HW3_Solutions

HW3_Solutions - Homework Solutions ECE302 Spring 2007 HW 3...

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Unformatted text preview: Homework Solutions ECE302 Spring 2007 HW 3 1 3.56 (a) 5 = 10 4 I D + V D | V D = 0 I D = 0.500 mA | I D = 0 V D = 5 V Forward biased - V D = 0.5 V I D = 4.5 V 10 4 = 0.450 mA (b) - 6 = 3000 I D + V D | V D = 0 I D = - 2.00 mA | I D = 0 V D = - 6 V In reverse breakdown - V D = - 4 V I D =- 2 V 3 k = - 0.667 mA (c) - 3 = - 3000 I D + V D | V D = 0 I D = - 1.00 mA | I D = 0 V D = - 3 V Reverse biased - V D = - 3 V I D = 1 2 3 4 1 mA 2 mA (a) Q-point v D 6 5-1-2-3-4-5-6-1 mA-2 mA (b) Q-point i D (c) Q-point 3.65 The one-volt source will forward bias the diode. Load line: 1 = 10 4 I D + V D | I D = 0 V D = 1 V | V D = 0 I D = 0.1 mA 50 m A , 0.5 V ( 29 Mathematical model: f = 1- 10- 9 exp 40 V D ( 29- 1 [ ] + V D 49.9 m A , 0.501 V ( 29 Ideal diode model: I D = 1V/10k = 100 A; (100 A, 0 V) Constant voltage drop model: I D = (1-0.6)V/10k = 40.0 A; (40.0 A, 0.6 V) Homework Solutions ECE302 Spring 2007 HW 3 2 3.66 Using Thvenin equivalent circuits yields and then combining the sources +- +- +- V I 1 k 1.2 k 1.5 V 1.2 V 0.3 V +- +- V I 2.2 k (a) Ideal diode model: The 0.3 V source appears to be forward biasing the diode, so we will (a) Ideal diode model: The 0....
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This note was uploaded on 03/30/2008 for the course ECE 302 taught by Professor Barlage during the Spring '07 term at N.C. State.

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HW3_Solutions - Homework Solutions ECE302 Spring 2007 HW 3...

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