Basic circuit analysis - City U

1 Prof. Tse: Basic CircuitAnalysis1 EIE209 Basic ElectronicsBasic circuit analysisProf. Tse: Basic CircuitAnalysis2 Fundamental quantities Voltage potential difference bet. 2 points across quantity analogous to pressure between two points Current flow of charge through a material through quantity analogous to fluid flowing along a pipeProf. Tse: Basic CircuitAnalysis3 Units of measurementnVoltage: volt (V)nCurrent: ampere (A)nNOT Volt, Ampere!!Prof. Tse: Basic CircuitAnalysis4 Power and energyWork done in moving a charge dq from A toB having a potential difference of V isW = V dqPower is work done per unit time, ,ABdqProf. Tse: Basic CircuitAnalysis5 Direction and polaritynCurrent direction indicates the direction of flow of positive chargenVoltage polarity indicates the relative potential between 2 points:+ assigned to a higher potential point; and assigned to alower potential : Direction and polarity are arbitrarily assigned on circuitdiagrams.

2 Actual direction and polarity will be governed by thesign of the Tse: Basic CircuitAnalysis6 Independent sourcesnVoltage sourcesnCurrent sourcesIndependent stubborn! never change!Maintains avoltage/current(fixed or varying)which is notaffected by anyother independent voltage source can never be independent current source can never be Tse: Basic CircuitAnalysis7 Dependent sourcesnDependent sources values depend on some other variablesProf. Tse: Basic CircuitAnalysis8 CircuitnCollection of devices such as sources and resistors in whichterminals are connected together by conducting wires converge in NODESnThe devices are called BRANCHES of the circuitCircuit analysis Problem:To find all currents andvoltages in the branchesof the circuit when theintensities of thesources are Tse: Basic CircuitAnalysis9 Kirchhoff s lawsnKirchhoff s current law(KCL)nThe algebraic sum of thecurrents in all brancheswhich converge to acommon node is equalto s voltage law(KVL)nThe algebraic sum of allvoltages betweensuccessive nodes in aclosed path in the circuitis equal to Tse: Basic CircuitAnalysis10 Overview of analysisnAd hoc methods (not general)nSeries/parallel reductionnLadder circuitnVoltage/current divisionnStar-delta conversionnMore generalnMesh and nodal methodsnCompletely generalnLoop and cutset approach (requires graph theory)Done inBasicElectronics!

3 }NEWProf. Tse: Basic CircuitAnalysis11 Series/parallel reductionnSeries circuit eachnode is incident tojust two branches ofthe circuitKVL gives=Hence, theequivalentresistance is:Prof. Tse: Basic CircuitAnalysis12 Series/parallel reductionnParallel circuit one terminal ofeach element is connected to a nodeof the circuit while other terminals ofthe elements are connected toanother node of the circuitKCL givesHence, theequivalentresistance is:Prof. Tse: Basic CircuitAnalysis13 Note on algebranFor algebraic brevity and simplicity:nFor series circuits, R is preferably parallel circuits, G is preferably example, if we use R for the parallel circuit , we get theequivalent resistance aswhich is more complex than the formula in terms of G:G = G1 + G2 + .. + GnProf. Tse: Basic CircuitAnalysis14 Ladder circuitnWe can find the resistancelooking into the terminals 0and 1, by apply the series/parallel reduction , lumping everything beyond node 2 as G2, we haveThen, we focus on this G2, which is just G20 in parallelwith another subcircuit, ,We continue to focus on the remainingsubcircuit.

4 Eventually we getProf. Tse: Basic CircuitAnalysis15 Voltage/current divisionFor the series circuit , we can find thevoltage across each resistor by theformula:Note the choice of R and G in theformulae!For the parallel circuit , we can find thevoltage across each resistor by theformula:Prof. Tse: Basic CircuitAnalysis16 Example (something that can bedone with series/parallel reduction)Consider this circuit , which is createddeliberately so that you can solve it usingseries/parallel reduction technique. Find :Resistance seen by the voltage source isHence,Current division gives:Then, using V2=I4R4, we getProf. Tse: Basic CircuitAnalysis17 Oops!Series/parallel reductionfails for this bridge circuit !Is there some ad hocsolution?Prof. Tse: Basic CircuitAnalysis18 Equivalence of star and a star circuit , find the delta equivalence. That means,suppose you have all the G s in the star. Find the G s in thedelta such that the two circuits are equivalent from theexternal reverse (star)D (delta)Prof.

5 Tse: Basic CircuitAnalysis19 Star-to-delta conversionFor the Y circuit , we considersumming up all currents into thecentre node: I1+I2+I3=0, whereY (star)D (delta)Thus,, andProf. Tse: Basic CircuitAnalysis20 Star-to-delta conversionFor the D circuit , we haveY (star)D (delta)Prof. Tse: Basic CircuitAnalysis21 Star-to-delta conversionNow, equating the two sets of I1, I2 and I3, we getThe first problem is Tse: Basic CircuitAnalysis22 Delta-to-star conversionThis problem is moreconveniently handled in termsof R. The answer is:Prof. Tse: Basic CircuitAnalysis23 Example the bridge circuit againWe know that the series/parallelreduction method is not useful forthis circuit !The star-delta transformation maysolve this question is how to apply thetransformation so that the circuitcan become solvable using theseries/parallel reduction or other achoc Tse: Basic CircuitAnalysis24 Example the bridge circuit againAfter we do the conversion from Y to D, we can easily solve thecircuit with parallel/series Tse: Basic CircuitAnalysis25 Useful/important theorems Th venin Theorem Norton Theorem Maximum Power Transfer TheoremProf.

