To amplify a given signal using an operational amplifier. The amplification is determined by the gain resistors. The math is pretty simple.
Half of the power rating of a 1/4 W resistor,
Rx = (V^2)*8 = (12^2)*8 = 1152 Ω.
Ry = R_x/11 = 104.7 Ω.
Rth = Rx*Ry/(Rx + Ry) = 96 Ω. Vth = 1 V.
Ri = 20*96 = 1.92 kΩ.Rf = 19.2 kΩ.
Data
Component
|
Nominal Value
|
Measured Value
|
Power or Current Rating
|
Ri
|
2000 Ω
|
1957 Ω
|
1/8 W
|
Rf
|
20000 Ω
|
19760 Ω
|
1/8 W
|
Rx
|
1152 Ω
|
1786 Ω
|
1/4 W
|
Ry
|
104.7 Ω
|
7020 Ω
|
1/4 W
|
V1 = Vcc
|
12 V
|
12.12 Ω
|
2 A
|
V2 = Vee
|
12 V
|
12.08 Ω
|
2 A
|
Vin
|
Vout (Measured)
|
GAIN (Calculated)
|
VRi (Measured)
|
IRi (Calculated)
|
VRf (Measured)
|
0.0 V
|
0.00 V
|
0.00
|
0.00 V
|
0.00 A
|
0.00 V
|
0.25 V
|
-2.55 V
|
-10.20
|
0.25 V
|
0.125 mA
|
2.57 V
|
0.50 V
|
-5.07 V
|
-10.14
|
0.50 V
|
0.250 mA
|
5.11 V
|
0.75 V
|
-7.62 V
|
-10.16
|
0.75 V
|
0.375 mA
|
7.65 V
|
1.00 V
|
-10.16 V
|
-10.06
|
1.01 V
|
0.505 mA
|
10.18 V
|
I_v1 = 2.31 mA
I_v2 = -1.645 mA
Pv1 = V*I = 28.00 mW
Pv2 = -19.87 mW
Iv1 + I_v2 = 0.665 mA
If = 0.509 mA
Error = 30.6%
Conclusion
To emulate a sensor's signal we use potentiometer to adjust the voltage to the amplifier. The result will amplify the signal ten times the potential input.
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