Christian_Andrade(lab4EE221)

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EE 22100

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Electrical Engineering

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Jan 9, 2024

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The City College of New York Grove School of Engineering EE 22100 Electrical Engineering Laboratory I Spring 2022 Lab Report Experiment #4 Superposition Student Name: Christian Andrade Instructor: Saurabh Sachdeva 04 /01 /2022 Objective:

The purpose of this experiment is to test whether or not superposition holds true in circuits containing various circuit elements. Superposition is when a circuit can be broken into multiple circuits with one source. Since theoretically superposition holds true for all linear elements, the circuits will be constructed and measured. The circuits will include simple resistor circuits, and circuits with diodes. Introduction: In this lab, we tested the superposition law by analyzing circuits which were built using resistors, capacitors, and diodes. We compared output voltage that all sources contributed together to the sum of the output voltage that each source contributed alone. In the end, by analyzing our results, we concluded if the law superposition was obeyed for our circuit. Apparatus Used: The DC Power Supply, The Function Generator, The Digital Multimeter, The Digital Oscilloscope, and the Protoboard, Resistors: 1kΩ, 22kΩ, 10kΩ, and 0.1 �� F Capacitor Software used: LabView, Multisim, and MATLAB Lab Procedures and Measurements: I) Superposition with Resistors: In this part of the lab, the circuit shown in figure 1 was constructed on the Protoboard with two input voltage V 1 set to 4V and V 2 set to 8V. Then the V out was measured using the Digital Multimeter with both input sources on. Measured Calculation with both V 1 V �� 2 :

V out = −0.7060 V V1 R 1 4 3 22kΩ R 3 R 2 22kΩ 4V 10kΩ R 4 5 0 10kΩ Figure 1a: Circuit 1 by shorting the 8V input voltage After this, the 8V input voltage was short by replacing the source with a wire and the circuit would look like as shown in previous figures. Then the output voltage V out1 was measured using the Digital Multimeter with only the 4V as the input voltage. Measured Calculation with V 1 : V out1 = 0.7056 V R 1 3 22kΩ R 2 22kΩ 2 4 R 4 10kΩ R 3 10kΩ 0 V2 8V Figure 1b: Circuit 1 by shorting the 4V input voltage Finally, I removed the 4V source and placed the 8V source in the circuit. Then I measured the output voltage V 2 with the 8V input source again using the Digital Multimeter. Measured Calculation with V 2 : V out2 = −1.4124 V Here we can clearly see the superposition principle holds for this circuit since: V out = V out1 + V out2 = 0.7056 V + (−1.4124 V) =

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−0.7068 �� II) Superposition with Capacitors: Color Code Value Measurement %Error % ( ) �� ≡ 100 ∗ �� − �� 1 = 10 kΩ 9.9969 kΩ 0.03 Table 2: Resistance measurements of Circuit 2 C1 1 2 3 R1 V1 4V 10kΩ 0 0.1µF V2 0.5Vpk 10kHz 0° Figure 2: Circuit 2 with capacitors In the second part of the lab, the circuit shown in figure 2 was constructed on the Protoboard. On the Oscilloscope, the DC supply was set to 4V, frequency to 10 KHz and the amplitude to be 1 V pp . Then the output voltage V out was measured using the Oscilloscope and the waveform was saved using the VI “Export” option provided in the WaveForm software. Again, the Oscilloscope trace was saved as a text file using the VI “Export” to be later plotted in MATLAB. Using the DMM, we measured the output voltage. Measured Calculation with both V 1 V �� 2 : V out = 3.99933 V

Questions: 1. The signal at �� should be of the form: �� + ( ) �� . By looking at the oscilloscope trace on the screen, determine the value of �� . Ans: From Figure 7 in the pot-lab section, we can observe that V out takes the form of a sine signal with an amplitude of 1.9V that is shifted up by V DC . Since the volts per division is 2 and our signal starts from the second box, then this must mean that V DC is 4V. Therefore, by looking at the oscilloscope A=1.92 V and V DC =4V. R 1 1 2 10kΩ V1 4V 0 C 1 0.1µF Figure 2a: Circuit 2 by shorting the AC Voltage Next, the AC Voltage V 2 was taken out and the only V 1 was set to 4V. Then by the use of the Oscilloscope, the image and data of the output voltage V out1 was saved. Using the DMM, we measured the output voltage. Measured Calculation with V 1 : V out1 = 3.9997 V R 1 10kΩ 2 V1 0.3535Vrms 10000Hz 0° 0 C 1 0.1µF 3

