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ENG 1450 Introduction to Electrical and computer Engineering On-line Lab 8 Resistor-Capacitor Filter, AM Receiver, Diode Rectifier Name: _______________________________ Student Number: _______________________________ Lab Section: _______________________________ Instructor: _______________________________ Date: _______________________________ Part A: Resistor-Capacitor Filter In this part of the lab you will use a capacitor to build a low pass filter. Figure 1 shows the circuit of a low pass filter. Build the circuit with the 160 pF capacitor and a 10 kΩ resistor. Figure 1. Low pass filter. Apply a 10 kHz sinusoidal signal from a function generator to your circuit (this is the voltage Vin) with an amplitude of 5 V and no DC offset. 1. In Table 1, record the value of Vout (amplitude) as you increase the frequency to 100 kHz. In Table 2, record the value of Vout as you increase the frequency from 120 kHz to 200 kHz. Table 1: Values of frequency and Vout Frequency (Hz) 10 kHz 100 kHz Vout (V) Table 2: Values of frequency and Vout Frequency (Hz) 120 kHz 200 kHz Vout (V) (Note: simulator automatically adjusts the voltage scale of the oscilloscope) 2. Plot Vout vs. frequency on the graph of Figure 2, using the results of Tables 1 and 2. (Note: you can use Excel or other software to make this plot. Make sure to keep the same units for axes) Figure 2. Vout (V) vs. frequency (kHz) The equation relating Vout to Vin for your low pass filter is: where f is the circuit frequency, and Analysis of this equation will show that when Vout = 0.7 x Vin , the magnitude of the two components in the denominator of the right hand side are equal. And so: 2πf x RC = 1 yielding, or 3. Let’s check the above concept (it also can be used to find a value of the unknown capacitor). Adjust the frequency so that the output voltage Vout = 0.7 x Vin and calculate C. C = ___________________ F Take a screenshot of the constructed circuit (with Vout = 0.7 x Vin signal shown) and insert it in the space below. In practice, a capacitor can be easily build using two metal plates and a piece of paper separating them. The capacitance of a capacitor constructed from two parallel metal plates can be found from: where ε = ε0 εr is the effective dielectric constant of the material between the metal plates, A is the overlap area of the metal plates, and d is the distance separating the plates. 4. Use the above equation to find the thickness of the piece of paper separating the two metal plates of the capacitor if it has a capacitance of 1 nF. - For the paper separating your metal plates, assume ε = 9 pF/m. - Assume that square metal plates with a side of 10 cm were used. d = ___________________ m 5. In Lab 6, you used a 100 µF capacitor to store energy for a simple LED flashlight. Assuming that you used two square metal plates with a piece of paper separating them, how large would your capacitor be (i.e. its side length) in order to equal 100 µF? 6. Suggest how you could shrink the size of this capacitor down to the ~1 cm (a typical size of 100 µF capacitor). Part B: AM Receiver Circuit You will build and investigate an AM demodulation circuit that can produce an audio signal. Assume that the AM signal is transmitted on a 200 kHz carrier signal, and an audio signal of 1 kHz is amplitude modulating the carrier signal. You will use a function generator for these signals. Rectifier and low pass filter circuit: For this circuit you will use a regular (silicon) diode. Figure 3. Schematic of a simple AM demodulator circuit. Start by building the circuit of Figure 4. Use R1 = 10,000 Ω. Figure 4. Diode rectifier circuit. Since the TinkerCAD simulator can not produce a true AM signal, we will use the carrier (200 kHz) and the modulation (1 kHz audio) signals separately for our virtual experiments. Set both signals to have 5 V amplitude with no DC offset. Connect the output of the function generator to Vin. Using the oscilloscope observe both Vin and V1 at the same time. Note: The time scales of the oscilloscope channels should be set to 2 µs for carrier and 1 ms for audio signals. Observe the carrier and audio signals one at a time. Vin should look something like Figure 5 (in our case it is just a full sine wave). For a real AM signal the amplitude of the carrier wave is changing rapidly in accordance with the audio signal (e.g. music). Figure 5. AM modulated signal. On the other hand, V1 should look something like Figure 6. Since the diode lets only positive signals pass, V1 is only the positive half of the signal Vin. Because the negative half of the sine wave was cut, it is said that the diode “rectified” the signal. Figure 6. Rectified AM modulated signal. Note: Verify that V1 is only the positive half of the signal Vin before you proceed any further. Since we should experiment with carrier and audio signals separately, verify that both of them are rectified (i.e. transmitted) properly. Take the screenshots of the rectified signals and put them side by side below. 7. Measure the amplitude of the positive part of Vin and compare it to V1 (use either signal). Vin = ______________ V V1 = ______________ V Now you will build the following low pass filter to remove the carrier wave, leaving only the envelope of the AM signal (i.e. the audio signal), as shown in Figure 7. Use R1 = 10 kΩ, and C1 = 1 nF. You will use only the 200 kHz carrier signal for this part. Figure 7. Diode rectifier with low pass filter. 