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Filename: DPKC_Mod06_Part04

Dynamic Presentation of Key Concepts Module 6 – Part 4 Special Cases and Approaches with First Order Circuits. Filename: DPKC_Mod06_Part04.ppt. Overview of this Part Special Cases and Approaches with First Order Circuits. In this part of Module 6, we will cover the following topics:

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Filename: DPKC_Mod06_Part04

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  1. Dynamic Presentation of Key Concepts Module 6 – Part 4Special Cases and Approaches with First Order Circuits Filename: DPKC_Mod06_Part04.ppt

  2. Overview of this PartSpecial Cases and Approaches with First Order Circuits In this part of Module 6, we will cover the following topics: • Finding Initial and Final Values • Sequential Switching • Negative Time Constants • Generalized Solution for First Order Circuits

  3. Textbook Coverage This material is introduced in different ways in different textbooks. Approximately this same material is covered in your textbook in the following sections: • Circuits by Carlson: Section #.# • Electric Circuits 6th Ed. by Nilsson and Riedel: Section #.# • Basic Engineering Circuit Analysis 6th Ed. by Irwin and Wu: Section #.# • Fundamentals of Electric Circuits by Alexander and Sadiku: Section #.# • Introduction to Electric Circuits 2nd Ed. by Dorf: Section #-#

  4. 6 Different First-Order Circuits There are six different STC circuits. These are listed below. • An inductor and a resistance (called RL Natural Response). • A capacitor and a resistance (called RC Natural Response). • An inductor and a Thévenin equivalent (called RL Step Response). • An inductor and a Norton equivalent (also called RL Step Response). • A capacitor and a Thévenin equivalent (called RC Step Response). • A capacitor and a Norton equivalent (also called RC Step Response). These are the simple, first-order cases. We have found the solutions for the inductive currents and capacitive voltages in the last two parts of this module.

  5. Finding Initial and Final Values In the step response solutions that we found in the last part, there were three parameters: • The time constant • The initial condition (at the time of switching) • The final value (a long time after the time of switching) The key in these solutions is finding these three items.

  6. Finding the Time Constant – 1 The time constant is L/R in an RL circuit, and RC in an RC circuit. If there is more than one inductor, we start by finding the equivalent inductance. Similarly, if there is more than capacitor, we start by converting them to a single capacitor. If we can’t take this step of converting to an equivalent with only one inductor or capacitor, we can’t use this method. The inductors L3 and L2 are in parallel, and that combination is in series with L1.

  7. Finding the Time Constant – 2 After converting to an equivalent with only one inductor or capacitor, we need to find the resistance seen by that inductor or capacitor. This means finding the Thévenin resistance seen by the inductor or capacitor. The inductor L sees an equivalent resistance, or Thévenin resistance, of RS1 in parallel with RS2. In some cases there may be dependent sources present. If so, we would use our test-source method to find the Thévenin resistance.

  8. Finding the Initial Condition – 1 In finding the initial condition, we usually are given the information that the circuit has been in some condition “for a long time”. In that case, the voltages and currents will stop changing. As a result, the inductor behaves like a short circuit, and the capacitor behaves like an open circuit. We are making an assumption here that should be kept in mind. The voltages and currents will only stop changing in the six circuits we are studying, or circuits that can be reduced with equivalent circuits to one of those six circuits.

  9. Finding the Initial Condition – 2 If we are told that the circuit has been in some condition “for a long time”, and the voltages and currents stop changing, then an inductor behaves like a short circuit, and a capacitor behaves like an open circuit. We can make this replacement, and then the circuit becomes a fairly straightforward circuit to solve. In the circuit shown here, we are told that it has been in this condition for a long time before t = 0. Therefore, for t < 0, the switch is closed, and we would replace the capacitor with an open circuit, and solve for vC. Note that this will have to be equal to vC(0), since it was valid just before zero, and can’t change instantaneously.

