## Answers

From Figure, V_E = I_E R_E Implies I_F = V_E/R_E I_E = 2.1/0.68 = 3.08823 mA I.E = 3.08823 mA But IE = (1 + beta) I_B beta = I_E - I_B/I_B = I_E/I_B - 1 beta = 3.08823 mA/20 muA - 1 = 153.4 beta = 153.4 I_C = beta I_B = 153.4 times 20 muA = 3.0682 I_C = 3.0682 mA Using KVL in collector to emitter, V_LC = I_C R_C + V_CE + V_E = 3.0682 times 2.7 + 7.3 + 2.1 V_CC = 17.6842 V \r\n Assume V_BE active = 0.7V Now using KVL in Base to emitter V_CC = I_B R_B + V_BE + V_E R_B = V_CC - V_BE - V_E/I_B R_B = 17.6842 - 0.7 - 2.1/20 muA R_B = 744.2 k Ohm (6) given circuit can be Redrawn as - V_BE = 0.7V (assume] V_th = V_CC times R_2/R_1 + R_2 V_th = 16 times 9.2/9.1 + 62 = 2.0478 V R_th = R_B = R_1 // R_2 = 62 // 9.1 = 7.9353 k Ohm Using KVL in Base Loop- V_th = I_B R_th + V_BE + (1 + beta) I_B R_E I_B = V_th - V_BE/R_th + (1 + beta) R_E I_B = 2.478 - 0.7/7.9353 + (80 + 1) 0.68 I_B = 21.388 muA

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