EE302-4-Advanced Topics

EE302-4-Advanced Topics - DC Circuit Theorems Superposition...

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Unformatted text preview: DC Circuit Theorems Superposition Superposition In this lecture, you will: Learn the principle of superposition Learn how to apply the principle of superposition to solve DC circuits Understand the tradeoffs in applying superposition versus other circuit analysis approaches 3 Linearity Linear operators: Add and subtract Multiply and divide by constants Differentiate and integrate Nonlinear operators: Exponents other than 1 • For example, squares and square roots Logarithms and many other functions. 4 Linear Circuit Elements Resistors are linear circuit elements: V = IR is clearly a straight line. Ideal capacitors and inductors are linear. Independent sources are linear. Dependent sources are linear provided that the output voltage or current is proportional to the first power of a specified voltage or current (or a sum of these). 5 Nonlinear Circuit Elements Semiconductor devices are nonlinear. Diodes Transistors Devices using ferromagnetic materials are nonlinear. Iron core inductors. Many other real world devices are nonlinear although their response may be approximately linear over a limited range. 6 Linear Circuits A linear circuit is a circuit composed exclusively of linear circuit elements. + 50 V ­ 5 Ω 10 Ω 40 Ω 3 A 7 Superposition Superposition is a property of all linear systems: f (ax + by ) = a[ f ( x )] + b[ f ( y )] 8 Superposition Definition: Any linear system obeys the principle of superposition, which states that whenever a linear system is excited or driven by more than one independent source of energy, the total response is the sum of the individual responses that result from each independent source acting alone. 9 What does this mean? Since we are dealing with linear circuits, we can determine the response of the circuit to each independent source acting alone and then sum the results. Of course, superposition only makes sense when there is more than one independent source. 10 Superposition Example 2A 3A 10Ω + VR ­ V = IR VR ( 2 A) = 2 × 10 = 20.0 V VR ( 3 A) = −3 × 10 = −30.0 V VR ( Total ) = VR ( 2 A) + VR ( 3 A) = 20 − 30 = −10.0 V 11 Superposition in Circuits The response in a linear circuit having more than one independent source can be obtained by adding the responses caused by the separate independent sources acting alone. How do I find the response of the circuit to one source acting alone? Turn off all of the other sources. 12 Turned Off Voltage Source I VS=0 VS V a Short Circuit I VS b a Voltage source always provides the specified voltage between its terminals. Turned off means VS = 0 volts + ­ b Current determined by the attached circuit. Behaves like a short circuit. 13 Turned Off Current Source I Current source always provides the specified current. IS V IS=0 a a + Open Circuit Voltage determined by the attached circuit. Turned off means IS = 0 amps V Behaves like an open circuit. IS ­ b b 14 Applying Superposition 1. Given a circuit, turn off all but one independent source. Replace a current source with an open circuit. Replace a voltage source with a short circuit. Leave dependent sources in place. 1. Solve for the requested voltage or current. 2. Repeat for each independent source. Must use a consistent reference directions! 1. Sum the results from each source to obtain the final answer. Example: three independent sources means you solve the circuit three times and then sum. 15 Example Find V1 + 50 V V1 5 Ω 10 Ω 40 Ω 3 A ­ 16 Example­­Step 1 Set the current source to zero V1 + 5 Ω 50 V 10 Ω 40 Ω ­ Req= 10 Ω ║ 40Ω = 8Ω Using a voltage divider approach, V1A = 50 Req/(Req + 5) = 50 (8/(8+5)) = 30.77 V 17 Example­­Step 2 Set the voltage source to zero V1 5 Ω 10 Ω 40 Ω 3 A The three resistors are now in parallel 1/Req = (1/5 + 1/10 + 1/40) Req = 3.077 Ω V1B = IR = 3(3.077) = 9.23V 18 Example­­Step 3 Sum the responses from each source V1 + 50 V ­ 5 Ω 10 Ω 40 Ω 3 A V1 = V1A + V1B = 30.77 + 9.23 = 40.0 V In this example, the circuit was solved using simple resistive manipulations Some problems may be more complicated using superposition if you need to use node voltage or mesh currents to solve 19 A few words of caution…. Suppose you are asked to find the power in the 10Ω resistor in Example 1 using superposition Can you use superposition to solve it? Yes, but only after finding the total voltage across the 10Ω resistor or the total current in the 10Ω resistor. Why can’t you find the power in the resistor from each source and then sum to get the answer? 