3.155J6.152J Microelectronic Processing Fall Term. 2003 Bob handley Martin schmidt Problem set 1 solutions Out Sept. 8. 2003 Due Sept 17. 2002 1. a) Mean free path 2= kBI d-p where k=1.38×102,T=293K,p=10mTor=133Pa,d=1A=100m 38×10-3(293 6.9cm ford=1A Jr(m)(332)017mfrd=2A b) Use the ideal gas law pV=nkB T 1.33Pa vk7(138×102X90)33×10m-33×10cm3 P c) Flux on a surface P 2IkBT where m=mass of particle =(28 amu)(1.67x 10-kg/amu)=4.76 x 10-kg J, =3.86x 1022 molecules 39×108cm2s- m2s d) We are trying to find the average molecular speed of n2 molecules impinging a surface in the chamber. two ways to do this 2.3=2.3×104 L 235
1 3.155J/6.152J Microelectronic Processing Fall Term, 2003 Bob O'Handley Martin Schmidt Problem set 1 Solutions Out Sept. 8, 2003 Due Sept.17, 2002 1. a) Mean free path l = kBT 2pd2 p where kB = 1.38 ¥ 10-23, T = 293 K, p = 10 m Torr =133 Pa, d ª 1A = 10-10 m l = 1.38 ¥10-23 ( )( ) 293 K 2p 10-10 ( ) m 2 ( ) 1.33Pa = 6.9cm 1.7cm Ï Ì Ô Ó Ô b) Use the ideal gas law pV = N kB T c) Flux on a surface: Jx = p 2pkBTm where m = mass of particle = ( 28 amu) ( 1.67 ¥ 10 –27 kg/amu) = 4.76 ¥ 10 -26 kg Jx = 3.86 ¥1022 molecules m2 s = 3.9 ¥1018 cm-2 s -1 d) We are trying to find the average molecular speed of N2 molecules impinging a surface in the chamber. Two ways to do this: vx = 2J n = 2.3 m s = 2.3¥104 cm s or vx = 2kBT pm = 235 m s for d = 1 A for d = 2 A N V = p kBT = 1.33Pa (1.38¥10-23 ) 293 ( ) K = 3.3¥1020 m-3 = 3.3¥1014 cm-3
some people calculated the average molecular speed in any direction, which is This is also accepted. (Note that the molecular speed is independent of pressure because both J and n are proportional to p. 2. SCCM stands for a flow of a"standard cubic centimeter per minute"'(standard cc is at temperature of 273 K and pressure of 760 Torr) L min 100 =0.001666 min人1000cm3人60s To change to 100 K and l mTorr L∥100KY760T 00066 =463 s八273K八10Tor CVD 3. Assume chemical equilibrium is established in a CVD reactor according to the quation Sih4 (g) (g)+H2(g) The temperature is maintained at 650C and the pressure at 0. 1 atm. If the equilibrium constant for the reaction is K(T)=2 x 10(Torr)exp[-1.8 eV/(kgT], find the partial pressure of each gas assuming P(H,) P(SiH,) 10 points. K(T=923K)=2×10°eX[-29×10J1ka×923 0.3 Torr Psi,X PH,-PsiHz=0.3 Torr (*)(assumption: PH2=Ps PH,+PH +PH=76 To in addition Psin, +2 76Tor(**) Substituting(**)into()
2 some people calculated the average molecular speed in any direction, which is v = 8kT pm = 476 m s This is also accepted. (Note that the molecular speed is independent of pressure because both J and n are proportional to p.) 2. SCCM stands for a flow of a “standard cubic centimeter per minute” (standard cc is at temperature of 273 K and pressure of 760 Torr). 100 cm3 min Ê Ë Á Á ˆ ¯ ˜ ˜ L 1000cm 3 Ê Ë Á ˆ ¯ ˜ min 60s Ê Ë Á ˆ ¯ ˜ = 0.001666 L s To change to 100 K and 1 mTorr, 0.001666 L s Ê Ë Á ˆ ¯ ˜ 100 K 273 K Ê Ë Á ˆ ¯ ˜ 760Torr 10-3 Torr Ê Ë Á ˆ ¯ ˜ = 463 L s CVD 3. Assume chemical equilibrium is established in a CVD reactor according to the equation: SiH4 (g) ´ SiH2 (g) + H2 (g) The temperature is maintained at 650 C and the pressure at 0.1 atm. If the equilibrium constant for the reaction is K(T)=2 ¥ 109 (Torr) exp[-1.8 eV/(kBT)], find the partial pressure of each gas assuming p(H2) p(SiH2). 10 points. K(T = 923K) = 2 ¥109 exp -2.9 ¥10-19 J / k [ ] ( ) B ¥ 923K = 0.3 Torr Also, K = pSiH2 ¥ pH2 pSiH4 = pSiH2 2 pSiH4 = 0.3 Torr (*) (assumption : pH2 = pSiH2 ) I in addition, pSiH4 + pSiH2 + pH2 = 76 Torr pSiH4 + 2pSiH2 = 76 Torr (**) Substituting (**) into (*):
