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# hw10sol - s The donor concentration in germanium is 2.88 ×...

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3.091 Fall Term 2004 Homework #10 solution outlines 1. Assume steady state. J = − D c x ; ∴ ∆ x = D c J = 3.091 × 10 4 cm 2 s × 1.5 × 10 19 atom cm 3 10 3 mol cm 2 hr × 1 hr 3600 s × 6.02 × 10 23 atom mol = 2.773 × 10 –2 cm 2. A solution to Fick’s second law for the given boundary conditions is: c c s = 1 erf 2 x Dt , from which we get erf 2 x Dt = 1 0.018 = 0.982 From the error function tables 0.982 is the erf of 1.67. This means that 0.002 2 Dt = 0.001 Dt = 1.67 D = 286 10 5 8.314 1253 0 e D = 13 6.45 10 2 cm sec t = 0.001 2 = 5.56 10 5 sec = 6.4 days 1.67 2 6.45 10 13 3. C s C c × = erfc x = erfc 3 10 3 = erfc (2.083) c s 2 Dt 2 Dt c = 1 erf ( . . 2 083), 1 c = 0 9964 c c s s 16 3 c = 3 6 × 10 3 , c = 2 88 × 10 cm

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Unformatted text preview: s The donor concentration in germanium is 2.88 × 10 16 /cm 3 . 3.091 Fall Term 2004 Homework #10 Solution Outlines page 2 × ( , c x = 5 × 10 16 = c erfc 25 10 − 4 4. c x t ) = − c erf x = c erfc s s s × 90 2 Dt s 2 Dt 2 7.23 10 − 9 × × 60 × ∴ c = 5 10 16 = 6.43 × 10 16 cm − 3 s 25 10 − 4 × erfc 2 7.23 10 − 9 × 5400 × erfc(0.20) = 1 – erf(0.20) = 1 – 0.2227 = 0.7773 5. c o /3 O x c o c ? c x t ) = c 0 e r f x ( , 2 Dt What is x when c = c o /3? x x c 0 = c 0 erf ⇒ 0.33 = erf ; erf (0.30) = 0.3286 ≈ 0.33 3 2 Dt 2 Dt x ∴ 2 × 10 = 0.30 ∴ x = × 0.30 × 3.091 10 − 6 × × 60 = 2.58 × 10 –2 cm = 258 µ m 2 Dt...
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hw10sol - s The donor concentration in germanium is 2.88 ×...

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