EE_331F_2008_-_HW4S

# EE_331F_2008_-_HW4S - EE331DevicesandCircuitsI...

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34 EE 331 Devices and Circuits I Autumn 2008 Problem #4 Solution 3.73 Diodes are labeled from left to right (a) D 1 on, D 2 off, D 3 on : I D 2 = 0 | I D 1 = 10 0 3.3 k Ω+ 6.8 k Ω = 0.990 mA I D 3 + 0.990 mA = 0 −− 5 () 2.4 k Ω I D 3 = 1.09 mA V D 2 = 5 10 3300 I D 1 =− 1.73 V D 1 : 0.990 mA, 0 V D 2 : 0 mA, 1.73 V D 3 : 1.09 mA, 0 V ( b ) D 1 2 off, D 3 I D 2 = I D 3 = 0 I D 1 = 10 0 V 8.2 k 12 k Ω = 0.495 mA V D 2 = 5 10 8200 I D 1 0.941 V I D 3 = 0 5 V 10 k Ω I D 1 = 0.005 mA D 1 : 0.495 mA , 0 V D 2 :0 A , 0.941 V D 3 : 0.005 m A V (c) D 1 2 3 on I D 1 = 0 10 8.2 k Ω V = 1.22 mA > I 12 K = 0 2 12 k Ω V 0.167 mA I D 2 = I D 1 + I 12 K = 1.05 mA > 0 I 10 K = 2 5 10 k Ω V = 0.700 mA I D 3 = I 10 K I 12 K = 0.533 mA > 0 D 1 : 1.22 mA V D 2 : 1.05 m A V D 3 : 0.533 m A V (d) D 1 2 off, D 3 I D 1 = 0, I D 2 = 0 I D 3 = 12 5 4.7 + 4.7 + 4.7 V k Ω = 1.21 mA > V D 1 = 0 5 + 4700 I D 3 0.667 V < 0 V D 2 = 5 12 4700 I D 3 1.33 V < 0 D 1 A 0.667 V D 2 A 1.33 V D 3 : 1.21 mA V 3.74 Diodes are labeled from left to right (a) D 1 2 off, D 3 I D 2 = I D 1 = 10 0.6 0.6 ( ) 3.3 k 6.8 k Ω = 0.990 mA I D 3 + 0.990 mA = 0.6 5 2.4 k Ω I D 3 = 0.843 mA V D 2 = 5 10 0.6 3300 I D 1 1.13 V D 1 : 0.990 mA, 0.600 V D 2 : 0 A, 1.13 V D 3 : 0.843 mA, 0.600V

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35 (b) D 1 on, D 2 off, D 3 off : I
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EE_331F_2008_-_HW4S - EE331DevicesandCircuitsI...

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