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Lecture 5. Nodal Analysis • Nodes and reference nodes • Steps of Nodal Analysis • Supernodes • Examples 1 Circuit 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. 2 Learning by Examples: A Summing Circuit 500W I1s + 500W V - 500W 1kW 500W I2s • 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? 3 Nodal Analysis: The Recipe 1. 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. 4 Step 1. Reference Node 500W 500W + I1s 500W V 1kW 500W I2s – The reference node is often called the ground node 5 Nodal Analysis: The Recipe 1. 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. 6 Step 2. Node Voltages V1 500W V 500W 2 1 I1s 2 500W V3 3 1kW 500W I2s V1, V2, and V3 are unknowns for which we solve using KCL. 7 Nodal Analysis: The Recipe 1. 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. 8 Step 3. Currents and Node Voltages V1 500W V1 V2 500W V2 V1 V1 500W 500W 9 Nodal Analysis: The Recipe 1. 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. 10 Step 4. KCL at Node 1 V1 I1s 500W 500W V2 V1 V2 V1 I 1s 0 500W 500W 11 Step 4. KCL at Node 2 V1 500W V2 500W V3 1kW V2 V1 V2 V2 V3 0 500W 1kW 500W 12 Step 4. KCL at Node 3 V2 500W V3 500W I2s V3 V2 V3 I 2s 0 500W 500W 13 Nodal Analysis: The Recipe 1. 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. 14 Step 5. Solving the Equations • Re-organize the Equations 1 1 1 V V2 0 V3 I1s 1 500W 500W 500W V1 V2 V1 I 1s 0 500W 500W V V V V V 2 1 2 2 3 0 500W 1kW 500W V3 V2 V 3 I 2s 0 500W 500W 1 1 1 1 1 V1 V3 0 V2 500W 500 W 1 k W 500 W 500 W 0 V1 1 1 1 V2 V3 I 2 s 500W 500W 500W • 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. 15 Matrix Notation • The three equations can be combined into a single matrix/vector equation. 1 1 500W 500W 1 500W 0 1 500W 1 1 1 500W 1kW 500W 1 500W V 1 I1 1 V2 0 500W 1 1 V3 I 2 500W 500W 0 • The equation can be written in matrix-vector form as Av = i • The solution to the equation can be written as v = A-1 i 16 Solving the Equation with MATLAB I1s = 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 1.5833 17 A General Solution 500W 500W + I1s 500W V 1kW 500W I2s – Solution: V = 167I1 + 167I2 • Can you prove this? 18 Another Example: A Linear Large Signal Equivalent to a Transistor 0.7V Ib 5V + – 1kW + – 50W 100Ib 2kW + Vo – 19 Nodal Analysis: The Recipe 1. 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. 20 Another Example: A Linear Large Signal Equivalent to a Transistor 0.7V 1 5V Ib V2 V1 + – 1kW 2 + – V3 V4 3 50W 4 + Vo 100Ib 2kW – Nodal Analysis: The Recipe 1. 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. 22 KCL @ Node 4 Node 1: V1 5 0.7V 1 5V Ib V2 V1 + – 1kW 2 + – V3 V4 3 50W 4 + Vo 100Ib 2kW – V3 V4 V4 100 I b 0 Node 4: 50W 2kW 23 How to Treat the Dependent Source • We must express Ib in terms of the node voltages: Ib V1 V2 1kW • Equation from Node 4 becomes V3 V4 V V V 100 1 2 4 0 50W 1 kW 2kW 24 How 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! 25 Supernode 0.7V 1 V2 V1 + – 1kW Ib + – V3 50W 100Ib V4 + 4 Vo 2kW V2 V1 V3 V4 0 1kW 50W And don’t forget – V2 V3 0.7 • If a voltage source is not connected to the reference node, then it is supernode! 26 Nodal Analysis: The Recipe 1. 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. 27 Step 5. Solving the Equations V1 5 Great, one variable is already known! V2 V3 0.7 V2 V1 V3 V4 0 1kW 50W V2 5 V3 V4 0 1kW 50W V3 V4 V1 V2 V4 100 0 50W 1 kW 2kW V3 V4 5 V2 V4 100 0 50W 1 kW 2kW • Write the equations for V2, V3 and V4: V2 V3 0V4 0.7 V V2 V 5 3 4 1kW 50W 50W 1kW 100 1 1 1 500 V2 V3 ( )V4 1 kW 50W 50 2kW 1 kW 28 Class Examples • Drill Problems P2-8 and P2-10 29