Question

Question 5 CIRCUIT C1 (2.5 Marks) Statement: Sketching relevant output (Vc-Ic) characteristics decide on Q-point coordinates and DC load-line details: That is, determine DC operating conditions of the NPN-BJT/Si circuit of Fig.5 (denoting the type Circuit C1). Assume the following: Vcc= 18 volt; Rc = 1.5 k; RB1 = 75 k; RB2 = 3.6 k; V CESat = 1 volt; and a = 0.99338. +Vcc VIC → Rc R81 1 IB Vc RB2 12 IE Fig. 5: CIRCUIT C1 Required Solutions In the multiple choice of answers supplied, choose the correct set of results on quiescent values representing: {VB, Base voltage; IE, Emitter current; Ic, Collector current; I1, Base- feeding current drawn from supply; I2, Base-biasing current thru' RB2; IB, Base current; Vc, Collector voltage}. (1) {VB = 1.7 volt; IE= 5.5 mA; Ic = 7.67 mA; I1 = 1.53 mA; 12 = 1.10 mA; IB = 584A, Vc = 10.5 volt} (2) {VB = 0.7 volt; IE= 5.7 mA; Ic = 5.67 mA; I1 = 0.23 mA; I2 = 0.19 mA; IB = 38ŅA, Vc = 9.5 volt} (3) {VB = 0.6 volt; IE= 3.7 mA; Ic = 3.68 mA; I1 = 1.23 mA; 12 = 0.29 mA; IB = 20ŅA, Vc = 10.5 volt}. (4) {VB = 1.5 volt; IE= 3.7 mA; Ic = 1.67 mA; I1 = 1.23 mA; I2 = 0.10 mA; IB = 30ŅA, Vc = 9.5 volt}. (5) {VB = 0.5 volt; IE= 2.7 mA; Ic = 3.67 mA; I1 = 0.23 mA; 12 = 0.15 mA; IB = 18uA, Vc = 9.5 volt}.

please choose one of the answer choices above.

Must show all work, thank you

Answer 1

Roi 95> R= 15k_2., 75 krd, RB - 3.6 Kate, vee- 18v = 1V, 2= 0.99338 0.99338 150. 057 1-0.99338 Veesat Vcc Vec fic Rc Roth B IR kon Loop! IE Vih Ž Ron - RB, // RB₂ = (75k-2) x (3.6k (75k_2)x (2.5k_2)=(3.4 3 5 kr (75 +3.6) Kr With = 3.6kr Roe LR RB, the Veca X 18 (3.6+75 kr Van = 0.8244 V IVEZOV V BE UBE VB TV = 0.IN 0.7v A VB (0.71) + VE = 0.7

Apply KVL an cloop- Vrh - (Rrei IB) - UBE → 0.8244 - ( 3.435 x 13 x Ie) 10.12 4427 Io- 3.622 3x105A 3.435 X 103 0.7 = 0 = IB = 36.22 de A Ic = ß Io 150.057X 36.22X16 17 Ic 5.435 mA LIE 5.44 mA tv I E = (x + 1) I Ve = vcc - IRC = 18-(5.495x16% )(1-5x603) Ver 9.8475V V 0.7V Iz = 1.944 8.20 4A RB2 3.6kr I 0.19 mA vi I Bt I (3.6-22 X16 9+(0.1982.) A 0.23 mA I, 2.306 8 0 4 1,- → Vo=o+V, L = 5.47mA, - 5.435 mA, I, – 0-2.3mA I= 36.22 MA, ve=9.8475V Anther values obtained are closest to option (2) So, option (2) is correct Answer. I. 0.19 MA >