  10. Finding the Initial Condition – 3 In this example, we have replaced the capacitor with an open circuit for t < 0. We solve then for vC(0). In order to find vC(0), we first solve for the current iX, which is Using this, we can get vC(0), which is

  11. Finding the Initial Condition – 4 It should be noted that the initial condition cannot always be found this way, since the circuit may not have been in a given condition for a long time. However, there must be some way that the initial condition can be found, if we are to find the solution. In some cases, an expression for the capacitive voltage or inductive current may be known, in which case we simply evaluate at the time of switching. Let’s imagine that in this circuit, we knew that We would plug in to get

  12. Finding the Final Value – 1 In finding the final value, we are looking for the value after “a long time”. Similar to the situation in finding the initial condition, this will be when the voltages and currents stop changing. As a result, again, this is when the inductor behaves like a short circuit, and the capacitor behaves like an open circuit. We are making an assumption here that should be kept in mind. The voltages and currents will only stop changing in the six circuits we are studying, or circuits that can be reduced with equivalent circuits to one of those six circuits.

  13. Finding the Final Value – 2 In this example, we have replaced the capacitor with an open circuit for a long time after the switching, or t >> 0. We might also refer to this as t = ¥. In this circuit, the value of vC(¥) is clear from the fact that the current through RS1 is zero, and

  14. Sequential Switching – 1 Now, we will consider the case where there is more than one switching event, which occur at more than one point in time. Remember that in general we define the time t = 0 as the time when the switching takes place. However, when there is more than one time involved, we can’t make time equal to zero at two different times in the same problem. Usually, we assign the time t = 0 to the first event, and then have all other events at times relative to that point in time. The question, then, is what to do with the general solution equation when this occurs.

  15. Sequential Switching – 2 The question, then, is what to do with the general solution equation when there is more than one switching event, which occur at more than one point in time. Conceptually, what we do is to apply a transformation of variables to the time variable. Specifically, we take a solution that would be valid with t' = 0 for that switching event, and then translate it back to the time variable t, which has another origin. In the example time line below, note that the time of the second switching was t = 2[s]. If we had a solution that would work for t' = 0, we can transform it to the time variable t, by replacing t' with t – 2[s], since t' = t – 2[s].

  16. Sequential Switching – 3 The general solution equation when there is more than one switching event, which occur at more than one point in time, applies a transformation of variables. The general solution has the quantity t – t0, wherever the variable t would otherwise occur. The other key is that the initial condition for this time period occurs at the time of the second switching, and holds for the time after the second switching. The general solution then becomes

  17. Sequential Switching – 4 Note that this new general solution includes, as a special case, the solution when the switching occurs at t = 0. In this case, t0 = 0, and this solution reduces to the solution that we had previously.

  18. Sequential Switching – Example – 1 Let's solve this example problem, below. The problem has two switches, called SWA and SWB. We are told that switch SWA was closed for a long time, and switch SWB was open for a long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t = 1[s], switch SWA opened. We wish to find the voltage across the capacitor, vC(t), for t³ 0.

  19. Sequential Switching – Example – 2 The problem has two switches, called SWA and SWB. We are told that switch SWA was closed for a long time, and switch SWB was open for a long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t = 1[s], switch SWA opened. We wish to find the voltage across the capacitor, vC(t), for t³ 0. Our first step is to redraw the circuit for t < 0. We replace the capacitor with an open circuit, since it has been in the condition for a long time, and the voltages and currents have stopped changing. We close switch SWA, and open switch SWB, since that was where they were for t < 0. We have the circuit that follows. Clearly, the voltage vC(0) = 9[V].

  20. Sequential Switching – Example – 3 The problem has two switches, called SWA and SWB. We are told that switch SWA was closed for a long time, and switch SWB was open for a long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t = 1[s], switch SWA opened. We wish to find the voltage across the capacitor, vC(t), for t³ 0. The next step is to redraw the circuit for 0 <t< 1[s]. This is the time period until the next switching occurs. During this time, the capacitor acts like a capacitor; we do not replace it with anything else. We close switch SWA, and close switch SWB. We have the circuit that follows. We want the time constant, which means we need the equivalent resistance seen by the capacitor.

