Professors HW2 Solutions

# Professors HW2 Solutions - EE323 Homework 2 Solution...

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EE323 Homework 2--- Solution, Au/2004 3.2 (20points) (a) Diode is conducting, and thus has a 0 drop across it. Consequently: V 3(3 ) 3, 0 .6 10 Vm K VI Ω −− =− = = A (b) Diode is cut off, so: 3 , 0 VA == (c) Diode is conducting, and thus has a 0 drop across it. Consequently: V ) 0 10 K Ω = A (d) Diode is cut off, so: 3 , 0 = 3.3 (20points) (a) D 1 is off and D 2 is on such that: ) 3 2 K Ω A − − = (b) D 1 is on and D 2 is off such that: 31 1, 1 2 K Ω A === 3.10 (20points) (a) Two resistors connected to the anode form a voltage divider such that the Thevenin voltage at the anode is 20 9 10 20 V anode V 6 = ⋅= + and two resistors are combined to 10 20 20 10 20 3 K Ω × ⎛⎞ = ⎜⎟ + ⎝⎠ . Therefore the current through the diode is 6 0.225 20 20/3 mA I + and the voltage is 20 64 . 5 20 20 /3 V V =⋅ = + . (b) Circuits connected to the cathode are simplified such that the Thevenin voltage at the cathode is 10 9 10 10 V cathode V = + 4 . 5 ; similarly, the voltage at the anode is 10 5 10 10 V anode V = + 2 . 5 . So the voltage applied to the diode is , the diode is off and 2.5 4.5 2 0 V V =−= < 0 A I = .

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Professors HW2 Solutions - EE323 Homework 2 Solution...

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