Slide 1Circuit Analysis – A Systematic ApproachLearning by Examples: A Summing CircuitNodal Analysis: The RecipeStep 1. Reference NodeSlide 6Step 2. Node VoltagesSlide 8Step 3. Currents and Node VoltagesSlide 10Step 4. KCL at Node 1Step 4. KCL at Node 2Step 4. KCL at Node 3Slide 14Step 5. Solving the EquationsMatrix NotationSolving the Equation with MATLABA General SolutionAnother Example: A Linear Large Signal Equivalent to a TransistorSlide 20Slide 21Slide 22KCL @ Node 4How to Treat the Dependent SourceHow to Deal With Nodes 2 and 3?Slide 26Slide 27Slide 28Class Examples• Nodes and reference nodes• Steps of Nodal Analysis• Supernodes• ExamplesLecture 5. Nodal Analysis12Circuit Analysis – A Systematic Approach•We’ve learned several tricks to perform circuit analysis: Single loop circuits, equivalent resistor, superposition etc.•Nodal Analysis is a rather general method that allows you to analyze virtually all the linear circuits via a well defined recipe.3Learning by Examples: A Summing Circuit•The output voltage V of this circuit is proportional to the sum of the two input currents I1s and I2s.•This circuit could be useful in audio applications or in instrumentation.•The output of this circuit would probably be connected to an amplifier.•Can you solve this problem using superposition method?+-V5005001k500500I1sI2s4Nodal Analysis: The Recipe1. Choose a reference node and assign 0 voltage to it.2. Assign node voltages to the other nodes.3. Express currents in terms of node voltages.4. Apply KCL to each node other than the reference node. 5. Solve the resulting system of linear equations.5Step 1. Reference NodeThe reference node is often called the ground node+–V 5005001k500500I1s I2s6Nodal Analysis: The Recipe1. Choose a reference node and assign 0 voltage to it.2. Assign node voltages to the other nodes.3. Express currents in terms of node voltages.4. Apply KCL to each node other than the reference node. 5. Solve the resulting system of linear equations.7Step 2. Node VoltagesV1, V2, and V3 are unknowns for which we solve using KCL.5005001k500500I1s I2s1 2 3V1V2V38Nodal Analysis: The Recipe1. Choose a reference node and assign 0 voltage to it.2. Assign node voltages to the other nodes.3. Express currents in terms of node voltages.4. Apply KCL to each node other than the reference node. 5. Solve the resulting system of linear equations.9Step 3. Currents and Node Voltages500V1500V1V250021VV5001V10Nodal Analysis: The Recipe1. Choose a reference node and assign 0 voltage to it.2. Assign node voltages to the other nodes.3. Express currents in terms of node voltages.4. Apply KCL to each node other than the reference node. 5. Solve the resulting system of linear equations.11Step 4. KCL at Node 1500500I1sV1V205005001211VVVIs12Step 4. KCL at Node 25001k500V2V3V10500k150032212VVVVV13Step 4. KCL at Node 305005002323sIVVV500500I2sV2V314Nodal Analysis: The Recipe1. Choose a reference node and assign 0 voltage to it.2. Assign node voltages to the other nodes.3. Express currents in terms of node voltages.4. Apply KCL to each node other than the reference node. 5. Solve the resulting system of linear equations.15Step 5. Solving the Equations•The left side of the equation is a sum of a linear combination of node voltages (variables to be determined).•The right side of the equation is a sum of currents from sources entering the node.•Re-organize the EquationssIVVV13210500150015001050015001k1150015001321 VVVsIVVV2321500150 0150010 05005001211VVVIs050 0k150032212VVVVV05005002323sIVVV16Matrix Notation•The three equations can be combined into a single matrix/vector equation.21321050 01500150 01050 0150 01k11500150 01050 0150 0150 01IIVVV•The equation can be written in matrix-vector form asAv = i•The solution to the equation can be written asv = A-1 i17Solving the Equation with MATLABI1s = 3mA, I2s = 4mA>> A = [1/500+1/500 -1/500 0; -1/500 1/500+1/1000+1/500 -1/500; 0 -1/500 1/500+1/500];>> i = [3e-3; 0; 4e-3];>> v = inv(A)*i v = 1.3333 1.1667 1.583318+–V 5005001k500500I1s I2sA General SolutionSolution: V = 167I1 + 167I2•Can you prove this?19Another Example: A Linear Large Signal Equivalent to a Transistor5V100Ib+–Vo50Ib2k1k+–+ –0.7V20Nodal Analysis: The Recipe1. Choose a reference node and assign 0 voltage to it.2. Assign node voltages to the other nodes.3. Express currents in terms of node voltages.4. Apply KCL to each node other than the reference node. 5. Solve the resulting system of linear equations.5V100Ib+–Vo50Ib2k1k0.7V12 3 4V1V2 V3V4+–+ –Another Example: A Linear Large Signal Equivalent to a Transistor22Nodal Analysis: The Recipe1. Choose a reference node and assign 0 voltage to it.2. Assign node voltages to the other nodes.3. Express currents in terms of node voltages.4. Apply KCL to each node other than the reference node. 5. Solve the resulting system of linear equations.23KCL @ Node 40k210050443VIVVb5V100Ib+–Vo50Ib2k1k0.7V12 3 4V1V2 V3V4+–+ –Node 1:Node 4:51V24How to Treat the Dependent Source•We must express Ib in terms of the node voltages:•Equation from Node 4 becomesk121VVIb0k2k11005042143VVVVV25How to Deal With Nodes 2 and 3? •The 0.7-V voltage supply makes it impossible to apply KCL to nodes 2 and 3, since we don’t know what current is passing through the supply.•We do know that V2 – V3 = 0.7 V•We need another equation!26100Ib+–Vo50Ib2k1k0.7V14V1V2 V3V4+–+ –050k14312VVVVSupernode0.732 VVAnd don’t forget •If a voltage source is not connected to the reference node, then it is supernode!27Nodal Analysis: The Recipe1. Choose a reference node and assign 0 voltage to it.2. Assign node voltages to the other nodes.3. Express currents in terms of node voltages.4. Apply KCL to each node other than the reference node. 5. Solve the resulting system of