HW#3 Solution

For this value of d every 2a increment of ib1 in the

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Unformatted text preview: drives current through the base- emitter junction of the BJT via R1, hence VBE = Vf = 0.7 V, and iB = (V2 – VBE)/R1 = (10 – 0.7)/1MΩ = 9.3 µA the Darlington pair becomes VfD = Vf1 +Vf2 = 1.2 V. When Q1 enters saturation, vBC of Q2 will still be positive, hence Q2 will remain the constant current region until vCE2 = v1 reaches the value Vsat. so that iC =...
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