Question

I keep getting the wrong answer from everyone else pls help me get the right answer

For the BJT bias circuit shown, what value of Rc in kilohms is needed to allow the maximum possible peak-to-peak signal swing on the collector without clipping? Use Vcc-9V, Vee--7V, Vb-1.2V, and Re -9.9k2. Assume that to keep the transistor in the forward-active region, the base-collector junction cannot be forward biased. Use B 20 and Vbe(on) 0.7V. Neglect the effects of base-width modulation. Vcc RE Q1 V. VEE Answer:

Answer 1

v_Cc = 9v v_b = 1.2 v Re = 9.9 k ohm Transistor is Active region betw = 2.0 V_B elementof = 0.7 v V_EE = -7 v R_c = ? Apply K.V.L from V_cc to base Vertical-bar V_E_B = -0.7 V_CC - I_E R_E - V_E B - V_B = 0 9v - I_E (9.9 k) = 8.5 I_E = 0.85 mA In active region I_C = beta I_B I_E = I_c + I_B = beta I_B = I_B = (beta + 1) I_B implies I_B = I_E/beta + 1 = 0.0408 mA I_B = 40.8 mu A I_C = 0.81 mA Apply K.V.L from V_ce to V_EE v_cc - I_E R_E - v_E_C - I_C R_c - v_EE = 0 the min value i.e V_EE > 0.2 v for the transistor is active regia i.e transistor is (n &n) But if p intersection p transistor v_CE < 0.2 for transistor active regia V_E_C > 0.2 v i.e V_E C = 0.3 v of if R_c is move V_E_C is min 9 - (0.8) x mA times 9.9 k) - 1 - I_c R_c + exit = 0 15 - (0.81 times 9.9) = I_C R_C R_C = 8.2 1 k ohm