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Response of First Order RL and RC Circuits

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Response of First Order RL and RC Circuits Response of First Order RL and RC Circuits First Order Circuits: Overview In this chapter we will study circuits that have dc sources, resistors, and either inductors or capacitors (but not both). Such circuits are described by first order differential equations. They will include one or more switches that open or close at a specific point in time, causing the inductor or capacitor to see a new circuit configuration. This in turn will cause a time-dependent change in voltages and currents. We will find that the equations describing the voltages and currents in these circuits (i.e., the circuit responses) are exponential in time, and characterized by a single time constant. In other words, we will have responses of the form / ( ) ( ) − For the natural response, and ∝ / ( ) ( ) (1 ) − for the step response, where τ is the time constant ∝. These− are single time constant circuits. Natural response occurs when a capacitor or an inductor is connected, via a switching event, to a circuit that contains only an equivalent resistance (i.e., no independent sources). In that case, if the capacitor is initially charged with a voltage, or the inductor is initially carrying a current, the capacitor or inductor will release its energy to the resistance. The circuit below shows an inductor that was initially connected to a current source, which establishes a current in the inductor. The switching event at t = t0 results in the inductor being connected to only a resistance. In this case, the inductor current iL(t) will decrease exponentially in time. t = t0 R i (t) S L L R iS We next consider a capacitor initially connected to a voltage source, which establishes a voltage across the capacitor. The switching event at t = t0 results in the capacitor being connected to only a resistance. In this case, the capacitor voltage vC(t) will decrease exponentially in time. RS t = t0 a) b) vS + R C vC(t) - In the circuits above, L and C may be the result of combining inductors or capacitors in series or parallel. The resistance RS is interpreted as the Thevenin Equivalent resistance seen by the inductor or by the capacitor. Note that the circuit after the switching event may contain dependent sources, which are part of the Thevenin resistance. Note that the circuit connected to the inductor before t = t0 is a Norton equivalent, and the circuit connected to the capacitor is a Thevenin equivalent. Here is an example of the use of Thevenin and Norton equivalent circuits to help us think in general terms about circuit analysis. We could have had considerably more complex circuits connected to the inductor or to the capacitor – and later we will have that - but it won’t matter because we can always reduce those circuits to a Norton or Thevenin equivalent. Note also that we could have chosen the Norton equivalent for the capacitor example, and the Thevenin equivalent for the inductor example. These are equivalent by the source transformation theorem, but making these choices now will make our analysis later easier. Step response occurs when an inductor or capacitor is connected, via a switching event, to a circuit containing one or more independent sources. Examples for the inductor and for the capacitor are shown below. t = t RS 0 t = t0 iL(t) L RS C vS iS Again, we are using Thevenin and Norton equivalent circuits to represent what gets connected to the inductor and capacitor. Later we will consider more complex circuits. The Natural Response of an RL Circuit The circuit below shows the natural response configuration we introduced earlier. We now specify that the switch had been closed for a long time, and then opened at t = t0. After the switch opened, the inductor was connected to the resistance R. We want to know what happens to the inductor current after the switch is thrown. Terminology When we write “t”, we mean the variable representing time. When we write t0, we mean a particular moment time, e.g., t0 = 0, or t0 = 5 [ms]. The “natural response” is one in which the inductor, with current flowing through it, undergoes a switching event that connects it to a resistance only. As we pointed out above, that resistance can be a single resistor, or it can be an equivalent resistance that arises from a circuit containing multiple resistors and/or dependent sources. t = t0 RS iL(t) L R iS A Long Time? What do we mean that the switch had been closed “for a long time”? We will be specific later about what that means, but it is important to know that before the switch opens, enough time had passed that voltages and currents are no longer changing. Recall that earlier we said that voltages and currents will be decaying exponentially in time for circuits like this. Therefore, after enough time has passed, voltages and currents will stop changing. Once that happens, we say the circuit is at steady state. Analysis We begin the analysis at t < t0, i.e., before the switch opened. The circuit for this time domain is shown below: the switch is closed. Since we know the circuit was in steady state (because it has been like this a long time), we have di L =⇒=00v . dt L From the current-voltage relationship for an inductor, we know that at constant current, the inductor acts like a short. Thus, the resistors R and RS have no current flowing through them, so the only current flowing is through the inductor, and iL = iS. Now we look at the situation for t > t0. The switch has been thrown, and the inductor sees only a resistance. We don’t need to worry about the current source or the resistor RS at + this point. iL(t) We are no longer in steady state: currents and voltages now L R begin to change. To find out how, we do a KVL: vL - di() t LL += i() tR 0 dt L Note that the inductor current and voltage are in the active sign relationship. You should convince yourself that the signs in this equation are correct. This is a first-order differential equation for iL(t). Solution it( ) dit R ∫∫L = − dt it( 00) t iLL it( ) R ln L =−−tt ( 0 ) itL ( 0 ) L −−R tt L( 0 ) ∴=itLL( ) ite( 00) tt ≥ We have integrated over time from the switching time t0 to an arbitrary time t, and we have integrated over current from its value at t0 to its value at t. Initial Condition For a complete solution, we need the initial condition itL ()0 . We found earlier that before the switching event, the current in the inductor was iS. But that was the current flowing in the inductor before the switch was thrown. We need to know the inductor current after the switch is thrown, which is want the differential equation refers to. To do that, we make use of the inductor property that we cannot have an instantaneous change in current through an inductor. That means the inductor current flowing immediately before − the switch moved, itL ()0 is equal to the current flowing in the inductor immediately after the + − − switch moved, itL ()0 . But itL ()0 is the current we found above, that is, itLS()0 = i, and + + itL ()0 is what we have called itL ()0 in the equation above. So, itLL()00= it () = i S, which is the initial condition we need in order to solve the equation. Valid Time Range To be clear about the validity of the solution, we indicate that tt≥ 0 , meaning that the solution as written is valid at time t0 and later. It is not valid for tt< 0 . Typically, L and t0 will be known, and we need to find the initial condition and the resistance seen by the inductor, R. As noted above, R will be the Thevenin equivalent resistance of whatever the inductor is connected to. Further, we have a natural response problem if the Thevenin equivalent voltage is 0, that, the Thevenin equivalent is a resistance only. To simplify notation, we will define the initial value of the current as itL ()00≡ I. So in this case we have, as an initial condition, I0 = iS (the value of the current source). The Time Constant The form of the response can be simplified by noting that the units of R/L are s-1 (inverse seconds): R ohm ⋅ Amp − = = s 1 . L volt ⋅s We can then write −−(tt0 ) τ i t= IeL t≥ t . L ( ) 00 L τ = L R Very often we have t0 = 0, in which case −t τL iL ( t) = Ie0 t ≥ 0 . We can interpret this equation to mean that the inductor current starts at the initial value I0, and then decays exponentially in time once it is removed from the current source and connected to the resistor. In the figure below we show a plot of itL ( ) for a long enough time that the exponential has decayed to 0 (or at least to a very small value). The initial current iS is chosen to be 10 [mA] and the time constant is 10 [mS]. The plot also shows the point at which the time t is equal to the time constant, i.e., t = τL . At this point, the current has decayed to a value given by −τL τL −1 iLL( t=τ=) Ie00 = e I . The numerical value of e-1 is approximately 0.368, or roughly 3/8. -1 iL(t = τL) = e I0 We note that after five time constants, or t = 50 [ms], the current has decayed almost to 0.
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