EECS40 RLC Lab guide Introduction Second-Order Circuits Second order circuits have both inductor and capacitor components, which produce one or more resonant frequencies, ω0. In general, a differential equation for the circuit can be written in the form: )()()(2)(d2022tftvdttdvdttv=++ωα (Eq. 1) Where α and 0ω depend on values for R, L, and C present in the circuit, and the forcing function f(t) depends on these values as well as voltage sources present. This differential equation follows the same form as that for a damped harmonic oscillator (ie. A mass oscillating due to a spring (resonance), but experiencing a damping force (friction), and possibly a driving force (on the mass to counteract the spring)). In the case of our second order circuit, the damping is produced by a resistor burning energy, resonance occurs between the inductive and capacitive components, and driving force would be a voltage source to force the circuit against the damping by the resistor. For a series RLC circuit, the damping factor LR2=α and resonant frequency LC10=ω Figure 1: Series RLC circuit The solution to the differential equation (Eq. 1) depends on these parameters α and 0ω, in that if: 0ωα> The circuit is overdamped, and the solution is: tstseKeKtv2121)(+= 0ωα= The circuit is critically damped, and the solution is: tststeKeKtv2121)(+= 0ωα< The circuit is underdamped, and the solution is: Lab 4.5 - guide - 1))sin()cos(()(21tKtKetvnntωωα+=− The roots and of the over- and critically damped solutions are given by: 1s2s 2021ωαα−+−=s 2022ωαα−−−=s The natural frequency nωof the underdamped case is given by: 220αωω−=n In this lab, we will observe the step response of a series RLC circuit. The step response is how the circuit behaves in response to a forcing function that is a step function (f(t) = 0 for t <0, and f(t) = 1 if t≥0) which we apply periodically to our circuit with a square wave. The solution is shown in Figure 2.6, where the values are normalized (amplitude of the forcing function divided out). Figure 2: Normalized Step Response of 2nd order circuit Quality Factor and Bandwidth Lab 4.5 - guide - 2Figure 3: Series RLC Bandpass Filter, with AC source The circuit shown in Figure 3 has a resonance frequency0ω, which has the same value as shown on page 1. At resonance, the reactance of the capacitor cancels out the reactance of the inductor so they must be equal in magnitude. The quality factor , of a series RLC filter is defined as the ratio of the inductive reactance to the resistance, at the resonant frequency: Q RLfRLQ002πω== Figure 4: Plots of Transfer Function magnitude, for different values of Qs, the quality factor of a series RLC circuit. Note that this plot is not on a log-log scale (which Bode plots feature). Lab 4.5 - guide - 3The bandwidth of a bandpass filter is the region between which the output is above half the maximum power. This is also the -3dB point, because in decibels, 10 log 0.5 ≈ -3, where 0.5 comes from the power ratio, or 2)(ωH. The bandwidth, B, of a series bandpass filter is related to quality factor, Q, by the equation: Qf0=B ___ The voltage measured at the half-power frequency should be ~0.707, or √0.5 of the maximum voltage, because power is proportional to the square of voltage. Phasors and Complex Impedance Phasors can be used to represent the complex amplitude of a sinusoidal function. For example: ()φφω∠=+AtAcos Impedance measures the opposition of a circuit to time varying current. The units of impedance are ohms. A resistor’s impedance is simply it’s resistance. Capacitors and inductors have complex impedances: Capacitor impedance =Cjω1 Inductor impedance = Ljω An RLC circuit that is driven by a sinusoidal voltage source can be analyzed using KVL, KCL, and Ohm’s law. The rules that apply to resistors apply to the complex impedances of the elements of the circuit. For example, Vout in Figure 3 can be found easily using the voltage divider equation with the circuit element’s impedances. (Doing this is part of a prelab problem) Lab 4.5 - guide - 4Hands On LC Circuit For this part of the lab, you will need to set your function generator to generate pulses at regular intervals. First, create a square wave with a frequency of 100 Hz, 3 Vpp, and a Voltage offset of 1.5V (the offset makes the square wave range between 0V and 6V). Hit the shift key, then the key with “burst” written in small blue letters above it. Use the oscilloscope to make sure the output your function generator looks something like this: F igure 5: pulses created by the function generator Now, Attach a 10 µF capacitor, as shown in figure 2. (make sure to check if the capacitor uestion 1: Sketch the voltage across the capacitor. ow, add a 1 mH inductor in parallel with the capacitor, as shown in figure 3. is polarized) Figure 6: Capacitor attached to a pulse generator A1 ScopeA1 Ground Q N A1 Scope Figure 7: inductor and capacitor attached to a pulse generator A1 GroundLab 4.5 - guide - 5You may have to move the trigger level up or down to get a still picture. If that doesn’t uestion 2: Sketch the voltage across the capacitor and inductor in parallel, and explain crease the frequency of the function generator to 2 KHz. uestion 3: Sketch what you see when the inductor is removed (like Figure 6). Sketch se the two time cursers (hit the curser key, then use the time softkey) to estimate the e uestion 4: Use the information from the time cursors to estimate the frequency of the emove the 1 mH inductor and replace it with a 10 mH inductor. uestion 5: Sketch the new waveform. eries RLC Circuit elow with a 1 nF Capacitor, 100 mH inductor, and a 2.2 KΩ work, you can always hit the stop button to freeze the image. DON’T rescale the image, you will need to graph the new output on the same graph as question 1. Qwhat you see. (Use the same graph as question 1) In Qwhat you see when the inductor is present (like Figure 7). Uperiod of the oscillation you see on your oscilloscope. (You may have to extrapolate thperiod from a trough to peak (1/4th of a period) to get the most accurate answer) Qoscillation. R Q SConstruct the circuit bresistor. Use a sinusoidal input with 6 Vpp and 0V offset. Determine the resonancefrequency, f0, and set the function generator to that
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