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# solution2 - ECE 442 Spring 2009 HW#2 Solutions 1 Note that...

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ECE 442 Spring 2009 HW#2 Solutions 1. Note that in all four cases I diode = I resistor (since they are in series). a) Using the ideal diode model: Circuit a): assume diode is ON V = - 3 V, I = 3 - V 10 k = 0 . 6 mA check assumption: I > 0 OK Circuit b): assume diode is OFF I = 0 , V = 3 - I · 10 k = 3 V check assumption: V diode = V anode - V cathode = ( - 3) - V = - 6 V < 0 OK Circuit c): assume diode is ON V = 3 V, I = V - ( - 3) 10 k = 0 . 6 mA check assumption: I > 0 OK Circuit d): assume diode is OFF I = 0 , V = - 3 + I · 10 k = - 3 V check assumption: V diode = V anode - V cathode = V - (+3) = - 6 V < 0 OK b) Using the constant voltage-drop diode model (with V D = 0 . 7 V ): Circuit a): assume diode is ON V = ( - 3) + V D = - 2 . 3 V, I = 3 - V 10 k = 0 . 53 mA check assumption: I > 0 OK Circuit b): assume diode is OFF I = 0 , V = 3 - I · 10 k = 3 V check assumption: V diode = V anode - V cathode = ( - 3) - V = - 6 V < V D OK Circuit c): assume diode is ON V = (+3) - V D = 2 . 3 V, I = V - ( - 3) 10 k = 0 . 53 mA check assumption: I > 0 OK Circuit d): assume diode is OFF I = 0 , V = - 3 + I · 10 k = - 3 V check assumption: V diode = V anode - V cathode = V - (+3) = - 6 V < V D OK Comparing the answers in a) and b): For circuits in which diode is off, the answers are the same, since both models coincide for V diode < 0; For circuits in which diode is on, the voltages V differ only by the amount of voltage drop across the diode (here V D =0.7V), and currents differ by V D 10 k = 0 . 07 mA .

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