3.155J/6,152J Microelectronic Processing Technology Fall Term. 2003 Quiz 2 With SOLUTIONs Dec.3,2003 vaporation and sputter deposition [151 1. a)In the plasma of a sputter deposition system, consider an argon atom that is ionized during a collision in the plasm 1) Express the ratio of the acceleration of the liberated electron to that of the argon ion in terms of m and m, the mass of the ion and electron; give the numerical value of this ratio, and say in which direction each species accelerates )31 11) Over a given mean free path, one of the particles will achieve a much higher speed. Explain how that affects the potential (voltage)of the plasma relative to that of either electrode? [31 111) In argon(atomic number A =40)sputtering from a target composed of equal parts of tungsten(A= 184)and aluminum(A 27), would the deposited layer more likely turn out richer in Wor AI?[3 b)Answer these items with either"" or"evaporation"or"both 1) Which deposition method affords better step coverage? [21 11) Which deposition method is more likely to have ballistic transport ng deposited? [21 111) Which method is more likely to be used for deposition of an epitaxial single crystal film? [2] 1. a) i)a=F/m, hence alecto aon_= M/m; electron accelerates toward anode, ion 11) Over a given mean free path, electron achieves a much hi gher speed and thus travels much farther than the ion. Since most of these particles are created in the plasma(some a generated at the cathode), more electrons than ions leave the plasma and as a result, the plasma is at a positive potential relative to either electrode 111) Argon ions are much more effective at transferring energy to species having similar masses(mM/m+M, so the sputtering yield for Al is much greater than that for tungsten in argon ion sputtering b) 1) sputtering; species approach substrate from a greater distribution of angles i1)evaporation; much lower pressure gives longer mena free path chan ii) evaporation; film can be cleaner because of lower background pressure I mber, and hence fewer defects available to upset periodicit Interconnects 25 2. Two different metals, tungsten(W) and copper(Cu) are to be used for interconnects on given IC. [12 a) Which of these metals is more likely to be used for local interconnects? [21
1 3.155J/6.152J Microelectronic Processing Technology Fall Term, 2003 Quiz # 2 With SOLUTIONS Dec. 3, 2003 Evaporation and sputter deposition [15] 1. a) In the plasma of a sputter deposition system, consider an argon atom that is ionized during a collision in the plasma. i) Express the ratio of the acceleration of the liberated electron to that of the argon ion in terms of M and m, the mass of the ion and electron; give the numerical value of this ratio, and say in which direction each species accelerates.) [3] ii) Over a given mean free path, one of the particles will achieve a much higher speed. Explain how that affects the potential (voltage) of the plasma relative to that of either electrode? [3] iii) In argon (atomic number A = 40) sputtering from a target composed of equal parts of tungsten (A = 184) and aluminum (A = 27), would the deposited layer more likely turn out richer in W or Al? [3] b) Answer these items with either “sputtering” or “evaporation” or “both”. i) Which deposition method affords better step coverage? [2] ii) Which deposition method is more likely to have ballistic transport of the species being deposited? [2] iii) Which method is more likely to be used for deposition of an epitaxial single crystal film? [2] 1. a) i) a =F/m, hence aelecton/aion = M/m ; electron accelerates toward anode , ion toward cathode. ii) Over a given mean free path, electron achieves a much higher speed and thus travels much farther than the ion. Since most of these particles are created in the plasma (some a generated at the cathode), more electrons than ions leave the plasma and as a result, the plasma is at a positive potential relative to either electrode. iii) Argon ions are much more effective at transferring energy to species having similar masses (mM /[m + M]), so the sputtering yield for Al is much greater than that for tungsten in argon ion sputtering. b) i) sputtering; species approach substrate from a greater distribution of angles. ii)evaporation; much lower pressure gives longer mena free path. iii) evaporation; film can be cleaner because of lower background pressure I chamber, and hence fewer defects available to upset periodicity. Interconnects [25] 2. Two different metals, tungsten (W) and copper (Cu) are to be used for interconnects on a given IC. [12] a) Which of these metals is more likely to be used for local interconnects? [2]
