# solution1 - ECE 442 Spring 2009 HW#1 Solutions 1 Thevenin...

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Unformatted text preview: ECE 442 Spring 2009 HW#1 Solutions 1. Thevenin equivalent of the given voltage divider is: V th = 10 Â· 1 k 1 k + 9 k = 1 V Rth = 1 k || 9 k = 1 k Â· 9 k 1 k + 9 k = 0 . 9 k Î© After loading the output with 1kÎ© , we can calculate the output voltage either directly (by placing load resistor in parallel with 1k resistor to ground): V out = 10 Â· 1 k || 1 k (1 k || 1 k ) + 9 k = 10 Â· . 5 k 9 . 5 k = 0 . 526 V or using the Thevenin equivalent (by placing load resistor in seres with the equivalent Thevenin resistor): V out = 1 Â· 1 k 1 k + 0 . 9 k = 0 . 526 V 2. The succession of Thevenin equivalent circuits, along with the calculated values, is as follows: V th 1 = 10 Â· 1 k 1 k + 1 k = 5 V, R th 1 = 1 k || 1 k = 0 . 5 k Î© V th 2 = 5 Â· 1 k 1 k + 1 . 5 k = 2 V, R th 2 = 1 . 5 k || 1 k = 0 . 6 k Î© 1 V th 3 = 2 Â· 1 k 1 k + 1 . 6 k = 0 . 77 V, R th 3 = 1 . 6 k || 1 k = 0 . 62 k Î© Finally, the output voltage and output resistance are given by: V o = 0 . 77 Â· 2 k 2 k + 0 . 62 k = 0 . 588 V R o = 2 k || . 62 k = 2 k Â· . 62 k 2 k + 0 . 62 k = 0 . 47 k 3 . Since the diodes are identical and connected in series, the voltage across each diode is equal to....
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solution1 - ECE 442 Spring 2009 HW#1 Solutions 1 Thevenin...

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