Steps:
we assume the Vin= 1V and Vout= -10V, so A= -(-10)/1=10. In addition, we must make the input current less than 1mA, so the input resistor =1V/1mA= 1kΩ and output resistor= 10*1k=10kΩ.
The 5 volt power source to get 1v, we make the Rx=286Ω and Ry=57.6Ω
Nominal value | Measured value | Rate | |
Ri | 1kΩ | 981 | 0.25 |
Rf | 10kΩ | 9830 | 0.25 |
Rx | 288Ω | 286 | 0.125 |
Ry | 58Ω | 58.7 | 1 |
V1 | 12V | 12.15 | |
V2 | 12V | 12.14 |
Then measure the value of Vout,Vin, and Vrf
step5A:
Vin | Vout(M) | Gain(C) | Vin(M) | Iri(C) | Vrf(M) |
0 | -0.32 | null | 0.02 | 0.0000204 | 0.34 |
0.25 | -2.91 | 11.64 | 0.27 | 0.275 | 2.85 |
0.5 | -5.31 | 10.62 | 0.52 | 0.53 | 5.33 |
0.75 | -7.8 | 10.4 | 0.77 | 0.785 | 7.86 |
1 | -9.97 | 9.97 | 0.99 | 1.01 | 10.02 |
Iv1=1.51mA
Iv2=0.49mA
Pv1= 12.15*1.51m=18.35mw; Pv2= 12.14*0.49m=5.95mw
It satisfies the power supply constraint to supply no more than 30mW.
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