1 UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences C. SPANOS Special Issues in Semiconductor Manufacturing EECS 290H Fall 0 PROBLEM SET No. Official Solutions 1. A process engineer is about to begin a comprehensive study to determine the effects of five variables on the etch rate of polysilicon. a) If a 2 factorial design were used, how many runs would be made? 2 = 32 b) If σ 2 is the experimental error variance of an individual observation, what is the variance of the main effect? Assuming 32 runs, then the variance of the main effect will be 2σ 2 /16 = σ 2 /8. c) What is the simple formula for a 99% confidence interval for the main effect of variable 1? (Simple here means that you can assume that the σ of the process is known.) Assuming that we know sigma, then the 99% interval is given as: Effx 1 ± 2.67σ/2.83 d) On the basis of some previous work it is believed that σ = nm/min. If the experimenter wants 9% confidence intervals for the main effect and interactions whose lengths are equal to nm/min, (i.e. the upper limit minus the lower limit is equal to nm/min), how many replications of the 2 factorial design will be required? The width of the 9% intervals will be = 2 x 1.96 * * Sqrt(4/N), where N=32, 64, 96,... In order to get as close as possible to satisfying this equation when N is a multiple of 32, I choose N=992, or 31 replications of the 32 run experiment. 2. Consider the following data representing non-uniformity (in %) of Films Grown by CVD, where A, B, C and D represent different chambers and (1), (2) and (3) represent different recipes. (1) 0.31 0.4 0.46 0.43 A B C D 0.82 0.43 1.10 0.4 0.88 0.63 0.72 0.76 0.4 0.71 0.66 0.62
2 (2) 0.36 0.29 0. 0.23 (3) 0.22 0.21 0.18 0.23 0.92 0.61 0.49 1.24 0. 0.37 0.38 0.29 0.44 0.3 0.31 0. 0.23 0.2 0.24 0.22 0.6 1.02 0.71 0.38 0. 0.36 0.31 0.33 Carry out analysis of variance first using the data as is and then using the transformation Y = y -1. Plot the residuals, and consider whether in the new response metric there is evidence of model inadequacy. Compare the treatment averages for the two different representations of the response. The basic ANOVA shown below reveals that the interaction term is insignificant for either the straight or the transformed data. The transformed data analysis however, does give me stronger significance and much more reasonably distributed residuals. In both cases the treatment averages are significantly impacted by the treatment choice. In the next pages I repeat the analysis in greater detail and by exluding the inisgnificant interaction term Response Non-Uniformity Model 11 2.4362 0.0396 9.0097 Error 36 0.8007 0.022242 Prob > F C. Total 47 3.000813 <.0001 Effect Tests Source Nparm DF Sum of Squares F Ratio Prob > F Recipe 3 3 0.92162 13.806 <.0001 Chamber 2 2 1.0312 23.2217 <.0001 Recipe*Chamber 6 6 0.137 1.8743 0.1123 Response Y^-1 Model 11 6.862181.16929 21.310 Error 36 8.6483 0.9 Prob > F C. Total 47 6.0264 <.0001 Effect Tests Source Nparm DF Sum of Squares F Ratio Prob > F Recipe 3 3.414289 28.3431 <.0001 Chamber 2 2 34.8771 72.6347 <.0001 Recipe*Chamber 6 6 1.70772 1.0904 0.3867
3 Response Non-Uniformity, excluding interaction term 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0.0.2.4.6.8 1.0 1.2 1.4 Non-Uniformity Predicted P<.0001 RSq=0.6 RMSE=0.182 Non-Uniformity Actual Model 1.942187 0.390844 1.69 Error 42 1.00862 0.021 Prob > F C. Total 47 3.000813 <.0001 Lack Of Fit Lack Of Fit 6 0.137 0.041690 1.8743 Pure Error 36 0.8007 0.022242 Prob > F Total Error 42 1.00862 0.1123 Max RSq 0.733 Effect Tests Source Nparm DF Sum of Squares F Ratio Prob > F Recipe 3 3 0.92162 12.2727 <.0001 Chamber 2 2 1.0312.6433 <.0001 Non-Uniformity Residual 0. 0.4 0.3 0.2 0.1 0.0-0.1-0.2-0.3.0.2.4.6.8 1.0 1.2 1.4 Non-Uniformity Predicted Residuals do NOT appear to be IIND. A transformation is NEEDED here
