Introduction to AC Circuits - Experiment #3 | ENGR 4300, Study notes of Engineering

Download Introduction to AC Circuits - Experiment #3 | ENGR 4300 and more Study notes Engineering in PDF only on Docsity! Electronic Instrumentation ENGR-4300 Spring 2000 Section ____________ Experiment 3 Introduction to AC Circuits Purpose: In the following exercises, we will investigate the properties of the other two passive circuit components -- inductors and capacitors. Equipment Required: · Oscilloscope (HP 54603B 2 Channel 60 MHz Oscilloscope) · Function Generator (HP 33120A 15 MHz Function/Arbitrary Waveform Generator) · Instrumented Beam · Battery (Each group should bring a battery to class) · Parts Kits Background on the Simple Pendulum – An Example of an Harmonic Oscillator Before we address capacitance and inductance, we will review some of the properties of the simple pendulum. The instrumented beam is a very good example of a simple pendulum, even though it looks more like a small diving board. Let us assume that the end of the beam moves in the x-direction. Obviously, this is a simplification, since it really travels along the arc of a circle. When the beam is stationary, we will assume that it is horizontal and at x=0. Again, this is an approximation because the beam must bend downward slightly due to its own weight. When the beam is bent, it experiences a restoring force like a spring F = -kx where k is the spring constant. From Newton’s Law F = ma = m dv/dt = m d2x/dt2 which, when combined with the previous expression , gives us m d2x/dt2 + kx = 0 the harmonic oscillator equation. In standard form, the harmonic oscillator equation is d2x/dt2 + w2 x = 0 where w is the frequency of oscillation. Thus, the beam will oscillate at w = (k/m)1/2 = 2pf The solution to this equation is x = xo cos(wt + fo) where xo is the initial deflection of the beam and fo is the initial phase. For simplicity, there is no need to include fo. We have also not included any damping in this model. Thus, we get an oscillation that will go on forever, rather than decaying slowly away. (If you do not recall that this is the solution, plug the expression for x into the differential equation and you will see that it works.) K. A. Connor Revised: 12/4/2020 Rensselaer Polytechnic Institute Troy, New York, USA 1 Electronic Instrumentation ENGR-4300 Spring 2000 Section ____________ The kinetic energy of the beam is KE = (1/2) m v2 while the potential energy is PE = (1/2) k x2 Upon initial deflection, the energy of the beam is W = PE = (1/2) k xo2 Since we have assumed no dissipation (no friction or other damping force), this total energy will be conserved and W = (1/2) m v2 + (1/2) k x2 = constant To summarize, the pendulum or any harmonic oscillator works by exchanging energy between two different forms. Not all forms of energy can be easily converted to another state and then back again, but we know this is trivial with the kinetic and potential energy of a mass. We can start at this expression of energy conservation to determine the equations of motion of the beam or any other simple pendulum. Since the total energy is a constant, we can take the time derivative of the entire expression and set it equal to zero. dW/dt = (1/2) m 2 v dv/dt + (1/2) k 2 x dx/dt = 0 Since v = dx/dt, we can write (m dv/dt) v + (k x) v = 0 or m dv/dt = -k x which is the original equation of motion. Thus, once we have a conservation law, we can use it even to find out how things change with time. Part A The Instrumented Beam as a Harmonic Oscillator In the last experiment, you should have measured the oscillation frequency of the unloaded beam using the strain gauge and the bridge circuit. This time we will use the output from the coil at the moveable end of the beam as the magnet clamped to the end of the beam moves through it. This signal is quite a bit larger than that from the bridge. However it is not proportional to the position of the beam, rather, it is sensitive to the beam velocity. Since the velocity and the position oscillate at the same frequency, we can use either signal to find the frequency. Measure the frequency again, since you probably do not have the same beam. Then, measure the frequency two more times using additional masses of your choice. Make sure that the mass is at least 50 grams. More than 100 grams is even better. There is a scale available in the studio to measure any mass you choose to add to the beam. Write down the masses you used, the corresponding frequencies and the location of the masses (it is difficult to place the masses at exactly the same location) in the table below. The location should be measured from the pivot point. Load Used Mass (kgm) Frequency (Hz) Location (m) None mbeam (unknown for now) fo = leff (unknown for now) m1 = f1 = l1 = m2 = f2 = l2 = K. A. Connor Revised: 12/4/2020 Rensselaer Polytechnic Institute Troy, New York, USA 2 Electronic Instrumentation ENGR-4300 Spring 2000 Section ____________ or in series. Also assume that the capacitor has been charged up to some voltage V at time t = 0, at which time it is connected to the inductor. The charge will begin to flow creating a current through the inductor. After a short time, all the charge that was originally on the capacitor plates will be gone and the current through the inductor will reach its maximum value. Thus, we began with all the energy stored in the capacitor and none in the inductor and end with the opposite condition. The current flowing through the inductor will then charge the capacitor back up and the process will begin again. The energy is traded back and forth between the two storage elements. The total energy of the system must remain constant because there is no dissipative element (no resistor). The total energy is given by W = (1/2) C V2 + (1/2) L I2 = constant Since this is a constant, we can take the time derivative of this expression and set it equal to zero. dW/dt = (1/2) C 2 V dV/dt + (1/2) L 2 I dI/dt = 0 or (L dI/dt) I + (C dV/dt) V = 0 which we can separate into L dIL/dt = VL and C dVC/dt = IC if we note that V = -VC = VL and I = IC = IL. V is the voltage at the top of the circuit and I is