EE 242 EXPERIMENT 8: CHARACTERISTIC OF PARALLEL RLC CIRCUIT BY USING PULSE EXCITATION 1

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1 EE 242 EXPERIMENT 8: CHARACTERISTIC OF PARALLEL RLC CIRCUIT BY USING PULSE EXCITATION 1 PURPOSE: To experimentally study the behavior of a parallel RLC circuit by using pulse excitation and to verify that the circuit theory accurately predicts the observed circuit behavior. This experiment complements the previous experiment 7 (parallel resonance) LAB EQUIPMENT: 1 Agilent 33120A Function Generator (FG) 1 Agilent 34410A Digital Multimeter 1 Agilent 54621A Oscilloscope 1 Impedance Bridge (1 per lab) 1 Resistance Decade box, 10 Ω/step (use the Clarostat 1 Ω/step decade box on your bench) 1 5 mh (nominal) standard or highq inductor 1 Bag of short leads 5 BananaBanana leads 3 BNCBanana leads DISCUSSION Natural response is the behavior of an electrical or mechanical system due to internal energy storage. Forced response is the behavior of a system due to an external energy source. Natural responses always decay with time due to finite energy storage while forced responses can be maintained indefinitely by a continuous input of energy. The simplest electrical systems have one energy storage element, either a capacitor C or inductor L, or their equivalent in a mechanical system. They also have a resistance R which dissipates energy. These systems can be described by a firstorder differential equation; hence they are called firstorder systems. With two storage elements, the circuit behavior is described by a secondorder differential equation. Systems that move a mass from one position to another usually exhibit second order characteristics similar to a series or parallel RLC circuit. Examples include the motion of a simple pendulum, a mass connected to a spring after it is displaced and released, shock absorbers in automobiles, and D Arsonal displacement (a permanent magnet moving a coil mounted on a needle thin axle used in many metering devices such as analog voltmeters). Such devices are designed to be critically damped so that they can provide a quick response time with minimal overshoot and oscillation. 1 Original experiment 7B, amended and revised 03/03//09, John Saghri 1

2 Consider the circuit shown (which is similar to the circuit R 1 you will be building in this lab). Via a source transformation, this circuit can be made equivalent to a parallel RLC circuit (where R = R 1 R 2 ) driven by a current source I s = v s / R 1. Application of a sudden DC current I (e.g., by setting v s s in the original circuit to a V s R 2 L C V square wave), causes the voltage v across the capacitor to exhibit the natural response of a parallel RLC circuit. This natural response is characterized by the following second order differential equation. d 2 v 1 d v 1 d t 2 RC d t LC v = 0 I S Equivalent circuit R L C V (1) The characteristic equation whose roots define the natural response of the circuit is: s 2 ( 1 / RC ) s ( 1 / L C ) = 0 The roots s 1 and s 2 of the characteristic equation can be expressed as: (2) s 1,2 = α ± α2 ω o 2 (3) where the resonant frequency (in radians/second) is : ω o = 1 / L C and the damping rate (in nepers/second) is: (4) (5) α = 1 / ( 2 RC ) Depending on the values of α and ω o, the natural response of the circuit will be overdamped, criticallydamped, or underdamped. If α < ω o the natural response will be underdamped as shown in the following figure. The response will have an oscillatory component with an angular frequency (called damping frequency ω d ) of: 2 2 ω d = ω o α (6) and will be of the form: v = e t / τ [ k1 cos (ω d t ) k 2 sin (ω d t ) ] v ( t ) (7) 2

3 The exponential decay e t / τ with the time constant of τ = 1/ α characterizes the envelope of the oscillatory functions. In this experiment, you will be applying a step function (using a square wave) to a parallel RLC circuit. For the specified values of R, L, and C, your output will be damped oscillatory (underdamped response) as shown above. Note that V( ) = 0 for a parallel RLC circuit (since the inductor acts as a short at t = ). You will be measuring the damping frequency ω d and damping rate α from the scope s display. To find ω d you will need to measure the period of oscillation T as shown. To determine damping rate α you need to measure the time constant τ of the decay representing the envelope of oscillation. The time constant of an exponential decay of the form k e t / τ is the time required for the decay to reach 1/e of its maximum (i.e., k / e). As shown in the diagram above, for the underdamped case the envelope of the oscillation is an exponential decay that can be modeled as: Env (t) = v ( ) [ A v( ) ] e t / τ (8) where Env (t) Envelope of oscillation for underdamped case A= Env (0) Initial value ( a constant) V( )= Env ( ) τ Final value Time constant By measuring two point (t 1, v 1 ) and (t 2, v 2 ) on the Env ( t ) curve, we can calculate the time constant τ from the following formula: (assuming t 2 > t 1 ) : (9) 3

4 Procedure 1 Establish the circuit shown below. Note that R 1 represents an external 280 Ω resistor and Z Th represents the 50 Ω internal output resistance of the function generator. 2 Set the function generator for a square wave output. Set the frequency at 50 Hz, peaktopeak voltage at 10 V, and the DC offset at 5 V. 3 Adjust the scope to clearly display the decaying sine wave. Note if your waveform is unstable, press Auto scale and then Run/Stop to freeze the waveform on the scope. Then you can scale the frozen waveform as desired. Capture and print the display in your report. 4 Obtain the damping frequency ω d from the graph. Note that ω d = 2π / T where T is one period of oscillation. To obtain a more precise measurement of T from your oscillatory decaying waveform, measure the total time corresponding to three periods and then divide that time by three. 5 Obtain the time constant of the envelope, τ, from the graph using the equation 9. Note that it is advisable to choose the two points (t 1, v 1 ) and (t 2, v 2 ) far from each other in order to increase the accuracy of measurement of τ. For this case, it is advisable that you select t 1 and t 2 to correspond to first and third peaks of oscillation. Note again that V ( ) = 0 for a parallel RLC circuit.. 6 Determine the damping coefficient α = 1/ τ 7 Use equation 6 to calculate the resonance frequency ω o 8 Determine the bandwidth B = 2α and the quality factor at the resonance frequency Q o = ω o / B 4

5 9 Compare your measured values of the damping frequency ω d and damping rate α (as described above) to their theoretical values defined in formulas 5 and 6. Compute the percent differences and explain the reason for any difference. Notes: For theoretical values of α and ω o, you must use the actual values (not nominal) of the R, L, and C. To find R, you must find the parallel model resistances of the capacitor and the inductor, i.e., R PC and R PL. Use the following formulas to find R PC and R PL. o R SC = D / (ω C S ), where R SC is the series model resistance of the capacitor, D is the dissipation factor of the capacitor (measured via the impedance bridge), ω is equal to 2πf (where f is the frequency set on the bridge, i.e., 1000 Hz), and C S is the series capacitance value of the capacitor as measured by the bridge. o Use R PC = R SC ( 1 1 / D 2 ) to obtain R PC o To find R PL, use R PL = Q ω L P, where R PL is the parallel model resistance of the inductor, Q is the quality factor of the inductor (measured via the bridge) and L P is the parallel inductance value of the inductor as measured by the bridge. R is then equal to the parallel sum of R PC, R PL, and the 330 Ω effective load resistive value (i.e., 50 Ω function generator output resistance plus the 280 Ω resistor). 5

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