Thursday, April 27, 2017

Lab:Summing and difference Amplifiers(April 6, 2017)

Date: April 6, 2017
Non inverting op Amp; Summing Amplifier; Difference Amplifier
Activity 1: Inverting op amp graphs
            The first activity done today involved revisiting inverting op amps to determine practice our knowledge of the op amp.


The above picture shows the equivalent circuit of a basic inverting op amplifier. Since  we assume the op amp is ideal we can assume infinite open loop gain, A, infinite resistance at Ri, and zero resistance at, Ro. Therefore, we only look at R1 and R2. Since the R2 is connected to the Vo section of the op amp it will act as the feedback resistor. The R1 will be the initial resistor. We follow inverting op amp equation to obtain a gain of -3. The graphs display input voltage on top and output voltage on the bottom.

Activity 2: Noninverting op amp example
            The second activity is an example of how to solve a noninverting op amp circuit.


            The above picture shows the circuit we are working with along with the incorrect approach we took to solving it on the left of the picture. The correct method of solving it included doing nodal analysis at Va giving us the first equation. The second thing was realizing that Vs is equal to Vb which is equal to zero for this example. Then doing nodal analysis at Vb giving the second equation. We simplified both equations and got the Vo to be -1.64. An important observation of this example is that because Vs was equal to zero the circuit acted like an inverting amplifier. If Vs had any voltage the results would not have been a negative number rather a positive Vo.
Lab 1: Summing Amplifier
Pre-lab:
            In order to get this summing circuit to preform the addition of the two incoming voltages, Va and Vb, the three resistance must all be equal to each other, R1=R2=R3.     


The picture above is the summing circuit we set up using all three resistors that are equal to each other.


In the above picture is the results of the changing the voltage at Va and keeping Vb constant. Since all three resistors equal each other that means that this circuit follows the equation, Vo=-(Va+Vb).
Va
Vb
Vo theory
Vo exper
percent error
-4
1
3
2.96
1.33
-2
1
1
0.98
2.00
-1
1
0
0
0.00
0
1
-1
-0.97
3.00
1
1
-2
-1.96
2.00
2
1
-3
-2.05
31.67
3
1
-4
-3.43
14.25
5
1
-6
-3.43
42.83
The table above shows that for the most part the circuit acts the way we want predicted except at one point where the data is skewed, Va=3. However, the rest makes sense because we have observed that op amps saturate at high voltages and the saturation is higher when the Vo is negative.
Lab 2: Difference Amplifier
Pre-lab: First we determined the relationship between Va and Vb which is that they equal each other because they are both connected to the op amp from the positive and negative terminals and we assume ideal op amp; therefore, the equal each other and they both equal zero. But both of them equally zero does not go into deriving the difference amp equation.


In the Picture above we derived the relation ship between Vo and V1 and V2. The circuit is called a difference amplifier because if the all the resistances equal the same value then relationship becomes: Vo=(V2-V1). Resistances are on the bottom left of the board. We let R1=R3 and R2=R4. Furthermore, the results of when Vb=1 are on the bottom right and is the left of the two sets of data. The one on the very bottom right is when we changed the Vb to -1V.
V1
V2
Vo theory
Vo experiment
percent error
-4
1
10
4.27
57.3
-2
1
6
3.28
45.3
-1
1
4
3.27
18.3
0
1
2
2.01
-0.5
1
1
0
0
0.0
3
1
-4
-3.97
0.7
5
1
-8
-4.54
43.3
The table above( is for the results of when Vb=1 or V2=1) shows that high saturation occurs at higher voltages but this is to be expected. However, within smaller ranges of it does what we expect; it follows the equation, Vo=R2/R1(V2-V1) or Vo=2(V2-V1).

The graph above shows the relationship between the input and outputs of this circuit.


The above picture is our difference circuit.
In conclusion:

            We went over inverting op amps again. We continued our study of op amps by covering: non-inverting, summer, and difference op amps. In the experiments, we tested out the summing op amps and were able to determine That if all three resistors are the same the input voltages are added up then made negative. The Difference op amp circuit showed us that if the resistors R2/R1=R4/R3 we get a gain of the ratio of the two different resistors we also determined that if all the resistors were the same the input voltages would subtract which is why this is called a difference op amp. In both experiments we saw that at higher voltages there is a large saturation that occurs. There also appears to be more saturation when the output is negative.

Wednesday, April 26, 2017

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.