Response of First Order RL and RC Circuits 1

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 𝑣𝑣(𝑡𝑡) 𝑜𝑜𝑜𝑜 𝑖𝑖(𝑡𝑡) ∝ 𝑒𝑒−𝑡𝑡/𝜏𝜏 𝑣𝑣(𝑡𝑡) 𝑜𝑜𝑜𝑜 𝑖𝑖(𝑡𝑡) ∝ (1 − 𝑒𝑒−𝑡𝑡/𝑡𝑡) 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 RS iS iL(t) L R 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) vS b) + C vC(t) R In the circuits above, L and C may be the result of combining inductors or capacitors in series or parallel. 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. The resistance Rs may be interpreted as a source resistance. 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 we can always reduce those circuits to a Norton or Thevenin equivalent. After the switching event, the inductor and the capacitor are connected only to a resistance. That resistance should be interpreted as the equivalent Thevenin equivalent resistance seen by the capacitor. In more complex problems, there may be several resistors in series or parallel as well as dependent sources, all of which combine to a single Thevenin equivalent resistance. 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 later analysis 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. RS t = t0 t = t0 iL(t) iS RS C vS L 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 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 iS iL(t) L R 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 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 diL =0 ⇒ vL =0 . dt 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. We are no longer in steady state: currents and voltages now begin to change. To find out how, we do a KVL: L + iL(t) R L vL - diL (t ) 0 + iL (t ) R = dt 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 t R diL = − ∫ dt i t0 ) i t0 L L i( t ) ∫( ln iL ( t ) R = − ( t − t0 ) iL ( t0 ) L = ∴ iL ( t ) iL ( t0 ) e −R L( t − t0 ) t ≥ t0 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 iL (t0 ) . 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, iL (t0− ) is equal to the current flowing in the inductor immediately after the iL (t0+ ) . But iL (t0− ) is the current we found above, that is, iL (t0− ) = iS , and iL (t0+ ) is what we have called iL (t0 ) in the equation above. So, i= iL= (t0+ ) iS , which is L (t0 ) switch moved, 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 t ≥ t0 , meaning that the solution as written is valid at time t0 and later. It is not valid for t < t0 . 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. 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 iL (t0 ) ≡ I 0 . 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): We can then write  R  ohm ⋅ Amp −1  L  = volt ⋅ s = s . = iL ( t ) I 0e − ( t − t0 ) τL τL = Very often we have t0 = 0, in which case L R t ≥ t0 . iL ( t ) I 0e = −t τL 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 iL ( t ) 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 = τ L . At this point, the current has decayed to a value given by iL ( t = τL ) = I 0e −τ L τL = e −1I 0 . The numerical value of e-1 is approximately 0.368, or roughly 3/8. iL(t = τL) = e-1I0 Long Time We note that after five time constants, or t = 50 [ms], the current has decayed almost to 0. What is left is less than 1% of the initial value: t = 5τ L ⇒ iL ( t ) = 0.0067iL (t0 ) . For that reason, we can define a “long time” as five time constants. The Inductor is a Short in Steady State Earlier we defined steady state as the condition in which voltages and currents are no longer changing. We said that in switching problems we will need to know that the switch has been in its initial position for a long time so that we know the circuit is in steady state. Looking at the solution for the inductor current, we see that after a long time, the inductor current becomes 0, and again there is no longer any change in voltage or current. In other words, the circuit has returned to steady state. We also said above that in steady state, the inductor acts like a short circuit because although there is current flowing in it, there is no voltage across it. The inductor behaves like a short while the switch is in its initial position, and later returns to a short a long time after the switching event. In between those times, the inductor current and voltage are changing exponentially. These are important ideas that we will return to many times in this course and in future courses. Power and Energy Delivered