Ci it IICircuits IIEE221EE221Unit 2Instructor: Kevin D DonohueInstructor: Kevin D. DonohueReview: Impedance Circuit Analysis with nodal, mesh, superposition, source transformation equivalent circuits and transformation, equivalent circuits and SPICE analyses.Equivalent CircuitsEquivalent CircuitsCircuits containing different elements are Circuits containing different elements are equivalent with respect to a pair of terminals if and only if their voltage and 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.qEquivalent Circuit (Example)Equivalent Circuit (Example)Find and compare the voltages and currents Find and compare the voltages and currents generated in 3 of the following loads across terminals AB:A open circuit resistance RLsh t i it RthIsANorton short circuitVRthABThé iVsRthBThéveninResults -Equivalent CircuitResults Equivalent CircuitCtSVoltage So rceVABIABCurrent SourceNorton CircuitVABIABVoltage SourceThévenin CircuitVABIABOpen 0Sh0thsRIIVABIABOpen 0Sh0sVVShort0RLsIthLthLsRRRRIthLthsRRRIShort0RLthsRVthLLsRRRVthLsRRV1What is the Norton equivalent for the Thévenin circuit?What is the Thévenin equivalent for the Norton circuit?Finding Thévenin and Norton Equivalent CCircuitsIdentify terminal pair at which to find the 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 CCircuitsThe equivalent circuits can then be The equivalent circuits can then be expressed in terms of these quantitiesAAscocthIVR  RthIscAVoc RthABBSource TransformationSource TransformationThe following circuit pairs are equivalent wrtto terminals AB. Therefore these source and resistor combinations can be Therefore, these source and resistor combinations can be swapped in a circuit without affecting the voltages and currents in other parts of the circuit.AAA RthIsABIs RthRthABVs RthAB RthABthsRVBBBB RAA AAIsRthBVs BVs BBIsSource TransformationSource TransformationSome equivalent circuits can be determined by Some equivalent circuits can be determined by transforming source and resistor combinations and combining parallel and serial elements around a terminal of interestterminal of interest. This method can work well for simple circuits with psource-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Use source transformation to find the phasorvalue cVˆ3k ˆ3 k-j3.5 k05cVˆ6 kShow = VVˆ3082Show = VcV308.2Nodal Analysis Identify and label all nodes in the system.Slct n nd s f nc nd (V0)Select one node as a reference node (V=0). Perform KCL at each node with an unknown voltage, expressing each branch current in g, p gterms of node voltages. (Exception) If branch contains a voltage source One way: Make reference node the negative end of the l d d l h i i d l ygvoltage 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 diff r nc b t n th n d v lt s n ith r nd f th 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.ExampleExampleFi d th t dtt l f (t) i Find the steady-state value of vo(t) in the circuit below, if vs(t) = 20cos(4t): 10 1 Hi+vs2 ix0.5 H0.1 Fix+vo-Show: v0(t) = 13.91cos(4t-161.6º)Show 0(t) .9 cos(t6.6)Loop/Mesh AnalysisLoop/Mesh AnalysisCreate loop current labels that include every 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 p(ypin common branches). Perform KVL around each loop expressing all voltages in terms of loop currents.gp 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)A th L t th l th h d t 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 ExampleAnalysis ExampleFind the steady-state response for vc(t) when vs(t) Find the steadystate response for vc(t) when vs(t) = 5cos(800t) V3k+ vc(t) - 114.86 nF6k3 kvs(t)C b d i d ith h d l l i t f ti6 kCan be derived with mesh or nodal analysis or source transformation:V 6800cos8868.2)( 302.8868 j1.4434 - 2.5000 ˆttvVcc6Linearity and SuperpositionLinearity and SuperpositionIf a linear circuit has multiple independent sources then a 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.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 ExampleAnalysis ExampleFind the steady-state response for vc(t) when Find the steadystate response for vc(t) when vs(t) = 4cos(200t) V and is(t) = 8cos(500t) A.10 5 mF6 4vs(t)+vc(t)-10 mis(t)C b d i d ith itiCan be derived with superposition:V )94.4-t.92cos(200 135500cos1.2)(  ttvcSPICE SolutionSteady-State Analysis in SPICE is performed using the AC (f ) ti i th i l ti t .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