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Given RB = 240 k., Rc = 1.2 k., and = 120. Determine: a) IB b) Ic c) le d) VBE e) VCE f) Mode of operation (cutoff, active, s

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VCC & ROVIC SB RB JE VIE [YBE = 0.70 a) By KVL at input VOC - IB.RB -VBE = 0 IB - VCC-VBE RB 1B: -0.0304 mA 8-0.7 240K 16 = 3RB = 47K, RC = -2K, B = 120 13 = 8-0.7 = 0.1553 ma 1B 47K IC BIB - (120)/0.1553 in) = 18.636 mA VCE = VCC - IC.RO = 8 - 18-63VEEQ = = VCC-ICO RC 16-12.952m)(2K) VCEQ - 10.096 v linear active and IC 70 50 BJT is in AS VCE > 0.2 region Load line: Ica (

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