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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+// Exa 6.14
+clc;
+clear;
+close;
+format('v',6)
+// Given data
+R1 = 100;// in ohm
+R2 = R1;// in ohm
+R3 = 3.9;// in k ohm
+R3 = R3 * 10^3;// in ohm
+R_F = R3;// in ohm
+Vx = -3.2;// in V
+Vy = -3;// in V
+// output voltage due to Vx, Vox = -(R_F/R1)*Vx and due to Vy, Voy = (R3/(R2+R3)) * (1+(R_F/R1))*Vy
+// Vo = Vox + Voy = -(R_F/R1)*Vx + (R_F/R1)*Vy (as R1=R2 and R3=Rf)
+//So, Aod = Vo/(Vx-Vy) = -R_F/R1;
+Aod = -R_F/R1;
+disp(Aod,"The closed loop differential gain is");
+Vo = (-R_F/R1)*(Vx-Vy);// in V
+disp(Vo,"The output voltage in V is");