Question

Q.A. 2 -Statement: DC biasing of a CE amplifier circuit is as in Fig.2 Assume the parameters, VCC = 12 volt, RE = 450 ohm;  = 100; RL = 2250 ohm, RC =1000 ohm and source resistance and Rs = 2500 ohm. Also, the device emitter resistance re is negligible.

The amplifier is excited by a source, vs(t) = Vm × sin(2ft) with Vm = 500 mV peak and f = 5500 Hz; and, coupling and by-pass capacitors can be assumed to be of zero reactance at the working frequency.

Required Solutions: Determine the following: Input resistance Rin (as seen by the ac signal source at the input); value of biasing resistance, RB, ac power dissipated (PL) in the load, RL and quiescent parameters: (VCQ; ICQ; IBQ; IEQ).

Fig. 2 +V CC IBQ RC Ica RE Cc2 V. in Ca in CQ R +V BO R. in R. + +V EO RB2 TEQ

Multiple-choice answers

1. (Rin ≈ 6.61 Mohm; RB ≈ 7.1 Mohm; quiescent parameters: (VCQ ≈ 10.1 volt; ICQ= 3.9 mA; IBQ = 0.45 microA; IEQ = 6.91 mA; and, PL (microwatts) = 100.4)

2. (Rin ≈ 5.42 Mohm; RB ≈ 37.8 Mohm; quiescent parameters: (VCQ ≈ 10.7 volt; ICQ= 4.65 mA; IBQ = 0.23 microA; IEQ = 3.72 mA; and, PL (microwatts) = 80.5)

3. (Rin ≈ 4.55 Mohm; RB ≈ 67.8 Mohm; quiescent parameters: (VCQ ≈ 10.7 volt; ICQ= 3.65 mA; IBQ = 0.33 microA; IEQ = 4.02 mA; and, PL (microwatts) = 102.5)

4. (Rin ≈ 4.42 Mohm; RB ≈ 57.8 Mohm; quiescent parameters: (VCQ ≈ 9.7 volt; ICQ= 5.65 mA; IBQ = 0.30 microA; IEQ = 4.74 mA; and, PL (microwatts) = 69.2)

5. (Rin ≈ 3.61 Mohm; RB ≈ 17.1 Mohm; quiescent parameters: (VCQ ≈ 7.1 volt; ICQ= 4.9 mA; IBQ = 0.49 microA; IEQ = 4.94 mA; and, PL (microwatts) = 113.4)

Answer 1

Given circuit diagram is TV cc ER & Ica RB GRLA IL Rs + RB2 VIEQ To simplify the circuit, take a part of it IB Ix VIE IE =(3+1) Ix since IE = IB tic = IB + BIB=(3+1) IB Here IB i Ix ) Io * Se put=> [p+372, IIE = (+1]"IB Pretot = (B+1]? Bnew = BP+1+2B-1 Bnew = Brtaß Given, ß= = 100 2 Bnew 100"7200 = 10 200.

can Here, We simple method for calculating propose resistance Rin : input Step 1: az RL Rs Rina The Rg RB2 IR BE Ring Rin ac circuit [ All De vo voltage sources are grounded Ring =rt (ß +1) RB2 (3+1) is the multiplying factor when emitter resistance from Base we see Rinz =[[+4)[Ren] + rms] (p+1) + Von NOTE : Tog is the resistance of the transistor seen en through Base, AT Bre or (Btiore gn and given that re =0 → Y =0. Rinz (1 +B)[(4 +p) Rea] (B+13² RB2 = (Pnew +1) RB2 = (10201) RB2 (1020)(450) Rin [Rsll Rine (I) 45904502

Dc Analysis : Vcc=12v Rc = 1k2 RB Prew ’+2B = 10200 IB VIE RB2: Re = 450.12 Mesh Equation : Vcc - IBRB - VBE - IERE =0 UBE =0.77 xasince re=o. 12 IB RB 340.7) -(4+5) Ig RE =O IB[45450 + Rs] = 11.3-0.7 - 10.6V, Also Vcc Ic Rc - VCE TERE =0 from Eq. Ⓡ. We could not get RB value. In Eq.②, We couldnot get Vce value, So, only procedure remained عنا to verify through given choices. From We knew > Rin= Roll Rine Rin - RBII (4.59 M.2) By logic AllR2 = less than both R1 and R₂. Thus Roll 4.59M LS less than 4.59 M2 Choices satisfying this condition are 9 and ②.

Take 5): Given Rin = 3.61 M 2 [less than Ring Getting Back RB and verifying. Rg. Rina RB (4.59x104) Rin 3.61X10 RBT Rina RiB +4.59x106 » RB = 3.61 4.59 R8 +4.59x106) » Rol Rol ? 3.61 4.59 ? ] 3.61 x16 → RB = (49] [3.64 xlo“) = 46.9 m2 = 17M2. 17 MR. Also, RB given in choice (5) is also around 5 Thus An Answer Option is