Chapter 4 Techniques of Ckt Analysis Ch1

Chapter 4: Techniques of Circuit Analysis This chapter gives us many useful tools for solving and simplifying circuits. We saw a few simple tools in the last chapter (reduction of circuits via series and parallel combinations of resistances, for example) but in this chapter we take circuit simplification much farther… 4.1 Terminology Planar Circuits can be drawn on a plane with no crossing branches. Non-Planar Circuits cannot be drawn on a plane without branches crossing one another somewhere. Figure 4.1 (left) shows a circuit that has crossing branches but that can be re-drawn without any branches crossing. This is a planar circuit. Figure 4.2 (right) shows a circuit that is not planar: there is no way to draw it without branches crossing. In Circuit Analysis we will consider planar circuits only. Many of the techniques we learn here cannot be applied to non-planar circuits. Definitions A node is a point in a circuit where two or more circuit elements meet. The number of nodes in a circuit is n. A path is formed when adjoining (connected) circuit elements are traced, in order, without passing through any node more than once. A closed path is a path whose starting and ending points are the same. A branch is a path connecting two nodes. A mesh is a closed path that does not contain any other closed paths The number of meshes in a circuit is m. Looking at the circuit in Figure 4.1b (shown again below), we can identify: 4 meshes; 8 closed paths; 6 nodes. Two of the four meshes are shown in green; two nodes are shown as red dots; one possible closed path which is not a mesh is shown in blue. The blue closed path contains two meshes. (The arrows on the paths are just for show: we can trace a path in either direction.) Essential Node: A useful kind of node is the essential node, which is a node at which at least three circuit elements meet. In the circuit above there are four of these; one of them is labeled as a green dot. One of the red dots is also an essential node; the other is not. An essential branch is a branch that connects two essential nodes. There are quite a few in the circuit above. Where are we going with all this?? This terminology helps us figure out how many equations we will need to solve the circuit: We need as many equations as there are essential branches with unknown currents; call this number be. Then we can write (ne – 1) KCL equations so we need [be – (ne – 1)] KVL equations. Here, ne is the number of essential nodes. We will shortly look at two very important and very powerful circuit analysis techniques:  Node Voltage Method (NVM): allows us to solve a circuit with (ne – 1) KCL equations.  Mesh Current Method (MCM): allows us to solve a circuit with [be – (ne – 1)] KVL equations. Using either method, we save quite a few equations, since without them we would need (ne – 1) KCL equations and [be – (ne – 1)] KVL equations. 4.2 Introduction to the Node Voltage Method To use the Node Voltage Method (NVM) we define new circuit variables called the node voltages. We find the node voltages by solving simultaneous equations called the node voltage equations. These are generally much fewer in number than we need with basic KVL, KCL, Ohm’s Law techniques. If we have the node voltages, we can find any other current or voltage in the circuit with a single equation (i.e., no more simultaneous equations will be necessary). Node Voltage Method Algorithm Basic steps in using NVM (we will clarify and expand on this shortly):  Find and label all essential nodes. o There are ne essential nodes, and we will need to solve ne – 1 node voltage equations.  Choose one of the essential nodes as a reference and label it with the symbol . o This is where the “- 1” arises in “ne – 1 equations”: we do not need a node voltage equation at the reference. o Typically we will choose the reference node to be the one with the largest number of circuit elements connected to it. But we do not have to do this, and we will see there may be reasons for choosing a different node as the reference.  Define the non-reference node voltages. o The non-reference nodes are the ones you did not choose as the reference node. The non-reference node voltages are the voltage drops from each non-reference node to the reference node (an example or two will clarify this). These are labeled with the positive sign at the node, and the negative sign at the reference node.  Apply Kirchhoff’s Current Law (KCL) to each non-reference node, writing currents at each node in terms of the node voltages and any sources present. o Although there is more than one way to do this, in this class we will always set the sum of the currents leaving the node equal to zero.  