the packages it uses have eight legs
In this experiment, we assume that we have a sensor which has an output ranges from 0 to 1V with negligible impedance. The specification for the Signal conditioning needs to be 0 to -10v
and must draw no more than 1mA from the sensor and the op-amp supplies should supply no more than 30mW of power each.
here is the schematic of the circuit
first, we need to determine the value of R1
since the max input voltage is 1 and we want it be lower than 1mA
we can determine the value R1 =(1V)/1mA = 1kΩ
next, we need to determine the feedback resistor
since the max output voltage is 10
10V=Rf/1000
Rf=10KΩ
Next we need to determine Rx
since Rx is to be 1/4W resistor and we want to operate at half of its rated power
The worst case voltage across Rx occurs when Ry is set to zero
(1/8)W=(6^2)/Rx
Rx=1152Ω
Determine Ry
Assuming the value of Rx determined above and no loading on the divider
Ry that will yield 1V across it
1=[Ry/(288+Ry)]6
Ry=57.6Ω
The Thevenin equivalent of the divider circuit from the inverting amplifier
Rth=48Ω
and the thevenin resistance is at least a factor of 20 times smaller than the resistance value for Ri
Next, we devise the experiment
we record the values of the components
Component | Nominal Value | Measured Value | Power or Current Rating |
Ri | 1K Ω | 984Ω | 1/4W |
Rf | 10k Ω | 9.9KΩ | 1/4W |
Rx | 288 Ω | 287Ω | 1/4W |
Ry | 57.6 Ω | 58.8Ω | 1/4W |
V1 | 12V | 12.13V | 12W |
V2 | 12V | 12.06V | 12W |
Next we build the circuit on the breadboard
Next we record the data from the output by looking at the multimeters we set up for the experiments
Vin (V) | Vout(Measured)(V) | GAIN(Calculated) | V_Ri(mA) | I_Ri(Calculatd)(mA) | V_Rf(Measrued)(V) |
0 | -0.05 | 0 | -0.007 | -0.007113821 | 0.06 |
0.25 | -2.54 | -10.16 | 0.269 | 0.273373984 | 2.57 |
0.5 | -5.08 | -10.16 | 0.536 | 0.544715447 | 5.12 |
0.75 | -7.56 | -10.08 | 0.797 | 0.80995935 | 7.62 |
1 | -9.91 | -9.91 | 0.946 | 0.961382114 | 9.91 |
Power supply current
Iv1=2.32mA
Iv2=1.49mA
P_V1=12.13*2.32mA=28.14mW
P_V=12.06*1.49mA=17.97mW
Both power supply are less than 30 mW.
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