Pm2=0.3psm,=0.3(76-2Psmn2) 28-06 →Psi,+0.6p 28=0 PiH,=4.48 Torr H3 Ps 67 Tor Check that the sum of the partial pressures, p, =76 Torr=0. 1 atm 4. Assume a CVd process based on the reaction: 2AB(g)+ 2A(S)+B(g) a) Sketch and briefly describe the individual steps that control the reaction B 7 6 B 1)Bulk transport governed by gas velocity outside boundary layer, u 2) Diffusion across boundary layer CR -Cs) S(x)is the thickness of the boundary layer, a function 8(x)U 3)Adsorption involves sticking 4)Dissociation of AB due to temperature and possibly catalyzed by interaction of A with surface b may remain adsorbed or desorb upon dissociation 5)A(s) actually bonds with a site on substrate surface(film growth) 6)B, 8)must diffuse across boundary layer. JB. =B(cB-cB) 7) Bulk transport of B, (g) under carrier gas velocity u
3 pSiH2 2 = 0.3pSiH4 = 0.3(76 - 2pSiH2 ) = 22.8 - 0.6pSiH2 fi pSiH2 2 + 0.6 pSiH2 - 22.8 = 0 pSiH2 = 4.48 Torr pH2 ª 4.48 Torr pSiH4 = 67 Torr Check that the sum of the partial pressures, pi = 76 Torr = 0.1 atm. 4. Assume a CVD process based on the reaction: 2AB(g) ´ 2A(s) + B2(g). a) Sketch and briefly describe the individual steps that control the reaction. 15 points. 1) Bulk transport governed by gas velocity outside boundary layer, u•. 2) Diffusion across boundary layer J µ D d(x) ( ) Cg -Cs , d(x) is the thickness of the boundary layer, a function of x. 3) Adsorption involves sticking. 4) Dissociation of AB due to temperature and possibly catalyzed by interaction of A with surface. B may remain adsorbed or desorb upon dissociation. 5) A(s) actually bonds with a site on substrate surface (film growth). 6) B2(g) must diffuse across boundary layer. ( ) 2 2 2 2 ( ) B s B g B B C C x D J = - d 7) Bulk transport of B2(g) under carrier gas velocity u•. Cs 4 3 A B 2 Cg 1) A B A B B 5 B B 6 • u B B 7 A B A
b) How would you distinguish between i) the reaction-limited and ii)a transport- limited cases 5 p In the reaction-limited regime, the temperature dependence of the film growth ate exhibits arrhenius behavior △G △G In the transport-limited regime, the deposition rate is given by 32v, pu,x AN xv n kaT and入 p T 2md2potal so that P√T e. Thus, temperature dependence is the biggest differentiator between the two c)Sketch the variation of the log of the Cvd growth rate as functions of the square root of the gas flow velocity and as a function of 1/T 5 points. Reaction regim Transport regime Reaction regime Transport regime 4
4 b) How would you distinguish between i) the reaction-limited and ii) a transportlimited cases? 5 points. In the reaction-limited regime, the temperature dependence of the film growth rate exhibits Arrhenius behavior: ˜ ˜ ¯ ˆ Á Á Ë Ê - D µ ˜ ˜ ¯ ˆ Á Á Ë Ê - D = = k T G v k T G k k N kC v B B o g exp , exp In the transport-limited regime, the deposition rate is given by: v = 3lvxCg 4Nf x ru• x h , where vx = 2kBT pm , Cg Pg = 1 kBT , and l = kBT 2pd2 Ptotal so that v µ Pg T u• Thus, temperature dependence is the biggest differentiator between the two regimes. c) Sketch the variation of the log of the CVD growth rate as functions of the square root of the gas flow velocity and as a function of 1/T. 5 points. ln(v) u Transport regime Reaction regime 1/T ln(v) Transport regime Reaction regime