  21. Sequential Switching – Example – 4 The problem has two switches, called SWA and SWB. We are told that switch SWA was closed for a long time, and switch SWB was open for a long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t = 1[s], switch SWA opened. We wish to find the voltage across the capacitor, vC(t), for t³ 0. To find the equivalent resistance seen by the capacitor, we remove the capacitor, and find the resistance seen by the two terminals of the capacitor, which we will name A and B. We next set the independent sources equal to zero, which in this case means that they act like short circuits. We have the circuit that follows. The equivalent resistance seen by the capacitor, that is, seen by terminals A and B, will be RS1||RS2, or 5[W]||8[W] which is 3.1[W]. This gives us the time constant for this time period, t = REQC = 6.2[s] for 0 < t < 1[s].

  22. Sequential Switching – Example – 5 The problem has two switches, called SWA and SWB. We are told that switch SWA was closed for a long time, and switch SWB was open for a long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t = 1[s], switch SWA opened. We wish to find the voltage across the capacitor, vC(t), for t³ 0. The next step find the final value for the circuit for 0 £t£ 1[s]. This will be the circuit valid when the voltages and currents would have stopped changing. Note that this will not happen for this time period, since it is less than a time constant. The voltage never reaches this final value. However, we need it to get the equation for the solution. We have the circuit that follows. We solve for this voltage in two parts. First, we find the current, iX,

  23. Sequential Switching – Example – 6 The problem has two switches, called SWA and SWB. We are told that switch SWA was closed for a long time, and switch SWB was open for a long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t = 1[s], switch SWA opened. We wish to find the voltage across the capacitor, vC(t), for t³ 0. Now we can write the expression for the voltage vC(t) for 0 £t£ 1[s]. We have found the initial condition, the time constant, and the final value. We can write the solution as

  24. Sequential Switching – Example – 7 The problem has two switches, called SWA and SWB. We are told that switch SWA was closed for a long time, and switch SWB was open for a long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t = 1[s], switch SWA opened. We wish to find the voltage across the capacitor, vC(t), for t³ 0. Now we want to find the solution for t ³ 1[s]. To do this, we need to again find the initial condition, time constant, and final value, this time for the new time period, during which switch SWA is open. That circuit is shown below. The portion of the circuit to the left of the capacitor does not affect the rest of this solution, and we will remove it.

  25. Sequential Switching – Example – 8 The problem has two switches, called SWA and SWB. We are told that switch SWA was closed for a long time, and switch SWB was open for a long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t = 1[s], switch SWA opened. We wish to find the voltage across the capacitor, vC(t), for t³ 0. Now we want to find the initial condition of the solution for t³ 1[s]. Here, the solution for 0 £t£ 1[s] has not had time to reach a steady state, thus the voltages and currents do not stop changing. However, we can get the initial value for the capacitive voltage, since it cannot change instantaneously. Substituting the time t = 1[s] into the solution for 0 £t£ 1[s], we get

  26. Sequential Switching – Example – 9 The problem has two switches, called SWA and SWB. We are told that switch SWA was closed for a long time, and switch SWB was open for a long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t = 1[s], switch SWA opened. We wish to find the voltage across the capacitor, vC(t), for t³ 0. Next, we want to find the time constant of the solution for t³ 1[s]. Here, the equivalent resistance seems fairly clear, and we can see that the equivalent resistance will be 8[W], after we have set the independent source equal to zero. The time constant will be RS2C, which will be

  27. Sequential Switching – Example – 10 The problem has two switches, called SWA and SWB. We are told that switch SWA was closed for a long time, and switch SWB was open for a long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t = 1[s], switch SWA opened. We wish to find the voltage across the capacitor, vC(t), for t³ 0. Next, we want to find the final value of the solution for t³ 1[s]. Here, the solution seems fairly clear. For t = ¥, the final value can be obtained when the capacitor is replaced by an open circuit, since the voltages and currents will have stopped changing. We get the circuit below, and we can solve, to get

  28. Sequential Switching – Example – 11 The problem has two switches, called SWA and SWB. We are told that switch SWA was closed for a long time, and switch SWB was open for a long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t = 1[s], switch SWA opened. We wish to find the voltage across the capacitor, vC(t), for t³ 0. Now we can write the expression for the voltage vC(t) for t³ 1[s]. We have found the initial condition, the time constant, and the final value. We can write the solution as

