Lab 3 – First and Second-Order Circuits | ECEN 3010, Lab Reports of Electrical and Electronics Engineering

Download Lab 3 – First and Second-Order Circuits | ECEN 3010 and more Lab Reports Electrical and Electronics Engineering in PDF only on Docsity! ECEN 3010 Lab Circuits and Electronics University of Colorado at Boulder Lab #3 First- and Second-Order Circuits Names of All Participating Lab Partners: Lab Section (Circle One): M T W Th F Learning Objectives This lab will introduce you to first- and second-order circuits. You will build high-pass and low-pass filters and analyze their effects on signal amplitude at different frequencies. You will also build RLC circuits with three different response types. Minimum Parts Required: Various resistors and capacitors; inductors or the laboratory “magic box.” Part 1 - Transient Response of an RC Circuit Resistor-capacitor (RC) circuits are first-order systems. Consider the circuit diagram shown in Figure 1. When the capacitor is charging, the voltage drop across the resistor (using KVL) is nonzero and is given by Vin  Vout iR . (1) The variable i is the current through the resistor. Since the output terminal has no load connected to it, the variable i is also the current through the capacitor (by KCL). The current through the capacitor is given by i C dVout dt . (2) Substituting Eq. (2) into Eq. (1), yields ECEN 3010 Lab 3 Page 1 of 14 Rev. 2008-08-24 This material is subject to copyright notice on last page. inout out VV dt dV  . (3) For this lab, the voltage Vin will be a square wave. The square wave can be represented as a time-domain step function that either rises from 0V to voltage V1, or falls from voltage V1 to 0V. The time constant of the RC circuit is τ = RC. For a rising step function input voltage, the solution to equation (3) takes the form e t out VVtV    11)( . (4) OutputR = 1.5kInput C = 0.1uF i+ – Vin Vin Vout Figure 1. A low-pass filter is a common and useful RC circuit. Figure 2 shows the response (output voltage) of the RC circuit when the input is a rising step function (from 0V to V1). The output voltage reaches 63% of the final value (V1) in one time constant. After five time constants (5), we say that the response has converged to its final or steady-state value (V1). 0.632 V1 Vout(t) V1 0 0 1 2 3 4 5 6 7 t /  t =  0.368 V1 Figure 2. The first-order circuit response to a step input (0 to V1 step) is an exponential that approaches the input step voltage V1. The current through the resistor charges the capacitor. ECEN 3010 Lab 3 Page 2 of 14 Rev. 2008-08-24 This material is subject to copyright notice on last page. Thus, for a low-frequency sinusoid, the output amplitude is approximately the same as the input amplitude. At high frequency, the output amplitude decreases approximately inversely with frequency. The transition between passing signals and attenuating signals occurs at the cut-off frequency (ωc) of the filter, which is the reciprocal of the time constant (c=1/τ). On a graph of amplitude ratio versus frequency, using a log-log scale, τ). ). On a graph of amplitude ratio versus frequency, using a log-log scale, the ‘corner’ at the cutoff frequency can be seen. This plot is called a Bode plot (pronounced BOH-dee). 2.1) Configure your function generator to be a sinusoidal input to the low-pass filter. Create a 0 to 5 volt sine wave (5V peak-to-peak amplitude, 2.5V DC offset). Change the frequency to the values listed in the Table 1, measure the peak amplitude of output voltage Vout, and record your measurements in Table 1 (Vout column). Table 1. Frequency response measurements. Voltages are peak values. Freq. Vout Vout/Vin HdB 50 100 200 500 1000 2000 4000 8000 16000 2.2) Enter the measurement data from the table into an Excel spreadsheet and use it to calculate the ratio Vout/τ). On a graph of amplitude ratio versus frequency, using a log-log scale, Vin. Plot the data using an Excel scatter plot with Vout/τ). On a graph of amplitude ratio versus frequency, using a log-log scale, Vin on the y-axis and the frequency in Hz on the x-axis. Set the plot properties to make the x-axis a logarithmic scale. 