HW#3 Solution

827 k 49 v just upon entry into saturation vce vsat

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Unformatted text preview: µA 5 8.8 10 vCE (V) 5.103 Assume the BJT to operate in the constant- current region. In this case, iB = i1 = 20 µA, hence iC = βFiB = (50)(20 µA) = 1 mA; vCE = v1 – iCR1 = 10 V – (1 mA)(5 kΩ) = 5 V. 5.106 If iB = i1 = 15 µA, then iC = βFiB = (120)(0.015 mA) = 1.8 mA, and the drop across R1 becomes (1.8)(2.7 kΩ) = 4.9 V. Just upon entry into satura...
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