Question

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-Ic) 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 Rload with respect to the Q-point. Assume the input signal at the base is a sine function with o= 21 * frequency (f) and other circuit details are as follows: Vcc = 15 volt; Vc = 8.25 volt; VE = 1.5 volt; Rc= 2.0 k; Rı= 1.0 k; Rs = 2 k; |Av1 = 200; VI = 0.026 volt; 1B1 = Current through RBi taken as 15% of the collector current; Frequency bandwidth (BW) response: 50 Hz to 10000 Hz, and a=0.98958. +Vcc Rc Ibi + Vc RB1 HE Rs HE Ссі Cc2 + vs(t)( s Rload vo(t) RB2 RE CE FIG. 10: CIRCUIT Y

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Answer 1

tvec Rc V ce=154 IC CC2 IBI th Vecsv Rs www ㅔ NPN Ve=1.5V Cal Rout ww Vout Rc=aku RL= l.okn 17 CE Rs = 2x RE laul=doo VI-006 D-0.989538 from given a 2 1-2 = 94.6 ~ 95 295 By DC analyus (opening all the Gpaciloss) +15V Ill1 RL & Skin Vla8.254 YIL Nha V cxRBO Rb Rait RB2 Bags VBE RB= RBill Roz Roba Venlisy Ie NIC RE By kul -15+ (Ic+I8, Da'+ Vce + Je pe=0 Giren Ibi 0.15'Ic 26.15) Io tic Re IS- 8.25 -1,5 0

Ve= Vcc Vec- c AIR)RE 9:35 - 15- (is) Ic (3) 3 TC 6.15 Ica 2.95mA Ica Ico -1.935mA I.Do = 0.01087004 95 I BO - 0 15xIco IBI o uumA = put Ic in (2.3 +RE) (2.935 x Sx10-1)- sas :. RE=0.51kL Vco = 8.250 Ie = 2.95 MA (uchoman is obtained when it's in Cut-off Gc=o) : -Vcc + CIC +Io.) Rc (uc man lao At Ic=0 IBI=0 tuc Ncc = Ncc mon = isu Vc (man) 15 v Saturation (Ich man is obtained when BJT is in ov Consider Cuce) sat ideally -15+ (1.153c Ja +(Ic) (0.51)=o

El c) mons 5.4MA Sane Close to 7.5rips [vc mang mam, Iccom] - [isv, sumo] VOE = 0.7V B-Ve=0.7 =) VB = Vth =d&V IBI IBL + IB o.uYMA = IB2 to 03089 mA IB2 = o. 4o9lMA Van IB2 = RB2 Riz ku 4091 5.378 kul RB2 Vc-\th Psa 5-2 IBI = =o.uy MA kn RBI ouy Riba 13.75kch RDS R32) = (13.75x kul, 5.37kW 134 kus) RB RB, RBL = 3.86242 1495 HP factor (s) Stability RE 01511 HB 1495 2.511 RBTRE 13.8 to:51

11.986 10 match in option Values Catulated All the ** OS = (75) huarlay Horng sama waning Given lower - 50 Hz so 2T XRx GE R CE seen by Resistane = Rs tra Rs = Roll RB ll = RE Ithfe axl3.80 = 1:3H GUKA gro.o IRO 0030891 ܡܛܠ 0.841 tku R= 1.376 to. But I (o.silk) as. 6 o odskus JUSIEJ) . 50 ato xo.odd x 10 x ce option Dis answer
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