ee606_s04_soln10 - B(0 ≈ p B W ≈(1-α p B ≈ 11.12...

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Solution to Selection Problems Homework 9 and 10 (iv) and (v) – similar to (ii) and (iii) respectively except semiconductor regions interchanged. 1
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(c) 11.12 Configuration (a) (a) I = I E , V EB = V A , and I C = 0. From Eq(11.47), we have I = (1 - α R α F ) I F 0 ( e qV A /kT - 1) (b) Δ p B (0) /p B 0 = ( e qV EB /kT - 1) = ( e qV A /kT - 1) Δ p B ( W ) /p B 0 = ( e qV CB /kT - 1) = ( α F I F 0 /I R 0 )( e qV A /kT - 1) (c) Simplifying, Δ p B (0) /p B 0 = ( e qV A /kT - 1) Δ p B ( W ) /p B 0 = α ( e qV A /kT - 1) 2
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(d) If V A >> kT/q , p B (0) = p B 0 e qV A /kT p B ( W ) αp B 0 e qV A /kT (e) If - V A >> kT/q
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Unformatted text preview: B (0) ≈ p B ( W ) ≈ (1-α ) p B ≈ 11.12 Confguration (b) (a) I = I E , V EB = V A , and V CB = 0. From Eq(11.47a), we have I = I F ( e qV A /kT-1) (b) Δ p B (0) /p B = ( e qV EB /kT-1) = ( e qV A /kT-1) 3 Δ p B ( W ) /p B = ( e qV CB /kT-1) = 0 (c) There is no simplifcations/ 4...
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