d) If you wanted to increase the growth rate of a transport-limited CVd process what processing variables would be most effective?(List them in decreasing order of efficacy, e.g. v o exp(x)first, etc 5 nts For the fixed length, in the transport-limited regime, in decreasing order of efficacy, we have Pg, yu,, and 5. Phosphorus-doped polysilicon is produced by CVd based on the following reactions Sih(g)e Si(s)+ 2H,(g) 2PH3(g)“2P(s)+3H2(g) Deposition occurs at 1000oC, Ns=5x 10cm", the Si deposition is reaction limited (cn=108cm3,k。=6×10°cm/s),△G=1.0eV. and the deposition of the P is transport limited(c=108cm 3, D=10-xT cm/s, and(pu/n)=0.01 cm.) a)What range of P concentration in the poly-silicon might you expect across a 30 cm diameter wafer? Explain the assumptions you must make to answer this question 10 points. The silicon deposition is reaction limited k=6×105ex 8617×10eV/K×1273K C=10cm Namn=5×102cm3 1.32x10 he phosphorus deposition is transport limited DCR pLu n D=10-3×(1273)2=0.45cm2/s 3×10 phosphorus
5 d) If you wanted to increase the growth rate of a transport-limited CVD process, what processing variables would be most effective? (List them in decreasing order of efficacy, e.g. v µ exp(x) first, etc.) 5 points. For the fixed length, in the transport-limited regime, in decreasing order of efficacy, we have Pg, u• ,and T . 5. Phosphorus-doped polysilicon is produced by CVD based on the following reactions: SiH4 (g) ´ Si (s) + 2H2 (g) 2PH3 (g) ´ 2P (s) + 3H2 (g). Deposition occurs at 1000o C, NSi = 5 ¥ 1022 cm-3, the Si deposition is reaction limited (cg = 1018 cm-3, ko = 6 ¥ 105 (cm/s), DG = 1.0 eV), and the deposition of the P is transport limited (cg = 1018 cm-3, Dg = 10-5 ¥ T3/2 cm/s, and (ru/h) 0.5 = 0.01 cm-0.5). a) What range of P concentration in the poly-silicon might you expect across a 30 cm diameter wafer? Explain the assumptions you must make to answer this question. 10 points. The silicon deposition is reaction limited: vsilicon = kCg N k = 6 ¥105 exp -1eV 8.617 ¥10-5 eV /K ¥1273K Ê Ë Á ˆ ¯ ˜ = 66 cm /s Cg =1018 cm-3 Nsilicon = 5 ¥1022 cm-3 vsilicon =1.32 ¥10-3 cm /s The phosphorus deposition is transport limited: v phosphorus = DCg LNP rLm h D =10-5 ¥ (1273) 3 2 = 0.45cm 2 /s v phosphorus = 3¥10-6 L cm /s
We assume that a boundary layer develops at l=0 cm (0) But the phosphorus will not dope the silicon beyond its solid solubility limit, about 1.5 x 02cm(Fig 2.4 in Campbell) 1.5×10=3% v=1.32×10°cm/s phosphor÷547×10°cm/s ma547×107cm/s 415×10+=41ppm 132×10-3cm/s b) If the temperature at the far end of the wafer is constrained to be 10 C hotter than that at the near end of the points With the hotter temperature, both deposition rates should increa 1283 3×10-7cm/sec=10ly° 1273 Thus, the silicon deposition rate has increased by 7%, while the phosphorus deposition rate increased by only 1%. The increase in temperature actually leads to a less 6
6 We assume that a boundary layer develops at L = 0 cm: vP (0) = • But the phosphorus will not dope the silicon beyond its solid solubility limit, about 1.5 ¥ 1021cm-3 (Fig 2.4 in Campbell): 3% 5 10 1.5 10 22 21 = ¥ ¥ At L = 30 cm: vsilicon =1.32 ¥10-3 cm /s v phosphorus = 5.47 ¥10-7 cm /s Cphosphorus Csilicon = 5.47 ¥10-7 cm /s 1.32 ¥10-3 cm /s = 4.15 ¥10-4 = 41ppm b) If the temperature at the far end of the wafer is constrained to be 10 C hotter than that at the near end of the reactor, how does this affect your answer? 10 points. With the hotter temperature, both deposition rates should increase: vsi =13.3mm s vsi 0 124 34 ¥ e 1eV kB 1 1273- 1 1283 Ê Ë Á ˆ ¯ ˜ =1.4 ¥10-3 cm /s =1.07 ¥ vsi 0 v p = v p 0 ¥ 1283 1273 Ê Ë Á ˆ ¯ ˜ 3 2 = 5.53¥10-7 cm /sec =1.01v p 0 Thus, the silicon deposition rate has increased by 7%, while the phosphorus deposition rate increased by only 1%. The increase in temperature actually leads to a less uniform doping
