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243S2008-04-soln - Sp 2008 EE243 HW#4 Solution Problem 1...

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Sp 2008 EE243 HW#4 Solution Problem 1 C max = C (x=0,y →∞ ,t) = Q π Dt N B = Q π Dt e - x j 2 4Dt = Q 2 π Dt [1 + erf( - y j 2 Dt )] or N B = C max e - x j 2 4Dt = C max [ 1 + erf( - y j 2 Dt ) 2 ] = C max [ 1 - erf( y j 2 Dt ) 2 ] x j 2 Dt = [ - ln( N B C max )] 1/2 y j 2 Dt = erf -1 [ 1 - 2N B C max ] y j x j = erf -1 [ 1 - 2N B C max ] [ - ln( N B C max )] 1/2 0.2 0.4 0.6 0.8 0 10 -5 -4 -3 -2 -1 10 10 10 10 x x x x y j / x j log( N B C max - ) Problem 2 (i) Injection of Si interstitials from the SiO2/Si interface during oxidation ( to release the mechanical stress due to oxide volume expansion). Law of Mass Action [C I C V =constant] implies more interstitial conc C I will yield less vacancy conc C V . (ii) D= D I C I +D V C V . Boron has enhanced diffusion with more C I implies its diffusion constant is affected dominantly by insterstitials. (iii) D= D I C I +D V C V . Antimony has reduced diffusion with more C I implies its diffusion constant is affected dominantly by vacancies. Problem 3 At 1000 °C , n i = 1 × 10 19 /cm 3 D h × D i + p n i At low dopant concentration <<n i , n = p= n i , and h 1. Therefore D = 1 × 10 -14 cm 2 /sec implies N B C max y j x j 10 -5 0.89 10 -4 0.87 10 -3 0.83 10 -2 0.77
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D i + = 1 × 10 -14 cm 2 /sec At a dopant concentration of 10 20 /cm 3 , h 2 D h × D i + p n i = 2 × 1 × 10 -14 cm 2 /sec × 10 20 10 19 = 2 × 10 -13 cm 2 /sec Problem 4 (i) Intrinsic carrier concentration, n i . D = D i o + D i - n n i + ...... Since n i T 3/2 exp[-E g /2kT] , E g implies n i implies D (ii) Charged vacancies .
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