2.3) Use equation (9) to create a column of the theoretical Vout/τ). On a graph of amplitude ratio versus frequency, using a log-log scale, Vin values. Plot the theoretical values on the same graph as the measured values. Use 50 Hz as a starting frequency, and then use frequency increments of 100 Hz for the 100 Hz to 2 kHz range, and use 1 kHz increments from 2kHz to 16kHz. The goal is to produce a theoretical curve for comparison to your experimental results in Table 1. Adjust the Excel plot properties so that theoretical result appears as a line without data points (you will have more points on the theoretical curve than you have on the measured curve). Do your measured data points fall on the theoretical line? Print the graph and include it with this lab report. 2.4) Set the plot properties so that both plot axes are on a log scale. With log-log axes applied to your plot, you now have a Bode plot. Determine the cutoff ECEN 3010 Lab 3 Page 5 of 14 Rev. 2008-08-24 This material is subject to copyright notice on last page. (“corner”) frequency? The cutoff frequency occurs where 2/τ). On a graph of amplitude ratio versus frequency, using a log-log scale, VinVout  . Print the graph and include it with this lab report. 2.5) Calculate the theoretical cutoff frequency (ωc) in rad/τ). On a graph of amplitude ratio versus frequency, using a log-log scale, sec and fc in Hz. How does it compare with the value you determined from the plot? 2.6) The magnitude of the transfer function in decibels (HdB) is calculated using in out dB V V H log20 . (10) Determine HdB for your data points in Table 1. Complete Table 1 and include it with this lab report. 2.7) Set the function generator frequency to the theoretical cutoff frequency fc that you calculated. At this frequency, what is the measured Vout/Vin? What is HdB ? Note: Decibel (dB) notation is commonly used to express amplitude ratios over a wide range. The performance of an electronic filter is often specified by the cutoff frequency and the attenuation in dB per decade (the amplitude ratio change in dB for each factor of ten change in frequency) or in dB per octave (the amplitude ratio change in dB for each factor of 2 change in frequency). The RC filter has an attenuation of 6 dB per octave. More sophisticated active electronic filters can have a much larger roll-off such as 24 dB per octave or more. Part 3 – Frequency Response of a High-Pass Filter The RC circuit shown in Figure 4 is a high-pass filter. In a high-pass filter, high frequency input is ‘passed’ to the output but low frequency input is attenuated (lowered in amplitude). Output R = 1.5k Input C = 0.1uF + – Vin Vin Vout Figure 4. This RC circuit is a high-pass filter. ECEN 3010 Lab 3 Page 6 of 14 Rev. 2008-08-24 This material is subject to copyright notice on last page. The amplitude of the high-pass filter output is given by              221 2 fRC fRC VV inout   . (11) The variable f is the frequency of the sinusoidal signal. Thus, for a high-frequency sinusoid, the output amplitude is approximately the same as the input amplitude. At low frequency, the output amplitude decreases approximately inversely with frequency. The transition between passing signals and attenuating signals occurs at the cut-off frequency (ωc) of the filter, which is the reciprocal of the time constant (c=1/τ). On a graph of amplitude ratio versus frequency, using a log-log scale, τ). ). Just like the low-pass filter, on a graph of amplitude ratio versus frequency, using a log-log scale, the ‘corner’ at the cutoff frequency can be seen. 3.1) Construct the circuit in Figure 4. 3.2) Configure your function generator to be a sinusoidal input to the high-pass filter. Create a 0 to 5 volt, 100 KHz sine wave (5V peak-to-peak amplitude, 2.5V DC offset). Measure the maximum and minimum voltage of the output sinusoid at 100 KHz. Record the maximum and minimum voltage. Why is the lowest voltage now below ground (a negative node voltage value)? 3.3) Change the frequency to the values listed in the Table 2, measure the peak amplitude of output voltage Vout, and record your measurements in Table 2 (Vout column). Table 2. Frequency response measurements. Voltages are peak values. Vin Freq. Vout Vout/Vin HdB 50 100 200 500 1000 2000 4000 8000 16000 3.4) Enter the measurement data from the table into an Excel spreadsheet and use it to calculate the ratio Vout/τ). On a graph of amplitude ratio versus frequency, using a log-log scale, Vin. Plot the data using an Excel scatter plot with Vout/τ). On a graph of amplitude ratio versus frequency, using a log-log scale, Vin on the y-axis and the frequency in Hz on the x-axis. Set the plot properties to make the x-axis a logarithmic scale. ECEN 3010 Lab 3 Page 7 of 14 Rev. 2008-08-24 This