HW3solution

# HW3solution - ∆ R P = 2.35 x 10 17/cm 3 From Eqn(5.9 with...

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5.4 First, calculate the total oxide thickness needed to ensure that the implanted impurity concentration is less than 10 15 /10 = 10 14 /cm 3 at the Si-SiO 2 interface. We know that R P = 0.05 μ m, the thickness of the oxide. For arsenic, this requires E = 80 keV from Fig. 5.3, and R p = 0.017 μ m. N P = Q/

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Unformatted text preview: ∆ R P = 2.35 x 10 17 /cm 3 . From Eqn. (5.9) with N P /N B = 235, X O = R P + 3.94 ∆ R P . X O = 0.05 + 3.94(0.017) = 0.117 μ m of oxide. The additional oxide required is 3.94(0.017) = 0.067 μ m. However, X nitride = 0.85 X O so only 0.057 μ m of silicon nitride is required. 5.8...
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## This note was uploaded on 10/24/2011 for the course EEE 5322 taught by Professor W.r.eisenstadt during the Fall '10 term at University of Florida.

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HW3solution - ∆ R P = 2.35 x 10 17/cm 3 From Eqn(5.9 with...

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