Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11EE201 Lecture 7 P. 1Nodal AnalysisNodal AnalysisA systematic technique for computing the node voltages of a circuit. Once the node voltages are known, the CE voltages can be found. This yields all branch currents if resistances are known. +_veq_+Req vin(t)iin(t)AFrom KCL @A: iin = vA / Req This is the simplest use of nodal analysis. Once vA is known, current through the resistor can be calculated.EE201 Lecture 7 P. 2Nodal Analysis - guidelines1) Draw a clear circuit diagram indicating nodes.2) Define a reference node (ground). This should usually be the node with the most branches.3) Assign node voltages to all nodes other than ground.4) If circuit contains voltage sources, include them in a ‘supernode.’ This reduces the number of independent variables.5) Apply KCL at each node to obtain node voltages.6) Define no current variables.EE201 Lecture 7 P. 3Given: I, R1, R2, R33 nodes in this circuit. Node C is ground.Voltages across resistors:V1 = VAC = VA - 0V2 = VAB = VA - VBV3 = VBC = VB - 0CBAR1R2R3Circuit solution using nodal analysis_V1+V2V3_+_+IEE201 Lecture 7 P. 4Apply KCL (basis of nodal analysis) at each node:@A: I = I1 + I2 = V1 / R1 + V2 / R2@B: I2 = I3 = V2 / R2 = V3 / R3Insert nodal voltages into the above equations.@A: I = VA / R1 + (VA - VB) / R2 @B: 0 = (VB - VA) / R2 + VB / R3Which are 2 equations with 2 unknowns (VA, VB)Solve these simultaneous equations to yield VA and VB, and the circuit behavior is known.EE201 Lecture 7 P. 5Example: Find vx and iy. +_+_+_22225vx3vx5V5Aiy10V3A2iyEE201 Lecture 7 P. 61) Clear circuit diagram with nodes labelled.+_+_+_22225vx3vx5V5Aiy10V3A2iy2) Define reference node (ground).3) Assign node voltages: VA, VB, VC, VDTrouble: how do you find current through a voltage source?AB CDEE201 Lecture 7 P. 7+_+_+_22225vx3vx5V5Aiy10V3A2iy4) Define supernode to eliminate difficulty with finding current through voltage sources. Supernodes enclose two or more nodes and are an extension of the Guassian surface concept.AB CDSN1EE201 Lecture 7 P. 85) Use KCL at each node to obtain set of equations that will yield node voltages.How many unknown node voltages are there?2 unknowns: VB and VC (or VD)KCL@B: -5 - 3vx + iy + 2 vx = 0-5 - 3(VB - 5) + 2 (VB - VC) + 2(VB - 5) = 0-5 - (VB - 5) + 2 (VB - VC) = 0 VB - 2VC = 0KCL@SN1: -3+ 5+ 2(VC–VB) + 5VC + 2VC – 2iy +2(VC–10) = 0Insert iy = 2(VB - VC) and reduce, -6VB + 15 VC = 18vx = (VB - 5) = 7 Viy = 2(VB - VC) = 12 AEE201 Lecture 7 P. 9Node Analysis with Dependent Voltage SourcesGenerally, dependent voltage sources are included in a supernode.+_30V800 _+110V50 100 500 ix+_AB CD40 ixEEE201 Lecture 7 P. 10+_30V800 _+110V50 100 500 ix+_AB CD40 ixDefine supernode to include independent and dependent source.EEE201 Lecture 7 P. 11KCL@supernode:(VA - VE)/50 + VB/500 + VC/800 = 0But VE = 110 V; VC = VB + 30 (VA - 110)/50 + VB/500 + (VB + 30)/800 = 0Current through 100 resistor:(VA - VB)/100 = ix ; VA - VC = 40 ix(VA - VB)/100 = (VA - VB - 30)/40Simultaneous equations to solve (VA - 110)/50 + VB/500 + (VB + 30)/800 = 0 (VA - VB)/100 - (VA - VB - 30)/40 = 0Solve these for VA and VB, then substitute into equation for current
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