2. The BJT Circuit show, Please calculate 2.1 VB, VRB, VCE, VCC and VRE 2.2 RB, RC and RE 2.3 IB and IC IC VRC = 3 V RC VC = 9 V VB ? = 125 RB IB VE = 8 V VCC VBB = 20V RE IE = 4.5mA
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Since the base-emitter junction is forward-biased, we can assume a voltage drop of about 0.7V across it (typical for silicon transistors). Therefore, \( V_B = V_{BE} + V_E \), where \( V_E \) is given as 8V. \( V_B = 0.7V + 8V = 8.7V \) Show more…
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