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.

3 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!}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.

4 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. 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!

5 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.

6 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.

7 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.

8 Tse: Basic CircuitAnalysis28 Equivalent modelsTh venin equiv. cktVoltage source in series with , V + IRT = VTwhich is consistent with case1 equationNorton equiv. 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.

9 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).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.

10 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.