Solve the node voltage equations. The resulting node voltages are considered the solution to the circuit. If you have those, you can find any current or voltage in the circuit with one equation (no more simultaneous equations needed). To see how to express the current leaving a branch in terms of the node voltages, we look at the following example. i12 The essential nodes are labeled 1, 2, and (reference node). The node voltages are defined as described above: positive at the non-reference node and negative at the reference. The current i12 is leaving node 1 and heading from node 2. It can be written in terms of node voltages as i12  v1  v2 . 2[] Where did we get that equation for i12? From KVL: v1  v2  i12 (2[])  0 Solving this equation for i12 gives the previous equation. We continue in this way to set up KCL at each non-reference node. So, at node 1, the sum of the currents leaving node 1 is v1 v1  v2 v1  10   0 5 2 1 The sum of the currents leaving node 2 is v2  v1 v2  2 0 2 10 If the third term of the first equation is confusing, do a KVL around the loop containing the 10 [V] source, the 1 [] resistor, and the 5 [] resistor. Solve this KVL for the current leaving node 1, and you will get the third term in that equation. Eventually, you will get to the point where these terms can be written down quickly without having to examine each KVL individually. The third term in the second equation is easy, since we know the current in that branch: it’s 2 [A] entering node 2, so the current leaving is – 2 [A]. In this case, we don’t try to express it in terms of the node voltages. Note also that we have summed the currents leaving each node. This is an arbitrary choice, but we must choose something, and be consistent. We have two equations for two unknowns. The solution is… 𝑣1 = 9.091 [𝑉] 𝑣2 = 10.91 [𝑉] Now that we have the node voltages, we can find any other voltage or current in the circuit. To do that we probably need to solve one equation, but we will not have to solve simultaneous equations anymore. For example, the current in each branch is easy – we have already shown how to calculate i12 in the equation above. Also, the voltage across any individual component can be found with one KVL. As one more example, the voltage across the current source in the circuit above is nothing but v2. Notation Rules (you knew we were going to have this…) All node voltages must be labeled just as they are in the diagram above. Those node voltages have a ‘+’ sign at the non-reference node, and a ‘-‘ sign near the reference node. It is not sufficient to simply label v1, for example, near node 1. It is certainly not acceptable to leave out the node voltages from your diagram. If you do, significant credit will be subtracted from quizzes and exams. 4.3 The Node Voltage Method and Dependent Sources If we have dependent sources, we will need one equation for each controlling variable. We call these auxiliary equations. We will need to write these equations in terms of the node voltages. This is because we are trying to find the node voltages (which are the circuit variables), and we don’t want to introduce any extra variables. 4.4 The Node Voltage Method: Some Special Cases Special cases arise when we have a voltage source connected either (1) between an essential node and the reference node, or (2) between two non-reference essential nodes. (1) Voltage source between an essential node and the reference node. In the circuit below the 100 [V] source is connected from essential node 1 to the reference node. 1 10  + vs  100V + - v1 - 2 + 25  50  v2 5A - This is a simple case. Remember that we are trying to find the node voltages, which are the voltage values between each of the essential nodes and the reference node. But we already know that in this case !! v1  vs  100V This is the node voltage equation for node 1. We have still have the same number of node voltage equations, but one of them is trivial. This works for dependent sources, too, although we of course still need the auxiliary equation. The rest of the circuit is solved as follows: 𝑣2 − 𝑣1 𝑣2 + −5=0 10 50 This gives v1 = 100 [V]; v2 = 125 [V]. (2) Voltage source between two non-reference essential nodes. In the circuit below the dependent source is connected between two non-reference essential nodes. We are using a dependent source for illustration, but our remarks are valid for independent sources. 5 1 + 50 V + - v1 iQ + 40  10 i Q -+ 2 50  v2 - i - 3 + 100  v3 4A - Node voltage v1 is already known: 𝑣1 = 50 [𝑉] . Let’s write a node voltage equation at node 2: v2  v1 v2  i 0 5 50 How do we write the current in the dependent source? There is no way to write a ‘v/R’ term there. But there is a current there that we have to account, for so we will call it ‘i’. Now at node 3: v3 4i 0 100 What should we do about this unknown current ‘i’? Let’s add the last two equations together: v2  v1 v2 v   3 40 5 50 100 Hmmm…this last equation is what we have got if we had treated node 2 and node 3, with the dependent source in between, as one big node…it’s as if it were a … a… Supernode!! 