Chapter 5: Circuit Theorems

Chapter 5: Circuit Theorems This chapter provides a new powerful technique of solving complicated circuits that are more conceptual in nature than node/mesh analysis. Conceptually, the method is fairly straightforward to write down, however, they are complicated to apply in practice. One will find that the amount of work increases dramatically using these circuit theorems to solve a circuit. There are four theorems we will look at in this chapter: Source Transformations, Superposition Principle and Thévenin & Norton Equivalents, and Maimum Power Theorem. There are two useful ideas to take away from this chapter.. Thévenin-Norton source transformations. In certain situations, there is a way to convert a voltage source into a current source and vice-versa. Source transformations are going to change how you solve circuits and in some cases, a series of source transformations will have a dramatic affect in making the circuit calculation quite straight forward. Source transformations will become an additional technique in your arsenal in solving circuits regardless of the techniques one uses (T, Node, Mesh ).. Thévenin-Norton Equivalents. The techniques of Chapter 4 (Mesh and Node) are very powerful and fairly straight forward to use. When we start studying nonlinear circuits (circuits with capacitors and inductors), mesh/node techniques will not be applicable. The way we will solve these nonlinear circuits are similar to solving irreducible dependent source circuits with KVL and KCL equations. The Thévenin-Norton Equivalents will help us to simplify circuits as they get more and more challenging. Picture wise, we will break-up the circuit into two subcircuits: the (i) load circuit and (ii) subcircuit to be simplified via the Thévenin-Norton Equivalent. That is, we will rewrite it into an equivalent form that is either a simple series circuit or a simple parallel circuit. Thévenin Norton Source Transformation There are transformations that convert current sources into voltage sources (and vice versa). That is, there is an equivalent circuit transformation between a series resistor and voltage source into a parallel resistor and current source: In order for these two circuits to be equivalent, the two separate subcircuits must have the same terminal current i and the terminal voltage v. In order to have the same terminal current and voltage, certain conditions must be met. These are derived by applying KVL to the series circuit and KCL to the parallel circuit. Here is what we get: KVL: v = v ir v v v i = i = R = R and v = i R KCL: i = i v/r R R R Th Th solving for i Th N Th N Th N Th N N N Th Th necessary conditions for equivalance That is, a Thévenin-Norton Source Transformation converts a voltage source with a resistor in series into a current source and the same resistor in parallel: 5.

Eample 5. Use source transformations to solve for the current through the 3Ω-resistor. Solution There are several ways one can solve for the current in this circuit: simple circuits or mesh/node. With the help of source transformations, you will see how useful these can be in reducing the circuit down to one loop before you can change a baby s diaper. Important Point: do not apply a source transformation to the 3Ω element. If a source transformation changes the 3Ω resistor to a parallel one, that specific information on the current through that 3Ω resistor is lost. Leave it alone! Let s convert this circuit using three different source transformations: Applying KVL/Mesh to the last loop, (3 4) i 6 = 0 i =.9A 3Ω 3 3Ω Superposition Theorem In Chapter, we briefly addressed the issue of a linear circuit. We stated then that for a circuit to be linear it meant that () the elements of the circuit are linear themselves and () the total response of an element (the voltages or currents) can be determined by the sum of the individual responses from each independent source in the circuit. This last statement is the Superposition theorem. Analogy: Newton s nd law (F net = F F F n) Superposition Principle The current i R through an element is equal to the algebraic sum of the currents (i, i i n) produced independently by each individual source. That is, i = = R i in isource all sources current produce by source- current produce by source-n Solving Strategies Step : Isolate one source and deactivate all other independent sources. To deactivate an independent current and voltage sources replace a voltage source with a short circuit (a short has v short = 0) current source with an open circuit (an open has i open = 0) With only one active source in the circuit, calculate the current/voltage of the element of interest. Repeat this process for each independent source in the circuit. Step : If there are n-sources, then there are n-circuits to solve using superposition. Eample 5. Use superposition to find the current i through the 0kΩ-resistor. Solution Superposition states that to calculate the current i 0kΩ, this current is the sum of all of the individual currents produced by the V, 3mA and 9mA-sources: i = i i i i i i 0k V 3mA 9mA 3 5.