to R We can find expressions for the power delivered to the resistor and the energy dissipated by it as a function of time. Note that these equations are valid only for 2 L 2 pabs by= i= (t ) R I 0 R e R (t ) w= abs by R (t ) t pabs , R dt ∫= t0 t ≥ t0 −2( t −t0 ) . τL −2( t −t0 )   1 τL L I 0 2 1 − e  2   As we expect, after a long time, the total energy dissipated in the resistor is 1 L I 0 2 , i.e., it is the 2 total energy delivered by the inductor, which was initially carrying a current iL (t0 ) = I 0 . Inductor Voltage We can also find the voltage across the inductor as follows (recall we have an active sign relationship between vL and iL. diL d  − ( t − t0 ) τ L  −L = − L  I 0e vL ( t ) =  t > t0 . dt dt   Then,  1  − ( t − t0 ) τ L vL ( t ) =− L I 0 ⋅  −  e  τL  − t −t / τ = R I 0e ( 0 ) L . Valid Time Range for the inductor voltage Although the current through the inductor cannot change instantaneously, the voltage across the inductor can change instantaneously, and in this case it does. Note that immediately before the switch was moved, the voltage across the inductor was 0. However, at t = t0 in the expression above, the voltage across the inductor is changed instantaneously. vL ( t0 ) = I 0 R . Therefore, the inductor voltage has Because the voltage changes instantaneously, we cannot say what its value is at t = t0. The inductor voltage is indeterminate at t = t0, meaning it is not possible to say what it is. Therefore, we have indicated the valid time range as being t > t0. This implies that the solution for the voltage across the inductor is not valid at t = t0. Find the Inductor Current First We found the inductor current, and from that we found the inductor voltage. There was an important reason for doing it in this order: it is straightforward to find the initial condition for the differential equation that describes the current, which we have done. Although it is possible to find the initial condition for the equation that describes the inductor voltage, this is less straightforward. Therefore, it is a good idea to find the inductor current first, and then if necessary, differentiate to find the inductor voltage. The Natural Response of an RC Circuit We now go through a similar analysis for the RC circuit. The circuit below shows the natural response configuration for the RC circuit. In this case, a switch is moving from position a to position b. We specify that the switch had been in position a for a long time, and then moved to position b at time t = t0. After the switch moved, the capacitor was connected to the resistance R, and we want to know what happens to the capacitor voltage after the switch is moved. The “natural response” is one in which the capacitor, with a voltage across it, undergoes a switching event that connects it to a resistance only. As for the inductor, 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. Analysis We begin the analysis at t < t0, i.e., before the switch opened. The circuit for this time domain is shown next: the switch is in position a. Since we know the circuit was in steady state (because it has been like this a long time), we have dvC =0 ⇒ iC =0 . dt This follows from the current-voltage relationship for the capacitor. Since currents and voltages are no longer changing, the capacitor is in steady state, and the capacitor acts like an open circuit, because there is a voltage across it but no current flowing through it. Thus, the voltage across the capacitor is vC = vS. Now we look at the situation for t > t0. The switch has been moved to position b, and now the capacitor sees only a resistance. We don’t need to worry about the current source or the resistor RS at this point. We are no longer in steady state: currents and voltages now begin to change. To find out how, we do a KCL: C dvC (t ) vC (t ) + = 0 dt R Note that the capacitor 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 vC(t). Solution This equation has the same form as that for the inductor current, so we can write down the solution directly. The only difference is that the constant in the exponent changes to the 1/RC. = vC ( t ) vC ( t0 ) e − ( t − t0 ) RC t ≥ t0 . Valid Time Range As for the inductor current, the capacitor voltage is valid at the time t = t0, and at times greater than this value. Initial Condition For a complete solution, we need the initial condition vC (t0 ) . We