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Solutions_1 - EE 310 Problem Set No 1 Solutions 1.41(a I...

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EE 310 Problem Set No. 1 Solutions 1.41 (a) 1 2 1 D D I I mA (ii) ± ² 617 . 0 10 5 10 ln 026 . 0 14 3 1 ¸ ¸ ¹ · ¨ ¨ © § u ³ ³ D V V ± ² 557 . 0 10 5 10 ln 026 . 0 13 3 2 ¸ ¸ ¹ · ¨ ¨ © § u ³ ³ D V V (b) 2 1 D D V V (ii) 10 . 0 10 5 10 5 13 14 2 1 2 1 u u ³ ³ S S D D I I I I So 2 1 10 . 0 D D I I 1 1 . 1 2 2 1 ´ D D D I I I mA So 909 . 0 2 D I mA, 0909 . 0 1 D I mA Now ± ² 554 . 0 10 5 10 0909 . 0 ln 026 . 0 14 3 1 ¸ ¸ ¹ · ¨ ¨ © § u u ³ ³ D V V ± ² 554 . 0 10 5 10 909 . 0 ln 026 . 0 13 3 2 ¸ ¸ ¹ · ¨ ¨ © § u u ³ ³ D V V 1.42 (a) ± ² 426 . 2 026 . 0 635 . 0 exp 10 6 14 3 ¸ ¹ · ¨ © § u ³ D I mA 635 . 0 1 635 . 0 R I mA 061 . 3 635 . 0 426 . 2 2 1 ´ D D I I mA ± ² 641 . 0 10 6 10 061 . 3 ln 026 . 0 14 3 2 1 ¸ ¸ ¹ · ¨ ¨ © § u u ³ ³ D D V V V ± ² 917 . 1 635 . 0 641 . 0 2 ´ I V V
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EE 310 Problem Set No. 1 Solutions 1.43 (a) Assume diode is conducting. Then, 0.7 D V V V J So that 2 0 7 23 3 30 R . I . A P 1 1 2 0 7 50 10 R . . I A P ³ Then 1 2 50 23 3 D R R I I I . ³ ³ Or 26 7 D I . A P (b) Let 1 50 R k : Diode is cutoff. 30 (1 2) 0 45 30 50 D V . . V ´ Since , 0 D D V V I J µ 1.45 (b) ± ² 1 i O I V ; 0 D I ; for 7 . 1 0 d d i I mA 7 . 1 O V V; ± ² 7 . 1 ³ i D I I mA; for 7 . 1 t
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