  29. Sequential Switching – Example – 12 The problem has two switches, called SWA and SWB. We are told that switch SWA was closed for a long time, and switch SWB was open for a long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t = 1[s], switch SWA opened. We wish to find the voltage across the capacitor, vC(t), for t³ 0. So, our complete solution, which has two parts, is the following,

  30. Note 1 – Sequential Switching Notice, as you go through the previous slides in this example, that we switched back and forth between < and >, and £ and ³. This has been done carefully. Note that when we talk about the capacitive voltage, vC, we used £ and ³. When we talk about anything else, such as the circuit as a whole, we use < and >, since there can be step changes in any other quantity.

  31. Negative Time Constants – 1 A circuit that illustrates what happens with a negative time constant is shown below. The switch was closed for a long time, and then opened at t = 0. We will solve this circuit to find vX(t) for t > 0. Before we start, though, let us consider what we should expect from a negative-valued time constant.

  32. Negative Time Constants – 2 Let's start by remembering what happens when we have a positive-valued time constant. We get a solution that moves from an initial condition, towards a final value. It moves exponentially, meaning that it approaches the final value, getting closer and closer but never actually reaching the final value, which occurs when things stop changing. The final value is an steady-state value, and is found with all derivatives equal to zero, which means that inductors become short circuits, and capacitors become open circuits.

  33. Negative Time Constants – 3 Now, notice what happens when we extend the range of time for this solution to t < 0. This is the same solution, but plotted for negative values of time. Note that the time constant is still positive. The solution is decreasing rapidly as time goes more and more negative, that is, moving to the left.

  34. Negative Time Constants – 4 Next, see what happens when we change the time constant to a negative value. We have also changed the time period to that from –300[ms] to +100[ms]. This looks just like our last plot, but rotated around the ordinate (vertical axis). Note in particular that we have the same value at t = 0, and we have the same steady state value, which is 10[mA].

  35. Negative Time Constants – 5 The key, then, is that when we have a negative time constant, our solution technique really does not change. The final value is no longer a final value, but it is still the steady-state value, and is found in the same way as with a positive time constant. That is, we replace the inductor with a short circuit, and the capacitor with an open circuit. We should really refer to the final value as the steady-state value, which happens to occur at t = ¥ for the case where the time constant is positive.

  36. Negative Time Constant Example – 1 We will start, as always, by defining the inductive current. Then, our next step is to find the initial condition. For this circuit, that means drawing the circuit for t < 0. Remember that the circuit had been in the given condition for a long time. Let's solve this circuit, which will have a negative-valued time constant. The switch was closed for a long time, and then opened at t = 0. We will solve this circuit to find vX(t) for t > 0.

  37. Negative Time Constant Example – 2 We start by finding the initial condition. We redraw the circuit for t < 0. The switch was closed. Remember that the circuit had been in the condition for a long time, so we replace the inductor with a short circuit. When we see this circuit, it is clear that vX is zero. In addition, there is no current through the 30[W] and 50[W] resistors. Thus, we can say that Let's solve this circuit, which will have a negative-valued time constant. The switch was closed for a long time, and then opened at t = 0. We will solve this circuit to find vX(t) for t > 0.

  38. Negative Time Constant Example – 3 Next, we need to find the time constant. We redraw the circuit for t > 0. The switch is now open. Let's solve this circuit, which will have a negative-valued time constant. The switch was closed for a long time, and then opened at t = 0. We will solve this circuit to find vX(t) for t > 0.

  39. Negative Time Constant Example – 4 With the circuit redrawn for t > 0, we can find the time constant. We want the equivalent resistance seen by the inductor. Thus, we remove the inductor, and then set the independent source equal to zero. Then, since we have a dependent source present, we apply a test source to the terminals of the inductor, A and B. Let's solve this circuit, which will have a negative-valued time constant. The switch was closed for a long time, and then opened at t = 0. We will solve this circuit to find vX(t) for t > 0. Writing KCL at top node, we get

  40. Negative Time Constant Example – 5 We want the equivalent resistance seen by the inductor. Solving for vD, we get Let's solve this circuit, which will have a negative-valued time constant. The switch was closed for a long time, and then opened at t = 0. We will solve this circuit to find vX(t) for t > 0. Then, we write KVL around the loop, and we get The equivalent resistance is

  41. Negative Time Constant Example – 6 From this result, we can find the time constant, which is Let's solve this circuit, which will have a negative-valued time constant. The switch was closed for a long time, and then opened at t = 0. We will solve this circuit to find vX(t) for t > 0. Next, we need to find the steady-state value of the current through the inductor.