20 Superposition and Power Consider the power in the 10­ohm resistor of Example 1: V1(50V) = 30.77 volts • Power in 10 Ω resistor = (V1)2/10 = (30.77)2/10 = 94.7W V1(2A) = 9.23 volts • Power in 10 Ω resistor = (V1)2/10 = (9.23)2/10 = 8.52W If we then add the results, • Power in 10Ω resistor = 94.7 + 8.52 = 103.2 W However, if we use the final calculated value: V1 = 40 V, • Power in 10 Ω resistor = (V1)2/10 = (40)2/10 = 160. W Power is a nonlinear function: (VA)2 + (VB)2 ≠ (VA+VB)2 160. W is the correct answer Superposition applies to voltage and current, not power! 21 Superposition and Power Suppose we use superposition on a circuit containing two independent sources to find the voltage V across a resistor R We solve using each independent source and obtain two voltages: VA and VB V = VA + VB If we tried to use superposition to compute power: P = VA2/R + VB2/R Correct approach: Power in the resistor = V2/R V2/R = (VA + VB)2/R = VA2/R + VB2/R + 2VAVB/R We would come to the same conclusion if we had chosen to solve for current instead of voltage 22 Superposition and Power Conclusion: Superposition may be used to find any current or voltage in a circuit With this current or voltage, power can be computed You cannot use a superposition approach to calculate power directly 23 When do I use superposition? Superposition is required only when the independent sources in a circuit are fundamentally different Example: a combination of DC and AC sources In the DC circuits we are analyzing, superposition is never required For some simple circuits, superposition may offer a reasonable alternative to node voltage or mesh current techniques. For most circuits, superposition is more difficult because you must solve the circuit multiple times. 24 Why learn to apply superposition? Superposition is an important and powerful principle that will be used on more complex linear systems that are driven simultaneously be different types of energy sources. EE302 is preparing you with this fundamental understanding of how to apply superposition 25 Standard Example Find the power dissipated by the 8­ohm resistor. 20 Ω 10 Ω + 1A 20 Ω 8Ω 10 V - 26 4.5 W In­class Exercise #1 Find v using superposition 4 A 2Ω + 5 Ω 110 V + 10 Ω v 12 Ω ­ ­ v = 50.0 V 27 In­class Exercise #2 Find v using superposition 8 Ω + 128 V ­ 18 Ω 10Ω + 70 V ­ + 48 Ω v 20 Ω ­ v = 96.0 V 28 In­class Exercise #3 Find ix using superposition 3ix 24 V + ­ ix = 1.00 A 2 Ω 4A 8 Ω ix 29 Superposition Summary Learn the principle of superposition. Learn how to apply the principle of superposition to solve DC circuits. Understand the tradeoffs in applying superposition versus other circuit analysis approaches. 30 Superposition Questions? Source Transformations Source Transformations In this lecture, you will: Learn to transform voltage sources and series resistors into current sources with parallel resistors and vice versa Learn to apply this technique as another tool in solving circuit problems 33 One Port Networks Sometimes we only care about the way a linear network behaves at its + V output terminals ­ We can represent this as a black box with a pair of terminals This black box is called a one-port network I Linear network 34 One Port Network The behavior of this one­port network can be completely + described by the relationship V between voltage and current at ­ the terminals We don’t care about the contents of the black box as long as the I­V characteristic is correct I Linear network 35 One Port Network­­Example 10Ω 10Ω 5Ω To an outside observer, these two networks are identical 36 Source Transformations We can use a similar approach to convert between voltage and current sources to simplify circuits We need to find the relationship between Is, Vs, and R that guarantees the two configurations are equivalent + ­ Vs R R Is 37 Add a Load Resistor IL + ­ R Vs RL IL IL Is R RL IL Vs = R + RL R = I s( ) R + RL 38 The Load Currents Must Be Equal If the “boxes” are equivalent, the load resistor can’t tell the difference. In order for the two “boxes” to be equivalent, the load currents must be equal for all values of load resistor. IL Vs R = = I s( ) R + RL R + RL Is Vs = R 39 Compare The Two Boxes IL Open circuit voltage: + ­ R Vs RL Vs Short circuit current: Vs/R Voltage across RL: IL R Is= Vs/R RL VsRL/(R+RL) Open circuit voltage: (Vs/R)*R = Vs Short circuit current: Is = Vs/R Voltage across RL: VsRL/(R+RL) 40 I­V