Figure 2b: Circuit 2 by shorting the DC Voltage Finally, the DC Voltage V 1 was taken out and the only V 2 was set with Freq=10kHz and Amplitude = 1 V pp . Then using the Oscilloscope, the image and data of the output voltage V out2 was saved. Using the DMM, we measured the output voltage. Measured Calculation with V 2 : V out2 = 0.00013 V Here we can clearly see the superposition principle holds for this circuit since: V out = V out1 + V out2 = 3.9997 V + (0.00013 V) = 3.99983 �� Question 2: If you were to sum the two oscilloscope images that you obtained in part 4 would y ou get the same trace that you got in part 3? Does superposition hold when a circuit has both capacitors and resistors? Ans: If we summed Figure 7 with Figure 8 from the post-lab section, we would get the graph shown on Figure 9. Every point of our sine signal on Figure 9 would be shifted up by the value of the signal shown on Figure 8 which is 4V. Since the AC signal is a sine wave with a peak of 1.92 V, then our next shifted peak would be 4. Therefore, superposition does hold for a circuit with both capacitors and resistors. III) Superposition with Diodes: 1 3 R1 1kΩ 2 R2 1kΩ V2 4V 0 D1 1N4001G V1 4V

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Figure 3: Circuit 3 with diode In this part of the lab, the circuit shown in the figure 3 was constructed on the Protoboard with the diode in a forward biased position. We were asked to consider the voltage source polarities. Forward Biased: 2 0 R1 1kΩ R2 1kΩ D1 1N4001G 3 V1 4V Figure 3a: Circuit 3 with Diode by shorting V 2 ( ) �� Using the DMM, we measured the output voltage. Measured Calculation with V 1 : V out1 = 0.63827 V 1 2 V2 4V 0 R1 1kΩ R2 1kΩ D1 1N4001G Figure 3b: Circuit 3 with Diode by shorting V 1 ( �� ) �� Using the DMM, we measured the output voltage. Measured Calculation with V 2 : V out2 = 0.63778 V

As shown in figure 3a and 3b, we have, two resistors ( �� 1 = �� 2 = 1 kΩ ), two DC source ( �� 1 = �� 2 = 4V). The output voltage is taken across the diodes. Output voltage with both source on = V out = 4V Output voltage with V 2 turned off and V 1 on = V out1 = 0.63782V Output voltage with V 1 turned off and V 2 on = V out2 = 0.63778V To hold the superposition principle, the following condition must be satisfied: V out = V out1 + V out2 = (0.63782 + 0.63778) = 1.2756 �� But V out = 4V Thus, this proves that the principle of superposition does not hold for the circuits containing diodes (forward biased). R everse Biased: R1 1kΩ 2 V1 R2 1kΩ D1 1 1N4001G 0 4V Figure 3c: Circuit 3 with Diode by shorting

V 2 ( ) �� Using the DMM, we measured the output voltage. Measured Calculation with V 1 : V out1 = 3.9997 V 1 V2 4V R1 1kΩ 2 R2 1kΩ D1 1N4001G 0 Figure 3d: Circuit 3 with Diode by shorting V 1 ( �� ) �� Using the DMM, we measured the output voltage. Measured Calculation with V 1 : V out2 = 2.01 V As shown in figure 3c and 3d, we have, two resistors ( �� 1 = �� 2 = 1 kΩ ), two DC source ( �� 1 = �� 2 = 4V). The output voltage is taken across the diodes. Output voltage with both source on = V out = 4V Output voltage with V 2 turned off and V 1 on = V out1 = 1.9988V Output voltage with V 1 turned off and V 2 on = V out2 = 2.01V To hold the superposition principle, the following condition must be satisfied: V out = V out1 + V out2 = (1.9988 + 2.01) = 4.0088 ��

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But V out = 4V Thus, this proves that the principle of superposition does hold for the circuits containing diodes (reverse biased). Question 3: Determine if superposition hold for the diode in the reversed direction? Ans: No, it may seem as if it does, but a reversed diode allows no current to pass through and makes the voltages across the diode 0V. When you sum the voltages from each source you also will get 0V, but this is as if you were measuring nothing. Superposition will still not work on a diode regardless of orientation. Conclusion: Throughout the lab, the elements used have different properties, and the concept of superposition applies to those circuits. Basically, this superposition principle is not limited to circuit analysis, but also applicable in many fields where cause and effect allow a linear relationship to one another, this is also known as linearity property. Since superposition rests in a linearity property, a diode could not be analyzed with this principle, and this is because the diode is the most fundamental nonlinear element. This was seen in this experiment, where the (ideal) diode acts as a short circuit in the forward bias and as an open circuit in the reversed bias. As for the errors of the theoretical and measured values, the reason for could be human error. Post Lab: 1. In your lab report redraw the circuits 1, 2, and 3. Ans: V1 R 1 4 3 22kΩ R 3 R 2 22kΩ