  42. Negative Time Constant Example – 7 The steady-state value of the inductive current can be found by redrawing the circuit, for the switches in the t > 0 positions, and the inductor replaced by a short circuit. We have the circuit shown below. Let's solve this circuit, which will have a negative-valued time constant. The switch was closed for a long time, and then opened at t = 0. We will solve this circuit to find vX(t) for t > 0. Solving, we have

  43. Negative Time Constant Example – 8 Thus, our solution for the inductive current is Let's solve this circuit, which will have a negative-valued time constant. The switch was closed for a long time, and then opened at t = 0. We will solve this circuit to find vX(t) for t > 0. Since this is the current through the 10[W] resistor, we can say that

  44. Note 1 – Negative Time Constants Note that the voltage vX made a step jump in this problem. Just before t = 0 (at t = 0-) the voltage vX was zero, because of short across it. Just after t = 0 (at t = 0+) the voltage vX was 10[mV]. This is one of the reasons why it is a good idea to solve for the inductive currents and capacitive voltages first.

  45. Note 2 – Negative Time Constants Some students are concerned about the practical implications of a voltage which increases exponentially with time. This concern is reasonable. In practice, such a circuit cannot continue like this for a very long time. Note also that this circuit is an exception to our rule that the six circuits we are studying always move to steady-state conditions “after a long time”. Such circuits are useful, however. They require an amplifier (the dependent source) to be able to provide the energy required. This kind of situation typically results from positive feedback, which is considered “unstable”. More about all of this may be covered in your future electronics courses.

  46. Complete Generalized Solution We are now ready for our most generalized version of the solution of first order circuits. We use xss for the steady-state value, which is also the final value when the time constant is positive. We use t0 for the time of switching, so that it can be at times other than t = 0. Using all this, we get the following general solution, In this expression, we should note that t is L/R in the RL case, and that t is RC in the RC case. The expression for greater-than-or-equal-to (³) is only used for inductive currents and capacitive voltages. Any other variables in the circuits can be found from these.

  47. Generalized Solution Technique – Step Response To find the value of any variable in a Step Response circuit, we can use the following general solution, • Our steps will be: • Define the inductive current iL, or the capacitive voltage vC. • Find the initial condition, iL(t0), or vC(t0). • Find the time constant, L/REQ or REQC. In general the REQ is the equivalent resistance as seen by the inductor or capacitor, and found through Thévenin’s Theorem. • Find the final value, iL,SS, or vC ,SS. • Write the solution for inductive current or capacitive voltage using the general solution. • Solve for any other variable of interest using the general solution found in step 5).

  48. 6 Different First-Order Circuits The generalized solution that we just found applies only to the six different STC circuits. • An inductor and a resistance (called RL Natural Response). • A capacitor and a resistance (called RC Natural Response). • An inductor and a Thévenin equivalent (called RL Step Response). • An inductor and a Norton equivalent (also called RL Step Response). • A capacitor and a Thévenin equivalent (called RC Step Response). • A capacitor and a Norton equivalent (also called RC Step Response). If we cannot reduce the circuit, for t > 0, to one of these six cases, we cannot solve using this approach.

  49. Isn’t this situation pretty rare? • This is a good question. Yes, it would seem to be a pretty special case, until you realize that with Thévenin’s Theorem, many more circuits can be considered to be equivalent to these special cases. • In fact, we can say that the RL technique will apply whenever we have only one inductor, or inductors that can be combined into a single equivalent inductor, and no capacitors. • A similar rule holds for the RC technique. Many circuits fall into one of these two groups. • Note that the Natural Response is simply a special case of the StepResponse, with a final value of zero. Go back to Overview slide.

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