Question

7.) In the in the cascaded summing / inverting / noninverting operational amplifier circuit shown in Figure 7, find the output AC signal voltages V1 V2 V3 and Vo. The values of input AC signal voltages Vsi, Vs2 and V sa are -0.25 V, 0.40 V and 0.80 V, respectively. (160 pts) V1 = V2 = Vos V3 =

Answer 1

Solution as the given Circuit consists of different types of operational amplifiers first we will consider Single Cor) complifier at a time ão I, 25k2 I ~ (Loveting Auplifier - VS lokn A A - kel at Node +8=0 Vgg -VA + V - VA =0 lokn 2 arka A Becomes Short Circuit Voltage at node Due to Virtual approximately approximately 30" o v oco chtitok is above Subetitute in above Equation quanto & given Veq = -0.25V =0 -0.25-0 + V₂ -0 oknack-n 0.25 xarks . V - cks V = o-barv (0) Hor Preventing complice vo = chai x

at note B KEL Ve = vs due to Virtual Stort aka Vsa + Va t Vsa zo oksz ke Given V = 0.40V je VB +ve vo =0 AĞ 36 0.40 V2 +0.4 =0 6k 36kg 0-4 + 6x04 + V2 = 36 + V2 = +7x0.4 = 729v ../V = 128v Loker 5k t & (sonverting Ampletion) a s Explained above (to Feding ve) V₂ = - RFX V2 Giren 4= 0.8 :: V2 = - 40 x 0.8 [V2 = -64 v

14 Short c an at mode c = 2 due to Virtual " Apply ker at Node "C" is I Ve +6 -14 = 0 la 7 Watt V V V y = 0 + 64. X = 0 6 V4 = -3x6.4 la Vy=-3x2. . Jy = -9.64 all the values of V, Vgg Vz, Vy the total as follows. is i On Substituting Gresit becomes 30k_2 lokn Joker - - M 15kn "4 Now By Can End applying ket at Node "B" (p=0) we output Village (vo).

=0 VnV+ kovot v vo + Vo - Vp Toka 20kan Trka 30kn & V, s. 300 Subetitute as we got re , &o,& in above Squation Vp + 2 + 4 + t 10 do 15 30 = 0 Vo = -6v, -3V2 - 4/4 30 OOPPE Von – by - 3V - Yry Vo= ~3V, 1.5 Va - 2 Vy No = -3x (0-bas) – 1,5(2-3) --(-9.6) Vor – 1.875-42 + 19.2 T.Vo = 13.125 Va = 2.8V u. V = o.basv =-9.6V V – – 6.4V