Characteristic for Both Networks I­V Characteristic graphically describes the terminal behavior Vs/R of both networks Vs/R R + ­ Vs IL (Amps) Slope = ­1/R RL R RL Vs 41 VL (Volts) Example #1 4Ω + 6V - 6Ω 30 Ω 5Ω 20 Ω + 40V - 10 Ω Find the power associated with the 6V source Three options: Node voltage method: 3 equations with three unknowns Mesh current method: 3 equations with three unknowns Source transformation Since we are only asked to find the power in the 6V source, it may be easier to use source transformation 42 Example #1 ­ Step 1 5Ω + 40V - 8A 5Ω Is= Vs/5 = 8A + 6V ­ 6 Ω 4 Ω 20 Ω 30 Ω 10 Ω 5 Ω 8 A 20 ║5 = 4Ω 43 Example #1 ­ Step 2 Vs= IsR = 8(4)=32V 4Ω + 32V - 8A 4Ω + 6V ­ 6 Ω 4 Ω 4 Ω 30 Ω + 32 V ­ 10 Ω Req = 6+4+10=20Ω 44 Example #1 ­ Step 3 20 Ω Is= Vs/R = 32/20=1.6A + 32V - 20 Ω 1.6 A 30 ║ 20 = 12 Ω 4 Ω + 6V ­ 30 Ω 20 Ω 1.6 A 45 Example #1 ­ Step 4 Vs= IsR = 1.6(12) = 19.2V + 19.2V - 1.6 A 12 Ω 12 Ω 12 Ω 4 Ω + 6V + I 19.2 V ­ ­ ­6 + (4+12)I + 19.2 = 0 I = ­.825A P6v = ­VI = ­(­0.825) x 6 = 5.00 watts absorbed 46 Compare The Two Boxes + ­ R Vs RL IL The resistor R has the same value in both circuits. But it is not the same resistor. • What is the power dissipated in R in the open circuit and short circuit cases? IL R Is= Vs/R Only the terminal behavior is equivalent. RL Do not transform any resistor that is required for a dependent source or the problem solution. 47 Special Case Resistors in series with current sources have no impact on source transformations R1 Is= Vs/R R RL R RL Is= Vs/R Resistors in series with a current source may be neglected for source transformations, but MUST be included for power calculations 48 Special Case Resistors in parallel with a voltage source have no impact on source transformations. + Vs ­ R1 R RL + ­ Vs R RL Resistors in parallel with a voltage source may be neglected for source transformations, but MUST be included for power calculations. 49 Standard Example Find the power dissipated by the 8­ohm resistor. 20 Ω 10 Ω + 1A 20 Ω 8Ω 10 V - 50 4.5 W Practice Problem #1 Using a series of source transformations, calculate Vo 1A 5Ω 6A 12Ω 20Ω 10Ω 120Ω 40Ω + - 72Ω 20 V + Vo ­ 30Ω Vo = 16.0 V 51 Practice Problem #2 Find V and the power delivered or absorbed by the 120V source 1.6Ω 20Ω 120V + ­ + 60 V 5Ω 36A 6Ω + 8Ω Vx ­ Vx = 48.0V, P120V = 420. W delivered 52 Practice Problem #3 2.3KΩ 2mA Io 2.7KΩ 1KΩ 0.6mA 1. Find Io using a series of source transformations 2. Find Io using the node voltage method 3. Find Io using the mesh current method Io = ­1.00 mA 53 Questions? Source Transformations Thévenin Equivalent Circuits Thévenin Equivalent Circuits The objectives of this lecture are to: Define what is meant by an equivalent circuit. Learn how to define the two types of equivalent circuit: • Thévenin equivalent circuit • Norton Equivalent Circuit How to use equivalent circuits to simply the analysis of complex circuits. 56 Why are Equivalent Circuits Needed? We have learned a number of methods and tools for analyzing electrical circuits: KVL, KCL, and Ohm’s Law Current and Voltage Division Node Voltage and Mesh Current Analysis Superposition Source transformation Problems exists in analyzing circuits where part of the circuit is complex or unknown: Appliance plugged into a wall outlet. Attaching speakers to a stereo system. 57 Speakers for a CD Player CD Reader Digital to Analog Converter Amplifier Speakers I don’t care about the details of the CD player, I want to optimize the speaker design for maximum volume. Only the voltage and current at the speaker terminals are important. 58 Importance of Equivalent Circuits The equivalent circuit allows us to represent most of the circuit in a much simpler form. After determining the equivalent circuit, we know the current and voltage going to any load placed at that point. This allows us to do optimization more easily. 59 Transistor Amplifier i1 R1 vs + - + io vo 20,000 + - 40io Transistor A v1 5000 vo - + R3 + - 70i1 R2 R4 v1 R5 Transistor B We want to know how much power is going to resistor R5. What is important? Only the voltage and current for resistor R5. We are not interested in the voltage and current of every element, just R5. 60 Equivalent Circuits i1 R1 vs + - + io vo 20,000 + 40io - v1 5000 vo - + R3 70i1 R4 v1 R5 Transistor B The goal here is to create an equivalent I R5 - R2 Transistor A Complex Circuit + circuit for the complex one such that: + V ­ Voltage across R5 is the same. Current to R5 is the same. This resistor is traditionally called the LOAD Resistor. 