Op Amp 1

  

In this lab we are to have a sensor whose output ranges from 0 to +1V and whose Thevenin output resistance is negligible. the output range must be between 0 and -10V and must draw no more than 1mA from the senseor and the op amp power supplies should be no more than 30 mW.
From our calculations we determined Ri = 1 k ohms and Rf = 10 k ohms
We proceed by determining Rx = 8 * V^2 = 1152 ohms      RY / (RY + RY) * 12 =1     RY = 288 ohms
From this we find that RTH = 230.4 ohms which is at least 20 times smaller than the resistance value of R1

Applying our theoretical results we attain the following

Component Nominal Value Measured Value Power Rating

        Ri                1k ohm      984 ohm            0.25 W

        Rf                10k ohm      9.74 ohm            0.25 W

        Rx                1k ohm      982 ohm            0.25 W

        Ry                     x                    x                         x

        V1                  12 V      12.06 ohm               2 A

        V2                  12 V       12.1 ohm               2 A

Vin           Vout Gain          VRi         IRi           VRf

0.0V          0V             0           0V          0 mA      0

0.25V -2.48V -9.92 .249V .253 mA -2.48 V

0.5V -       4.97V -9.94 .5V         .508 mA -5.01 V

0.76V -7.48V -9.97 .751V .763 mA -7.52 V

1.004V -10.03V -9.98 1.005V 1.02 mA -10.01 V

IV1 = 1.737 mA                            IV2 = 1.276 mA

PV1 = IV1 * V1 = 20.9 mW            PV2 = IV2 * V2 = 15.4 mW

I_Ri = V_in / R_i =1.004 / 984 = 1.02 mA

I_Rf =  V_Rf / R_f  = 10.01 / 9740 = 1.02 mA

From our calculations we see that the power supply constraint of no more than 30 mW is maintianed. In order to reduce the power drawn we should increase the resistor values in a way that the ration Rf/Ri = 10.