Response of the V-source Deactivate both current sources (replace them with opens), and calculate the current i produced by the V source using a KVL/Mesh loop: 36 i = i = ma 3 Response of the 3mA-source Deactivate the other sources (short V and open 9 ma sources), and calculate the current i produced by the 3mA source using CDR: 6 i 4 = (3mA) = ma = i 6 0 3 Response of the 9mA-source Deactivate the other sources (short V and open 3 ma sources), and calculate the current i 3 produced by the 9mA source using CDR: i 3 = (9A) = 3 ma = i3 4 According to Superposition, the current i 0kΩ is the sum of all of these individual currents: i 4 0kΩ = i i i3 = 3 = ma = i 3 3 0kΩ Thévenin s Theorem History: this theorem was independently derived in 853 by Helmholtz and in 883 by Léon Charles Thévenin. Léon Charles Thévenin was a French telegraph engineer who etended Ohm's law to the analysis of comple electrical circuits. The goal of Thévenin s theorem is to identify and separate a portion of a comple circuit (called the subcircuit) and replace it with a Thévenin equivalent (series voltage source and resistor). In doing so, the solving of the circuit problem is greatly simplified. A typical application of a Thévenin theorem goes something like this: Step. Focus on a load and separate it out from the subcircuit to be replaced with a Thévenin equivalent. Step. Perform the necessary calculations to rewrite the subcircuit as a Thévenin equivalent circuit. Step 3. Reconnect the Thévenin equivalent circuit back onto the load circuit and now solve the simplified circuit. To accomplish this, one applies the Thévenin s Rules. Thévenin s Rules (Notation: v Th = v OC, R Th = R N, and i N = i Sc) a. Determine v oc with all sources activated If the open circuit contains sources about the open, apply KVL around the open circuit to determine v OC. 5.3

Use any method to solve for v OC b. Determine R Th. Method-: Independent Sources Only (Conceptual Approach) Deactivate all sources and determine R Th relative to the terminal points a-b. Method-: Independent & Dependent Sources (Calculational Approach) i. Place a short at the terminal points a-b and calculate i sc using any method. ii. Determine R Th by using R Th = v OC/i. c. Redraw the equivalent Thévenin circuit. If the Thévenin resistance is negative, subtract it from the load resistance. Dependent sources only the book does not give any problems related to only dependent sources. I will not cover them; however, if you are transferring to San Jose State and plan on majoring in EE, at the end of the semester I can teach you these in about 0-5 minutes. Independent Sources Only Because this eample circuit only has an independent source, I will not only demonstrate how a Thévenin equivalent is determine between points a-b, but also show that Method- and Method- are equivalent. Suppose the following circuit has its load identified and removed, so that we can determine the Thévenin equivalent between points a-b. Both Method- and Method- determine v oc eactly the same. Step : Determine v oc What does the voltage v oc mean? It is the effective voltage that the load sees between the points a and b with respect to the rest of the circuit. Identifying nodes and applying node analysis because one of the nodes is v oc, we get Node v oc : ( ) v 0 oc ( ) v 0 ( ) 8 = 0 v oc = vth = V Node v : ( ) v ( ) v ( ) 8 = 0 3 6 0 0 oc 6 What differentiates Method- and Method- is in the way that the Thévenin resistance is determined. Step : Determine R Th Method : Deactivating Sources There is only an independent voltage source in the circuit, and because of this, we can use Method-. Method- is NOT applicable if there are dependent sources in the circuit. We start by deactivating the voltage source by shorting it out (replace the 8V-source with a wire) and place an Ohmmeter at the terminal points. Effectively, one way to envision how an Ohmmeter measures resistance is by sending out a calibrated current which is then compared to the returning current, and reads the resistance. The R Th resistance is (0 3 6) R Th = (0 3 6) = = 6Ω= 0 3 6 What does this resistance physically mean? It is the effective resistance that the load sees between the points a and b with respect to the rest of the circuit. The Thévenin equivalent circuit has replaced a comple circuit with a Thévenin equivalent circuit with values of v Th = V and R Th = 6Ω: 5.4

