EE242 Lab 7 - lab report PDF

Title EE242 Lab 7 - lab report
Author Willaim Winter Ng
Course Electric Circuit Analysis Laboratory III
Institution California Polytechnic State University San Luis Obispo
Pages 7
File Size 231.3 KB
File Type PDF
Total Downloads 44
Total Views 165

Summary

lab report...


Description

EE 242 Electric Circuit Analysis Laboratory III

Winter 2018

Instructor – John Saghri

Experiment 7: Parallel Resonance

Due Date: 3/2/18

Bench #12

Alex Painter and James Kim

Objective: We hope to create and observe the characteristics of a circuit with parallel resonance.

Afterwards, we want to confirm that our theoretical formulas and circuit theory apply to these circuits as well.

Parts and Equipment: ● ● ● ● ● ● ● ● ●

1 Agilent 33120A Function Generator (FG) 1 Agilent 34410A Digital Multimeter 1 Agilent 54621A Oscilloscope 1 Impedance Bridge (1 per lab) 2 Resistance Decade boxes, 100 Ω/step 1 5 mH (nominal) standard or high-Q inductor 1 Bag of short leads 5 Banana-Banana leads 3 BNC-Banana leads

Procedure Summary: 1. Find the values of Cs and the dissipation factor D using the impedance bridge and Rs using given formulas. 2. Convert from series-model to parallel-model using given formulas. 3. Find the values of Lp and Q using the impedance bridge and Rp using given formulas. 4. Recreate the circuit shown in Figure 1.

5. Calculate wo, B, and Qo when changing R1 to infinity, 100, and 5 ohms. Find the values of L and C using the impedance bridge. 6. Change the frequency to 600-1200 and measure Zin.

7. Create a table showing frequency, magnitude of Zin, and phase of Zin. 8. Create graphs showing the magnitude of Zin vs frequency and the phase of Zin vs frequency.

Experimental Data and Sample Calculations: Capacitor Calculations: Cs = 4.944 uF D = 0.0085, Q = 1/D = 117.647 Rs = D/wCs = 117.647/(6283.2*4.944*10^-6) = 3787 ohms Cp = 4.944/(1 + 0.0085^2) = 4.9436 uF Inductor Calculations: Lp = 4.920 mH Q = 86.92 Rp = (86.92)(6283.2)(4.92*10^-3) = 2686.98 ohms

R1 (in ohms)

Wo (rad/s)

B (rad/s)

Qo



6411.77

0



500

6411.77

606.8

10.567

100

6411.77

90.91

70.529

R=100 ohms Frequency (hz)

|V1|

|Zin|

Phase of Zin

600

272 mV

108.8

N/A

700

335 mV

134.0

N/A

800

406 mV

162.4

N/A

900

491 mV

196.4

N/A

1000

549 mV

219.6

N/A

1100

519 mV

207.6

N/A

1200

458 mV

183.2

N/A

Frequency (hz)

|V1|

|Zin|

Phase of Zin

600

298 mV

119.2

N/A

700

381 mV

152.4

N/A

800

483 mV

193.2

N/A

900

640 mV

256.0

N/A

1000

1.29 V

516.0

N/A

1100

778 mV

311.2

N/A

1200

568 mV

227.2

N/A

Frequency (hz)

|V1|

|Zin|

Phase of Zin

600

304 mV

121.6

N/A

700

393 mV

157.2

N/A

800

541 mV

216.4

N/A

900

649 mV

259.6

N/A

1000

N/A

N/A

N/A

1100

N/A

N/A

N/A

1200

N/A

N/A

N/A

R=500 ohms

R=infinity ohms

LEGEND Series 1 (bottom line) : R1=100ohms Series 2 (middle line): R1=500ohms Series 3 (top line, ends early): R1=infinity ohms

