Operational Amplifier II

Statement of Purpose
To observe the signal produced by an inverting operational amplifier on a given load.

Procedure
Part 1.

Set up the operational amplifier with a sensor as the input, which is emulated by a variable resistor.

Although there is only two power supplies, we know that hooking the same potential can be attained by hooking the variable resistor in parallel. With this setup the following data is attained.

V_in (Desired)	V_in (Actual)	V_out (Measured)	V_RF (Measured)	I_op (Calculated)
0.25 V	0.24 V	-2.41 V	2.46 V	-0.0246 mA
0.5 V	0.50 V	-4.90 V	4.87 V	-0.0502 mA
1.0 V	1.00 V	-10.04 V	9.86 V	-0.1028 mA

I_cc = 0.874 mA
I_ee = -0.981 mA
I_cc + I_ee = -0.107 mA
P_cc = V*I = 10.488 mW
P_ee = 11.772 mW

Part 2.

This time a load resistor was connected to the output of the operational amplifier.

V_in (Desired)	V_out (Measured)	V_RF (Measured)	I_op (Calculated)	I_cc (Measured)	I_ee (Measured)
1.0 V	-9.99 V	9.72 V	-0.102 mA	0.887 mA	-0.987 mA

I_ee + I_cc = -0.1 mA
P_cc = 10.644 mW
P_ee = 11.844 mW


Part bonus

A useful purpose for an operational amplifier is its ability to change the amplitude of a signal by adjusting the values of the input and feedback resistors. We can take advantage of this fact that the gain is represented by the ratio of input and feedback resistors by making one of those resistors adjustable, instead of a fixed value.



The results are as shown below,

V_in (Desired)	V_out (Measured)	V_RF (Measured)	I_op (Calculated)	I_cc (Measured)	I_ee (Measured)
1.0 V	-5.03 V	4.99 V	-0.101 mA	0.885 mA	-0.985 mA

I_ee + I_cc = -0.1 mA

Conclusion 
We see that the current goes unchanged when adding a resistor to the output of the operational amplifier, since changes to the output of the operational amplifier have no affect to the input side, we see the many benefits of an operational amplifier. The use of this can be seen most obviously in the buffer operational amplifier.