243S2008-07-soln - = j j j j j j RESIST FROM Si b) The...

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EE243 HW#7 Solution Sp2008 Problem 1 T(t) = I out /I in = exp 0 z o - (Am + B) dz For t 0, m = 1 T(0) = exp(- (A + B)z o ) = 0.15 For t , m = 0 T( ) = exp(- Bz o ) = 0.85 z o = 2.2 µ m T( )/T(o) = exp(A z o ) = 0.85/0.15 Az o = ln(0.85/0.15) = 1.735 A = 0.79/ µ m B = 0.074/ µ m C = A+B AI o T(0)[1 - T(0)] dT dt t = 0 ( 1/ T 12 ) with T 12 factor = 0.8 I o = 30mJ/cm 2 -sec, dT dt t = 0 = 0.4 - 0.15 5 sec = 0.05/sec C = 0.0175 cm 2 /mJ Problem 2 Resist standing waves a) The period (min-to-min exposure) of the standing waves is λ /(2n r ) = 365/(2 x 1.7) = 107 nm at i-line b) () ( ) () () () 3 . 169 634 . 0 4 . 18 66 . 8 9 . 150 52 . 5 74 . 2 22 . 8 68 . 2 82 . 4 71 . 2 52 . 6 032 . 0 7 . 1 71 . 2 52 . 6 032 . 0 7 . 1 365 _ _ _ = = + = +
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Unformatted text preview: = j j j j j j RESIST FROM Si b) The magnitude of the reflection coefficient seen from within the resist that a bottom antireflection coating (BARC) layer must produce to make the vertical contrast C VERTICAL equal 0.2 is | | = 0.10 . This is found by finding I MAX = 1.5 I MIN , taking the square root, and setting 1+| | = sqrt(I MAX /I MIN ) (1 -| |) Problem 3 Resist Development Simulation: By adjusting the dose (trial and error) a) Dose at 80mJ/cm2 just cleared in about 60s. b) Dose of 144 mJ/cm2 cleared in about 20s. 80 mJ/cm 2 144 mJ/cm 2...
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This note was uploaded on 11/02/2010 for the course EECS 243 taught by Professor Ee243 during the Spring '03 term at University of California, Berkeley.

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243S2008-07-soln - = j j j j j j RESIST FROM Si b) The...

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