LAB 6_ RC CIRCUITS; PASSIVE FILTERS

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Jan 9, 2024

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LAB 6: RC CIRCUITS; PASSIVE FILTERS Lab Course: ECE291 Written By: Anna Chang, Abdallah Bereka, Shricharan Kulavanikerpuram Subramaniam Date of Submission: 10/20/2022 Instructor: Dr. Hongya Ge

1 Ⅰ . Introduction Capacitance and resistance are present in any electrical circuit, and to find out the characteristics of these components, it is important to understand how they respond to low pass and high pass filters. The objective of this lab is to measure and analyze the time response of an RC circuit to a step voltage. This will be done through observing the frequency response graphs of the amplitude and phase of the RC circuits. Ⅱ . Pre-Lab 1. What is the response of a simple low-pass RC filter (fig. 8 a) to a square-wave input? Sketch Vout assuming that the square wave period is larger than the circuit time constant. How will you measure the time constant (τ = RC) of this circuit using an oscilloscope? HINT: Vout is a voltage across a capacitor C which is charged and discharged from a source Vin through a resistor R. The scope gives you this voltage as a function of time. a. The oscilloscope has the convenient feature of calculating the time constant value, so we will be using that measurement to measure the time constant of the oscilloscope. [Figure 1] Graphs of response of RC circuit with time constant τ=RC 2. Draw approximate frequency responses for circuits (a) and (b). Indicate the filters characteristic frequency, called also the break frequency or the corner frequency. At that frequency the power drops to ½ of the power in the filter pass-band. Since the output voltage drops then by 3 dB, the frequency is also designated as f -3db. How does this frequency depend on R and C? a. The designated frequency at -3dB is the cutoff frequency. The frequency depends on R and C due to their inverse relation shown in the equation: . 𝑓 = 1 2Π𝑅𝐶

2 [Figure 2] Graphs of frequency gain and frequency response of RC circuit with high pass and low pass filter Ⅲ . Experimental Procedure: Equipment List: ● AC wave generator ● Multimeter ● oscilloscope ● Wires ● 10K Ohm resistor ● Breadboard ● Capacitor ● Inductor ● Multisim [Figure 3] Image of Circuit Schematic of a Low Pass Filter

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3 [Figure 3] Image of Circuit Schematic of a High Pass Filter LOW PASS FILTER (INTEGRATOR) 1. Using one capacitor and one resistor from your parts kit assemble the low pass filter schematic with its pass frequency (f -3db ) of a few kHz. Note the values of the selected components. 2. With a square wave at the filter input observe the output waveform. Using an oscilloscope determine the time constant and compare it to the value calculated from the measured resistance and capacitance used in the circuit. Check if the time constants on the rising and the falling parts of the waveform are the same. Make sure that you apply a square wave of low enough frequency so that the output waveform becomes flat before the input waveform changes polarity. a. HINT: To measure the time constant t, use the cursors on the digital scope screen. The scope may also show the signal "rise time" tr i.e. the time it takes for a signal to increase from 10% to 90% of its maximum value. ( tr ≅ τ ⋅ ln9). 3. Demonstrate that for a certain frequency range the circuit behaves as an integrator i.e. the output is proportional to the integral of the input signal. 4. Drive the circuit with a sine wave and measure its frequency response by recording the input and output voltages for different frequencies. Keep the generator voltage constant. Find the frequency f -3db at the -3 dB gain. Your data should include several points above f -3db , if possible within two decades above that frequency. Make preliminary graphs of the amplitude as a function of frequency in the laboratory; you will have a chance to repeat measurements that do not fit the expected distributions. 5. Measure phase shifts between input and output signals at 0.1f --3db , f --3db , and f = 10f --3db .. Measurement of the phase shift is an accurate method of determining the filter characteristic frequency (f --3db ). Try. HIGH PASS FILTER (INTEGRATOR) 1. Exchange positions of the capacitor and the resistor in your circuit. Show that the circuit acts as a high-pass filter. You do not need to make as many measurements as in the case low pass filter. Now that you know what to look for, obtain data at a few selected frequencies. Find the frequency f -3db .

4 2. Next, apply a triangular wave to the input. Observe the output while varying the generator frequency. Does the filter behave as a differentiator? If so, in what frequency range? CIRCUIT (MULTISIM) SIMULATION 1. Simulate measurements 1 a), 1 b), and 2: frequency responses of amplitude and phase, and the time dependence of current with a square wave input for low pass and high pass filters. Use the same values of R and C in simulations as in the laboratory. Compare results of simulations with measurements. IV. Results Capacitance Value Resistor (Ω) τ=RC 1 nF 10k 10*10^-6 [Figure 4] Table of value of components and time constant Capacitance Value Resistor (Ω) τ=RC 1 nF 10k 19*10^-6 [Figure 5] Table of value of measured components and time constant for square wave Frequency Range [Figure 6] Table of frequency range *rise and fall time are the same Having calculated that , we were now able to approach solving this using a guess ? 𝐼???? ? ?????? = 0. 86 and check method, such as the following: Input Voltage Output Voltage ? 𝐼???? ? ?????? Diff. to Actual Frequency 20 x log (V2 / V1) 1.02 1.04 0.980769 -0.12077 1K 0.99 1.04 0.951923 -0.09192 4K

5 0.91 1.04 0.875 -0.015 8K 0.894 1.04 0.859615 0.000385 8.376K 0.89 1.04 0.855769 0.004321 9K 0.85 1.04 0.817308 0.042692 10K 0.93 1.04 0.894231 -0.03423 15K [Figure 7] Table of Guess+Check Method, reformatted from Excel Calculated frequency Measured Frequency Phase Shift (degrees) (1/τ) / 2π = 8376.575 8.376k -26.05 V s 1.09 V - [Figure 8] Table of measured amplitude and phase shift for 1k Hz Frequency Phase shift 0.8k -2.44 8k -26.05 80k -73.76 [Figure 9] Table of frequency and phase shifts Ⅴ . Questions and Discussion In the beginning of the experiment we have calculated the time constant . We τ = 𝑅𝐶 have found that the time constant is 10 μs. Our experimental and theoretical values are quite similar, consistent with the way the sine wave behaves over time. The low pass filter does not dramatically alter the behavior of the input signal, showing rather similar waves. This explains the consistency in the time constants for both the rising and falling time.

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6 When taking a look at the frequency gain as frequency increases for the low pass filter, we see that the behavior looks similar to a line, with the frequency gain decreasing as the frequency of the input is increased. Using basic math to predict the slope of the frequency gain, we get a measurement of 19dB/decade (in reality, this value is negative due to the decreasing slope). If you divide this value by 3.32 (log 10 / log 2) you will get 5.7 dB/octave.This is consistent with the theoretical attenuation values of 20dB/decade and 6dB/octave. Measuring the slope from the 0.8kHz point to the 80kHz point, we get a value of 7dB/octave. Multiplying these values with (log 10 / log 2) gets the value of 23.34dB/decade. Again, these values are consistent with the theoretical values provided within this lab project itself. Ⅵ . Conclusion In conclusion, We have calculated the time constant within the high pass and low pass outputs to see its consistency with the formula and shown the decibel consistency between the high pass and the low pass filter. We have driven the different frequency from both circuits and the two have slightly different behavior outputs when given the same frequency and input. With the low pass filter, it “integrates” the wave input and essentially “reverses” its behavior. With the high pass filter, we saw how it differentiated the triangular wave and turned it into a square wave, consistent with the behavior of a triangular wave when it is tampered with.

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