Method : Determine i to calculate R Th = v OC/i Method- is applicable for both independent and dependent sources in the circuit. Place a short across the open and calculate current i. There are three loops in the circuit and will use mesh analysis to determine i. Loop - i : ( 0 6) i (6) i 0 i = 0 8 6 0 i 0 matri = form 6 9 3 i = 8 0 3 3 i = 0 Loop - i : (6 3) i (6) i 3 i 8 Loop - i : (0 3) i (0) i (3) i 0 solution i = However, I think that the most efficient way is using node analysis since there is only one node plus a KCL to get i. Applying node to the v y, we get y 3 6 0 y = 6 y Node v : ( ) v ( ) 8 0 v = 5V Now apply KCL at node a: Calculating R Th, we get : = 8Ω 0Ω 8 5 0 KCL i i i = = A = i = v = V = Ω OC 6 i A This agrees with our previous calculation of R Th and therefore, both methods are equivalent. A Eample 5.3 Determine the current through the 4Ω-resistor using a Thévenin equivalent circuit (Methods and ). Solution First, remove the 4Ω-resistor and convert the circuit into a Thévenin equivalent. Step : Determine voc = v Th. Focus on the open circuit and see how voc is defined. Since the 4Ω resistor is connected to an open loop, there is no current flowing through it and therefore, has no voltage drop. Applying a KVL loop around the open circuit loop, the 0V-source will still contribute to voc and we have v = 0 v v = v 0 0 oc oc 0 5.5

So to determine voc, my real task is first to determine the voltage across the 0Ω resistor. Question: what is the most efficient way to determine v 0Ω? For Carlos, I d say a source transformation on the series 8V and 9Ω, converting them into a parallel current source and resistor, then add the current sources together. The circuit I get is Applying CDR to get the current of the 0Ω resistor and Ohm s law to get the voltage v 0Ω, we get 9 0 Ohm's CDR : i 0Ω = A = A v 5 law 0Ω = R0Ωi = 0 5 = 4V 9 6 9 Using our KVL relation from above, we determine voc: v = v 0 = 4 0 = 4V = v = v oc 0Ω oc Th Step : Find R Th. Method : Deactivating Independent Sources Since the sources are all independent sources, we can use Method- and deactivate all of sources to determine the Thévenin resistance R Th at a-b. It is important to realize that in the previous voc-circuit, there was no 4Ω resistor; however, in the R Th-circuit the 4Ω resistor reappears and must be accounted for. 5 0 = 4 (6 9) 0 = 4 = 0Ω= 5 0 Now that we have both the Thévenin voltage and resistance, I can draw the equivalent Thévenin circuit, add back in the load 4Ω-resistor, and solve for the current i 4Ω. Using Ohm s law to determine the current i 4Ω, we see that the current flows from b to a, not a to b and therefore, I assign a negative value to this current: 4V i 4 Ω = = A = i 6 4 4 Ω Interpretation of results. Effectively, when we calculate the Thévenin equivalent, the 4Ω resistor sees an effective voltage of -4V and an effective resistance of 6Ω from the view point of terminal points a-b. That is, regardless of the load (whether it s a 4Ω resistor or some complicated load circuit), this load feels -4V and 6Ω from the rest of the circuit. It does not see the individual 0V, 8V and A-sources but a combination of these that effectively reads the Thévenin voltage of -4V between a-b. It does not see some combination of all of the resistors in the circuit, but the effective Thévenin resistance of 6Ω between a-b. 5.6

. It clearly is a lot more work to calculate the current i 4Ω using Thévenin Equivalent circuits, whereas using Node or Mesh would have given us this current more efficiently. So what s all the fuss about learning these Thévenin equivalent techniques then if there are other techniques more efficient? To put this into prospective, resistive circuits are only half of this course, with the other have being capacitive and inductive circuits. Node and Mesh only apply to linear resistive circuits (as well as to ac phasor circuits). When we start capacitive and inductive circuits (nonlinear circuits), we will not be able to solve these circuits using Node or Mesh and we will be forced to use Simple Circuit Techniques (KVL & KCL) along with Thévenin equivalents to guide us. Thévenin equivalents are a powerful ally in solving these considerably tougher circuits. Method : Finding i to determine R Th = v OC/i When a short is placed across a-b to calculate i, note that i = i 4Ω. I epect that you are most likely thinking of calculating i using mesh analysis (3 loops = loop equations plus a ). Even though this is a different circuit, I already have reduced the left-hand portion circuit with a source transformation that combined the 8V-source with the A-source. I want to continue using this simplified circuit and make one more source transformation on the 0/9A-source to convert it into a voltage source. This is what I get: I immediately see that there is only one node voltage to solve for and a solver equation for i. 5 0 4 5 4 Node v : ( ) v ( ) ( 0) ( ) (0) = 0 v = 0.4V 0 v Solver : i = =.4A = i 4 4Ω Using a combination of voc and i, the Thévenin resistance is determined to be v 4V = = = Ω OC 0 i.4a Interpretation of results. Both methods ( & ) determine the Thévenin resistance, however, Method- is only valid if there are ONLY independent sources. If there are any dependent sources, Method- is NOT valid and ONLY Method- can be used.. The combination voc/i, is a measure of the effective resistance (R Th) that the load sees relative to the terminal points a-b. Norton Equivalent Circuits 5.7