8. Use the following equation to calculate the cutoff frequency of the fcuttoff for this low pass filter. fcuttoff = ___________________ Hz 9. How many times larger is fcarrier compared to fcuttoff ? = ___________________ 10. Observe both Vin and V1 at the same time on the oscilloscope. You should see that V1 now looks to be almost a constant DC voltage near the peak value of Vin, but you can still see small ripples in V1 where the peaks of Vin occur. Now observe both Vin and V2 at the same time on the oscilloscopes. You should see that V2 now looks to be almost a constant DC voltage near the peak value of Vin. This is because you have filtered the carrier wave from V2, leaving only the envelope signal. Since the amplitude of our carrier wave in the simulator was not modulated, i.e. it was constant, you should get a constant DC voltage after the low pass filter. Take the screenshots of the V1 and V2 signals and put them side by side below. 11. Measure the amplitude of the positive part of Vin and compare it to V2. Vin = ______________ V V2 = ______________ V A picture of the real filtered AM signal is shown in Figure 8. In this case V2 is not a constant voltage since it is an actual (varying) audio signal. Figure 8. Vin and V2 viewed at a slow enough time setting to see the audio modulation of the carrier frequency. V2 is the transmitted (and filtered) audio signal. Part C: Diode Rectifier Circuits In question 10 you saw that an AC signal (carrier wave) was converted to a DC voltage. This process is called rectification and is often used to convert the sinusoidal voltages into the constant ones. Rectifiers are at the heart of our smartphone, laptop, tablet chargers as well as DC power supplies, i.e. everywhere where a constant DC voltage is needed. The circuit of Figure 4 is called a half-wave rectifier because it removed the negative half of the sine signal, i.e. the diode passed the current only during the positive half of the sine wave. At the same time the resulting signal was not a DC-like voltage and changed from zero to a maximum value, i.e. it was pulsed. The output can be improved by adding a capacitor in parallel to the resistor R1. In this case the capacitor will smooth the output because of its charging during the voltage rise stage and discharging afterwards. The charging/discharging of a capacitor typically creates a voltage ripple (the difference between the max and min values of the rectified voltage) shown in Figure 9 that is not always acceptable/desirable when a truly constant voltage is needed. Figure 9. Output signal of a half-wave rectifier with a capacitor. Build the circuit of Figure 4. Use R1 = 10,000 Ω. Add a capacitor C1 = 0.5 µF in parallel to R1. Apply 60 Hz signal (i.e. wall plug frequency) with 4 V of amplitude and no offset. 12. Measure V1 using the oscilloscope with 10 ms time scale. Provide below the oscilloscope screenshots of the V1 for the cases of C1 = 0.5 / 1 / 5 / 10 µF. As can be seen, the value of the capacitor plays an important role in modifying the fluctuations of the output signal. This happens because the capacitance directly affects the RC time constant (i.e. charging/discharging speed of a capacitor) and the amount of stored energy. Measure the voltage ripple (approximately) and calculate the RC time constant for the cases above: Vr_0.5µF = __________ R1C1_0.5µF = __________ Vr_1µF = __________ R1C1_1µF = __________ Vr_5µF = __________ R1C1_5µF = __________ Vr_10µF = __________ R1C1_10µF = __________ 13. Based on the above results, what are the two strategies you can suggest to reduce the ripple voltage Vr? 14. What is the frequency of the ripple voltage? fripple = _____________ The other reason for voltage ripple is that only positive cycles of the wave are used to produce the output signal. The circuit that allows us to use both positive and negative cycles of a sine wave is called a full-wave rectifier. It uses four diodes and is shown in Figure 10 below. This circuit “inverts” the negative signal cycles thus making them positive at the output. This helps to reduce the ripple voltage. Figure 10. Full-wave rectifier with a smoothing capacitor and output waveform. Let’s investigate operation of the full-wave rectifier. Initially build the circuit of Figure 10 without the smoothing capacitor. Apply 60 Hz signal with 4 V of amplitude and no offset. 15. Measure the output voltage using the oscilloscope with 10 ms time scale. Provide below the oscilloscope screenshot of the output signal. Note: Verify that the output has both halves of the input signal Vin (and all are positive) before you proceed any further. 16. Indicate the states of the diodes (ONN/OFF) during the positive and negative cycles of the input signal. (Note: during the positive signal cycle the node a is positively biased and the node b is negatively biased. Diode operation regimes were covered in Lab 4). Positive cycle Negative cycle D1 = __________ D1 = __________ D2 = __________ D2 = __________ D3 = __________ D3 = __________ D4 = __________ D4 = __________ Now add the smoothing capacitor C to the circuit (as in Figure 10). Important note: do not use a polarized capacitor for this circuit. 17. Measure the voltage ripple (approximately) for the following cases (C = 0.05 / 0.1 / 0.5/ 0.9nF): Vr_0.05nF = __________ Vr_0.1nF = __________ Vr_0.5nF = __________ Vr_0.9nF = __________ Take the screenshots of the rectified signals for C = 0.05 and 0.9 nF and insert them side by side below. Take a screenshot of the operating circuit (for C = 0.5 nF) and insert it in the space below. 18. What is the frequency of the ripple voltage? fripple = _____________ How can you explain this result when compared with the ripple frequency of question 14? ENG 1450 - Lab 8 Summer 2020-v1 Page 14 V o u t V i n = 1 1 + j 2 p f R C j = - 1 C = 1 2 p f R f c u t o f f = 1 2 p R C C = e A d f c u t o f f = 1 2 p R 1 C 1 f c a r r i e r f c u t o f f

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