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ECE 3144 Lecture 13

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  1. ECE 3144 Lecture 13 Dr. Rose Q. Hu Electrical and Computer Engineering Department Mississippi State University

  2. Nodal analysis case 1: with independent current sources and resistors only. Reminder from Lecture 12 • Nodal analysis case 2: with dependent current sources • Nodal analysis case 3: with independent voltage sources • Scenario 1:The independent voltage source is connected to the reference node. • Any time an independent voltage source is connected between a reference node and a nonreference node, the voltage for the nonreference node is known. • Scenario 2:The independent voltage source is connected between the two nonreference nodes. • The supernode technique is introduced. • First circle the voltage source and the two connecting nonreference nodes to form a supernode • Write the equations that defines the voltage relationships between the two nonreference nodes as a result of the presence of the independent voltage source. • Write the KCL equation for the super node. • Nodal analysis case 4: with dependent voltage sources • Networks containing dependent sources are treated the same way as case 3.

  3. At node 1:  (1) At node 0:  (2)  (3)  Case 4/scenario 1: nodal analysis with dependent voltage sources connected between the reference node and a nonreference node 0 1 Problem 3.37: Find V0 in the circuit given The dependent voltage source is connected to the reference node Write the controlling equation for the dependent voltage source Put equations (1),(2) and (3) in matrix form

  4. (1) (2)  (3) Case 4/scenario 2: nodal analysis with dependent voltage sources connected between two nonreference nodes Problem 3.35: Find Vo in the circuit network given. 2 1 0 The dependent voltage source connected to two nonreference nodes 0 and 2. Thus node 0 and node 2 form a supernode. A voltage source is connected between the reference node and node 1. Apply KCL to the supernode: The controlling function for the voltage source is:

  5. Example 3: cont’d Put the three equations (1), (2) and (3) into matrix format: 

  6. Summary for nodal analysis • The circuit variables are node voltages • Always select one node as the reference node, whose voltage is zero. All other node voltages are measured with respect to the reference node. • We have introduced four cases for nodal analysis. • If only independent current sources are present, you can write the KCL equations for the N-1 nonreference nodes or you can establish the G matrix directly by taking the advantage of G matrix symmetry. • If dependent current sources are present, write the KCL equations for the N-1 nonreference nodes first; then replace the controlling equations for the dependent current sources. • If voltage sources are present in the network, they may be • connected between the reference node and a nonreference node. In this scenario, the voltage at the nonreference node is already known if the voltage source is independent. If the voltage source is dependent, the voltage at the nonreference node is expressed as the controlling function of the dependent voltage source. • connected between two nonreference nodes. The supernode technique is introduced.

  7. Loop analysis • In loop analysis, the unknown parameters are currents and KVL is employed to determine the unknowns. • If there are N independent loops, N independent equations are needed to describe the network.

  8. In nodal analysis, the unknown parameters are nodal voltages and KCL is used to solve the problem. If there are N nodes in the network, N-1 equations are required to solve the problem. KCL is applied to each nonreference node. In loop analysis, the unknown parameters are current and KVL is employed to determine the unknowns. If there are N independent loops, N independent equations are needed to describe the network. KVL is applied to each independent loop. Nodal analysis vs. nodal analysis We would like to consider nodal analysis and loop analysis as mirror scenarios:

  9. vS1 vS2 b c a + + - 4 V 12 V v3 2 v1 v2 W W W k 2 k 2 k - - + d e f => 4000I1 - 2000I2 = 12 (1) I I 2 1 => -2000I1 + 4000I2 = -4 (2) Loop analysis case 1: containing independent voltage source only • Let us first identify how many independent loops here: Loop 1 and loop 2 • The unknown variables are loop currents: I1 and I2 2 • Now we determine the physical currents through each branch. Branch d->a: I1 Branch a->b: I1 Branch b->c: I2 Branch c->f: I2 Branch f->e: I2 Branch b->e: I1-I2 Applying KVL to loop 1: v1 – vS1 + v2 = 0 => 2000I1 + 2000(I1-I2) -12 = 0 Applying KVL to loop 2: -v2 + vS2 + v3= 0 => -2000 (I1-I2) + 4 + 2000I2 = 0 There are two equations and two unknowns I1and I2. You can solve them. I1 =10/3 mA and I2 =2/3 mA

  10. Loop analysis case 1: cont’d The two equations are put into the matrix format: Again, we notice that the coefficient matrix for this type of circuits is symmetrical. We will explain later why this coefficient matrix is symmetrical and how to write the mesh equations by inspection. An important definition: what is mesh: A mesh is a special kind of loop that does not contain any loops within it. So when you traverse the path of a mesh, you do not encircle any circuit elements. The majority of our loop analysis will involve writing KVL equations for meshes, thus loop analysis usually is called mesh analysis. The following example derives the general interpretation on the symmetry of the coefficient matrix.

  11. v4 + v5 –v3 + vS2 = 0 => i2R4 + i2R5 – (i1-i2)R3 + vS2 = 0 => Loop analysis case 1: independent voltage source only There are two independent loops (meshes) in this circuit. Applying KVL to loop 1 v1 + v3 + v2 – vS1 = 0 => i1R1 + (i1-i2)R3 + i1R2 - vS1 = 0 => i1(R1 + R2 + R3) - i2R3 =vS1 (1) Applying KVL to loop 2: -i1R3 +i2(R3 + R4 + R5) =-vS2 (2) In the matrix format It is again the Ohm’s law in matrix format: RI = V. Remember in nodal analysis case 1 we have Ohms' law in matrix format GV = I

  12. Nodal analysis case 1: cont’d • the R matrix is symmetrical. • In the first equation, the coefficient of i1is the sum of all resistors through which mesh current i1 flows; the coefficient of i2 is the negative of the sum of the resistances common to mesh current 1 and mesh current 2. The right-hand side of the equation is the algebraic sum of the voltages sources in mesh 1. The sign of the voltage source is positive if it aids the assumed direction of current flow 1 and negative if it apposes the assumed flow direction. • In the second equation, the coefficient of i2 is the sum of all resistors through which mesh current i2flows; the coefficient of i1 is the negative of the sum of the resistances common to mesh current 1 and mesh current 2. The right-hand side of the equation is the algebraic sum of the voltages sources in mesh 2. The sign of the voltage source is positive if it aids the assumed direction of current flow 2 and negative if it apposes the assumed flow direction. • In general, we assume all of the mesh currents to be in the same direction (clockwise or counterclockwise). If KVL is applied to mesh j with mesh current ij, the coefficient of ijis the sum of all resistors in mesh j; the coefficients for other mesh currents ik (kj) are the negative sum of the resistors common to mesh k and mesh j.The right-hand side of jth equation is equal to the algebraic sum of the voltage sources in mesh j. These voltage sources have a positive sign if they aid the current flow ij and a negative sum if they appose it.

  13. Homework for lecture 13 • Problem 3.40, 3.41 • Due Feb 11