61 Thévenin Equivalent Circuit Any linear DC circuit can be replaced by an equivalent circuit consisting of an ideal voltage source and a series resistor with appropriate values. If the complex circuit contains a dependent source, then the control voltage or current must be within that circuit. RTh I Complex Circuit + V ­ VTh I + ­ Thévenin Equivalent Circuit The voltage and current at the terminals are identical. 62 + V ­ Thévenin Equivalent Voltage First consider the open circuit case: The open circuit voltage of the Thévenin equivalent must equal the open circuit voltage of the original circuit. RTh I = 0 Complex Circuit + VOC - I = 0 + VTh + VOC - ­ VOC = VTh 63 Thévenin Equivalent Resistance Next consider the short circuit case: The short circuit current of the Thévenin equivalent must equal the short circuit current of the original circuit. RTh ISC + V = 0 ­ Complex Circuit I SC VTh = RTh VTh ISC + + V = 0 ­ ­ VOC ⇒ RTh = I SC 64 Thévenin Equivalent Circuit RTH VTH VTh = VOC VOC RTh = I SC 65 Thévenin Equivalent Example Compute the Thévenin Equivalent for this circuit treating the resistor RL as the load. 6Ω 4Ω + 36 V 12 Ω R L = 16 Ω - 66 Thévenin Equivalent Example First, remove the load resistor from the circuit. It is not part of the Thévenin Equivalent. 6Ω 4Ω + 36 V - + 12 Ω VL R L = 16 Ω - 67 Thévenin Equivalent Example Now compute the open circuit voltage. 6Ω 4Ω + + + V12 - 36 V - 12 Ω VOC - Note that VOC = V12 Use the voltage divider equation to find V12: V12 = 36(12/18) = 24 volts = VOC 68 Thévenin Equivalent Example Now compute the short circuit current. 6Ω 4Ω + 36 V 12 Ω ISC - Use the current divider equation to find ISC: I SC 12 4 36 = = 0.75[ 4] = 3 A 4 (6 + 12 4) 69 Thévenin Equivalent Example RTH VTH VTh = VOC = 24 V VOC 24 RTh = = =8Ω I SC 3 70 Thévenin Equivalent Example How much power is delivered to the load resistor? 8Ω + 24 V RL = 16 Ω - PL = RLIL2 = 16(24/24)2 = 16 W 71 Thévenin Equivalent Example Let’s compute PL for the original circuit. 6Ω 4Ω + 36 V 12 Ω IL R L = 16 Ω - 12 20 36 7.5 36 IL = 20 = 13.5 20 = 1 A 6 + 12 20 2 PL = 16 I L = 16 W 72 What about the Value of RL? We didn’t use the value of the load resistor in determining the equivalent circuit. Does this mean that ANY resistor could be placed where the load resistor is? YES!!!!! 73 Why does this work? Resistors are linear elements V = I*R When we set up the equivalent circuit, we look at the extreme values for a resistor Infinite Open circuit Zero Short Circuit Since we know the response at the ends and a resistor is linear, the equivalent circuit works for ANY resistor!! 74 We can replace RL with any resistor Why is this important? The equivalent circuit provides information about the currents and voltage that could be provided to a point in the original circuit. We can either use the original circuit or the equivalent circuit to calculate these values We can then place any load across the output terminals and be able to predict the behavior of the circuit. • For example, the loads could be different kinds of speakers attached to a stereo system. 75 Thévenin Equivalent Summary Any linear DC circuit can be replaced by its Thévenin Equivalent (VTh and RTh). To compute VTh and RTh: 1. Remove the load resistor from the circuit. 2. 1. Helps to make sure you do not include it in the resistance calculation. • 1. • It is often necessary to use KCL to find ISC. Compute the open circuit voltage across the terminals where the load was removed. Compute the short circuit current where the load was removed. VTh = VOC and RTh = VOC/ISC This method works for any linear circuit that contains at least one independent source. 76 The Req Method for Computing RTh Must I calculate ISC and VOC every time to get RTh? NO! For circuits that contain NO dependent sources: Remove the load resistor from the circuit. • Helps to make sure you do not include it in the resistance calculation. Turn OFF all independent sources: • Voltage source become shorts. • Current sources become opens. Calculate the equivalent resistance seen looking into the circuit from the load terminals. RTh = Req Compute VOC (or ISC) in the normal way. 77 Thévenin Equivalent Example Let’s use the Req method on the example circuit: 6Ω 4Ω + 36 V - + 12 Ω VL R L = 16 Ω - 78 Thévenin Equivalent Example Turn off all independent sources 6Ω 4Ω Req 12 Ω Compute the equivalent resistance seen looking into the circuit from the load terminals: Req = 4 + 6 12 = 4 + 4 = 8 Ω = RTh 79 Example 2: Req Method 40 kΩ 0.2 mA 5 kΩ 10 kΩ RL