143f2010-07-soln - EE143, Fall 2010 HW#7 Solutions Problem...

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EE143, Fall 2010 HW#7 Solutions Problem 1 All etchimg rates have only vertical components ( completely anisotropic) Poly etech rate = 0.1 m /min Oxide etch rate = 1/5 poly etch rate = 0.02 m /min Resist etch rate = 1/ 2 poly etch rate = 0.05 m /min Slope of resist before etching = tan -1 (0.5/0.15) = 73.3 o After 5 minutes of etching (a) Poly-Si will be etched to a depth of 0.1 5 = 0.5 m Photo resist will be etched to a depth of 0.05 5 = 0.25 m Bottom of resist opening will increase by 2 x ( 0.05 5 cot 73.3 o ) = 0.15 m (b) Oxide slope angle = tan -1 ( 0.1/ 0.075) = 53.1 o Problem 2 i.)The poly etching is completely isotropic and the mask etching is amistropic Note that the lateral etching rate of poly > lateral removal rate of mask! (Ii) x max = 5 m iii) x min = (5 - 2) m = 3 m iv) mask etched = 2x = 2 (0.01 m/min 1 m 0.1 m/min cot 60ー = 2 0.058 m = 0.116 m mask max = 5 - 0.116 = 4.88 m v.) t mask (max) = 0.5 m - ( 1 m 0.1 m/min ) 0.01 m/min = 0.4 m poly mask 1 m x max X Y x min before after m 5 mask max x x mask
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143f2010-07-soln - EE143, Fall 2010 HW#7 Solutions Problem...

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