4 Response Y^-1, excluding interaction term 6 4 Y^-1 Actual 3 2 1 1 2 3 4 6 Y^-1 Predicted P<.0001 RSq=0.84 RMSE=0.4931 Model.2919 11.083 4.4723 Error 42 10.2138 0.2432 Prob > F C. Total 47 6.0264 <.0001 Lack Of Fit Lack Of Fit 6 1.70772 0.26179 1.0904 Pure Error 36 8.6483 0.86 Prob > F Total Error 42 10.2138 0.3867 Max RSq 0.8681 Effect Tests Source Nparm DF Sum of Squares F Ratio Prob > F Recipe 3 3.414289 27.9816 <.0001 Chamber 2 2 34.8771 71.7084 <.0001 Y^-1 Residual 1.0 0. 0.0-0. -1.0 1 2 3 4 6 Y^-1 Predicted
3. Consider a 2 8-4 fractional factorial design. (a) How many variables does the design have? This design has 8 variables. (b) How many runs are involved in the design? 2 8-4 = 2 4 = 16 runs. (c) How many levels are used for each of the variables? This is a 2-level factorial, so two levels are used. (d) How many independent generators are there? Since this is a 16 th fraction (the -4 in the exponent), we will need four independent generators. (e) Write the required generators that give you the maximum resolution. Here I cheated by looking them up at Box Hunter an Hunter: I=1248, I=138, I=2368, I=1237 (f) What is the resolution of your design? This is a resolution IV design: no main effects are confounded with two parameter interactions or with other main effects. 4. Construct a 2 7-1 fractional factorial design. Show how the design may be divided into eight blocks of eight runs each so that no main effect or two-factor interaction is confounded with any block effect. I looked up the answer on page 8 for the Box Hunter and Hunter textbook. I used one generator 7 = 12346, and the blocks were generated as (I need three blocking factors to generate 8 blocks (2 3 ) of 8 runs each for a total of 2 7-1 =64 runs): B 1 = 137 B 2 = 126 B 3 = 1234 Clearly on main factors or two factor interactions interact with those bock effects.
6. Fit the followind PECVD deposition data with a linear model using regression techniques, and perform analysis of variance to evaluate the quality of your model. Is the linear model sufficient? Observation (u) Time (x u, min) Thickness (y u, μm) 1 8 6.16 2 22 9.88 3 3 14.3 4 24.06 7.34 6 73 32.17 7 78 42.18 8 87 43.23 9 98 48.76 Formally I cannot talk about model sufficiency because without experimental replication I cannot directly test for it. Under these circumstances, my job is to find the simplest significant model, and to exclude all insignificant terms. Analysis shows that a linear model is clearly significant. This is true with or without the intercept term, which, on its own it is not significant. A quadratic term is clearly insignificant.
7 Response Thickness (full model) Thickness Actual 0 1 0 10 0 Thickness Predicted P<.0001 RSq=0.97 RMSE=3.241 Model 2 18.762 91.383 86.922 Error 6 63.1842 10.31 Prob > F C. Total 8 1893.9494 <.0001 Parameter Estimates Term Estimate Std Error t Ratio Prob> t Intercept 0.674766 2.82244 0.24 0.8190 Time 0.489698 0.03728 13.0 <.0001 (Time-.3333)*(Time-.3333) 0.00012 0.00121 0.10 0.9221 Thickness Residual 4 2 0-2 -3 - Time Thickness Leverage Residua 0 10 0 10 0 Thickness Predicted 0 10 0607080 90 110 Time Leverage, P<.0001 Time*Time Thickness Leverage Residua 0 1 27.7 27.8 27.9 28.0 Time*Time Leverage, P=0.9221
8 Response Thickness (Intercept and Linear Term Only) Thickness Actual 0 1 0 10 0 Thickness Predicted P<.0001 RSq=0.97 RMSE=3.007 Model 1 18.67 18.66 2.4624 Error 7 63.2937 9.04 Prob > F C. Total 8 1893.9494 <.0001 Parameter Estimates Term Estimate Std Error t Ratio Prob> t Intercept 0.8386661 2.10023 0.39 0.7081 Time 0.4891 0.03437 14.23 <.0001 4 Thickness Residual 2 0-2 -3 - Time 0 Thickness Leverage Residua 1 0 10 0 Thickness Predicted 0 10 0 60 70 80 90 Time Leverage, P<.0001
9 Response Thickness (Intercept forced to zero, Linear Term Only) Thickness Actual 0 1 0 10 0 Thickness Predicted P<.0001 RSq=. RMSE=2.8432 Model 1 8836.64 8836.64 1093.146 Error 8 64.669 8.08 Prob > F C. Total 9 8901.313 <.0001 Tested against reduced model: Y=0 Parameter Estimates Term Estimate Std Error t Ratio Prob> t Intercept Zeroed 0 0.. Time 0.00983 0.0112 33.06 <.0001 Thickness Residual 3 1-1 -3 - Time Thickness Leverage Residua 0 1 0 10 0 Thickness Predicted 0 10 0 60 70 80 90 Time Leverage, P<.0001