6 Tse: Basic CircuitAnalysis26Th venin and Norton theoremsCircuit in questionExternal apparatus(another circuit )Problem:Find the simplest equivalent circuit model for N, such that theexternal circuit N* would not feel any difference if N isreplaced by that equivalent solution is contained in two theorems due to Th venin Tse: Basic CircuitAnalysis27Th venin and Norton theoremsLet s look at the logic behind these theorems (quite simple really).If we write down KVL, KCL, and Ohm s law equations correctly, we willhave a number of equations with the same number of , we can try to solve them to get what we suppose everything is linear. We are sure that we can get thefollowing equation after elimination/substitution (some high schoolalgebra):Case 1: a 0 Case 2: b 0Th veninNortonProf. Tse: Basic CircuitAnalysis28 Equivalent modelsTh venin equiv. cktVoltage source in series with , V + IRT = VTwhich is consistent with case1 equationNorton equiv.

7 CktCurrent source in parallel witha , I = IN + V/RNwhich is consistent with case2 equationProf. Tse: Basic CircuitAnalysis29 How to find VT and INTh venin equiv. cktOpen- circuit the terminals(I=0), we get VT as theobserved value of ! VT is just the open- circuit voltage!Norton equiv. cktShort- circuit the terminals(V=0), we get IN as theobserved current ! IN is just the short- circuit current!= VTI = INProf. Tse: Basic CircuitAnalysis30 How to find RT and RN (they are equal)Th venin equiv. cktShort- circuit the terminals(V=0), find I which is equal toVT/RT. Thus, RT = VT / IscNorton equiv. cktOpen- circuit the terminals(I=0), find V which is equal toINRN. Thus, RN = Voc / both cases,RT = RN = Voc / IscI = Isc= VocProf. Tse: Basic CircuitAnalysis31 Simple exampleStep 1: open-circuitThe o/c terminal voltage isStep 2: short-circuitThe s/c current isStep 3: Th venin or Norton resistanceHence, the equiv. ckts are:Prof. Tse: Basic CircuitAnalysis32 Example the bridge againProblem: Find the current flowing in solution is by delta-star conversion (as done before).

8 Another simpler method is to find the Th venin equivalentcircuit seen from Tse: Basic CircuitAnalysis33 Example the bridge againStep 1: open circuitThe o/c voltage across A and B isStep 2: short circuitThe s/c current isStep 3: RT= VTProf. Tse: Basic CircuitAnalysis34 Example the bridge again= Current in R5 = VTR5+RTProf. Tse: Basic CircuitAnalysis35 Maximum power transfer theoremWe consider the power dissipated by current in RL isThus, the power isThis power has a maximum, when plottedagainst 0 gives RL = Tse: Basic CircuitAnalysis36A misleading interpretationIt seems counter-intuitive that the MPT theorem suggests a maximumpower at RL = t maximum power occur when we have all power go to theload? That is, when RT = 0!Is the MPT theorem wrong?Discussion: what is the condition required by the theorem?Prof. Tse: Basic CircuitAnalysis37 Systematic analysis techniquesSo far, we have solved circuits on an ad hoc manner.

9 We are ableto treat circuits with parallel/series reduction, star-delta conversion, withthe help of some about very general arbitrary circuit styles?In Basic Electronics, you have learnt the use of MESH and planar circuits only; solution in terms of mesh any circuit ; solution in terms of nodal THEY ARE NOT EFFICIENT!Prof. Tse: Basic CircuitAnalysis38 Mesh analysis (for planar circuits only)Meshes windowsPlanar or not?Prof. Tse: Basic CircuitAnalysis39 Mesh analysisStep 1: Define meshes and unknownsEach window is a mesh. Here, we have twomeshes. For each one, we imagine a currentcirculating around it. So, we have two suchcurrents, I1 and I2 unknowns to be 2: Set up KVL equationsStep 3: Simplify and solvewhich gives I1 = 6 A and I2 = 4 we know the mesh currents, we canfind anything in the circuit ! , current flowing down the 3 resistorin the middle is equal to I1 I2 ;current flowing up the 42V source is I1 ;current flowing down the 10V source is I2 ;and voltages can be found via Ohm s Tse: Basic CircuitAnalysis40 Mesh analysisIn general, we formulate the solution in terms of unknown mesh currents:[ R ] [ I ] = [ V ] mesh equationwhere [ R ] is the resistance matrix[ I ] is the unknown mesh current vector[ V ] is the source vectorFor a short cut in setting up the above matrix equation, see Sec.

10 Of thetextbook. This may be picked up in the Tse: Basic CircuitAnalysis41 Mesh analysis observing superpositionConsider the previous example. The mesh equation is given by:orThus, the solution can be written asRemember what 42 and 10 are? They are the sources! The above solution canalso be written asorSUPERPOSITIONof two sourcesProf. Tse: Basic CircuitAnalysis42 Problem withcurrent sourcesThe mesh method may run into trouble if the circuit hascurrent source(s).Suppose we define the unknowns in the same way, ,I1, I2 and I3 .The trouble is that we don t know what voltage isdropped across the 14A source! How can we setup the KVL equation for meshes 1 and 3?One solution is to ignore meshes 1 and 3. Instead welook at the supermesh containing 1 and , we set up KVL equations for mesh 2 and thesupermesh:Mesh 2:Supermesh:One more equation: I1 I3 = 14 Finally, solve the Tse: Basic CircuitAnalysis43 Complexity of mesh methodIn all cases, we see that the mesh method ends upwith N equations and N unknowns, where N is thenumber of meshes (windows) of the important point:The mesh method is over-complex when applied tocircuits with current source(s).