Question

4. (10) In characterization of a circuit as shown below, some accurate resistors are used and the capacitor Ce is considered large enough for its ac impedance to be neglected. It's measured that when the switch is shifted from node A to B, the voltage gain of Ayi = v/v; is reduced by a factor of 5. In either case, the DC voltage drop across Rg is 2.0 V. Determine the biasing current Ibius and the common-emitter current gain of the BJT. 200 L

Answer 1

Page 1 Solution I Ž RC Skr ovo RE Vi 200kr loonB IS -CE T Ibias - Given figure DC analysis quce IC IRC ovo RE NW IB for DC analysis AC signal source usig 18 replaced with short circuit to deactivate it during DC analysis. Capacitor CE will be Open circuited during DC analysis > Corresponding circuit as shown in fig. A. V HOA RE . Irios V IE OB. FigiA) :- Circuit for DC O analysis.

III Page 2 IB: Current flowing an base. Ic :- Current flowing an collector IE :- Current flowing an Emitter. Also, from given figure, TE = Ibias We know that, IE= IC TIB Ic=B IB IE = Ibias = (Hp) IB AC analysis for Ac analysis . DC source us 98 replaced & RC=5K2 - with short ciraut to deactivate Hovo It during ac analysis RB-200k Current source Ibias as replaced with open circuit. Bypass capacitor CE IS replaced with short circuit g during ac analysis. 1 L + Fig-CB) shows the correspent ding irait for ac Fig. (B) :- Circuit for ac analysis analisis usig 1000RE

According to Question Ease-l :- When switch 98 at circuit & a common bЯрах capace{6% . page-3 position A the given emitter amplifier with 1o. RE comes in parallel with short circuit and the equivalent circuit as shown an fig(c)e And terminal Ais directly grounded The ciruit of fig. (C) can be replaced with small signal circut as shown an figis). • Vout. RB-200k RB vi Tobo vi & mui &RC -) :- Circuit You Case 1: Fig. (D) : - Small signal circuit for CE amplifier with сурах capaciox : from fig. (D) bo--gmuirc o -- gm Rc (5) for common emittere amplifier Eq.(5) is a current gain with bypass capacitor!

Case 2 :- When switch as at position B the given circuit - is a common emither amplifier without bypass Capacitor lo Re is not short cirauted reather it provides negative fudback Equivalent circuit as shown an fig. (E) The circut of fig. (E) can be replaced with equivalent small &RC signal circuit as shown an fig (F) RB RB V RR hovo T + هم یا ERE Figlt): - Circuit for case a. Fig (F) :- Small signal circuit for CE amplifier without bypass capacitor . from fig (f). wo --gmRc V tem RE (6) Equation (6) is a current gain for common emither amplifier without bypass capacitore from eais) and (6) It can be concluded that voltage gain of eqes) is reduced by a facter of (temple).

Page o According to Question gain is reduced by a factor of 5. (Given) » Item RE=5. gmRE=4 > gm = 42 > gmte - soma lv Also, voltage drop across RB = av. (Given) → Voltage chrop across RB = IBRB = 2 ( 00 Rp=200k) → Is X200K= 2 IB = 10uA Also, gm = Ich Take Vr=26mV (approx) 7 Ic=mV1 = 40 x26= 1.84mA Ic = 1:04MAJ Also, Ic= BIB = ß= Ic = 1.04MA - 104 IB TONA. -104 L Required common-emitter Current gain ß. Also, Ie> Ibias = (HB) IB » Ibias= (1H1OA) XIGMA = 1050 MA Tricy = 1.05mA Ik Required bias current - UT EN