Lab: Inverting Amplifier (April 04, 2017): Activities: Nonideal op amp, ideal op amp,

Date: April 04, 2017

Nonideal Operational Amplifiers, Ideal Operational Amplifier, and Inverting Voltage Amplifier

First activity: Nonideal Op Amp

The first activity completed in class was an introduction to the operational amplifier question. This question takes into consideration the design of an operation amplifier and asks to rearrange to it equivalent circuit.

Then, the question asks to be solved for current, i.

In the picture above, we wrote down the circuit with the op amp in the circuit and the equivalent circuit without. The left of the picture is mesh analysis. 

The picture above shows the results of the first activity. The current, i, resulted in equaling about 0.2 mA. KCL was used on both nodes 1 and 0 in order to determine the values of v1 and the relationship between v1 and v0. Then the v1 determined at node 1 is plugged into the results of node 0 to give an approximate relationship of vs/v0. That relationship was then used to determine the values of all the v’s.  Finally, we plug them all into our current equation to get our results. The conclusion for the first activity was that using this method to solve nonideal operational amplifiers is difficult because many different relationships that must be considered.

Second Activity: Ideal Op Amp

Since working with nonideal operational amplifiers is difficult and tedious, in the second activity we work with an ideal model for an operational amplifier instead.

In the above pictures, we tried to solve the ideal operational amplifier using methods for the nonideal operational amplifier. It did not work in the slightest and we were very confused.

It became clear that it was a lot easier to solve the problem by turning it into an ideal op amp question. The picture above shows the reworking of the circuit to include the op amplifier. From there, I solved using ideal op amplifier methods; this includes assuming i1=0, i2=0, and v1=v2. 

First major step was using voltage division to determine the relationship between v1 and v0. Then implementing all ideal op amp methods previously stated to get v0/vs=9. Next, nodal analysis at node 0 was used to determine the current, i0, to be about 0.65mA. I say about because using nonideal op amp methods would reveal the current to be just slightly higher. The conclusion for the second activity was that ideal op amplifier method is much easier than nonideal op amplifier methods.

Lab: Inverting Amplifier

Pre-lab:

            Since Vout=-(R2/R1)Vin, to achieve a gain of 2 the R2 was 4k-ohm and R1 was 2k-ohm resistors.

Actual resistance values of R1=2.2k-ohm resistor and R2=4.7k-ohm resistor.

The above picture shows the input and output of the circuit in the table along with a drawn version of the circuit.

The above picture shows the built circuit.

The above picture shows a graph of the input and output graphed on Microsoft Excel.

input voltage	Theoretical output	Actual output	percent error
-3	6	4.21	29.83
-2.5	5	4.21	15.8
-2	4	4.22	-5.5
-1.5	3	3.22	-7.33
-1	2	2.14	-7
-0.5	1	1.06	-6
0	0	0	0
0.5	-1	-1.06	-6
1	-2	-2.14	-7
1.5	-3	-3.23	-7.67
2	-4	-3.43	14.25
2.5	-5	-3.42	31.6
3	-6	-3.42	43

In conclusion:

            We expected to get a output voltage with 2 times gain with this op amplifier along with it being negative what we put in the input voltage. This is what we got even though the graph looks backwards because we implemented a negative input voltage but we got out a positive output. It is important to note that the higher the input the more saturation occurs. This can be observed at the edges of the graph. Mathematically, the results should be an inverted linear function; however, the saturation experienced at higher voltages is design limits of the op amp used.  Every op amp will experience a degree of saturation. There also appears to be a favored direction as there is a larger saturation for a positive input/ negative output.