HW#3 Solution

HW#3 Solution

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Unformatted text preview: tion, vCE = Vsat = 0.1 V. Hence v1 must exceed Vsat + iCR1 = 0.1 V + 4.9 V = 5 V. 5.107 Assume that Vf = 0.7 V and that the BJT operates in the constant- current region. Taking KVL around the base- emitter loop yields iB = (VBB – Vf)/RB = (2 – 0.7)/20 kΩ = 65 µA, so that iC = βFiB = (50)(65µA) = 3.25 mA. Taking KVL around b. Assume that Vf = 0.7 V. Taking KVL around the base- emit...
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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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