9 Lesson Mesh Current Analysis and Linearity. (Sections 3-2 and 3-3)(CLO 3-1 and 3-2) Students generally do not have serious problems in understanding Mesh Current analysis. You might mention that it is the dual of Node Voltage analysis. Mesh Current analysis cannot be applied to nonplanar circuits and is often difficult to apply to Op-Amp circuits studied later. Node Voltage analysis does not have this limitation. Linearity is the basis of most circuits and electronic courses. Getting students aware of what constitutes linearity is important to their understanding and development as good engineers. Mesh-Current Analysis There are several areas where students might have some early difficulties with mesh current analysis. Loops versus meshes. If you were to draw a four-pane circuit and ask students how many different loops there were they might say five, but there are many more thirteen, in all. Loops are defined in the text as A closed path formed by tracing through an ordered sequence of nodes without passing through any node more than once. A mesh is a special type of loop that does not enclose any elements although we appear to violate this rule when we discuss super meshes. While one can write as many KVL equations as there are loops there are only as many independent KVL equations as there are meshes. Fortunately, students can readily determine the number of meshes by counting the window panes in the circuit. Each window pane hosts its own current. Tell the students ALWAYS to draw each mesh current clockwise this will help them recognize errors later on. Use different colored markers and claim that the electrons that travel around that mesh are so endowed with that color. Multiple currents in one branch. This brings us to the next area of early difficulty for some students. That is rationalizing how different currents can flow in one branch. There are two types of voltage drops in mesh analysis the voltage across an element with one mesh current flowing through that element and the voltage across an element with two branch currents flowing in opposite directions (hence the reason for always drawing the mesh currents clockwise). The first is simply applying the i-v characteristic to the element using the single branch current: V = IMESH R for resistors. The second is the same except that the current will be the difference of the two mesh currents, one positive and one negative, with the positive current being the direction of the KVL being written, e.g. for resistors: V = (I+ MESH I MESH) R Continuing the color analogy in some branches there are say blue electrons traveling in one direction and red ones traveling opposite. They do not change their minds and travel in the opposite direction. Both contribute to the voltage drop across that element but in opposite ways. At some point, you might tell them that this is only an analytical technique and that there really is only one actual branch current in each branch. They can speculate on the color. They should label the mesh currents IA, IBIN. Some of the mesh currents may be known, like node voltages when the node is connected to a voltage source and the other side of the voltage source is connected to ground. If a mesh flows through a branch with a current source IS and no other mesh current is flowing through that particular branch then, IA = IS or if the source current is pointing in the opposite of direction the mesh current, IA = IS. Writing mesh-current equations follows from KVL. Here the trick is to always assume that the clockwise direction is positive. Hence, we write equations at each mesh (IA and IB) _ I
1
R
1
+
V
3
_ V
S
R
2
IA
R
2
V +
3
IB
IS
Mesh A : VS + R1 I A + R2 ( I A I B ) = 0 Mesh B : I B = I S
I1 is simply IA. V2 = (IS-IA) R2 and V3 = IS R3.
DC
This brings us to a super mesh the dual of the supernode. Consider the circuit shown. It has three meshes. The only source is a current source, IS, but it is located between two meshes, IA and IB. To solve this by mesh analysis we declare a super mesh shown by VS the dashed line. The students should note that the super mesh is not another mesh current, rather it is a useful loop that will help IC us solve the problem. In writing mesh equations around the R1 R3 super mesh, we still use the pre-identified mesh currents. Here are the equations starting with the super mesh:
R2 IA IS IB R4
Super mesh
R2 I A + R1 ( I A I C ) + R3 ( I B I C ) + R4 I B = 0 VS + R3 ( I C I B ) + R1 ( I C I A ) = 0
This is well and good but it leaves us one equation short. We find the missing equation by relating the two mesh currents IA and IB. We obtain that equation by realizing that the source current IS can be written as IS = IA IB. This will give us the three equations needed to solve the three unknown mesh currents. Solution then proceeds like that used for finding Node Voltages. Linear Circuits. There are two properties of linear circuits Proportionality (K= VOUT/VIN) and Superposition. We will study both but spend most of the time with the latter. Proportionality is simply that for linear circuits given an input and an output one can find a relationship that often is called the Gain factor or simply K. Once K is known that output can be found for any input by using VOUT = K VIN. A good example is a Voltage Divider. V IN V OUT This is a useful first block in developing block diagrams. In the next chapter, we will see that |K| can be greater than 1, but in this chapter dealing with only passive circuits |K| must be 1.
K
R
1
+ V
IN
In the circuit on the left VOUT = (R2 VIN)/(R1 + R2). If we solve for VOUT/VIN we get K=R2/(R1 + R2). Knowing K we can find VOUT given any VIN. You can derive a similar one for a Current Divider. An application of the proportionality principle is the Unit Output Method for solving ladder circuits.
R
2
V
OUT
_
Consider the following ladder circuit. Suppose we want to find iO for a vS of 30 V. Using the Unit Output Method we assume an output of 1 A and work backwards to determine what input it would take to produce that output. Then using proportionality we find K = iO/vS. Once we have K then we can find iO for any vS. We start by letting iO = 1 A. vO then equals 20 1A i 3 30 10 i 1 = 20 V. The current through the 60- resistor then is 20/60 = 0.33 A. i1 is equal to 1 + 0.33 = 1.33 A The v3 iO i 2 v1 voltage v1 across the 10- resistor is 10 1.33 = 13.3 vS 60 20 V. The voltage v2 across the 100- resistor then is v2 100 vO 13.3 + 20 = 33.3 V. The current i2 through the 100- resistor is 33.3/100 = 0.33 A. The current i3 is 0.33 + 1.33 = 1.67 A. The voltage v3 is 30 1.67 = 50 V. So the voltage vS that will produce 1 A at the output is 50 + 33.3 = 83.3 V. K then is 1/83.3 = 0.012. Therefore, a 30 V input will produce iO = 0.01230 = 360 mA.