3.155J/6,152J Microelectronic Processing Fall Term. 2003 Bob handl Martin schmidt Problem set 2 Solutions Out Sept 17. 2003 Due Sept 24. 2003 Problem 1 Deal-Grove model states that the thickness of the oxide is related to a time constant r and two constants A and b by the relation A+A2+4B(t+) 2 where +ax T B A.B. and r can be obtained from the oxidation of si lecture notes For the initial 20 minutes of dry oxidation at 1000 C A=0.165m B=00117m2hr τ=0.37h Solving for t xox=0.040lum For the 90 minutes of wet oxidation, A=0.226 um and B=0.287 um/hr. We need to calculate a new t using the xox calculated in the previous step and the new a and b values that apply for the wet oxidation τ=0.037h After 90 minutes of wet oxidation xox= 0.5bl um The final 20 minutes of dry oxidation has the same A and b as the first step and we need to calculate a new t from the total xox grown in the previous two steps, 0.561 um 4.8h Solving for our final x
1 3.155J/6.152J Microelectronic Processing Fall Term, 2003 Bob O'Handley Martin Schmidt Problem set 2 Solutions Out Sept. 17, 2003 Due Sept.24, 2003 Problem 1 Deal-Grove Model states that the thickness of the oxide is related to a time constant t and two constants A and B by the relation † xox = -A + A2 + 4B(t + t) 2 where † t = x0 2 + Ax0 B A, B, and t can be obtained from the oxidation of Si lecture notes. For the initial 20 minutes of dry oxidation at 1000 C, † A = 0.165mm B = 0.0117mm 2 /hr t = 0.37h Solving for tox: xox = 0.0401mm For the 90 minutes of wet oxidation, A = 0.226 µm and B = 0.287 µm2 /hr. We need to calculate a new t using the xox calculated in the previous step and the new A and B values that apply for the wet oxidation: t = 0.037 h After 90 minutes of wet oxidation, xox = 0.561 mm The final 20 minutes of dry oxidation has the same A and B as the first step and we need to calculate a new t from the total xox grown in the previous two steps, 0.561 µm. t = 34.8 h Solving for our final xox:
x3=0.564um Using the charts, one can estimate that the oxide thickness should be around 0: 56 um. This shows that the Deal-Grove model is a good approximation for oxide growth Problem 2 The Deal-Grove coefficients for this problem are the same as in problem 1 A=0.165m B=00117m2 τ=0.37h Using the relation a+Va2+4B(+r) After 1 hour the oxide thickness is Problem 3 In general: Using the knowledge that you have regarding the oxidation process, explain qualitatively what you expect to happen when polycrystalline silicon is oxidized and also what is anticipated when Boron is present. You may use some assumptions about B and How will oxide growth rate be affected by poly versus single-crystal silicon tures A, but also speculate physically from what was discussed in the oxidation lec To the first order, the same set of models that model single crystal silicon oxidation can be used for polycrystalline silicon. Polycrystalline means that there are a variety of crystal orientations present in the grains, which will clearly have an effect on the parameters of the Deal-Grove model. No crystal orientation effects are expected on the coefficient b because it describes oxidant diffusion through the sio2. the ratio b/a is effected because it involves reaction at the Si/Sio2 interface. For polycrystalline silicon Plummer says to use an average b/A value. The average b/a value will be for the rate of oxidation of the orientation which is between the two extremes of and Orientation effects incorporated as stated in equation 6. 40 of Plummer. B A 00) How will oxide growth rate be affected by boron atoms? Highly doped substrates oxidize more rapidly than do lightly doped wafers. At lower temperatures, and for thinner oxides, the enhancement can be three to four times that for doped si
2 xx = 0.564 mm Using the charts, one can estimate that the oxide thickness should be around 0:56 mm. This shows that the Deal-Grove model is a good approximation for oxide growth. Problem 2 The Deal-Grove coefficients for this problem are the same as in problem 1. † A = 0.165mm B = 0.0117mm 2 t = 0.37h Using the relation † xox = -A + A2 + 4B(t + t) 2 After 1 hour, the oxide thickness is: xox = 0.0686 mm Problem 3 In general: Using the knowledge that you have regarding the oxidation process, explain qualitatively what you expect to happen when polycrystalline silicon is oxidized and also what is anticipated when Boron is present. You may use some assumptions about B and A, but also speculate physically from what was discussed in the oxidation lectures. How will oxide growth rate be affected by poly versus single-crystal silicon? To the first order, the same set of models that model single crystal silicon oxidation can be used for polycrystalline silicon. Polycrystalline means that there are a variety of crystal orientations present in the grains, which will clearly have an effect on the parameters of the Deal-Grove model. No crystal orientation effects are expected on the coefficient B because it describes oxidant diffusion through the SiO2. The ratio B/A is effected because it involves reaction at the Si/SiO2 interface. For polycrystalline silicon, Plummer says to use an average B/A value. The average B/A value will be for the rate of oxidation of the orientation, which is between the two extremes of and . Orientation effects incorporated as stated in equation 6.40 of Plummer. † B A Ê Ë Á ˆ ¯ ˜ 111 =1.68 B A Ê Ë Á ˆ ¯ ˜ 100 How will oxide growth rate be affected by boron atoms? Highly doped substrates oxidize more rapidly than do lightly doped wafers. At lower temperatures, and for thinner oxides, the enhancement can be three to four times that for undoped Si
B 1+2.62×103 A A kTC The boron dopant concentration is implied by the usage of the term"heavily-doped Heavily-doped means degenerately-doped, which is doping greater than 10cm(page 22 in Plummer) This equation and definition are provided for reference but to solve this problem, we're not expecting you to use the equation. Instead, using a physical understanding of the idation process, calculate the oxide thickness for typical silicon material and then knowledge that this is a lower bound, and then qualitatively explain how the polycrystalline nature and doping affect this bound. One could also look at the lecture notes that show how oxidation thickness is affected by doping and also how b/a and B are affected
3 † B A = B A Ê Ë Á ˆ ¯ ˜ i 1+ 2.62 ¥103 exp -1.1eV kT Ê Ë Á ˆ ¯ ˜ CV T CVi T -1 Ê Ë Á Á ˆ ¯ ˜ ˜ È Î Í Í Í ˘ ˚ ˙ ˙ ˙ The boron dopant concentration is implied by the usage of the term “heavily-doped”. Heavily-doped means degenerately-doped, which is doping greater than 1019 cm-3 (page 22 in Plummer). This equation and definition are provided for reference, but to solve this problem, we're not expecting you to use the equation. Instead, using a physical understanding of the oxidation process, calculate the oxide thickness for typical silicon material and then acknowledge that this is a lower bound, and then qualitatively explain how the polycrystalline nature and doping affect this bound. One could also look at the lecture notes that show how oxidation thickness is affected by doping and also how B/A and B are affected