Solutions_1

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) ( ) 617 . 0 10 5 10 ln 026 . 0 14 3 1 = × = D V V ( ) 557 . 0 10 5 10 ln 026 . 0 13 3 2 = × = D V V (b) 2 1 D D V V = (ii) 10 . 0 10 5 10 5 13 14 2 1 2 1 = × × = = 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 ( ) 554 . 0 10 5 10 0909 . 0 ln 026 . 0 14 3 1 = × × = D V V ( ) 554 . 0 10 5 10 909 . 0 ln 026 . 0 13 3 2 = × × = D V V 1.42 (a) ( ) 426 . 2 026 . 0 635 . 0 exp 10 6 14 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 ( ) 641 . 0 10 6 10 061 . 3 ln 026 . 0 14 3 2 1 = × × = = D D V V V ( ) 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 VV V γ = = So that 2 07 23 3 30 R . I .A µ = 1 12 07 50 10 R .. IA = Then 12 50 23 3 DRR III . =−= Or 26 7 D I = (b) Let 1 50 Rk = Diode is cutoff. 30 (1 2) 0 45 30 50 D V . .V = ⋅= + Since , 0 DD VVI <= 1.45 (b) ( ) 1 i O I V = ; 0 = D I ; for 7 . 1 0 i I mA 7 . 1 = O V V; ( ) 7 . 1 = i D I I mA; for 7 . 1
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Solutions_1 - EE 310 Problem Set No. 1 Solutions 1.41 (a) I...

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