Question

You have a sensor that has a maximum output of 25 Volt, and you want to feed this signal intoa data acquisition (DAQ) board with a maximum input in the range 0, +101 Volt. Design a circuit that matches the sensor to your DAQ board with the following design criteria: a) You need to attenuate the input signal by a factor of 2.5 b) The sensor cannot provide any current; therefore you need to make sure not to cause a loading error. c) You can only use a single OpAmp and two resistors (and everything else you need to make this work such as power supplies, wires etc.) You consider the following three options: For each option,I explain why each option does or does not work, and II give the value of two resistors if the option works (choose one, calculate the other). Also, insert photos of HAND-DRAWN images here that show your attempts; you need figures for ALL THREE OPTIONS. 1. Use a non-inverting OpAmp circuit with two feedback resistors. 2. Use an inverting OpAmp circuit with two feed back resistors. 3. Design your own circuit (again use only one OpAmp and two resistors)

Answer 1

Here is the solution.

Use a non-investing opAmp circuit with two feedback resistances Here v_in is the sensor voltage, we need Gain = 1/10, Gain = 1 + R_2/R_1 so 1 + R_2/R_1 = 10, R_2/R_1 = 1/10 -1 = -9/10 = -0.9 Take mod = Vertical-bar R_2/R_1 Vertical-bar = 0.9, if R_1 = 1k, R_2 = 0.9 R_2 = 0.9 k Ohm This circuit won't give gain of 1/10, so, it will not work 2) use an inverting op-Amp with two feedback resistor Gain = -R_2/R_1 and Gain should be = 1/10 so -R_2/R_1 = 1/10: Take mod: Vertical-bar R_2/R_1 Vertical-bar = Vertical-bar 1/10 Vertical-bar if R_1 = 10k, R_2 = R_1/10 = 1 k Ohm This circuit will work Own circuit Here V_out = V_in times R_2/R_1 + R_2 So, R_2/R_1 + R_2 = 1/10: so if R_2 = 10 K R_1 + 10K = 100K R_2 = 70 K

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