b) Give two reasons for this choice. [4 c)Which is more likely to be used for global interconnects? [2] d) Give two reasons for this choice. 4 2. a) Tungsten more likely for local or interlevel interconnects b) local interconnects lengths are short so higher resistivity of w is less of a problem, and high melting temperature of w means it can withstand subsequent processing steps c)Copper more likely used for global interconnects d) lower resisitivity, it cannot withstand higher temperatures of earlier steps but acceptable later in processing 3. Two global interconnects are fabricated to be 0.25 microns in both thickness and width. They are made of a metal having a resistivity of 2 micro-Ohm-cm. They are parallel to each other over a length of 5 mm and their nearest sides are separated by 0.5 microns filled with SiO, (relative dielectric constant, K=4).[13 a) What is the resistance of each of these two independent lines? [5] b) What is their relative capacitance ( ignore interactions with other lines)?[41 c) what is the resulting rc time constant? 4 3.a)R=pLA=5x103/{0.252x101}=160092 b)C=kEAd=0.25X10°X5X10/{0.5×10°}=885X104F )RC=1.42×10 Reliability [15] 4. a) Your company has developed a new chip fab line having 20 independent processes: 10 of them with a yield of 97%o each and 10 with a yield of 96% each. Your estimated break-even yield is 60%. If you were to go into production now, would you expect to make a profit? [31 b)Would electromigration be more of a problem with tungsten or aluminum interconnects? Why? 31 c)You need to double the current density in a given interconnect. Would your problems with electromigration double or could things be much worse? Explain. [4] d) Briefly explain how grain boundaries play a role in hillock and/or void formation in interconnects. 5 4. a)No profit. Your yield will be about 44%(0.97 x0.95) b) Electromigration is more of a problem with Al because of its smaller mass relative W, even though W may have a higher resistivity c)Doubling the current density would result in a very large increase in electromigration by increasing the temperature, which is in the exponential of the grain boundary diffusion term. Increasing resistivity with increasing temperature is also a valid factor d) Hillocks and voids form at point along a conductor where electromigration flux converges or diverges. respectively. Hence hillocks are more likely where more grain
2 b) Give two reasons for this choice. [4] c) Which is more likely to be used for global interconnects? [2] d) Give two reasons for this choice. [4] 2. a) Tungsten more likely for local or interlevel interconnects. b) local interconnects lengths are short so higher resistivity of W is less of a problem, and high melting temperature of W means it can withstand subsequent processing steps. c) Copper more likely used for global interconnects. d) lower resisitivity, it cannot withstand higher temperatures of earlier steps but is acceptable later in processing. 3. Two global interconnects are fabricated to be 0.25 microns in both thickness and width. They are made of a metal having a resistivity of 2 micro-Ohm-cm. They are parallel to each other over a length of 5 mm and their nearest sides are separated by 0.5 microns filled with SiO2 (relative dielectric constant, k = 4). [13] a) What is the resistance of each of these two independent lines? [5] b) What is their relative capacitance (ignore interactions with other lines)? [4] c) What is the resulting RC time constant? [4] 3. a) R = rL/A =5 x 10-3/{0.252 x 10-12} = 1600 W. b) C = ke0A/d =0.25 x 10-6 x 5 x 10-3/{0.5 x 10-6} = 8.85 x 10-14 F c) RC = 1.42 x 10-10 s. Reliability [15] 4. a) Your company has developed a new chip fab line having 20 independent processes; 10 of them with a yield of 97% each and 10 with a yield of 96% each. Your estimated break-even yield is 60%. If you were to go into production now, would you expect to make a profit? [3] b) Would electromigration be more of a problem with tungsten or aluminum interconnects? Why? [3] c) You need to double the current density in a given interconnect. Would your problems with electromigration double or could things be much worse? Explain. [4] d) Briefly explain how grain boundaries play a role in hillock and/or void formation in interconnects. [5] 4. a) No profit. Your yield will be about 44% (0.9710 x 0.9510). b) Electromigration is more of a problem with Al because of its smaller mass relative to W, even though W may have a higher resistivity. c) Doubling the current density would result in a very large increase in electromigration by increasing the temperature, which is in the exponential of the grain boundary diffusion term. Increasing resistivity with increasing temperature is also a valid factor. d) Hillocks and voids form at point along a conductor where electromigration flux converges or diverges, respectively. Hence hillocks are more likely where more grain
