Question

Solve 11 and 12

1. (10 pts.) A base current Is of 0.048 mA is applied to the transistor in the Figure below. A voltage of 3.6 V is measured across resistor Re. Determine the emitter current le and the transistor pox 100 kn Van IE: 12.(15 pts.) For the NPN transistor circuit bhown,cakculate la, l and Va says that pa-oTV when the transistor is ON. (Given ?. 150) Use the usual approximation, which cC CE 200 kQ IB: Ic: IE: Va:

Answer 1

is given rightarrow I_B = 0.04 6 mA , V_Bc = 3.6 V a) so I_c. R_c = V_RC I_x = 3.6 v/1 km I_c = 1 mA therefore I_b + R_c = 1.040 mA I_B = 1.048 mA b) I_c = beta I_B beta = 1/0.04 theta beta_AC = 20.83 \r\n Apply kvl of Base V_BB - I_R R_B - V_BB = 0 5 - I_B times 0.7 = 0 I_B = 4.3/200 = 0.0215 mA therefore I_C = 3.225 mA So I_E = I_c + I_b = 3.225 + 0.0215 = 3.2465 mA Again Apply kvl from collector to emitter V_cc - R_c times R_c - v_CB = 0 V_cB = 3.55 V is because V_cB > V_oE 3.55 v > 0.2 v so transmitter is in active region