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

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Unformatted text preview: EE 310 Problem Set No. 1 Solutions 1.41 (a) I D1 (ii) V D1 0.026 ln VD 2 10 3 5 10 14 0.026 ln 0.617 10 3 5 10 13 I D2 1 mA V 0.557 V (b) V D1 V D 2 (ii) I D1 I D2 I S1 IS2 5 10 5 10 14 13 0.10 So I D1 0.10 I D 2 I D1 I D2 1.1I D 2 1 mA So I D 2 Now V D1 VD 2 0.909 mA, I D1 0.0909 mA 0.0909 10 5 10 14 0.909 10 5 10 13 3 3 0.026 ln 0.026 ln 0.554 0.554 V V 1.42 (a) I D 3 IR I D1 V D1 VI 6 10 14 exp 0.635 0.026 2.426 mA 0.635 0.635 mA 1 I D 2 2.426 0.635 VD 2 2 0.641 0.026 ln 3.061 mA 3 3.061 10 6 10 14 0.641 V 0.635 1.917 V EE 310 Problem Set No. 1 Solutions 1.43 (a) Assume diode is conducting. Then, VD V 0.7 V 0.7 23.3 A 30 1.2 0.7 I R1 50 A 10 Then I D I R1 I R 2 50 23.3 So that I R 2 Or I D 26.7 A (b) Let R1 50 k VD Diode is cutoff. Since 30 (1.2) 0.45 V 30 50 VD V , I D 0 1.45 (b) VO VO ; I D 0 ; for 0 I i 1.7 mA 1.7 V; I D I i 1.7 mA; for I i 1.7 mA Ii 1 1.47 (a) (ii) I (b) (ii) I (c) (ii) I 5 0.6 20 5 0.6 40 0.220 mA, VO 5 0.6 V 0.235 20 5 0.30 0.235 mA, VO V 2 0.6 25 8 0.376 mA, VO 2 0.376 5 0.12 V (d) (ii) I 0 , VO 5 V EE 310 Problem Set No. 1 Solutions 1.50 (a) I R 2 I D1 ID2 ID2 0.65 0.65 mA I D1 1 2(0.65) 1.30 mA VI 2Vr V0 5 3(0.65) 1.30 R1 R1 R1 2.35 K (b) I R 2 ID2 I D1 I D1 0.65 0.65 mA 1 8 3(0.65) I D 2 3.025 mA 2 I D 2 I R 2 3.025 0.65 2.375 mA 1.53 a. diode resistance rd VT I vd rd rd RS vS VT I vS VT RS I vo vd VT vs VT IRS b. I I I RS 260 v0 vS v0 vs v0 vS VT VT IRS 0.026 0.026 0.026 (1)(0.26) 0.026 0.1 0.26 v0 vS v0 vS v0 vS 0.5 0 0.909 0.0909 1 mA, 0.1 mA, 0.01 mA. 0.026 0.026 (0.01)(0.26) 1.60 For VD = 0, ISC = 0.2 A For ID = 0 VD VT ln VD VDC 0.754 V 0.2 1 5 10 14 EE 310 Problem Set No. 1 Solutions 2.3 (a) S 120 2 O 1 10 16.97 V (peak) peak 16.27 V (b) i D peak (c) O 16.27 8.14 mA 2 16.97 sin t 0.7 sin t 1 0.7 16.97 0.04125 t 2 1 2.364 t % 177.64 2.364 360 100% 180 2.364 177.64 48.7% (d) O avg 1 2 0.9869 16.97 sin x 0.7 dx 0.01313 1 2 16.97 cos x 0.9869 0.01313 0.7 x 0.9869 0.01313 O avg O 1 16.97 2 5.06 V 0.99915 0.99915 0.7 0.9738 (e) i D avg avg 2 5.06 2 2.53 mA 2.4 (a) R t 15 sin t 0.7 9 15 sin t 9.7 t 1 sin 1 9.7 15 40.29 0.2238 rad rad t R 2 180 40.29 139.71 0.7762 avg 1 2 0.7762 15 sin x 9.7 dx 0.2238 1 2 R 15 cos x 0.7762 0.2238 9.7 x 0.7762 0.2238 1 2 15 0.7628 0.7628 9.7 0.5523 avg 0.9628 V 0.8 0.9628 R R 1.20 i D avg (b) % 139.71 40.29 360 100% 27.6% ...
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This note was uploaded on 04/26/2010 for the course E E 310 taught by Professor Wharton,markjedwards,perrysto during the Spring '10 term at Pennsylvania State University, University Park.

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