# hw3_sol - ECE 442 HW3 Solutions 1 a The two iterative...

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ECE 442 HW3 Solutions 1. a) The two iterative equations we will use are derived as follows: I C = I S e V BE V T V BE = V T ln I C I S (1) In Forward Active mode, the V CE voltage is not well-de±ned. Therefore, we will use the B & E circuit. Writing the KVL over the contour L: 5 50 k · I B V BE 3 k · I E = 0 5 50 k 100 I C V BE 3 k · 101 100 I C = 0 I C =( 5 V BE ) · 100 353 k (2) We start with the initial guess of V BE 0 = 0 . 7 V . I C 1 5 0 . 7 ) 100 353 k = 1 . 218 mA V BE 1 = 25 m · 1 . 218 m 10 16 = 0 . 753 V I C 2 5 0 . 753 ) 100 353 k = 1 . 203 mA V BE 2 = 25 m · 1 . 203 m 10 16 = 0 . 753 V V BE 1 I C = 1 . 203 mA , V BE = 0 . 753 V , I B = I C β = 12 . 03 μ A , I E = β + 1 β I C = 1 . 215 mA b) V BE = 0 . 7 V I C = I S e V BE V T = 0 . 145 mA I B = I C β = 1 . 45 μ A , I E = β + 1 β I C = 0 . 146 mA The issue here is the sensitivity of the exponential relation to changes in V BE , so if starting from that relation even a relativelly small error in the initial guess would cause huge errors in resulting currents and cause the iterative process to diverge.

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## This note was uploaded on 02/21/2010 for the course ECE 442 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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hw3_sol - ECE 442 HW3 Solutions 1 a The two iterative...

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