HW#3 Solution

2 v10 k 5 k 132 ma taking kvl around the loop

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Unformatted text preview: βFiB = (50)(9.3 µA) = 0.47 mA and vCE = v1- iCRC = 10 V –(0.47 mA) (3.3 kΩ) = 8.47 b. For βF = 200, the collector current become (200)(9.3 µA) = 1.86 mA, and vCE = 10 V – (1.86 mA)(3.3 kΩ) = 3.86 V. c. The V2 voltage source can be eliminated by connecting...
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This note was uploaded on 02/13/2014 for the course EE 2230 taught by Professor Staff during the Spring '08 term at LSU.

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