Question

6th Edition. (Bart) Electen'c Devices and cucuts The Ideal operational Amplifier (chap 8) → Assume that the feedback resistance in Figure. I is doubled and the input resistance is halved Find the output for each of the following input signals. (a) Vi = -60.5 mV de (6) Vi - 500 sin wt u V (5) Vi = -0.16 tsin wt v (d) Vi = -0.2 sin (wt - 30°) V 39kn Aw we 15 kn Figure 1

Answer 1

Important concept : - ideas vo Vo = ve due to virtual ground concept te = TE = OA due to very high inpul impedance of ориме ve_te RE . given circuit vo vi VBVO = OV toCS COM O -V o O- V Ri Reso Kel at node Vi standard T resut Vo Viº - RS - RE given, RF = 39ks, Ri=15151 o in quetion saying that RE RE = 2 RF and 1. Then "ro" become as Ivo = – PRE Vi = -4. Ref. Vo T Rila I now put RF = 39k2, Ri= 15k)

fake care o ) v° = - Go: mv dc Vo = +4.x. 160.6mudo urult omv.dc - 629.2 m dc [vo = 0.6292 rata ve = 500 sinut MV Vo = -4 * 3.9 10w.sinwt MV Vo = -5.2 mv sinut Vo = -52 sinut mutt Vº = -0.16+ Sinwe v © Vo - - UX390.16 15 -19 15 @ Vo = 1,664 – 10.4 sinut Via = -0.2 Sin (wt-30°) V Vo = -4x2 (0.2) sin (wt-30°) V Vo = 2.08 Sin (wt-30°) V Tvoz 2016 in (wt-30) * Ivo = 2:08 sin (wt-30) Voy