boundaries that are more nearly parallel to the carrier flow direction enter a point than those types of grain boundaries leaving that point. Similar arguments for thermal gradient-induced migration stal growth [15 The change in Gibbs free energy during a transformation is dG= dH - Tds Answer the following 3 questions about crystallization of Si from the melt using increases","decreases or "remains the same 1) What happens to the enthalpy(heat content) of the material [21 11) What happens to the entropy? [2 i11) What happens to the Gibbs free energy?[21 b) In the depiction of the of crystal growth shown belo 1 indicate with arrows the direction of convection in the melt. [2] 11) Why is there a maximum velocity for growth of a crystal from the melt? 4] i)What processing conditions most affect the appearance of dislocations in the crystal, and how would they be removed? 31 see tang NVAC ry 1500°C 5. a) 1) Enthalpy decreases on crystallization of Si from melt 11) Entropy, ds,(disorder) decreases; -TdS increases 111) Gibbs free energy decreases if the temperature is below the solidification temperature. D 1 See figure. 11) The thermal gradient in the solid must be than that in the liquid in order to also remove the heat of fusion ted at the solid/liquid interface. The amount of heat generated due to solidification increases with growth rate 111) Dislocations arise due to stress in crystal. Stress in crystal growth is most often due to thermal gradients in the crystal. The thermal gradients can re reduced by slower growth or smaller boule radius
3 boundaries that are more nearly parallel to the carrier flow direction enter a point than those types of grain boundaries leaving that point. Similar arguments for thermal gradient-induced migration. Crystal Growth [15] 5. a) The change in Gibbs free energy during a transformation is dG = dH - TdS. Answer the following 3 questions about crystallization of Si from the melt using “increases”, “decreases” or “remains the same” i) What happens to the enthalpy (heat content) of the material [2] ii) What happens to the entropy? [2] iii) What happens to the Gibbs free energy? [2] b) In the depiction of the of crystal growth shown below i) indicate with arrows the direction of convection in the melt. [2] ii) Why is there a maximum velocity for growth of a crystal from the melt? [4] iii) What processing conditions most affect the appearance of dislocations in the crystal, and how would they be removed? [3] 5. a) i) Enthalpy decreases on crystallization of Si from melt. ii) Entropy, dS, (disorder) decreases; -TdS increases. iii) Gibbs free energy decreases if the temperature is below the solidification temperature. b) i) See figure. ii) The thermal gradient in the solid must be larger than that in the liquid in order to also remove the heat of fusion generated at the solid/liquid interface. The amount of heat generated due to solidification increases with growth rate. iii) Dislocations arise due to stress in crystal. Stress in crystal growth is most often due to thermal gradients in the crystal. The thermal gradients can re reduced by slower growth or smaller boule radius. 1500°C seed tang VAC crystal
Some useful equations Constants:k=1.38×102J/K=8.62×103eV/K,e=1.6×101°Cou atm =760 Torr= 10 Pa Ideal gas: PV=NKBT, N/V=n=C(concentration), J nv Gas kinetics: v Mean free path: 2 kBT d= molecular diameter v2TdP 2.tk, Tm, -SP 4M.M Energy transfer in collision of M, and Mi: 4- (M, +M, oS Reliability: mean time to failure: MTTF=t. f(odt or o-ne*E /g Electromigration: -9-0DE-,g (number/m-x sec) Ja is flux of species A, having a concentration of Ca, where flux is due to current in a medium of electrical resistivity is p. at temperature, T, by grain-boundary diffusion Da= Dexp/-E/(kBT)/ Change in concentration due to electromigration and temperature gradients d a dT dh dx dt dx Electrical: Resistivity, R=PlA(Ohm) mobility Capacitance, C=KE A/d(Farad), E=8.85 x 10(F/m) Crystal growth: vacancy concentratio p-E/ E=2.6ev =5x1032si/cm3 Oxygen concentration: Cor,=2x10exp[-1.03eV/kBT] Maximum crystal growth velocity: Vmas LPm a Isolid
4 Some useful equations Constants: kB = 1.38 ¥ 10-23 J/K = 8.62 ¥ 10-5 eV/K, e = 1.6 ¥ 10-19 Coul. 1 atm = 760 Torr = 105 Pa Ideal gas: PV = NkBT, N/V = n = C (concentration), † Jx = nv x 2 . Gas kinetics: † v x = 2kBT pm , Mean free path: † l = kBT 2pd2 P , d = molecular diameter. † J = P 2pkBTm , m = species mass. Energy transfer in collision of M1 and M2: Reliability: mean time to failure: Electromigration: (number/m2 x sec) JA is flux of species A, having a concentration of cA, where flux is due to current J in a medium of electrical resistivity is r.. at temperature, T, by grain-boundary diffusion obeying: Change in concentration due to electromigration and temperature gradients: Electrical: Resistivity, R =r l/A (Ohm), mobility, Capacitance, C = ke0A/d (Farad), e0 = 8.85 x 10-12 (F/m) Crystal growth: vacancy concentration: Eg = 2.6 eV n0 = 5 x 1022 Si/cm3 . Oxygen concentration: Maximum crystal growth velocity: JA = cAvA = cA DAF RT = cA DAqZA * Jr RT DA = DA o exp[-Ea /(kBT)] † DE = E1 4M1M2 (M1 + M2) 2 cos2 q † MTTF = t ⋅ f (t)dt 0 • Ú or µ J-n e+Ea / kBT dcA dt = - ∂JA ∂x - ∂JA ∂T dT dx † nvac = n0 exp -Eg /k [ BT] † Coxy = 2 ¥1022 exp -1.03eV /k [ BT] † m = v E = s ne = et m * † vmax = ks Lr m ∂T ∂z ˘ ˚ solid