Independent & Dependent Sources There are two important changes when dealing with independent and dependent sources in circuit where Thévenin s theorem is going to be applied:. Without dependent sources, zeroing all the independent sources leaves a resistive circuit so that the Thévenin resistance can be calculated directly. However, dependent sources typically alter the Thévenin resistance so those can't be zeroed. That is, Method- is NOT VALID whenever dependent sources are in the circuit and the only way to determine R Th is by using Medthod-; determine the ratio v OC/i = R Th.. Thévenin circuit with dependent sources can have positive and negative Thévenin resistance. It is not like there is an actual negative resistor that one can actual purchase at the store or something like that, but has implications that circuits have a impedance matching and can be resonated. Eample 5.4 Obtain the Thévenin equivalent for the following circuit. Solution The fact that there is a dependent source in the circuit immediately tells us that Method- cannot be applied to this circuit. Only Method- is valid. Step : Determine voc Focusing on the open circuit where voc is defined, the open immediately tells us that the kω resistor is a dead resistor since there is no current flowing through it and has no voltage drop. So the voltage across the 3kΩ resistor is equivalent to voc: v = v = 3i oc Clearly, to determine voc we shift our attention to first determining i. Thinking in terms of source transformations first, converting the.5ma source into a voltage source allows us to reduce the circuit to one loop. 3kΩ X : X X X oc X KVL (6 4 3)i = 5 i i = ma v = 3i = 3V Step : Determine i sc such that R Th = voc /isc. Since I ve already modified the left side of the circuit, I will continue using it. Placing the short across a-b, i sc is the current through the kω resistor. To determine i sc we can either use (i) mesh analysis which will produce 3 equations and 3 unknowns (i sc, i, i z), (ii) node analysis which also has 3 equations and 3 unknowns (i sc, i, v y) or (ii) simple circuits that has equations and unknowns (i sc, i ). Let s do simple circuits! 5.8

I will use VDR to determine the voltage 3i (and current i ) and use Ohm s law to get i sc: 3 VDR: v3kω = vkω = 3i = (5 i ) 3 0 solving for i i = 3 ma solving vkω for i i = = ma = i kω Applying R Th = v OC/i, we find that voc 3V R = Th 3kΩ i = ma = Step 3 Redraw the equivalent circuit: sc Eample 5.5 Determine the current through the Ω-resistor at a-b using a Thévenin equivalent circuit. Solution Step : Determine v oc The simplest way to determine v oc is by applying KVL around the top loop. Why? Because I immediately know the dependent current i since it is in series with the 4A-source. i= 4A voc 8 i 6i = 0 voc = 4V Step : Determine i sc such that R Th = v oc/i sc. Placing the short at a-b sets up three loops (i, i sc, i ). However, this short really sets up an unusual situation where the loop current-i (which also has a dependent source controlled by i) is directly connected to i sc via a constant 4Asource. Note that I have completely ignored i. The trick is to see that doing a mesh loop for loop-i and applying KCL between a-b, there are equations and unknowns (i sc, i) that quickly determines i sc: solving for i Loop - i : 6i i = 8 i = A KCL at a-b: i = 4 i = 6A Applying R Th = v OC/i, we find that v 4 V = = = Ω oc 4 isc 6 A Step 3: Redraw the equivalent circuit and solve the circuit problem. 5.9