10i Q 2 3 -+ In the figure above, the red dashed line encircles the supernode, and the arrows show currents leaving it. Let’s apply our usual node voltage technique to those currents. At node 2: v2 v3  v1 v3   40 50 5 100 As we found above, node voltage v1 is already determined: 𝑣1 = 50 [𝑉] These are our node voltage equations. But we have four unknowns (v1, v2, v3, and iQ) and only two equations. We need two more equations. For one, note that we have not included any information about the dependent source. We include this by simply doing a KVL around a loop that includes the dependent source: v2  v3  10 iQ  0 This equation comes from the fact that the voltage between node 2 and node 3 is “constrained” by the voltage source. This is the constraint equation. Finally, since we have a dependent source, we need an auxiliary equation. Remember it has to be written in terms of the node voltages. iQ  v2  v1 5 Now we have four equations in four unknowns and we can solve for the node voltages. Important Note The supernode equation (the first of our four equations above) contains v2 AND v3, so we do not have any fewer node voltages to solve for. But the kind of equations is different from the case where we do not have a supernode: Instead of two node voltage equations, we have a supernode equation and a constraint equation. Counting Equations It is good practice to count essential nodes and determine how many equations, and of what type, you are going to need. We will do this in the examples in class. You will need…  …one node voltage equation for each non-reference essential node.  …one constraint equation for each supernode  …one auxiliary equation for each dependent source. Choosing the Reference Node You can choose any essential node as the reference node. Typically this will be the one with the largest number of branches connected to it, since then you will avoid having to write an equation with a lot of terms. One reason you might not want to choose the node with the largest number of branches is that you might be able to simplify the problem by choosing the reference node so that one or more of them is trivial. If you choose the reference so that a voltage source (dependent or independent) is connected to it, with the other end of the voltage source at another essential node, then that node voltage is trivial. 4.5 Introduction to the Mesh Current Method To use the Mesh Current Method (MCM) we define new circuit variables called the mesh currents. We find the mesh currents by solving simultaneous equations called the mesh current equations. These are generally much fewer in number than we need with basic KVL, KCL techniques. If we have the mesh currents, we can find any other current or voltage in the circuit with a single equation (i.e., no more simultaneous equations will be necessary). Consider the circuit below, where KCL is applied to the node at the top. R1 R2 i1 vs1 KCL: i3 + - i2 R3 + - vs2 i1  i2  i3 KVL around the left and right loops gives… 𝑣𝑆1 = 𝑖1 𝑅1 + 𝑖3 𝑅3 𝑣𝑆2 = −𝑖2 𝑅2 + 𝑖3 𝑅3 If we solve the KCL equation for i3 and substitute the result into the KVL equations we get: 𝑣𝑆1 = 𝑖1 (𝑅1 + 𝑅3 ) − 𝑖2 𝑅3 𝑣𝑆2 = −𝑖2 (𝑅1 + 𝑅3 ) + 𝑖1 𝑅3 We now have two equations, in the form of KVLs, in two unknowns. This substitution, and the resulting equations, can be done “automatically” using the mesh current method … To use the mesh current method to get the last two equations directly, we do the following. Mesh Current Method Algorithm 1. Find the meshes and label each with a current. Remember that we learned how to count meshes m, essential nodes ne, and essential branches be. Then the number of meshes will be 𝑚 = 𝑏𝑒 − (𝑛𝑒 − 1) . o The two mesh currents in the circuit below are labeled ia, ib. The direction shown (clockwise for both) is arbitrary – we could have chosen either one (or both) to go in the other direction. We will refer to the two meshes as “mesh a” and “mesh b”. o Note that the mesh currents are defined as the currents going “around” the perimeter of the loops. It is important to realize that these are not necessarily the same as branch currents i1, i2, i3. R1 vs1 + - ia R2 R3 ib + vs2 - 2. Apply KVL around each mesh, writing voltages across resistors in terms of the mesh currents. The result for the circuit above is: Mesh a Mesh b vS1  ia R1   ia  ib  R3  0 ib  ia  R3  ib R2  v2  0 3. Solve for the mesh currents. Then, find the branch currents in terms of the mesh currents. Going back to the original circuit where we defined the branch currents i1, i2, i3, is should be clear that i1 = ia and i2 = ib. What about i3? Looking at the circuit below you should be able to convince yourself that i3 = ia – ib. So we have found the branch currents in terms of the mesh currents. This is the reason for the statement above that the mesh currents are not necessarily equal to the branch currents.