143f2010-02-soln

# 143f2010-02-soln - EE143, Fall 2010 HW#2 Solutions Problem...

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EE143, Fall 2010 HW#2 Solutions Problem 1 From k (0.436 /NA ) =0.5 and 0.436 / [ 2(NA) 2 ]=1, we obtain NA=0.467 and k = 0.536 For Stepper A: R = 0.536 0.365 / 0.6 = 0.326um -- does not meet R requirement DOF = 0.365/ [2 (0.6) 2 ] = 0.507um > 0.4 um – o.k. For Stepper B: R = 0.536 0.248 / 0.5 = 0.266 um -o.k. DOF = 0.248/ [2 (0.5) 2 ]= 0.496um > 0.4um - o.k. Only Stepper B will meet both R and DOF requirements . Problem 2 (i) L =L  T ( glass - Si ) Maximum run in/out error = ｱ 0.5 m with respect to the alignment marks on wafer => L = ｱ1 m on photomask   T maximum = ｱ10 -4 cm 10 cm (9-2.3) 10 -6 = ｱ1.5ｰC (ii) L is due to Si expansion only. L = L   Si  T   T maximum = ｱ10 -4 cm 10 cm 2.3 10 -6 = ｱ 4.3ｰC Problem 3 The center only has translational errors: 0.5 m along x, -0.5 m along y. After substracting the translational error, we have: Top Right Center Left Bottom x -0.5 +0.2 0 -0.2 +0.5 y +0.2 +0.5 0 -0.5 -0.2 The run out error is 0.2 m The rotational error is 0.5 m (counter-clockwise) Problem 4 (a) Example 1

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## This note was uploaded on 03/03/2012 for the course EECS 142 taught by Professor Ee142 during the Spring '04 term at University of California, Berkeley.

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143f2010-02-soln - EE143, Fall 2010 HW#2 Solutions Problem...

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