Question

1. Two directly coupled common emitter amplifiers are shown below in Fig. 1. Since base current IBase2 of the transistor Qp is much smaller than the collector current of QN, simply ignore I Base 2 current while solving the problem. Vsig is a purely AC signal. Find DC operating points (Ic, VCE) of the transistors. ON I Qp VA= OV B = 150 VE(ON) = 0.7 V VESAT) = 0.2 V VA= OV B = 70 V(ON) = 0.7 V VESAT) = 0.2 V W LO! C & C & Cour = 0 Rss = 502 R = 100 k22 R = 50 k2 Rai = 2 k2 Ru=1k2 Rc = 2 k2 Re = 1.5kQ2 R = 5 k92 Vec 10 V ! Res = Fig. 1. Directly coupled common emitter amplifiers under consideration,

Answer 1

GOео ск€ Ucc & RE2 =1-5k look= R1 z Lop Mython Ht your P2 sok: DC cond Pepong , Tauec slou 2E. RC & zkr V 33.3k2 Bato BE1502 I tho Ist stage .. . Apply Kulon Ost stoga . Rub . RUTIR 2 = 100k | 150K - 100.5 150 Rth= 33-3) -utht Ib,Reth + UBCH TERESO Ie=C1+6) 18, IB,=. Uth - UBE Rth+ CI+B) REJ. utho. Ucc XR2 RIAR2 = 10x50 150 . uth=3-3v. =-3-3-0-7 33.3k +(151)x1ksa FB,= 14-1946) TE = (1+B) IB) = 151 x 13] Ic= die, = 150 2-13 TIer= 2-13m Al Ici = 2-11 mA)

VE = IERE VED ICIRC, =) 2-13.MAXIKO I VE=2-13 Ve= ucc-'ICRC = 100- 2-11 160X2kór vez 5-78. 'cc- ICRC VCE - Uc-VE = 5-78-2-13 VCE = 3.650 : e point.os CUCE, Ice Coo) CICO, UCE) => [ 2-117A, 3-650] -0. . it KUL on 2nd stage V8zruc, . = 5-780 ...-UB+UBE + IEZRĘ=O Iez= VBI UBEZ RE 5-78-0-7 akr Icz = 2-54 mA Ic= LIE B=70 Ich 70 IE2 VE2= vcc- FERRE TEZ = Ecuxa) 70 IOU - 2-57 MAXI.skor TIE= 2.57 mA • Vez Ica RC2 V6 = 6:14 Ve= USZ-vci. = 254m7 Axzlsac - 6-14-5-08 Ucz= 5-08v NEC2 = 1.060) . a point [2.54má, i-06070