If the fixed input resistor is 10 kΩ, then we need a 100 kΩ resistor to achieve a gain of -10.
If Vsen = 1 V, the current leaving the op-amp is IOP = 0.1 mA.
We constructed the following setup and made measurements
|
Vin
Desired
|
Vin
Actual
|
Vout
Measured
|
VRF
Measured
|
IOP
Calculated
|
ICC
Measured
|
IEE
Measured
|
|
0.25 V
|
0.248 V
|
-2.48 V
|
2.48 V
|
0.0248 mA
|
|
|
|
0.5 V
|
0.5 V
|
-4.81 V
|
4.83 V
|
0.05 mA
|
|
|
|
1.0 V
|
1 V
|
-10.03 V
|
9.95 V
|
0.1 mA
|
0.889 mA
|
-0.982 mA
|
Kirchhoff's Current Law is confirmed when we add ICC and IEE to yield IOP.
The power supplied by the 12 V supplies were
PCC = (12 V) (ICC) = (12 V) (0.889 mA) = 10.67 mW
PEE = (12 V) (IEE) = (12 V) (0.889 mA) = 11.78 mW
We added a 1 kΩ resistor to the output shown below
and measured the following
|
Vin
Desired
|
Vout
Measured
|
VRF
Measured
|
IOP
Calculated
|
ICC
Measured
|
IEE
Measured
|
|
1.0 V
|
-9.97 V
|
9.75 V
|
0.1 mA
|
0.887 mA
|
-0.987 mA
|
We can see that KCL is obeyed when we sum ICC and IEE.
Again, we calculated the power:
PCC = 10.64 mW
PEE = 11.84 mW
Extra Credit
In order to achieve a gain of -5 in our inverting amplifier, our value for the feedback resistor must be 50 kΩ.
Our measurements were:
|
Vin
Desired
|
Vout
Measured
|
VRF
Measured
|
IOP
Calculated
|
ICC
Measured
|
IEE
Measured
|
|
1.0 V
|
-5.04 V
|
4.98 V
|
0.1 mA
|
0.885 mA
|
-0.985 mA
|
The behavior of the circuit is exactly the same as the two previous measurements made in the experiment.
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