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EE_331F_2008_-_HW3S - Problem#3Solution 3.22 Autumn2008 V I...

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EE 331 Devices and Circuits I Autumn 2008 Problem #3 Solution 3.22 V D = nV T ln 1 + I D I S | I D = I S exp V D nV T 1 (a) V D = 1.05 0.025 V ( ) ln 1 + 7 x 10 5 A 10 18 A ⎟ = 0.837 V | (b) V D = 1.05 0.025 V ( ) ln 1 + 5 x 10 6 A 10 18 A ⎟ = 0.768 V (c) I D = 10 18 A exp 0 1.05 0.025 V 1 = 0 A (d) I D = 10 18 A exp 0.075 V 1.05 0.025 V 1 = − 0.943 x 10 19 A (e) I D = 10 18 A exp 5 V 1.05 0.025 V 1 = − 1.00 x 10 18 A 3.23 V D = nV T ln 1 + I D I S | I D = I S exp V D nV T 1 (a) V D = 0.025 V ln 1 + 10 4 A 10 17 A ⎟ = 0.748 V | (b) V D = 0.025 V ln 1 + 10 5 A 10 17 A ⎟ = 0.691 V (c) I D = 10 17 A exp 0 0.025 V ⎟ − 1 = 0 A (d) I D = 10 17 A exp 0.06 V 0.025 V
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