Question

Question 4 CIRCUIT B2 (2.5 Marks) +Vcc Rc IRC Vic RB Ів Vc VB VE RE IE FIG.4: CIRCUIT B2 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.4 (denoting the type Circuit B2). Assume the following: Vcc= 24 volt; Rc = 5.0 k; RE = 1000 ohm; RB = 1000 k and B = 150. Required Solutions In the multiple choice of answers supplied, choose the correct set of results on quiescent values representing: {VB, Base voltage Vce, Collector-Emitter voltage; IE, Emitter current; Ic, Collector current; Inc, Current drawn from supply, IB, Base current; Vc, Collector voltage; VE, Emitter voltage}. (1) {VB = 1.546 volt; VCE = 10.920 volt; IE= 0.845 mA; Ic = 0.834 mA; Irc = 2.846 mA; Vc = 15.760 volt; IB = 13.225uA, VE = 1.546 volt}. (2) {VB = 2.546 volt; VCE = 15.926 volt; IE= 1.946 mA; Ic = 1.734 mA; IRC = 2.846 mA; Vc = 20.770 volt; IB = 32.225uA, VE = 1.946 volt}. (3) {VB = 5.546 volt; VCE = 12.825 volt; IE= 1.946 mA; Ic = 2.834 mA; IRC = 3.846 mA; Vc = 4.780 volt; IB = 10.325uA, VE = 2.846 volt). (4) {VB = 2.546 volt; Vce = 12.925 volt; IE= 1.846 mA; Ic = 1.834 mA; Irc = 1.846 mA; Vc = 14.770 volt; 18 = 12.225uA, VE = 1.846 volt). (5) {VB = 6.546 volt; VCE = 22.924 volt; IE= 4.846 mA; Ic = 4.834 mA; Irc = 5.846 mA; Vc = 15.780 volt; IB = 11.225uA, VE = 3.846 volt}.

please choose one of the answer choices above.

Must show all work, thank you

Answer 1

Option 4 is correct

of Vec Given circuit ; Inch Rc Vcc - 24V Rc = 5kr REF loood - lkn RB = loooka & B = 150 IBV tic RB * from circuit IRC = IB + Ic = IE Vc V IRC SIE (+1) IB + 0.71 -151 IB VE IEL .RE * KUL at Base Emitter junction site: Vec o Inc. Rct IB.RB to.7 + TERE si transistor V200.70] 24 - 0.7 151 IB (sk) + IB (10ook) + 151 IB (IK) 23. 3 i 23.3 B [ısı (ok) +loook K] → IBP → Ig - 12.225 UH 1906k * Ic = ß. IB = X 12.2254 A Ic - 1.834 mA * IE = (B+1) IB = 151 x 12.225 MA te - IRC 1.846 mA * VE DE-RES 1.846x103 x 103 e VE - 1.846 velt * we know VBE 20.7 – VB-VES 0.7 o VB - 0.7 + VE VB = 0.7 t 1.846 1 VB U 2.546 volt T * Vc = Vcc IRC RC - 24 (1.846 xlo? xs *103) = 14.77 volt * VCE = Vc - VE - 14.77 1.846 VCE = 12.925 volt