4V 10kΩ R 4 5 0 10kΩ Figure 4: Circuit 1(redrawn) R 1 1 2 3 C 1 10kΩ V1 4V 0 0.1µF V2 0.3535Vrms 10000Hz 0° Figure 5: Circuit 2(redrawn) 1 2 V2 4V 0 R1 1kΩ R2 1kΩ D1 1N4001G Figure 6: Circuit 3(redrawn) 2. Include all your Oscilloscope waveforms in your report Ans: 3. For sections 3 and 4 (superposition with capacitors) use MATLAB to add the waveforms

in part 4 to show whether or not V out = V out1 + V out2 . Ans: MATLAB Code with both AC and DC voltage on: Vacdc=load( 'acdc.txt' ); Vac=load( 'aconly.txt' ); Vdc=load( 'dconly.txt' ); Vt1=Vacdc(:,1); Vout=Vacdc(:,2); plot(Vt1,Vacdc, 'r' ); xlabel( 'time' ) ylabel( 'Voltage ' ) grid on ; legend( 'voltage' );

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Figure 10: MATLAB Graph of both AC and DC Voltage Source on We see a sinusoid on the line y=4. The amplitude is 1V. Case 1: V 1 = 4 V and V 2 = 0 V In the first case we turn off the AC source and keep the DC source on. MATLAB Code for DC Source only: Vdc=load( 'dconly.txt' ); Vt3=Vdc(:,1); Vout_dc=Vdc(:,2); plot(Vt3,Vout_dc, 'b' ); xlabel( 'time' ) ylabel( 'Voltage ' ) grid on ; legend( 'voltage' );

Figure 11: MATLAB Graph of only DC Voltage Source on We can see on MATLAB also that the line is very close to 4. We can verify that the graph is of a straight line (DC source). Case 2: V 1 = 0 V and V 2 = 1 �� In the first case we turn off the DC source and keep the AC source on. MATLAB Code for AC Source only: Vac=load( 'aconly.txt' ); Vt2=Vdc(:,1); Vout_ac=Vac(:,2); plot(Vt2,Vout_ac, 'b' ); xlabel( 'time' ) ylabel( 'Voltage ' ) grid on ; legend( 'voltage' ); Figure 12: MATLAB Graph of only AC Voltage Source on We can see a sine wave on the x axis. Proof of superposition with capacitors: We plot the data from the circuit and also plot the data by adding both the separated circuits. MATLAB Code to proof the superposition principle: Vacdc=load( 'acdc.txt' ); Vac=load( 'aconly.txt' );

Vdc=load( 'dconly.txt' ); Vt1=Vacdc(:,1); Vt2=Vac(:,1); Vt3=Vdc(:,1); Vtc=(Vt1+Vt2)/2; Vout=Vacdc(:,2); Vout_dc=Vdc(:,2); Vout_ac=Vac(:,2); Vout_c=Vout_dc+Vout_ac; plot(Vt1,Vacdc, 'r' ,Vtc,Vout_c, 'b' ); xlabel( 'time' ) ylabel( 'Voltage ' ) grid on ; legend( 'voltage total' , 'voltage out' );

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Figure 13: MATLAB Graph of proving superposition of principle for Circuit 2 with capacitors Graphically we can see that V out = V out1 + V out2 . Thus, we can confirm that superposition holds in this case. 4. Using the data that you collected in the lab justify whether or not superposition holds for each circuit. If superposition does not hold speculate why. Ans: For circuit 1 and circuit 2, superposition principle holds because resistors and capacitors consist of linearity properties. For this reason, the input voltage and output voltage values are approximately similar. On the other hand, circuit 3 doesn’t hold superposition principle for having diode in the circuit. It is because reversed diode allows no current to pass through and makes the voltages across the diode 0V. When you sum the voltages from each source you also will get 0V, but this will be considered as measuring nothing. It concludes the fact that, superposition principle in circuit 3 will not work regardless of orientation of the diode.