Define the Thévenin equivalent circuit seen by the load resistor RL. 80 Calculate Req 40 kΩ Remove RL Turn off sources 5 kΩ 10 kΩ Req Voltages sources becomes short Current source becomes open RTH is the equivalent resistance seen by RL. RTH = 5kΩ + 10kΩ = 15.0 kΩ 81 Calculate Voc 40 kΩ 5 kΩ + 0.2 mA 10 kΩ Voc - Remove RL Calculate Voc using any technique VTH = 0.2mA ⋅10kΩ = 2.00 V 82 Final Thévenin Equivalent Circuit 40 kΩ 5 kΩ + 10 kΩ 0.2 mA Voc - 15 kΩ 2 V + - RL 83 What about the Value of RL? We never specified a value for RL. Does it matter? No!! 40 kΩ 0.2 mA Current flowing through RL = Voltage across RL = 5 kΩ 10 kΩ RL 10kΩ 0.2mA 15kΩ + RL 10kΩ 0.2mA ∗ RL 15kΩ + RL 84 Original vs. Thévenin Circuit: Current 40 kΩ 5 kΩ 10kΩ I = 0.2mA RL + 15kΩ RL 10 kΩ 0.2 mA 15 kΩ 2V + - I 2 = RL + 15kΩ RL These are equal for any value of RL!! 85 Importance of Equivalent Circuit The equivalent circuit allows us to represent most of the circuit in a much simpler form After determining the equivalent circuit, we know the current and voltage going to any load placed at that point This allows us to do optimization more easily. Although I did this presentation with the load as a resistor, in fact, the load can be anything. Only the circuit replaced by the Thévenin equivalent is required to be linear. The load can be non­linear! This is an important technique used in the analysis of electronic circuits with non­linear elements. 86 Thévenin Equivalent Circuits Summary of the Method: VTh To calculate the open circuit voltage: Remove the load resistor from the circuit Define the open­circuit voltage • If the resistor is vertical, make the positive reference on top • If the resistor is horizontal, make the positive reference to the left. Calculate Voc • Use the methods from before: KVL/KCL, Voltage and Current Division, or Node and Mesh. VTh = VOC 88 Summary of the Method: RTh Calculate RTh using short circuit current: Place a short circuit across the load terminals. Compute the current flowing from top to bottom through the short. • Use any method. You may have to invoke KCL. RTh = VOC/ISC This method works for any circuit with at least one independent source. Calculate RTh using Req: Turn off all independent sources Compute the equivalent resistance seen looking into the circuit from the load terminals. RTh = Req This method works for any circuit that has no dependent sources. 89 Standard Example Find the power dissipated by the 8­ohm resistor. 20 Ω 10 Ω + 1A 20 Ω 8Ω 10 V - 90 4.5 W Thévenin Equivalent: Sign Convention The positive reference for the Thévenin and/or open circuit voltage is at the top. If the load is horizontal, the positive terminal is on the left. The load or short circuit current flows from top to bottom (out of the Thévenin Equivalent circuit). Or, left to right. The load typically absorbs power. This convention is compatible with the passive sign convention. RTh I Complex Circuit + V ­ VTh I + ­ Thévenin Equivalent Circuit 91 + V ­ Thévenin Equivalent Circuits Norton Equivalent Circuits Reminder: Source Transformations + ­ R Vs R Is= Vs/R IL Open circuit voltage: V s RL Short circuit current: V /R s Voltage across RL: V R /(R+R ) sL L IL RL These circuits are equivalent!! Open circuit voltage: (Vs/R)*R = Vs Short circuit current: I = V /R s s Voltage across RL: V R /(R+R ) sL L 93 Types of Equivalent Circuits RTH IN VTH Thévenin Equivalent Circuit RN Norton Equivalent Circuit IN = VTH/RTH RN = RTH We will focus on the Thévenin Equivalent Circuit 94 Questions? Thévenin Equivalent Circuits Norton Equivalent Circuits Thévenin Equivalent Circuits Dependent Sources VOC and ISC Method This method applies for all circuits with at least one independent source – same method you have already learned: Remove load resistor, if present • Calculate the open­circuit voltage (Voc) using any method. Replace the load resistor with a short circuit (i.e., a wire) • Calculate the short circuit current (Isc) using any method. VTh = VOC and RTh = Voc Isc 97 Dependent Source Example A 10Ω IC + 10V ­ 1Ω B 5Ω 5IC 2A Treat the 5­ohm resistor as the load resistor and determine the Th évenin Equivalent. We solved this same circuit earlier. VB = 20 volts 98 Dependent Source Example: VOC A 10Ω IC + 10V 1Ω B + VOC ­ 2A 5IC ­ KCL at Node B: ­Ic + 5Ic = 0 ­­­ The only possible value of IC is zero. Therefore, VOC = 10V. 99 Dependent Source Example: ISC A 10Ω IC + ISC 10V ­ 5IC 2A 10V IC = =1A ∴ 10Ω KCL at Node B : - I C + 5I C + ISC = 0 5I C = 5 A ISC = - 4I C = - 4A R TH 1Ω B VOC 10 = = = ISC -4 - 2.5Ω Can RTh really be negative? What does this mean? 100 Dependent Source Example 10Ω IC + 10V ­ 1Ω 5Ω 2A ­2.5Ω + ­ 5 VL = 10 = 20 V 5 − 2.5 + VL 10V 5IC ­ 5Ω Negative RTh is possible with dependent sources. (More on this later.) 