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### lesson31

Course: MAE 140, Spring 2010
School: UCSD
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Word Count: 908

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31 Lesson RLC Series and Parallel Circuits (Sections 7-5 and 7-6) (CLOs 7-3 and 7-4) This is the first lesson on the behavior of RLC circuit. There are several key points that we want the cadets to learn in this and the next lesson (step response of RLC circuits). RLC circuits have two independent energy storage devices (C and L). RLC circuits are described by a second-order linear differential or...

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31 Lesson RLC Series and Parallel Circuits (Sections 7-5 and 7-6) (CLOs 7-3 and 7-4) This is the first lesson on the behavior of RLC circuit. There are several key points that we want the cadets to learn in this and the next lesson (step response of RLC circuits). RLC circuits have two independent energy storage devices (C and L). RLC circuits are described by a second-order linear differential or integro-differential equation the solutions of which require application of two initial conditions (vC(0) and its derivative dvC(0)/dt or iL(0) and diL(0)/dt ). RLC circuits result in a second-order characteristic equation. Natural responses to RLC circuits yield one of three different forms that depend on the parameters R, L and C. These are referred to as the overdamped, critically damped and under-damped cases. Start with a Thvenin circuit and add an L and a C in series. Using KVL well find a second-order integrodifferential equation if we choose to find iL(t) as our variable or a second-order differential equation if we choose to solve for vC(t). di (t ) 1 t L L + i L (t )dt + vC (0) + RT i L (t ) = v T (t ) dt C0 v L (t ) + v C (t ) + v R ( t ) = v T (t ) or LC d 2 v C (t ) dv (t ) + RT C C + v C (t ) = v T (t ) 2 dt dt v L (t ) + v R (t ) + v C ( t ) = v T (t ) Most students will not really know what to do with the former so we will focus our solution of the latter. However, I think it is useful to show both because in Laplace either will be equally easy to solve. At this point, we will solve the equation for the natural response only (zero input). It becomes important that they appreciate how different the three responses look. When solving the characteristic equation LCs 2 + RT Cs + 1 = 0 with roots s1 , s 2 = roles of each parameter on the roots. st Next, derive the zero-input response. If you substitute v C (t ) = Ke in the second-order equation, solve it for RT C ( RT C ) 2 4 LC 2 LC spend some time letting them see the the two roots you will get the following result: vC (t ) = K 1e s1t + K 2 e s2t , where s1 and s2 are the roots of the equation. In order to find the total solution we need to use the initial conditions the stored energy and the only energy driving the circuit. The two initial conditions are related to the solution as follows v C ( 0 ) = V0 = K 1 + K 2 dvC ( 0) I 0 = = K 1 s1 + K 2 s 2 dt C s 2V0 I 0 C s V + I0 C and K 2 = 1 0 Solving these, we get that 1 K = . s 2 s1 s 2 s1 These are then plugged back into the result for vC(t). There are three distinct results depending if the roots are real and different, real and identical, or complex. Example 7-15 or a variant thereof is a good example to do in class to show the effects of RTLC on the response. Their influence on the response becomes especially clear in the under-damped case. In a series RTLC circuit RT tends to control the damping while L and C control the rate of oscillation. vC (t) (V) 15 10 5 0 2 -5 4 6 8 t(ms) Overdamped - Case A Critically damped - Case B Underdamped - Case C The results of Example 7-15 are repeated here. You might ask them what the response would be if we could somehow remove all the resistance from the circuit the circuit would oscillate forever. In addition, ask them why that is not possible (parasitic R associated with wires, the source resistance and the inductor.) With an Op-Amp or other active device like a transistor, it is possible to use positive feedback to just cancel the effects of the parasitic resistance and have the circuit oscillate forever. This would result in a Case D. Such a circuit is called an Oscillator. Turn the class attention to a parallel RLC circuit using a Norton equivalent. You may choose to derive the equations but it is -10 easier to use the concept of duality. Either way you should end up with the following roots of the characteristic equation . Just like the series case, there are three distinct responses. The 2 LC biggest difference to point out is that in the series case the smaller RT the more towards underdamped the circuit becomes. In the parallel case, the opposite is true. The larger RN the closer towards underdamped the circuit becomes. Why? s1 , s 2 = In summary, the three distinct cases are as follows Series: Case A: v C (t ) = K 1e 1t + K 2 e 2t t 0 t t Case B: v C (t ) = K 1e + K 2 te t L RN ( L R N ) 2 4 LC t0 t Case C: v C (t ) = K 1e cos t + K 2 e sin t t 0 Parallel: Case A: i L (t ) = K 1e 1t + K 2 e 2t t 0 Case B: i L (t ) = K 1e t + K 2 te t t 0 Case C: i L (t ) = K 1e t cos t + K 2 e t sin t t 0 The and are found from the roots of the characteristic equation and K1 and K2 are found from the initial conditions. Just for information, a series RLC Case D would have the following response: Case D: v C (t ) = K 1 cos t + K 2 sin t t 0 Next lesson is on RLC step response.
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