EE40EE40Transients -First Order CircuitsFirst Order CircuitsProf Nathan CheungProf. Nathan Cheung09/22/2009RdiHblCh t 4Reading: HambleyChapter 4Slide 1EE40 Fall 2009 Prof. CheungFirst-Order Circuits• A circuit which contains only sources, resistors and a capacitor is called an RC circuit.• A circuit which contains only sources, resistors and an inductor is called an RL circuit.RLd RC i it ll dfi td•RL and RC circuits are called first-order circuits because their voltages and currents are described byfirstorder differential equationsdescribed by first-order differential equations.R R–+vsL–+vsCi iSlide 2EE40 Fall 2009 Prof. CheungFirst Order Circuits• RC and RL Circuits (one energy (gystorage element)•Natural Response, Step Response,Natural Response, Step Response, Forced Response•Time constant•Time constant• Particular Solution and Complementary SolutionComplementary Solution• Applications – Propagation Delay, Slide 3EE40 Fall 2009 Prof. CheungDRAMCircuit ResponseThe natural response of an RC or RL circuit is its behavior (i.e. current and voltage) when stored energy in the inductor or capacitor is released to the resistive part of the network (containing no independent sources).The step response of an RC or RL circuit is its behavior when a voltage or current sourcestepis applied to the circuit orsource step is applied to the circuit, or immediately after a switch state is changed.The force response of an RC or RL circuit is its behavior when a voltage or current source has a particular time Slide 4EE40 Fall 2009 Prof. Cheungdependence [e.g. sin t, exp(-t)].Example: Natural Response of an RC Circuit• Consider the following circuit, for which the switch is closed for t < 0, and then opened at t = 0:Rot = 0++CRVo+−+v–Notation:0–is used to denote the time just prior to switching0+i d t d t th ti i di t l ft it hi0+ is used to denote the time immediately after switching• The voltage on the capacitor at t = 0–is VoSlide 5EE40 Fall 2009 Prof. CheungSolving for the Voltage (t ≥ 0)• For t > 0, the circuit reduces to+Ri+vCRoRVo+−•Applying KCL to the RC circuit:–pp y g• Solution:RCtevtv/)0()(−=Slide 6EE40 Fall 2009 Prof. Cheungevtv)0()(=Sketch of RC/t Ce)0(v)t(v−=Slide 7EE40 Fall 2009 Prof. CheungSolving for the Current (t > 0)+RoiRCtoeVtv/)(−=v–CRVo+−• Note that the current changes abruptly:i−0)0(eVvtitiRCto==>=−)(0,for0)0(/VieRRtito=⇒>+)0()( 0,for Slide 8EE40 Fall 2009 Prof. CheungRi=⇒)0(Solving for Power and Energy Delivered (t > 0)+RoiRCtoeVtv/)(−=v–CRVo+−RCtoeRVRvp/222−==tRCxotdxeVdxxpwRR/22)(−==∫∫x= dummy variable for integration()RCteCVRp/220011)(−∫∫Slide 9EE40 Fall 2009 Prof. Cheung()oeCV12 −=Time Constant τ• In the example, we found thatRCtRCtV//)()(dRCtoRCtoeRVtieVtv//)( )(−−== and• Define the time constant–Att=τthe voltage has reduced to 1/e(~0 37)RC=τAt t= τ, the voltage has reduced to 1/e(0.37) of its initial value.–Att=5τthe voltage has reduced to less thanAt t 5τ, the voltage has reduced to less than 1% of its initial value.Slide 10EE40 Fall 2009 Prof. CheungNatural Response of an RL Circuit• Consider the following circuit, for which the switch is closed for t < 0, and then opened at t = 0:t = 0i +Notation:LRoRIov–Notation:0–is used to denote the time just prior to switching0+ is used to denote the time immediately after switchingt 0 th ti t i t t dtt dth id t•t<0 the entire system is at steady-state; and the inductor is Æ like short circuit• The current flowing in the inductor at t = 0–is Ioand v Slide 11EE40 Fall 2009 Prof. Cheunggo across is 0.Solving for the Voltage (t > 0)tLRoeIti)/()(−=LRoRIo+v–• Note that the voltage changes abruptly:v=−0)0(ReIiRtvttLRo==>+−)(0,for )/(Slide 12EE40 Fall 2009 Prof. CheungI0Rv=⇒+)0(Solving for the Current (t ≥ 0)• For t > 0, the circuit reduces toi +LRoRIov–• Applying KVL to the LR circuit:•v(t)=i(t)R•v(t)=i(t)R• At t=0+, i=I0, •At arbitrary t>0, i=i(t) and()()di tvt Ld=-y()Slti()dt=Ie-(R/L)ttLReiti)/()0()(−=Slide 13EE40 Fall 2009 Prof. Cheung•Solution:= I0e(R/L)teiti)()0()(=Solving for Power and Energy Delivered (t > 0)tLR)/(tLRoeIti)/()(−=+LRoRIov–tLRoReIRip)/(222 −==txLRotodxReIdxxpwp)/(22)(−==∫∫()tLRoeLI)/(220011−=∫∫Slide 14EE40 Fall 2009 Prof. Cheung()oeLI)(12 −=Time Constant τ• In the example, we found thattLRtLRReItveIti)/()/()()(−−==andLooReItveIti)()()( )(==and• Define the time constantRL=τ–At t = τ, the current has reduced to 1/e (~0.37) of its initial value.–At t = 5τ, the current has reduced to less than 1% of its initial value.Slide 15EE40 Fall 2009 Prof. CheungNatural Response SummaryRL Circuit RC Circuiti+RLv–RC• Inductor current cannot change instantaneously• Capacitor voltage cannot change instantaneously–gygy/)0()0( ii+−=)0()0( vv+−=ti t tLτ/)0()(teiti−=τ/)0()(tevtv−=RCSlide 16EE40 Fall 2009 Prof. Cheung•time constant• time constantR=τRC=τStep Response of 1st-Order CircuitsIl h li d t lt•In general, when an applied current or voltage suddenly changes, the voltages and currents in an RC or RL circuit will change exponentially aCoccu cageepoe aywith time, from their initial values to their final values, with the characteristic time constant τ:[]τ/)(00 )()(+−−+−+=ttffextxxtx• where x(t)is the circuit variable (v or i)()()xfis the final value of the circuit variablet0is the time at which the change occursSlide 17EE40 Fall 2009 Prof. CheungExample : v(t) across a capacitor* v is continuousv(t)v(t)tttttotoSlide 18EE40 Fall 2009 Prof. CheungProcedure for Finding Transient Response1. Identify the variable of interest• For RL circuits, it is usually the inductor current iL(t)yL()• For RC circuits, it is usually the capacitor voltage vc(t)2.Determine the initial value (att=t0+)ofthe2.Determine the initial value (at t t0) of the variable• Recall that iL(t)and vc(t)are continuous variables:L()c()iL(t0+) = iL(t0−) and vc(t0+) = vc(t0−)•Assuming that the circuit reached steady state beforeAssuming that the circuit reached steady state before t0 , use the fact that an inductor behaves like a short circuit in steady state or that a capacitor behaves like an open circuit in steady stateSlide 19EE40 Fall 2009 Prof. Cheungan open circuit in steady stateProcedure (cont’d)3. Calculate the final value of the variable (its value as t Æ ∞)()• Again, make use of the fact that an inductor
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