Question

Electronics1. It's a multiple choices question. use the formula sheet if needed (the last picture).

Question: 10 CIRCUIT Y/FIG.10 (5 Marks) Statement: Consider the RC-coupled amplifier circuit illustrated in Fig. 10 (CIRCUIT Y). Sketching relevant output (Vc-le) characteristics, decide Q-point coordinates and DC load-line details of the NPN-BJT/Si circuit of Fig.10 of Circuit Y). Further, establish an appropriate AC-load line and decide the maximum swing of the output voltage across Road with respect to the Q-point. Assume the input signal at the base is a sine function with = 21 x frequency (f) and other circuit details are as follows: Vcc = 15 volt; Vc = 8.25 volt; VE = 1.5 volt; R=2.0 k; R= 1.0 k; Rs = 2 k; AVI = 200; V1 = 0.026 volt; 1B1 = Current through Retaken as 15% of the collector current; Frequency bandwidth (BW) response: 50 Hz to 10000 Hz; and a = 0.98958. +Voc Rc 181 Vc RBI Rs HE Cei HE Cc2 | +9 + v.t) s Road v.(t) RB2 RE + CE A FIG. 10: CIRCUIT Y

Answer 1

Given circuit +Vec Ri Vcc = 15V Vc = 8.251 IBU Ve=1.25V I cca Vi RBU RS + Wt ΝΡΛΙ Vout Rc = 2ks RL = 1.0kr. Сс Rout VC RE TE Rg=2k TAvl= 200 V 0.026V From given p= = 94.6 9=0.989538 96 By DC anal analysis Copening all the capacitors) 415V IBI Ru 2K Vc =8.25V th VcxRB2 RBI +RB2 B:95 Yin 1 RBE RBill RB2 Vol.5V PBL TEIC RE -415V R$2K2 By KVL -15+ Ctc + Ipi)2 +YE PIERE = 0 Given IBI = 0.15 I 21:15) Ic + IERE = 15-8.25-1,5 (4.3 +Re) Ich -5.25 +0 RE

UC = Ncc - (IC & IB1) Rc 8.25 = 15-(1:15) Tc (2) IBQ IBI 0.15* Ice At £=0 => IB) = 0 4 Vc =Vcmanac 2.31% = 6.75 Ic = 0.935mA IcQ=0.935mA Ica - 0.03089mA 95 JBQ= 30.89UA | Bii 0.44ma Put Ic in (4:3+RE) (7.935x103) = 5.25 RE: 0:511122 VcQ = 8.25V TE = (1+8) IBQ = (96)[ 30.8944) TE 2.965mA (V)mar is obtained when it's in cut-Off CT=O) -Vect (Ic tIB) Retro Vec = Vimax -15V | (Vc)mar =151 (IC) mod is obtained when Bjt is in saturation consider (CE)sar FOV Cideally (1•151_)2 + (IC) (5:1) = 0 => Ichiman 5.4mA some close to 7.5mA - 15+

Ivemax Icmon ] = [15V, 5.uma] VBE = 0.7V VB-VE = 0.7 = V₂ = 4th = 2.2V IB IB 2t IB 0.44MA TB2 + 0.03089mA TB2 = 0.40911MA IB2 = Veh RB2 RB2 222 Kn .40911 RB2: 5:378 KB IBI = 0.44mA = Vc-4th RB1 RBI 8.25-22 .44 RBL = 13. 75 kn (RBI, RB2) = (13.75k12,5-37 kn) Rg = RBillR32 = 3.862 KN Stability Factor (5) (+95 HB RE lupe RETRE lt 95 6.517 3-8627571) 96 11986 S 10 (near option) All the calculated values match in option ).

Given lower cut-off frequency i fu) = 50Hz 1 -50 anx RXCE Ro Resistance seen by CE ,RS = Rs RB 2x1/3862 - 1.317648.C --- oni s V IBA 0-026 ΚΟ 0.03089 on = 0.8417661 lon RE [1.3176.8017 95.6 511K) R = 002260 56 204.022x10xCE CEL*5mF Option Ô is answer:

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