θ 0 and K r as T 450 K 1/T 0 0 θ 1 with K r as T 5 10 K n AS LiebermanMinicourse07 71

72 -ASSISTED ETCH KINETICS LiebermanMinicourse07 72

73 ION-ENHANCED ETCHING 1. Low chemical etch rate of silicon in XeF 2 etchant gas 2. Tenfold increase in etch rate with the addition of argon ion bombardment of the substrate, simulating plasma-assisted etching 3. Very low etch rate due to the physical sputtering of silicon by the ion bombardment alone LiebermanMinicourse07 73

74 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) LiebermanMinicourse07 74

75 SURFACE COVERAGE The steady-state surface coverage is found from dθ dt = K an OS (1 θ) K d θ Y i K i n is θ = 0 n OS is the neutral gas density near the surface n is is the ion density at the plasma edge K a is the rate coefficient for O-atom adsorption K d is the rate coefficient for thermal desorption of CO K i 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 E i E thr (as for sputtering) = θ = K a n OS K a n OS + K d + Y i K i n is LiebermanMinicourse07 75

76 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 K d K a n OS LiebermanMinicourse07 76

77 NORMALIZED ETCH RATES 10 8 Y i K i n is = 5 K d En C K d n Vertical (E v ) Horizontal (E h ) K a n OS K d High O-atom density highest anisotropy E v /E h = 1+Y i K i n is /K d Low O-atom density low etch rates with E v /E h 1 LiebermanMinicourse07 77

78 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 LiebermanMinicourse07 78

79 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) LiebermanMinicourse07 79

80 CONCLUSIONS 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 neutral radical balance determines the flux of radicals 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 LiebermanMinicourse07 80