Slide 1Equivalent CircuitsEquivalent Circuit (Example)Results - Equivalent CircuitFinding Thévenin and Norton Equivalent CircuitsFinding Thévenin and Norton Equivalent CircuitsSource TransformationSource TransformationSource Transformation ExampleNodal AnalysisExampleLoop/Mesh AnalysisAnalysis ExampleLinearity and SuperpositionAnalysis ExampleSPICE SolutionSPICE ExamplePlotting Frequency Sweep ResultsPlot of MagnitudePlot of PhaseSPICE ExampleCircuits IIEE221Unit 2Instructor: Kevin D. DonohueReview: Impedance Circuit Analysis with nodal, mesh, superposition, source transformation, equivalent circuits and SPICE analyses.Equivalent CircuitsCircuits containing different elements are equivalent with respect to a pair of terminals, if and only if their voltage and current draw for any load is identical. More complex circuits are often reduced to Thévenin and Norton equivalent circuits.Equivalent Circuit (Example)Find and compare the voltages and currents generated in 3 of the following loads across terminals AB:open circuitresistance RLshort circuit RthVsIsRthAABBThévenin NortonResults - Equivalent CircuitVABIABOpen 0Short 0RLthsRIsIthLthLsRRRRIthLthsRRRICurrent SourceNorton CircuitVABIABOpen 0Short 0RLsVthsRVthLLsRRRVthLsRRV1Voltage SourceThévenin CircuitWhat is the Norton equivalent for the Thévenin circuit?What is the Thévenin equivalent for the Norton circuit?Finding Thévenin and Norton Equivalent CircuitsIdentify terminal pair at which to find the equivalent circuit.Find voltage across the terminal pair when no load is present (open-circuit voltage Voc)Short the terminal and find the current in the short (short-circuit current Isc)Compute equivalent resistance as: Rth = Voc / IscFinding Thévenin and Norton Equivalent CircuitsThe equivalent circuits can then be expressed in terms of these quantities RthIscAB Voc RthABscocthIVR Source TransformationThe following circuit pairs are equivalent wrt to terminals AB. Therefore, these source and resistor combinations can be swapped in a circuit without affecting the voltages and currents in other parts of the circuit. RthIsAB Is RthRthAB Vs RthAB RthABthsRV IsRthABVs AB Vs ABABIsSource TransformationSome equivalent circuits can be determined by transforming source and resistor combinations and combining parallel and serial elements around a terminal of interest.This method can work well for simple circuits with source-resistor combinations as shown on the previous slide.This method cannot be used if dependent sources are present.Source Transformation ExampleUse source transformation to find the phasor valueShow = V cVˆcVˆ3 k6 k-j3.5 k05cVˆ 308.2Nodal AnalysisIdentify and label all nodes in the system.Select one node as a reference node (V=0).Perform KCL at each node with an unknown voltage, expressing each branch current in terms of node voltages.(Exception) If branch contains a voltage sourceOne way: Make reference node the negative end of the voltage source and set node values on the positive end equal to the source values (reduces number of equations and unknowns by one)Another way: (Super node) Create an equation where the difference between the node voltages on either end of the source is equal to the source value, and then use a surface around both nodes for KCL equation.ExampleFind the steady-state value of vo(t) in the circuit below, if vs(t) = 20cos(4t): vs10 2 ix1 H0.5 H0.1 FixShow: v0(t) = 13.91cos(4t - 161.6º)+vo-Loop/Mesh AnalysisCreate loop current labels that include every circuit branch where each loop contains a unique branch (not included by any other loop) and no loops “crisscross” each other (but they can overlap in common branches).Perform KVL around each loop expressing all voltages in terms of loop currents.If any branch contains a current source,One way: Let only one loop current pass through source so loop current equals the source value (reduces number of equations and unknowns by one)Another way: Let more than one loop pass through source and set combination of loop currents equal to source value (this provides an extra equation, which was lost because of the unknown voltage drop on current source)Analysis Example Find the steady-state response for vc(t) when vs(t) = 5cos(800t) VCan be derived with mesh or nodal analysis or source transformation: V 6800cos8868.2)( 302.8868 j1.4434 - 2.5000 ˆttvVcc 114.86 nF6 kW3 kWvs(t)+ vc(t) -Linearity and SuperpositionIf a linear circuit has multiple independent sources, then a voltage or current anywhere in the circuit is the sum of the quantities produced by the individual sources (i.e. activate one source at a time). This property is called superposition.To deactivate a voltage source, set the voltage equal to zero (equivalent to replacing it with a short circuit).To deactivate a current source, set the current equal to zero (equivalent to replacing it with an open circuit).Analysis Example Find the steady-state response for vc(t) when vs(t) = 4cos(200t) V and is(t) = 8cos(500t) A.Can be derived with superposition:  V )94.4-t.92cos(200 135500cos1.2)(  ttvc5 mF6 W4Wvs(t)+vc(t)-10 mHis(t)SPICE SolutionSteady-State Analysis in SPICE is performed using the .AC (frequency sweep) option in the simulation set up. It will perform the analysis for a range of frequencies.You must indicate the:1. Starting frequency2. Ending frequency3. Number of stepping increments and scale (log or linear)4. Scale for the results (linear or Decibel, Phase or radians)Sources in the AC analysis must be set up in “edit simulation model” menu to:1. Identify source as sinusoidal through the small signal AC and distortion tab.2. Provide a magnitude and phase and check the USE boxex16-Small Signal AC-2-TableFREQMAG(V(IVM)) PH_DEG(V(IVM))(Hz) (V) (deg)+100.000 +3.299 -8.213+200.000 +3.203 -16.102+300.000 +3.059 -23.413+400.000 +2.887 -30.000+500.000 +2.703 -35.817+600.000 +2.520 -40.893+700.000 +2.345 -45.295+800.000 +2.182 -49.107+900.000 +2.033 -52.411+1.000k +1.898 -55.285SPICE ExampleFind the phasor for vc(t) for vs(t)= 5cos(2ft) V in the circuit below for f = 100, 200, 300, 400, 500, …..1000 Hz. Note that 400 Hz was the frequency of the original example problem.VR2 6k R1 3k IVmC114.86nPlotting Frequency Sweep ResultsChoices for