found earlier that before the switching event, the voltage across the capacitor was vS. We then make use of the capacitor property that we cannot have an instantaneous change in voltage across a capacitor. That that means the capacitor voltage immediately before the switch moved, the voltage across the capacitor immediately after the switch moved, voltage we found above, that is, vC (t0− ) is equal to vC (t0+ ) . But vC (t0− ) is the vC (t0− ) = vS , and vC (t0+ ) is what we have called vC (t0 ) . So, + v= v= vS . This is the initial condition we need in order to solve the equation. C (t0 ) C (t0 ) Typically, C and t0 will be known, and we need to find the initial condition and the resistance seen by the capacitor, R. The resistance R will be the Thevenin equivalent resistance seen by the capacitor. 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 voltage as vC (t0 ) ≡ V0 . So in this case we have, as an initial condition, V0 = vS (the value of the voltage source The Time Constant The form of the response can be simplified by noting that the units of RC are in fact s-1 (inverse seconds): = [ RC ] Ohm = Farad We can then write = vC ( t ) V0e − ( t − t0 ) As for the inductor current, we indicate that at time t0 and later. It is not valid for t ≥ t0 . t ≥ t0 , meaning that the solution as written is valid t < t0 . If we simplify by setting t0 = 0, i.e., we have τC OhmVolt = s −1 . Amps = vC ( t ) V0e −t τL t≥0 . We can interpret this equation to mean that the capacitor voltage starts at some initial value V0, and then decays exponentially in time once it is removed from the voltage source and connected to the resistor. In the figure below we show a plot of vC ( t ) for a long enough time that the exponential has decayed to 0 (or at least to a very small value). The initial voltage vS is chosen to be 10 [mV] 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 = τC . At this point, the voltage has decayed to a value given by vC ( t = V0e τC ) = −τC τC e −1V0 . = The numerical value of e-1 is approximately 0.368, or roughly 3/8. 12 vC(t) [mV] 10 8 vC(t = τC) = e-1V0 6 4 2 0 0 10 20 30 40 50 60 t [ms] As for the inductor, after five time constants, or t = 50 [ms], the voltage has decayed almost to 0. What is left is less than 1% of the initial value: t = 5τC ⇒ vC ( t ) = 0.0067vC (t0 ) . As we pointed out above, we define a “long time” as five time constants. The Capacitor is an Open Circuit in Steady State Just as in inductor problems, we need to know that the switch has been in its initial position for a long time so that we know the circuit is in steady state. Looking at the solution for the capacitor voltage, we see that after a long time, the voltage becomes 0, and again there is no longer any change in voltage or current. In other words, the circuit has returned to the steady state. We also said above that in steady state, the capacitor acts like an open circuit because although there is a voltage across it, there is no current flowing in it. The capacitor behaves like an open circuit while the switch is in its initial position, and later returns to an open a long time after the switching event. In between these times, the capacitor current and voltage are changing exponentially. Power and Energy Delivered to R We can find expressions for the power delivered to the resistor and the energy dissipated by it as a function of time. Note that these equations are valid only for t ≥ t0 . 2 C 2 (t ) R V0 R e pabs= v= by R (t ) −2( t −t0 ) τC −2( t −t0 )   1 τC 2 w= ( t ) = p dt C V 1 − e   abs by R ∫t0 abs,R 2 0   t As we expect, after a long time, the total energy dissipated in the resistor is 1 C V0 2 , i.e., it is the 2 total energy delivered by the capacitor, which was initially had a voltage vC (t0 ) = V0 . We can also find the current through the capacitor as follows. dvC d  − ( t − t0 ) τ C  iC ( t ) C= C  V0e = . dt dt   Then,  1  −(t −t0 ) τC iC ( = t ) C V0 ⋅  −  e t > t0  τC  V0 −(t −t0 ) τC = e t > t0 . R We note that although the voltage across the capacitor cannot change instantaneously, the current through it can change instantaneously, and in this case it does. Immediately before the switch was moved, the capacitor current was 0. However, at t = t0 in the expression above, the current through the capacitor is iC ( t0 ) = V0 . Therefore, the capacitor current has in fact changed R instantaneously.