Using Ohm s law, the current through the -resistor is voc 4 iω = = = A = iω Ω 4 Because this current is positive, the current runs positive from terminal a to terminal b. Interpretation of Negative Thévenin Resistance Suppose now that the load resistor was replaced with a variable resistor (i.e., decade resistor bo) such that we adjusted the resistance. How does the circuit behave? One would think that if one sets R ab = 0Ω, then maimum current would flow through the load resistance R ab. Let s look at this: voc 4 iload = = = 6 A Rload 4 0 If I now adjusted the decade resistor bo to R load = Ω, the current now reads A, double the value of the short circuit current: 4 iload = = A 4 And as you can see, something very special happens as R load 4Ω, the current goes to infinity: 4 lim iload = lim Rload 4Ω Rload 4Ω 4 R load Physical interpretation The key phase is impedance matching. Negative resistance cannot physically occur in the case where the circuit is linear and contains only passive components (resistors, capacitors and inductors). But for active circuits, where dependent sources (amplifiers) are applied, a virtual negative resistance can be realized. The physical meaning of negative resistance is that power is absorbed by the circuit with zero phase shift rather than dissipated. Maimum Power Delivered to a Circuit The Maimum Power Transfer Theorem is not so much a means of analysis as it is an aid to system design. Simply stated, the maimum amount of power will be dissipated by a load resistance when that load resistance is equal to the Thévenin resistance of the circuit supplying the power. If the load resistance is lower or higher than the Thévenin resistance, its dissipated power will be less than maimum. This is essentially what is aimed for in radio transmitter design where the antenna or transmission line impedance is matched to final power amplifier impedance for maimum radio frequency power output (impedance matching). Impedance, the overall opposition to AC and DC current, is very similar to resistance, and must be equal between source and load for the greatest amount of power to be transferred to the load. A load impedance that is too high will result in low power output. A load impedance that is too low will not only result in low power output, but possibly overheating of the amplifier due to the power dissipated in its internal Thévenin impedance. Mathematically, one can determine the maimum power delivered to the load by calculating the critical point where dp load/dr load = 0: 5.0

Rload vload dpload d R load vload = vth Pload = = v Th = 0 Rload Rload drload dr load ( Rload ) After doing some algebra, we arrive d R load Rload Rload = = 0 R 4 Th = R dr load ( Rload ) ( Rload ) Eample 5.6 What is the maimum power that can be delivered to the load resistor R L? load Solution Step : Find v OC What is the most efficient way to solve for v OC? I first note that the controlling voltage v y is known since the Ω-resistor is in series with the 9A-source: v y = 8V. So I can replace the VCCS as a 36A-source instead. So if I do the numbers, this is what I get: 4 nodes sources = node eqs [ node (v ) & SN eq (v,v ) (v,v ) DC (i ) OC eqs, unknowns (v OC,v,v,i ) Writing out the equations: node v : v v 0 = 9 OC OC SN: ( )v 3i v v 3i = 7 : v v = OC DC: v 3i = i In matri form and solution: 38 v = = 79.3V OC 3 0 0 v 9 OC 90 v 7 v = = 63.3V Step : Find i. Let s do the numbers: 3 9 solutions = 3 0 0 v 84 i 0 0 0 5 3 v = = 6.3V 38 i = =.7V 3 5.

4 loops sources = loops eqs ( regular (i ) & SL (i,i,i ) DC (i,v ) y Solver (i ) 3 6 eqs, 6 unknowns (i,i,i,i,v,i ) Writing out the equations, Loop : i i = 3i ( ) 3 y SL: i i 0 i = 3i : loop loop i 3 3 y y 3 i i = 9, i i = v DC : v = i i Solver : i = i i The matri and solutions are 3 0 3 0 0 i 0 i = 3.58A 0 5 0 0 i i =.6A 0 0 0 0 i i 3 9 3 5.4A solutions = = 0 0 0 i i 0 = 0.9A 0 0 0 v v y 0 y =.65V 0 0 0 i 0 i = 7.67A Step 3: Find R TH and solve the circuit problem Applying R Th = v OC/i, we find that v 79.3V = = = Ω oc 0.3 isc 7.67A R = 0.3 Ω v = 79.3V Th Th maimum power theorem v L (79.3 / ) R ma Th = RL = 0.3Ω P = = = 53 W = P R 0.3 L ma 5.