Question

VUMI JUNY SUULE No-Sp- Paragraph Styles (A) For the Op Amp circuit in Fig. 2.1. Infer the expression of you (s) / vin (s). (B) For the Op Amp in Fig. 22. Evaluate the Av R 40 KOR340K 200 KO R w 50K Ко ко) Fig. 2.1 Fig 2.2

solve it neatly on paper

Answer 1

A. Giwen circuit in s-domain is, Isc HE R mm X Vints) + • Vout (S) 1 Vina(s) R ./sc We need to find your (s) by using Superposition theorem Case-i: Vinils) is acting aline, Vinge is short circuited. (- : Ving(9)=0v) Then positive termince node voltage is zero. Therefore, according to Ideal op-amp negative terminal voltage is zero (at node-x) Apply Ked at node-x, we get o-Vino(s) + 0- Voutis) (/sc) (Isc) -0 R Vout(s) Vinics R SCR Case-ii =) Vout (s) = -/ Vinics) Maquina() is acting alene, Vini() jo shoot circuited. (.. Vini (5)=0) we need to find node voltage at 'y' by using voltage dividen mule. Vy = Ving(s) X (Ysc) R+Ysc Vy= Ving(s) x I HRSC =) Yy= Vina(s) Ving(s) ITSCR HSCR Vy-o + +vx According to Ideal op-amp; Vy=vy = Apply kce at node-x, we get Vy-yout (S) (Ysc) Vу Vout(s) R (Ysc) Youtis)= (1 + SCR) VX = (1+ sch Vinals) HSCR IIISER Vina(s) (yso) SCR I USER Yout(s) = Vina(s) a ScR

By adding & we get vaut (s) ie, Vout (S) = – Vinics) Vinals) + SCR SCR SCR See (Vingis) – Vini (9) (Vincs) = Vinals) - Vinils) Vout (5) = Q - Vin(s) SCR ㅗ SIR Vout (s) = Vin(s) Ginen cirait ip, :40KR § 40kr M Cooker MW IB SOKA • Yout(s) A Vin - 1 350kn 1 According to Ideal op-amp, V=VA=ov Apply kce at node-A, we get -O + ovo HOR vo -50k vo uok 16 o-VB SOK VB sox B = Zvo (or) UB= Now Apply KCl at node B, we get VB-vo +VB-O + + VB- Vin VB SOK uok SOK Rook

vz-vintuvB +5(UB-Uo)+44B =0 2ook VB-vin +4UB +5VB - 5%+ 4 VB = 14VB - 5% = vin >) 141-1 o) - 5%= vin - - 500 (14 + 1) - -vin = -22.5V = vin vo vin -0·04494 → Av= -0.0004)