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243S2009-m1-soln

# 243S2009-m1-soln - EE243 Sp 2009 Midterm Exam Solutions...

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- EE243 Sp 2009 Midterm Exam Solutions Problem 1 (a) Let δ E be the peak E-field contributed by an isolated square transparent point defect ( d < 0.6 λ /NA) with δ E proportional to the defect area (d 2 ). The maximum intensity is δ E 2 . The opaque isolated square defect will have a minimum E-field = (1- δ E) with intensity = (1- δ E) 2 The cross-over implies (1- δ E) 2 = δ E 2 δ E=0.5 or corresponding intensity=0.25. (b) (i) ) P x 2 cos( E E 2 E E ) e e ( E E E E ) e E e E )( e E e E ( EE ) x ( I e E e E e E e E ) x ( E 1 o 2 1 2 o x i 2 x i 2 1 o 2 1 2 o x 2 i 2 1 x 2 i 2 o x 2 i 2 1 x 2 i 2 o * x 2 i 2 1 x 2 i 2 o x ) ( i 2 1 x i 2 o m m m m m m m m m π + + = + + + = + + = = + = + = ν π ν π ν π + ν π ν π ν π ν π ν π ν ν π ν π α α I max = 0.6724 I min =0.0324 C= 0.908 (ii) E(x) is a pure sinusoid already and I(x) will have higher harmonics. Therefore, one cannot reproduce I(x) of part (i). (c) m(z, t exp ) = exp [ -C I(z) t exp ) = exp [ -C I o t exp ] exp[e -Bz ] and C I o t exp =2 ( i) Region 2 : I (z=0) is only 0.1 × I o . The resist surface will have m(z=0) = exp [-2 × 0.1 × 1] = 0.82 > m= 0.7. Note that m ( at z>0) will even be larger. Therefore the minimum intensity region will have

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• Spring '03
• ee243
• Semiconductor device fabrication, Crystallographic defect, Shallow trench isolation, clear field region, shallow junction formation, CMOS PROCESS FLOW

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243S2009-m1-soln - EE243 Sp 2009 Midterm Exam Solutions...

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