COMPUTING

Chapter 5 Operational AmplifiersIn the circuit of 5.60, the total current iswhereAlsoThen,i 1i = i 1 + i 2 + … + i Nv in1 v= --------- i in2vR 2 = --------- … i1R N = ---------- inN2R Nv out= – R f i = – R f ( i 1 + i 2 + … + i N )v out= – R f i = – R f ( i 1 + i 2 + … + i N ) =– R fR f-----vR in1-----v1 R in2 … R f+ + + ------v⎝⎛ 2 R inN N⎠⎞(5.49)If all input resistances are equal, that is, ifthe relation of (5.49) reduces toR 1 = R 2 = … = R N = Rv out=R f– ---- ( vR in1 + v in2 + … + v inN )(5.50)IfR f=R , relation (5.49) reduces further tov out = –( v in1 + v in2 + … + v inN )(5.51)and this indicates that the circuit of Figure 5.60 can be used to find the negative sum of any numberof input voltages.The circuit of Figure 5.60 can also be used to find the average value of all input voltages. Theratio R f ⁄ R is selected such that the sum of the input voltages is divided by the number of inputvoltages applied at the inverting input of the op amp.The circuit of Figure 5.61 shows the basic non−inverting summing and non−inverting averaging opamp circuit. In Figure 5.61 the voltage sources v in1 , v in2 , …,v inNand their series resistancesR 1, R 2, …,R Ncan be replaced by current sources whose values areand their parallel resistancesin Figure 5.62.v in1 ⁄ R 1, v in2 ⁄ R 2, …,v inN ⁄ R NR 1, R 2,…,R N. * The circuit of Figure 5.61 can now be represented as* For voltage source with series resistance to current source with parallel resistance transformation please refer to CircuitAnalysis I with MATLAB Applications, ISBN 978−0−9709511−2−05−38Electronic Devices and Amplifier Circuits with MATLAB Computing, Second EditionCopyright © Orchard Publications