Defects and crystal growth

Transcription

1 CRYSTL GROWTH CRYSTL GROWTH steps and questions 1. Reactants in molten form Si 2. Transport to S/L interface 3. dsorbtion: entropy decreases -TDS increases DH decreases (exo) 4. Critical nucleus size DS DH 5. Growth 6. Impurities, defects more stable at high T ; how grow pure crystal? 7. Segregation solid vs. liquid What do we need to know prior to crystal growth? 1 Defects and crystal growth Defects impurities, vacancies, dislocations T dependence growth techniques: float zone, ridgman, Czochralski Segregation during growth Segregation coefficients 2 1

2 Thermodynamics and phase diagrams H - H = DH = heat of formation of from Do all reactions that give off heat proceed? H H Configurations S - S = DS = Entropy (disorder) change from to Do all reactions that increase disorder proceed? S S nswer in Gibbs free energy: G = H-TS G - G = DG = DH - TDS G must decrease if reaction is to proceed. (From equilibrium, all changes increase G). G 3 Thermodynamics and phase diagrams H DH =offrom Do all exothermal reactions proceed? DS =from to DG = DH - TDS S G Configurations H Exothermic S < 0 more ordered H Configurations Endothermic H S S > 0 more disordered Does disorder always increase in reactions? G Will not go Will not go above T = DH /DS below T=DH /DS Examples: freezing of water melting of copper 4 2

3 Under what conditions will Si melt crystallize? T T m Liquid Si T = T m + T = T m - Si high S high H low S low H For solidification: DS = ST ( - ) final - ST ( + ) initial < 0 DH = HT ( - )- HT ( + ) < 0 DG =DH - T m DS < 0 >0 TDS must have smaller magnitude than DH for solidification; this defines solidification temp. 5 Note the relatively large solid solubility of s in Si.then decreases on approaching T melt rsenic solubility in Si increases with T Whereas this field is s + Sis 2 6 3

4 1-dimensional defects: We saw soluble impurities in Si. 1.0 T/T m From Phase diagram s P Impurity content (cm -3 ) S mix -TS mix Si Impurity Si Impurity 7 1-dimensional defects: More point defects Vacancy Interstitial impurity Self interstitial V-I pair = Frankel defect Substitutional impurity Strain field of vacancy interstitial Strain and surface energy can be reduced by agglomeration: Vacancy => void, Interstitial => precipitate when concentration > equilibrium 8 4

5 onding-antibonding orbital energy separation E Gp IV atom Semicon crystal s-p 3 antibonding Gp IV atom bonding/antibonding => energy gaps in semiconductors, insulators s 2 p 2 s 2 p 2 Conduction band s-p 3 bonding Gap Valence band 9 Vacancy concentration: Vacancy requires breaking 4 bonds n vac = n 0 exp -E g /k T [ ] E g =1.12eV Empirical: n vac = exp[ -2.6 ev /k T] Ê lná n vac ˆ =-E Ë a /k T n 0 Ê lná n vac ˆ Ë n 0 E a Conduction band Valence band rrhenius plot Equilibrium vacancy concentration: t RT: n vac = 3.4 x /cm /k T ( 300 km between vacancies) t 1273 K: n vac = 2.6 x /cm 3 ( 700 nm between vacancies) Vacancies abundant at high temperature 10 5

6 Oxygen impurities in Si: Observed to follow rrhenius [ ] C oxy = exp -1.03eV /k T 40 C oxy (ppm) 20 0 gglomeration => warpage, stress, dislocations Optimal No agglomeration (many isolated O 2- ions) 1414 T ( 0 C) 1200 High C oxy 3 hr anneal => 15 ppm Want about ppm (7 x /cm 3 ) which occurs at T C Oxygen 25 microns gglomeration nneal => denuded zone deeper than deepest feature Oxygen E g Vacancy Interstitial ctivation energies (ev): Dopants, impurities (substitutional, interstitial) t RT number of intrinsic carriers: n i = (n e n h ) 1/2 = n 0 exp(-e g /2k T) => n i =2x10 10 /cm 3 5 x So doping at 1 ppm => 10-6 = n D, /(5 x /cm 3 ) n D, =5x10 16 /cm 3 Very small doping concentration => large increase in carrier concentration What do dopants do? s = ne2 t m * J =se = ne < v > m = v E = s ne = et m * 12 6

7 CRYSTL GROWTH 1 Confined 2 CZ Meniscus controlled Normal freezing (ridgman) Zone melting Pull from solution 3 Floating zone feed rod horizontal high-purity crystal poly LEC vertical poly 2 O 3 Si 13 7

8 1) Reactants: first need high-purity Si Making high-purity Si: C boat SiO 2 Si + CO(g) SiHCl 3 (l) Si + HCl MGS -98% pure (Metallurgical grade Si) HCl + H 2 distill H C poly EGS 9 9s pure (Electronic grade Si) VC seed tang crystal 1500 C 16 8