101 Test Source Method for RTh This method can be used to find the Thévenin Equivalent for ANY linear circuit. Test source method: Remove the load from the circuit. Turn OFF all independent sources. Attach a test source to the output terminals: • Either a voltage source or a current source. • A one amp current source is often convenient. Compute the resulting current or voltage. RTh = Vtest/Itest 102 Thévenin Equivalent Example Let’s use the test source method on a simple circuit (same example as before): 6Ω 4Ω + 36 V - + 12 Ω VL R L = 16 Ω - First, turn off the independent sources and replace the load with a test source. Let’s use a 1 amp current source: 103 Test Circuit Example Now, compute the voltage across the test source: VTest = 1[ 4 + 6 12] = 8 V 6Ω 4Ω + 12 Ω 1 A VTest - RTh = VTest/ITest = 8V/1A = 8 Ω Same result as before. 104 Test Source Example 2 A 10Ω IC + 10V ­ 1Ω B 5Ω 5IC 2A Treat the 5­ohm resistor as the load resistor and determine the Th évenin Equivalent. We solved this same circuit earlier. VB = 20 volts 105 Test Source Example 2 10Ω 1Ω IC + 1A 1 + I C = 5I C VTest 5IC ­ ⇒ I C = 0.25 A VTest = −10( 0.25) = −2.5 V VTest RTh = = −2.5 Ω 1 For this problem the test source method is a reasonable approach. 106 The Special Case Circuits with dependent sources, but no independent sources are a special case. 10Ω ­ 0.5VC VC 5Ω + With no independent energy source: VOC = 0 ISC = 0 RTh = VOC 0 = = undefined I SC 0 The test source method is required. 107 The Special Case Circuits with dependent sources, but no independent sources are a special case. IS 0.5VC 10Ω 5Ω ­ + VC V5 + 1A ­ VTest = 30 + 10 = 40 V + VTest = = = ­ VTest RTh = = 40 Ω 1 The test source method is required. 108 Thévenin Equivalent Circuits­­Summary Finding VTH Remove the load resistor if present Define the open­circuit voltage • If the resistor is vertical, make the positive reference at the top • If the resistor is horizontal, make the positive reference to the left Calculate Voc = VTH using any method: node voltage, mesh, source transformation, voltage and current division 109 Summary: Finding RTH Method #1: Equivalent resistance method. No dependent sources are present Remove load resistor if present Turn OFF all independent sources • Replace voltage sources with short circuits and current sources with open circuits. Calculate equivalent resistance seen looking into the circuit from the load terminals. 110 Summary: Finding RTH Method #2: Short­circuit current method. Independent sources are present and dependent sources may be present Remove load resistor if present Calculate VTH using any method Place a short circuit across the terminals where the load resistor was connected Using any method, calculate the short circuit current • Downward for vertical load resistor • Left­to­right for horizontal load resistors RTH = VTH/ISC 111 Summary: Finding RTH Method #3: Test Source Method Dependent and/or independent sources are present. This method always works. Remove load resistor if present Turn OFF all independent sources Apply a test source to the terminals where you want to find RTH • Can be either a current or voltage source. • A one amp current source can be a good choice. Use any circuit analysis method to calculate resulting Vtest (or Itest). RTH = Vtest/Itest 112 Example Problem Find the Req, with respect to RL, for this circuit when V1=10V, R1=2Ω , R2=3Ω , R3=5Ω , and R4=6Ω : R1 + - R2 R4 ix R3 V1 RL ix/3 Method 1 using Req is not applicable due to dependent source. 113 Method #2: Short Circuit Current Method Remove load resistor, if present Calculate the open­circuit voltage (Voc) using any method R1 + - R2 R4 ix Note that ix = 0! R3 V1 ix/3 R3 5 Voc = V 1 = 10 = 5.00 V = VTh R1 + R 2 + R3 2 + 3+ 5 114 Method #2 Continued Replace the load resistor with a short circuit (i.e., a wire) Calculate the short circuit current (Isc) using any method • Downward for vertical resistor • Left­to­right for horizontal resistors R1 + - R2 R4 ix R3 V1 ix/3 Isc 115 Method #2 Con’t R1 + - A R2 R4 B ix R3 V1 ix/3 1 1 i V1 1 + VA − VB − x = R1 R 2 R2 3 R1 1 1 1 1 − VA + + + VB = 0 R2 R 2 R3 R 4 V ix = B R4 Isc 1 1 V1 1 1 + VA − + VB = R1 R 2 R1 R 2 3R 4 1 1 1 1 − VA + + + VB = 0 R2 R 2 R3 R 4 116 Method #2 Con’t R1 + - A R2 R4 B ix R3 V1 ix/3 Isc V1=10V, R1=2Ω , R2=3Ω , R3=5Ω , and R4=6Ω 1 1 V1 1 1 + VA − + VB = R1 R 2 R1 R 2 3R 4 1 1 1 1 − VA + + + VB = 0 R2 R 2 R3 R 4 10 1 1 1 1 + VA − + VB = 2 3 2 3 18 1 1 1 1 − VA + + + VB = 0 3 3 5 6 117 Method #2 Con’t R1 A R2 B R4 ix + - R3 V1 ix/3 Isc Results VB = 3.673V ix = 0.612A isc = 0.612A Req = 5V/0.612A = 8.17Ω 118 Open and Short Circuit Circuits are Different! R1 + - R2 A R4 B ix R3 V1 VB = 5.00 V ix = 0 A ix/3 R1 A R2 B R4 ix + - R3 V1 ix/3 VB = 3.67 V ix = 0.612 A You cannot use a node voltage from one to solve the other! 