the current flowing around the circuit. The convention for circuit elements is that the current flows into the end with the highest voltage and thus we have to reverse the capacitor voltage to get the correct expression for V and I. These equations relating the current and voltage of capacitors and inductors are analogous to V = IR for a resistor and can be derived or determined empirically. (Please see Chapter 2, sections 1 and 2 of Gingrich and the electronic textbooks listed on the course webpage.) We can combine the two current-voltage relationships for the capacitor and inductor of our simple circuit to obtain d/dt (C dVC/dt) = d/dt IC = dIL/dt = VL/L = -VC/L or d2VC/dt2 + (LC)-1 VC = 0 which is the harmonic oscillator equation. The frequency can be trivially determined since the form is the same as we saw with the pendulum. Thus, w = (LC)-1/2 If we add a resistor to make a series RLC circuit, we will have a damped oscillator, the circuit for which is shown below. K. A. Connor Revised: 12/4/2020 Rensselaer Polytechnic Institute Troy, New York, USA 5 Electronic Instrumentation ENGR-4300 Spring 2000 Section ____________ Part B RC and RLC Circuits – Examples of AC Circuits We have used the LC oscillator circuit to develop the equations we need to analyze AC circuits. Passive AC circuits can consist of any combination of R, L, or C (as discussed in Chapter 2 of Gingrich). We will begin with a combination we have seen before, an RC circuit. 0 C1 0.68uF R1 1k V1 + + Set up this circuit using PSpice, with a 100mV source voltage and a frequency of 1kHz. Run a transient simulation, using the parameters we have applied previously for a 1kHz source. The capacitor and the resistor form a voltage divider combination in the same manner as one configured with two resistors. From the Probe plot, we want to show that, just like the resistor divider, the source voltage (V1) is equal to the sum of the voltages across the resistor and the capacitor (VR + VC). To determine V1, place a voltage level marker at the top of the voltage source. To determine VR and VC, use voltage differential markers. Place the positive voltage marker at the end of each component marked with a +. This follows the convention that voltage is positive at the end where the current enters. When you run your simulation, you will obtain three voltages. You will have to add a trace that is the sum of the two differential voltages. When you have done this correctly, you should see that the sum will equal the source voltage. Print out a plot of your results. Repeat for frequencies of 10 Hz and 10 kHz. Change the transient analysis appropriately to obtain a useful plot. As a rule of thumb, it is best to display about three cycles of a sinusoidal signal. Be sure that you change your maximum time step or you might have to wait a long time for your 10Hz results. It is not necessary to make a copy of these plots. Verify that the voltages add as expected. Describe the qualitative differences you observe in these three cases. The configuration we have been addressing, where the input voltage is V1 and the output voltage is taken across the capacitor is a kind of filter. Run an AC sweep analysis of this circuit in decades (log scale) from 1Hz to 100kHz to determine what kind of filter it is (low pass or high pass). Do not use the differential voltage markers. Plot the voltages at the top of the voltage source and at the top (marked with a +) of the K. A. Connor Revised: 12/4/2020 Rensselaer Polytechnic Institute Troy, New York, USA 6 Electronic Instrumentation ENGR-4300 Spring 2000 Section ____________ capacitor. You will then be plotting these voltages with respect to ground. Modify your plot so that you are displaying the magnitude of the capacitor voltage divided by the magnitude of the source voltage. Produce a plot your results. This plot is the frequency response for this filter. Now, reverse the order of the resistor and capacitor in the circuit and take the output voltage across the resistor. Perform the same AC sweep and determine what kind of a filter you get when you take the output voltage across the resistor. Produce a plot of the frequency response for the second type of filter . That is, plot the magnitude of the output voltage divided by the magnitude of the input voltage. Set up the low pass RC filter on a protoboard. Connect the function generator as the input AC source. Monitor the input and output voltages with the ‘scope. Measure the relative amplitude ½Vout½/½Vin½ at the three frequencies 100 Hz, 1 kHz, and 10 kHz. You cannot do this directly, since this is not an option with the ‘scope. However, you can measure the input and output voltages and then divide the latter by the former. If you have time, also do the measurement at 10 Hz. This is such a low frequency, that the display on the ‘scope is somewhat harder to work with. Add these experimentally measured points to the PSpice plots. Repeat for the high pass RC filter. If you are having any trouble figuring out which configuration is which, check the figures in Chapter 3 of Gingrich where he discusses filters. We will now consider an RLC circuit, with all three kinds of passive components. Here we have used two switches to change the input voltage from zero to 10 volts at time t = 0. The switches are in the parts list. Use PSpice to simulate the transient response of this circuit for a total time of 1msec. Describe the voltages plotted by Probe. What features of the voltages reminds you of the instrumented beam? 0 C1 .068uF L1 10mH R1 50U1 0 1 2 U2 0 1 2 V1 10v Report and Conclusions · Write out the mathematical expressions for the three voltages for the first RC circuit case you considered. · .Determine the range of frequencies for which the input voltage V1 is within 5% of the output voltage (either VR or VC) for the two configurations. This tells you the range of frequencies that pass through the filter more-or-less unchanged. Signals at other frequencies are attenuated by the filter. · Determine the resonant frequency of the RLC circuit you analyzed with PSpice. K. A. Connor Revised: 12/4/2020 Rensselaer Polytechnic Institute Troy, New York, USA 7