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5. Below is a process flow that involves multiple high temperature steps (2, 3, 4 and 6). a) Determine the Dt value associated with each of the individual steps, #3, 4 and 6, where D is Phosphorus Diffusivity at the indicated temperature. b) Find the Dt value for step #1, so that the thermal budget (sum for all steps), (Dt)ro =1.2 E-9 cm c) Determine the time of the Phosphorus implant anneal in step #1. Starting material: (100) Silicon, doped with Boron, at concentration, CB = 1E14 cm Assume a final Phosphorus profile (after step 6) given by: C(.1) Dexp ID -C(O)explain where C(0,0) = 4E17 cm Rs = 500 2/0, and Dt = (Dt) = 1.2 E-9 cm Note: This is the n-well design example in Lecture 14! (because Di >>AR) For both oxidation steps, the initial oxide thickness, x = 0. Use the oxidation charts to find times. Process flow: Step Process Phosphorus implant time Temp Dt (cm) Spec. Q=2.45E13 cm E = 170 keV X = 2 um (after step 6) Xox = 400 nm Phosphorus implant anneal Field oxidation, Steam 1100 C 1000 C Xox = 50 nm 1000 C Implant mask oxidation, dry02 Boron implant Q=5E14 cm, E=20 keV X; = 0.4 um 6 Boron implant anneal 1000 C Thermal Budget, (Dtrol 1.2 E-9 Use the equation given below to find t in step #6, using D(1000 C) for Boron, x = 0.4 pm and C(x, t) equal to the concentration of Phosphorus at x=0.35 pm, found using the final Phosphorus profile giver above. You will need to iterate the equation below to solve for the Boron anneal time. Cx,t)- expl_(x-R)? 20(Ar& +201) "1Ar +20) To find the Dt contribution from step #6 to the thermal budget you need to multiply D(1000 C) for Phosphorus times the Boron anneal time. This is an example of how the process design is complicated by interdependency of steps. Plotting the resulting profiles of Boron and Phosphorus may help you to comprehend this problem. 5. Below is a process flow that involves multiple high temperature steps (2, 3, 4 and 6). a) Determine the Dt value associated with each of the individual steps, #3, 4 and 6, where D is Phosphorus Diffusivity at the indicated temperature. b) Find the Dt value for step #1, so that the thermal budget (sum for all steps), (Dt)ro =1.2 E-9 cm c) Determine the time of the Phosphorus implant anneal in step #1. Starting material: (100) Silicon, doped with Boron, at concentration, CB = 1E14 cm Assume a final Phosphorus profile (after step 6) given by: C(.1) Dexp ID -C(O)explain where C(0,0) = 4E17 cm Rs = 500 2/0, and Dt = (Dt) = 1.2 E-9 cm Note: This is the n-well design example in Lecture 14! (because Di >>AR) For both oxidation steps, the initial oxide thickness, x = 0. Use the oxidation charts to find times. Process flow: Step Process Phosphorus implant time Temp Dt (cm) Spec. Q=2.45E13 cm E = 170 keV X = 2 um (after step 6) Xox = 400 nm Phosphorus implant anneal Field oxidation, Steam 1100 C 1000 C Xox = 50 nm 1000 C Implant mask oxidation, dry02 Boron implant Q=5E14 cm, E=20 keV X; = 0.4 um 6 Boron implant anneal 1000 C Thermal Budget, (Dtrol 1.2 E-9 Use the equation given below to find t in step #6, using D(1000 C) for Boron, x = 0.4 pm and C(x, t) equal to the concentration of Phosphorus at x=0.35 pm, found using the final Phosphorus profile giver above. You will need to iterate the equation below to solve for the Boron anneal time. Cx,t)- expl_(x-R)? 20(Ar& +201) "1Ar +20) To find the Dt contribution from step #6 to the thermal budget you need to multiply D(1000 C) for Phosphorus times the Boron anneal time. This is an example of how the process design is complicated by interdependency of steps. Plotting the resulting profiles of Boron and Phosphorus may help you to comprehend this