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243S2008-02-soln

# 243S2008-02-soln - N CHEUNG Sp2008 EE243 Homework#2...

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N. CHEUNG, Sp2008 EE243 Homework #2 Solutions Problem 1 When the trench is completely filled with oxide, the oxide width will be 1.85 um: Soln : 0.46 µ m Si gives 1 µ m SiO 2 . Let width of SiO 2 trench be x µ m then x - 1 0.46 = x x = 1.85 µ m The problem now becomes to solve time t such that x ox (110) heavliy doped + x ox (100) =1.85 µ m with x ox (t) = 1 2 A [{1+ 4Bt A 2 } 1/2 -1 ] (can ignore the 20nm initial oxide correction for D-G model since the two oxide thickness of interest are much larger than that) From Table 6.2 of PDG , steam oxidation at 1100C, B(100) = 3.86 × 10 2 × exp [ -0.78/ (8.62 × 10 –5 × 1373 )]= 0.53 µ m 2 /hr B/A (100) = (1.63 × 10 8 )/(1.68) × exp [ -2.05/ (8.62 × 10 –5 × 1373 )]= 2.91 µ m/hr Therefore A(100) = 0.182 µ m B (110) heavily doped = B(100) =0.53 µ m 2 /hr B/A (110) heavily doped = 4 × 1.2 B/A (100) = 4.8 B/A (100) = 13.98 µ m/hr Therefore A(110) = 0.038 µ m By iteration, oxidation time t is found to be 1.8 hr with x ox (100) = 0.89 µ m and x ox (110) = 0.96 µ m Note: Since the expected x

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243S2008-02-soln - N CHEUNG Sp2008 EE243 Homework#2...

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