material is subject to copyright notice on last page. L CR 20     . (22) The form of the complementary solution takes on one of three forms depending on the damping ratio. Case 1: Overdamped (ζ > 1, α > ω0) tsts out eKeKtV 21 21)(  . (23) Case 2: Critically Damped (ζ = 1, α = ω0) tsts out teKeKtV 11 21)(  . (24) Case 3: Underdamped (ζ < 1, α < ω0) )sin()cos()( 21 teKteKtV n t n t out     . (25) Where the natural frequency ωn is given by 22 0  n , (26) 4.1) Using L = 3.8 mH, C = 0.001 µF, and R = 12 Kohm, determine the damping ratio (). What damping case is this (underdamped, overdamped, critically damped)? Assume that the input is a voltage source that switches from 0 to 5V at t = 0. Given this input, the initial response voltage across the capacitor will be 0V, and the final (steady-state) response voltage will be 5V. Based on the initial voltage, final voltage, and damping case, roughly sketch the output of the circuit (the voltage across the capacitor) that you expect. Should the response converge to steady state quickly or slowly? Should the response show overshoot and ringing? 4.2) Using L = 3.8 mH, C = 0.001 µF, and R = 3.9 Kohm, determine the damping ratio (). What damping case is this (underdamped, overdamped, critically damped)? Assume that the input is a voltage source that switches from 0 to 5V at t = 0. Given this input, the initial response voltage across the capacitor will be 0V, and the final (steady-state) response voltage will be 5V. Based on the initial voltage, final voltage, and damping case, roughly sketch the output of the circuit (the voltage across the capacitor) that you expect. Should the response converge to steady state quickly or slowly? Should the response show overshoot and ringing? ECEN 3010 Lab 3 Page 10 of 14 Rev. 2008-08-24 This material is subject to copyright notice on last page. 4.3) Using L = 3.8 mH, C = 0.001 µF, and R = 390 ohms, determine the damping ratio (). What damping case is this (underdamped, overdamped, critically damped)? Assume that the input is a voltage source that switches from 0 to 5V at t = 0. Given this input, the initial response voltage across the capacitor will be 0V, and the final (steady-state) response voltage will be 5V. Based on the initial voltage, final voltage, and damping case, roughly sketch the output of the circuit (the voltage across the capacitor) that you expect. Should the response converge to steady state quickly or slowly? Should the response show overshoot and ringing? Part 5 – Measured Response of an RLC Circuit 5.1) Build the RLC circuit in Figure 5 using L = 3.8 mH, C = 0.001 µF, and R = 12 Kohm (case 1). You can use the lab’s “magic box” (see your TA). Use your function generator as an input to the RLC circuit; create a 0V to 5V square wave (5V peak-to-peak amplitude, 2.5V DC offset) that has a frequency of 4 kHz. View the input voltage (from the function generator) on channel 1 of the oscilloscope, and view the output voltage (voltage across the capacitor) of the RLC circuit on channel 2 of the oscilloscope. Sketch a full period of the input and output of the circuit. Label values on the voltage and time axes. 5.2) Compare your experimental results with your expected response (step 4.1). Is the response what you expect? 5.3) Change the resistor to 3.9 Kohm (case 2), again using L = 3.8 mH and C = 0.001 µF. Use a 4 kHz, 0V to 5V square wave (5V peak-to-peak amplitude, 2.5V DC offset). View the input voltage (from the function generator) on channel 1 of the oscilloscope, and view the output voltage of the RLC circuit on channel 2 of the ECEN 3010 Lab 3 Page 11 of 14 Rev. 2008-08-24 This material is subject to copyright notice on last page. oscilloscope. Sketch a full period of the input and output of the circuit. Label values on the voltage and time axes. 5.4) Compare your experimental results with your expected response (step 4.2). Is the response what you expect? 5.5) Change the resistor to 390 ohms (case 3), again using L = 3.8 mH and C = 0.001 µF. Use a 4 kHz, 0V to 5V square wave (5V peak-to-peak amplitude, 2.5V DC offset). View the input voltage (from the function generator) on channel 1 of the oscilloscope, and view the output voltage of the RLC circuit on channel 2 of the oscilloscope. Sketch a full period of the input and output of the circuit. Label values on the voltage and time axes. ECEN 3010 Lab 3 Page 12 of 14 Rev. 2008-08-24 This material is subject to copyright notice on last page.