119 Method #3: Test Source Method Find the Req, with respect to RL, for this circuit when V1=10V, R1=2Ω , R2=3Ω , R3=5Ω , and R4=6Ω : R1 + - R2 R4 ix R3 V1 RL ix/3 120 Method #3: Test Source Method Remove load resistor, if present Turn off any independent sources Voltage sources become shorts Current sources become opens Put a 1­A test source at the place where you want to calculate Req R1 R2 R4 ix R3 1A ix/3 121 Method #3 Con’t R1 A R2 B R4 ix R3 1A ix/3 ix = ­1 A 1 1 1 1 + VA − VB = − R1 R 2 R2 3 1 1 1 − VA + + VB = 1 R2 R 2 R3 5 1 1 VA − VB = − 6 3 3 1 8 − VA + VB = 1 3 15 122 Method #3 Con’t R1 A R2 B 6Ω + ix R3 1A Vtest ix/3 ­ ix = ­1 A Results VB = 2.17 V Vtest = 8.17 V ix = ­1 A Req = 8.17 Ω = RTh 123 Practice Problem #1 Find the Thévenin equivalent circuit for the following Where R1=5kΩ , R2=500Ω , and R3=1500Ω : 60V R2 2mA a R3 R1 b Answer: ­45­V source in series with a 1.2kΩ resistor 124 Practice Problem #2 Find the Thévenin equivalent circuit for the following circuit. Use R1 = 20Ω , R2 = 10Ω , R3 = 5Ω , R4 = 4Ω , R5 =5Ω , R6 = 20 Ω , and R7 = 10 Ω : + R3 R1 + R2 8.5 V vx R4 RL - + - R5 Vx/5 - 18i1 R6 i1 125 One Last “Method” Sometimes you can obtain the Thévenin Equivalent by applying repeated source transformations and resistor manipulations. This approach does NOT work reliably for circuits containing dependent sources. 126 Source Transformation Example Let’s use the source transformation method on the example circuit: 6Ω 4Ω + 36 V - + 12 Ω VL R L = 16 Ω - First, transform the 36­volt source and 6­ohm resistor into a current source. 127 Source Transformation Example 6║12 = 4 Ω 6A 4Ω 12 Ω 6Ω 4Ω 4Ω + 24 V - VTh = 24 V and RTh = 8 Ω (as before) 128 Source Transformation Example Will the source transformation method work on this circuit? R1 + - R2 R4 ix R3 V1 RL ix/3 V1=10V, R1=2Ω , R2=3Ω , R3=5Ω , and R4=6Ω VTh = 5 V and RTh = 8.17 Ω 129 Summary: RTh Methods Test Source Any circuit – No limitations! • But often difficult to apply Turn OFF the independent sources and compute the test voltage divided by the test current. Open circuit voltage and short circuit current Any circuit with at least one independent source RTh = VOC / ISC Equivalent Resistance Any circuit with NO dependent sources Turn OFF the independent sources RTh is the equivalent resistance seen looking into the circuit from the load terminals. Repeated source transformations Any circuit with NO dependent sources is a candidate. Not always a helpful approach – depends on circuit topography. 130 Think before you Act! Always remove the load resistor. Always need either the open circuit voltage or the short circuit current. Which is easier to compute? Use source transformation to get the other. Dependent source: Test source or VOC / ISC method. No dependent source: Req method is often easiest. Using Req or test source method? Remember to turn OFF all independent sources. Never turn off a dependent source with any method. 131 Questions? Thévenin Equivalent Circuits with or without Dependent Sources Thévenin Equivalent Circuits Maximum Power Transfer Maximum Power Transfer As stated earlier, the Thévenin circuit allows us to calculate the voltage and current for any load resistor. Power delivered to the load is RTH important What is the most power that can be delivered? What value of load resistance will receive the most power? VTH + - RL 134 Power to the Load RL VL = VTH ⋅ RL + RTH RTH VTH + - + VL ­ RL IL VTH IL = RL + RTH 2 VTH RL PL = VL ⋅ I L = ( RL + RTH ) 2 If RL = 0 or RL = ∞, no power is delivered. As RL is increased from zero, the load current decreases and the load voltage increases. What value of RL absorbs the maximum power? 