Examples of Transient RC and RL Circuits. The Series RLC Circuit Impulse response of RC Circuit. Let’s examine the response of the circuit shown on Figure 1. The form of the source voltage Vs is shown on Figure 2. VsRCvc+- Figure 1. RC circuit tVptp0Vs Figure 2. We will investigate the response as a function of the ()vc tpτ and Vp. The general response is given by: () 1 0tRCvc t Vp e t tp−⎛⎞=−≤⎜⎟⎝⎠≤ (1.1) If the capacitor voltage at tp RCttp= is equal to Vp. Therefore for times t the response becomes tp> ()()ttpRCvc t Vp e tp t−−⎛⎞=≤⎜⎟⎝⎠ (1.2) 6.071/22.071 Spring 2006, Chaniotakis and Cory 1A general plot of the response is shown on Figure 3 for 1sec, 6sec, 10RCtpVpV== =olts Figure 3 If the pulse becomes narrower, the value of vc will not reach the maximum value. By expanding the exponential in Equation (1.1) we obtain, 2311() 1 1 026tt tvc t Vp t tpRC RC RC⎛⎞⎡⎤⎛⎞ ⎛⎞=−−+ − + ≤≤⎜⎟⎢⎜⎟ ⎜⎟⎜⎟⎝⎠ ⎝⎠⎢⎥⎣⎦⎝⎠…⎥ (1.3) When the higher order terms may be neglected resulting in RC t () 0tvc t Vp t tpRC≤≤ (1.4) At the end of the pulse (at ) the voltage becomes ttp= ()Vptpvc t tpRC=  (1.5) 6.071/22.071 Spring 2006, Chaniotakis and Cory 2For the response becomes ttp> ()ttpRCVp tpvc eRC−−⎛⎞=⎜⎝⎠⎟ (1.6) The product Vp is the area of the pulse and thus the response is proportional to that area. As the pulse becomes narrower (i.e. as ) equation (1.6) simplifies to tp0tp → tRCVp tpvc eRC−⎛⎞⎜⎝⎠⎟ (1.7) If we constrain the area of the impulse to a constant AVp tp=, then as the pulse becomes narrower, the amplitude Vp increases, resulting in an impulse of strength A. Therefore the response of an impulse of strength A is tRCAvc eRC−= (1.8) Figure 4. Impulse response of RC circuit 6.071/22.071 Spring 2006, Chaniotakis and Cory 3The spark plug in your car (a simplified model) Consider the circuit shown on Figure 5. The battery Vb corresponds to the 12 Volt car battery. The spark plug is connected actors the inductor and current may flow though it only if the voltage across the gap of the plug exceeds a very large value (about 20 kV). +-VbRLsparkplugvL+- Figure 5 When the switch is closed, the current through the inductor reaches a maximum value of . The equation that describes the evolution of the current with the switch closed is /Vb R /() 1tLRVbit eR−⎛⎞=−⎜⎝⎠⎟ (1.9) And the corresponding voltage across the inductor is given by /()tLRvL t Vb e−= (1.10) When the switch is opened, the current path is effectively broken and thus the time rate of change of the current becomes arbitrarily large. Since the voltage is proportional to , the voltage developed across the inductor could become very large. /di dt As an example, let’s consider a system with a resistance of 5Ω, a solenoid with an inductance of 10mH connected to a 12 Volt battery. How long does it take for the solenoid to reach 99% of its maximum value? If the switch is opened in 1µs, what is the voltage developed across the solenoid? The time constant of the system is 0.010.002sec5LR== The maximum current that can flow in the system is 122.45AA=. The time to reach 99% of the maximum value is given by 6.071/22.071 Spring 2006, Chaniotakis and Cory 40.0020.99 1te−=− The voltage across the coil when the switch is opened is 62.40.01 24110ivL kVt−∆== =∆× 6.071/22.071 Spring 2006, Chaniotakis and Cory 5Response of RC circuit driven by a square wave. Let’s now consider the RC circuit shown on Figure 6(a) driven by a square wave signal of the form shown on Figure 6(b). VsRCvc+-+ vR - (a) VstVp-VpT (b) Figure 6 The response vc(t) is given by []response = final value + initial value - final valueteτ− (1.11) By assuming that the initial value of the voltage across the capacitor is –Vp the response during the first half cycle of the square wave is []() - - 1 - 2tRCtRCvc t Vp Vp Vp eVp e−−=+⎡⎤=⎢⎥⎣⎦ (1.12) During the second half cycle the initial condition is 6.071/22.071 Spring 2006, Chaniotakis and Cory 6/2( / 2) 1 - 2TRCvc T Vp e−⎡⎤=⎢⎥⎣⎦ (1.13) And the complete response during the second half of the first cycle becomes /2() - 1 - 2 + TtRC RCvc t Vp Vp e Vp e−−⎡⎤⎡⎤=+⎢⎢⎥⎢⎥⎣⎦⎣⎦⎥ (1.14) Similarly the response during the first part of the second cycle starts with the value of vc at t=T and evolves towards the value Vp. If the time constant is small compared to the period of the square wave, the response will reach the maximum and minimum values of the square wave as shown on Figure 7, where and thus T/2=10RC. 4110 secRC−=× Figure 7 As the time constant RC increases, it takes longer for the response to reach the maximum value. Figure 8 shows a plot of the response for T/2=RC. Note that the response does not reach the maximum values of the input signal and the average value of the response is equal to the average value of the input signal. 6.071/22.071 Spring 2006, Chaniotakis and Cory 7Figure 8 Figure 9(a) and Figure 9(b) show the system response for RC=5T/2 for a square wave with a duty factor of 50% that varies between 0 and 5 Volts. Notice that the average value is reached within a certain number of oscillations and that there is a variation of the response “ripple” about the average value. The magnitude of this ripple is inversely proportional to the time constant RC. This is the first step that one must take when an AC signal is converted to DC. Next week, when we learn about the diode, we will explore this circuit further. 6.071/22.071 Spring 2006, Chaniotakis and Cory 8(a) (b) Figure 9 6.071/22.071 Spring 2006, Chaniotakis and Cory 9Second Order Circuits Series RLC circuit The circuit shown on Figure 10 is called the series RLC circuit. We will analyze this circuit in order to determine its transient characteristics once the switch S is closed. VsRCvc+-+ vR -L+ vL -S Figure 10 The equation that describes the response of the system is obtained by applying KVL around the mesh vR vL vc Vs++= (1.15) The current flowing in the circuit is dvciCdt= (1.16) And thus the voltages vR and vL are given by dvcvR iR RCdt== (1.17) 22di d vcvL L LCdt dt== (1.18) Substituting Equations (1.17) and (1.18) into Equation (1.15) we obtain