Discussion and Answers to Questions: 1. If you were to plot ⎪V1⎪ versus the frequency for a given value of R1, how would it compare with the corresponding shape of your plot of ⎪Zin⎪ versus the frequency? Explain. If we were to plot the magnitude of V1 versus frequency for R1, it would have an identical shape to the plot of the magnitude of Zin with respect to frequency. This is because the value of V1 and Zin are perfectly proportional, that is Zin = V1 *400. We got this factor of 400 by using ohms law. Zin = V1/I and we can find I by using ohms law at the 200 ohm resistor: I = 0.5V/200 ohms = 2.5mA. Thus, Zin = Vin * 0.0025^-1 = Vin * 400. Given this constant proportionality, the two graphs would have the exact same shape. 2. For each value of R1, obtain the resonance frequency from your graphs of ⎪Zin⎪ versus the frequency and compare it with its expected value as determined in step 6. Explain the difference. We solved for w0 and got 6411.77 rad/s. Dividing this value by 2pi gives a resonance frequency of 1020 Hz. As shown in the graph above, our resonance frequency occurs right around 1000 Hz, showing that our data matched our hypothesis. Small discrepancies in the data can be accounted to a small experimental margin of error due to equipment error, and human error in assembling the circuit and measuring data points.

3. From your graph of ∠Zin .versus the frequency, determine ∠Zin at resonance for each value of R1. Compare these values with the theoretical values and explain the difference. N/A - no phase of Zin

4. At resonance, Zin is purely resistive and its magnitude should equal the true value of R in the parallel RLC circuit (with pure components). Your values of ⎪Zin⎪at resonance should be close to Req⎪⎪R1. For each value of R1, determine the percent difference between your measured ⎪Zin⎪at resonance and calculated Req⎪⎪ R1. For R1 = 100 ohms: True R = 100||1571.77 = 94.02ohms Observed Zin at resonance = 219.6 ohms Percent Error = 133.6% For R1 = 500 ohms: True R = 500||1571.77 = 379.33 ohms Observed Zin at resonance = 516 ohms Percent Error = 36.03% For R1 = infinity ohms True R = infinity ohms Observed Zin at resonance = ? We could not find the resonance frequency for this case as no amount of voltage would maintain a steady drop across R2. If we had more time with this lab, we would have tried to obtain values for a smaller voltage drop across R2.

5. For each value of R1, measure the bandwidth B from the corresponding plot of Zin(ω). Compare the measured values of B with the theoretical values (B = 1 / RC) and explain the differences. Unfortunately, we only had one of our three curves provide an accurate band of data so I will perform calculations for R1 = 500 ohms. A rough estimate of f1 and f2 where the Zin value was at a factor of 0.707 of its max would be around 920 and 1100 ohms. This gives a band width of f=180 ohms and B=1131 rad/s. This is about double what we expected in our early calculations. Likely, if we took more data around our resonant frequency, we may have gotten a more accurate data value. 6. How would you describe the characteristic of your parallel RLC circuit as a filter (i.e., lowpass, high-pass, band-pass, or band-stop)? Our filter was a band pass filter as it allowed a central range of frequencies through, creating a band of admittance. We can see this band around f =1000 Hz, as there is a hump in the graph,

representing the admittance band.

Conclusions: James Kim: Although the formulas and concepts introduced in this lab were initially foreign, I am glad I took the time to understand what they meant and how they worked, especially the ones to transfer from series-model to parallel-model and vice versa. Furthermore, I was able to see how the frequency, magnitude of the input transfer impedance, and phase of the input transfer impedance changed as a result of a resistor changing in a circuit with parallel resonance. Alex Painter: This lab was difficult for me as I had little exposure to the concepts and formulas in the lecture. Our group spent most of our time dealing with equipment issues as the oscilloscope did not behave as planned as trying to measure both V1 and V2 at the same time was not possible. As a result, the data we did acquire was rushed, and we could not find phase differences between these voltages. I did learn about parallel resonance and learned that inductors and capacitors have parallel and series resistances....


Similar Free PDFs