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EE2005-Tutorial02-Solution

# EE2005-Tutorial02-Solution - NATIONAL UNIVERSITY of...

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NATIONAL UNIVERSITY of SINGAPORE Department of Electrical and Computer Engineering EE2005 – Electronics Tutorial 2 Solution Q1. ] 025 . 0 737 . 1 /[ 8 ) /( ) 10 937 . 7 ( × × = D T D V nV V S D e mA e I I (a) Using the constant-voltage-drop model, V R = (5 – 1 – 0.7) V = 3.3 V Ω = k V I R D 2 = 1.65 mA (b) By iteration, If I D = 1.65 mA, the exponential diode equation evaluates V D to be ) ln( S D T D I I nV V = ) 10 937 . 7 65 . 1 ln( ) 025 . 0 )( 737 . 1 ( 8 mA mA × = = 0.73171 V If V D = 0.73171 V, V R = (5 – 1 – 0.73171) V = 3.26829 V Ω = k V I R D 2 = 1.63415 mA If I D = 1.63415 mA, the exponential diode equation evaluates V D to be ) ln( S D T D I I nV V = ) 10 937 . 7 63415 . 1 ln( ) 025 . 0 )( 737 . 1 ( 8 mA mA × = = 0.73129 V If V D = 0.73129 V, V R = (5 – 1 – 0.73129) V = 3.26871 V Ω = k V I R D 2 = 1.63436 mA If I D = 1.63436 mA, the exponential diode equation evaluates V D to be ) ln( S D T D I I nV V = ) 10 937 . 7 63436 . 1 ln( ) 025 . 0 )( 737 . 1 ( 8 mA mA × = = 0.73129 V

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The converged I D and V D are 1.6344 mA and 0.7313 V, respectively. (c) V R = (5.1 – 1 – 0.7) V = 3.4 V Ω = k V I R D 2 = 1.7 mA By iteration, If I D = 1.7 mA, the exponential diode equation evaluates V D to be ) ln( S D T D I I nV V = ) 10 937 . 7 70 . 1 ln( ) 025 . 0 )( 737 . 1 ( 8 mA mA × = = 0.73300 V If V D = 0.73300 V, V R = (5.1 – 1 – 0.73300) V = 3.367 V Ω = k V I R D 2 = 1.68350 mA If I D = 1.68350 mA, the exponential diode equation evaluates V D to be ) ln( S D
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EE2005-Tutorial02-Solution - NATIONAL UNIVERSITY of...

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