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HW#3 Solution

So that ic fib 5093 a 047 ma and vce v1 icrc

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Unformatted text preview: s of Fig. 5.35(a) over the constant- current region (e.g., at points A or B), reveals that βF = iC/iB = 50. The base current will equal iB = (VBB – Vf)/RB = (2 – 0.7)/20 kΩ = 65 µA, yielding iC = βFiB = (50)(65 µA) = 3.25 mA in the constant- current region, with vCE = VCC – iCRC = 10 V – (3.25 mA)(2 kΩ) = 3.5 V. This vCE is greater than Vsat, confirming that Q1 does operate in the constant- current region. 5.112 a. The circuit configuration...
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