9 Growth rate µg s -G L DG DG = 0 fi Equilibrium at DH = TDS, DS =DH /T eq. DG = DH - TDS È DG =DH T eq -T Í ÎÍ T eq DH < 0 \T interface < T eq Must decrease for solidification to occur H s - H L larger (latent) heat content 17 1) Pulling crystal from melt VC seed tang crystal z 1500 C Q t S-L interface: If growth velocity too large, solid cannot dissipate heat (Typical v = 1 mm / min.) k S T ˆ z S Thermal conductivities k s ª1.5 W /(cm - C) k s T v max = Lr m z T = k L T ˆ + L m z L t 0 m t = vr m Heat of fusion Liquid ÆSolid L ª 340 cal mole () S If Tz too large solid fi thermal stress 18 9

10 Czochralski growth of single crystals: stress, dislocations Seed Dead boundary layer dt/dx C/cm Melt For large temperature gradients, e.g. dt/dx C/cm., and given a = 2.6 x 10-6 / 0 C, then Dl/l = adt => strains of 0.6%, which exceeds the yield stress of Si, => dislocations 19 Line defects: dislocations Dislocations originate in shear strains, mostly induced by thermal gradients during growth. couple of dislocations/wafer is typical. Why so few? 1) Tang (neck at beginning of xtl) allows dislocations to move to surface 2) Large number of atoms are involved in a dislocation, => high energy, U Dislocation has low entropy (most atoms are in unique place) G = H - TS large small is very positive 20 10

11 Czochralski growth of single crystals: impurities oundary layer keeps impurities away from solid-liquid interface Impurities must diffuse across still boundary layer. field suppresses convection of ions (reactive) deflecting them from interface ƒ 21 ridgman growth of Gas Convection barrier Gas in quartz ampule: Gas s T m T C Vertical ridgman dt/dx < 10 0 C/cm => dislocation density < 10 3 /cm 2 but melt-ampule contact => lower resistivity. To correct contact problems, use vertical ridgman Gas in N ampule: T m v mm/hr. r 10 MWcm 2-5 x 10 3 dislocations /cm

12 Liquid encapsulated CZ or ridgman growth because Ga s vapor pressure at T m = C: atmospheres. Seed N Crucible (keep Si out) 2 O 3 liquid plug 2 O 3 minimizes loss of s; v 1 cm/hr. r 100 MWcm Gas + s 10 4 defects/cm 2 Pressure may 100+ dislocations /cm 2 build to 20 atm: HPLEC T melt k Crit. resolved shear stress Si Ge Gas Gas cannot dissipate heat; grow slowly Gas cannot take stress; grow slowly 23 Floating zone For very high purity Si (not used for Gas) Top seed (molten part seed supported by upper crystal) high-purity crystal Polycrystal feed rod ottom seed high-purity crystal seed 24 12

13 CRYSTL GROWTH reviewed 1 Confined 2 CZ Meniscus controlled Normal freezing (ridgman) Zone melting Largest crystals: 300 mm diameter Pull from solution 3 Floating zone feed rod horizontal poly vertical poly LEC highestpurity crystal; Larger radial variations than CZ 2 O 3 Si 25 Introduction of dopants via melt Different dopants have different solubilities in solid Define segregation coefficient, k, as ratio of dopant concentrations solid/melt: k = C s C l l s O P Sb k = k < 1 implies only a small fraction of dopant moves into solid; concentration builds up in melt as S/L interface advances This in turn drives up amount transferred to solid We can calculate the dopant/impurity concentration as a function of position in crystal (let x = fraction solidified) 26 13

14 Dopant or impurity concentration vs. Position assumptions: no solid state diffusion, perfect liquid mixing CZ growth Float zone growth C S L C liq = C 0 C S L S C 0 C 0 x=0 x=l [ C L (x) - C S (x)]dx = dc L (x) L - x [ ] C L (x)[ 1- k]dx = dc L (x)[ L - x] x dx C [ 1- k] Ú L (x ) dc = L (x) Ú 0 C L - x 0 C L (x) 1-k È L C L (x) = C 0 ÎL - x È L C S (x) = kc 0 ÎL - x 1-k x = 0 l o x=l [ C 0 - kc L (x)]dx = dc L (x)l 0 xdx C dc Ú L (x ) = L (x) Ú 0 C l 0 0 C 0 - kc L (x) x =- 1 [ l 0 k C C 0 - kc L (x)] L (x ) C0 - kx = ln C - kc (x) C È L (x ) 0 L l 0 Î Í C 0 (1- k) C0 C S (x) = C 0 [ 1-(1- k)exp(-kx / l 0 )] 27 Dopant/inpurity concentration vs. Position È L C S (x) = kc 0 ÎL - x CZ growth k = 0.1, L = 1 1-k C S (x) = C 0 [ 1-(1- k)exp(-kx / l 0 )] Float zone growth k = 0.1, l = 0.1L 10 C s /C x=0 x=l 28 14