Question

Answer is: (b) 6.2

12V Vcc 11k a(t) + UB 1k 10HF UE k NL 2N k Figure 5: Circuit for Questions 12-15 Question 14. (2 marks) Consider the circuit in Figure 5. Suppose v.(t) = cos(104t) volt. In addition, the voltage vN accross the nonlinear element NL is given by UN where UN is expressed in volt and in is expressed in mili-amp. Find the smallest value of Vcc (in volt) to ensure that the transistor always operates in the active region. Assume VCE(sat) = 0.2 volt a) 4.2 b) 6.2 c) 8.2 d) 10.2 e) 12.2 WW

Answer 1

fins Given that, Ns (t) = Cos (t0t) VCE (sat) =0.av IK C12)= tv Voc+ oheve Vpc Vi 1 tK+111K 1+ COS (L04t) 米 Vi = Op-Amp is ideal and nagative feed bock by virtual short Vp= VN= Vi ASsume So Up voltoge at non inverting terminal of Op-fimp. VNI Voltoge at inverting teminal of Gp-fmp VCC iN=Vi lK 1+CoS (104も) Vi VB VCE tN=1+COs (1o t) mA VE +0s (o VN = VN VN=1 200s (ot tcos (tot) iN Vi Vcc = VCE t VN tVi VCE t 1+2Os tot + Cos10t) + cos Cto4t) VCE t at 3caS (I04 t) +cos'(o t) At edge of Saturation Vre =0.a V, in active and cos function range Tegion VCE 6.2v in between - to.t1, Cos (to+) (V) Cc 2 0.a t 2 + 3 COs (tot) To nd Vcc mintmum aue COS Cto t) is at t UCC 2 G.a+a t 3(1) + C1) c Z 6.a V Smatlest Nalue of c=6.2V zI