135 Maximum Power to Load Resistor How do I determine the maximum power to the load resistor? Experiment: change RL and measure each value Calculation: change RL and compute each value Mathematics: Power Maximized when ∂PL 2 =VTH ∂RL [( R L ∂PL =0 ∂RL + RTH ) − 2 RL ( RL + RTH ) =0 4 ( RL + RTH ) 2 This equation is true when RL = RTH This is referred to as impedance matching… 136 Power Delivered to the Load Relative Power to the Load 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Load Resistance divided by Thevenin Resistance 137 5.0 Maximum Power to the Load Maximum power transferred when RL = RTh RTH VTH + - + VL 2 2 VTh RL VTh PL = = 2 ( RL + RTh ) 4 RTh RL = RTh ­ Note that: VL = 0.5 VTh PL = PRTh (V 2 ) = Th RTh 2 2 VTh = 4 RTh 138 Example Problem What is the maximum power that can be delivered to the load resistor? 40 Ω 25 Ω 15Ω 2A 10 V 5Ω 10 Ω RL + - 139 Find Req RL = RTH for maximum power 40 Ω 25 Ω 15Ω 5Ω 10 Ω Req = { 5 + [10 ( 25 + 15) ]} RTH = 13Ω 140 Find VTH 40 Ω 25 Ω 15Ω 2 A i1 10 V + ­ 5Ω + 10 Ω i2 VOC ­ 50i2 − 15i1 = 10 50i2 = −20 i2 = −0.4 A VTH = 10i2 = ­ 4 V 141 Final Thévenin Circuit 13 Ω -4 V Pmax= + - 2 VTH 4 RTH RL = 308. mW 142 Using Thévenin Equivalent Circuits I Complex Circuit + V ­ I RL Unknown + Circuit V RL ­ If I know 2 of the following, I can determine the Thévenin circuit I, V, or power to a known resistor Open­circuit voltage Short­circuit current Maximum power The Thévenin circuit can then be used to answer more advanced questions without knowing all the details of the original circuit. 143 Example A black box provides its maximum power (12.5 W) to an 8Ω resistor. What resistor value(s) will receive 5 W of power from the same black box Solution Since max. power goes to a 8Ω resistor, RTH = 8Ω Using the maximum power equation 2 VTH = 12.5W this leads to VTH = ± 20 V 4(8Ω) The power to a load resistor is now given by 2 VTH RL 400 RL = 2 2 = 5W ( RL + RTH ) (8 + RL ) Answers: 1.02 Ω or 63.0 Ω 144 Practice Problem #3 (We have two batteries, A and B. Battery A has an open­circuit voltage of 13.8 V and a short circuit current of 350 A. Battery B has an open­circuit voltage of 13.2 V and a short­circuit current of 220 A. The batteries may be connected in series or parallel to give power to a single resistor. What connection (series or parallel) has the most power producing capability and what is the power produced? 145 Practice Problem #4 Find the Thévenin equivalent circuit for the following Where R1=5kΩ , R2=500Ω , and R3=1500Ω : 60V R2 a R3 R1 2A b Answer: ­45.0­V source in series with a 1.18kΩ resistor 146 Always Maximize Power? Do I always want to design for maximum power to the load? RTH VTH + RL - 147 Questions? Thévenin Equivalent Circuits Maximum Power Transfer Negative Resistance Consider the circuit below… No independent sources. To find Thévenin Equivalent, you must apply an independent source to the output terminals. 10 Ω 10V1 50 Ω + ­ + 10 Ω V1 ­ 150 Insert a 1 volt test source 10 Ω Ia 10V1 Ib 50 Ω + ­ + + 10 Ω 1.00 V Therefore, ­ V1 V1 = 1/6 volts 10V1 = 1.667 volts Itest ­ Ia = (1.667­ 1.00)/10 = 0.0667 amps Ib = 1.00/60 = 0.0167 amps Itest = Ib – Ia = 0.0167 – 0.0667 = ­0.05 amps RTh = Vtest/Itest = 1/(­0.05) = ­20 Ω 151 Add a test source­­1A 10 Ω + 50Ω Itest=1 A 10V1 10Ω + V1 - Vtest Solve for Vtest: (Vtest - 10V1 ) Vtest + -1 = 0 10 60 10 V V1 = Vtest ( ) = test 10 + 50 6 Solving , Vtest = -20V R th Vtest Vtest - 20V = = = = -20Ω I test 1 1 152 The Thévenin Equivalent Circuit 10Ω 50Ω Rth = -20Ω 10V1 10Ω + V1 - 153 How Can this Be? Can only have negative resistance with dependent sources. The negative resistance describes the terminal behavior of the original circuit. When an independent source is attached to the terminals, the original dependent source supplies power. I= -.5A 10 V source applied Rth = -20Ω 10V I = 10/(­20) = ­.5A Power absorbed by 10V source= 5W 154 Using a Current Source… If we apply an independent current source, the original dependent source supplies power to it as well… Rth = -20Ω 2A + V = -40V - V = 2(­20) = ­40V Power supplied to independent source = 80W 155 A Quick Look Ahead… 10Ω The original circuit is a model for a circuit using an op­amp in a non­ inverting configuration. 50Ω 10V1 10Ω + V1 - + 1kΩ - 10Ω 50Ω 9kΩ 10Ω 156 Summary A negative Thévenin resistance can be realized using dependent sources. These represent models for actual circuits. A negative Thévenin resistance means the dependent sources in the original circuit supply power to any independent sources connected to its terminals. 157 Questions? Negative Resistance ...
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This note was uploaded on 02/26/2012 for the course ECE 302 taught by Professor Preston during the Spring '08 term at University of Texas at Austin.

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