Electronic Devices and Circuits Dr.K.Lal Kishore Ph.D, MIEEE,FIETE,MISTE,MISHM. Registrar andProfessor of Electronics & Communication Engineering, Jawaharlal Nehru Technological University,Kukatpally, Hyderabad - 500 072. SSPBS Publications 4-4-309, Giriraj Lane,Sultan Bazar, Hyderabad - 500 095 A. P. Phone: 040-23445688 Copyright2008by publisher Allrightsreserved.Nopartof thisbookorpartsthereof maybe reproduced, stored in a retrieval system or transmitted in any language or byany means, electronic,mechanical,photocopying,recording or otherwisewithout theprior writtenpermissionof thepublishers. Published by: SSP BS Publications - 4-4-309, Giriraj Lane,Sultan Bazar, Hyderabad - 500 095 AP. Phone: 040-23445688 Fax: 040 - 23445611 e-mail: contactus@bspublications.net www.bspublications.net Printed at Adithya ArtPrinters Hyderabad. ISBN:81-7800-167-5 CONTENTS Contents ............... .......................... ..... .... .................................................. ....... ..... . Symbols ............................................... ........... ....................................................... . BriefHistory ofElectronics ................................................................................... . Chapter 1 Electron Dynamics and CRO .............................................................1-39 1.1ElectronDynamics................................................................................................ 2 1.2Motion of ChargedParticlesinElectric andMagneticFields ............................... 2 1.3Simple ProblemsInvolving Electric andMagnetic Fields Only .......................... 24 1.4Principles of CRT................................................................................................ 26 1.5Deflection Sensitivity ........................................................................................... 29 1.6Applications ofCRO ............................................................................................ 36 Summary .............................................................................................................. 36 Objective Type Questions .................................................................................... 37 EssayTypeQuestions .......................................................................................... 38 Multiple Choice Questions ................................................................................... 38 Chapter2 JunctionDiodeCharacteristics .................................................... 39-134 2.1Reviewof SemiconductorPhysics..................................................................... 40 2.2EnergyBandStructures ....................................................................................... 61 2.3ConductioninSemiconductors ........................................................................... 62 2.4Conductivityof anIntrinsicSemiconductor ....................................................... 66 2.5DonorTypeorn-TypeSemiconductors............................................................. 67 2.6AcceptorTypeorp-TypeSemiconductors ......................................................... 68 2.7Ionization Energy ................................................................................................. 68 2.8Holes and Electrons............................................................................................. 68 2.9Mass ActionLaw ................................................................................................. 70 2.10Law of Electrical Neutrality ................................................................................. 70 2.11TheFermiDiracFunction................................................................................... 75 2.12TotalCurrent inaSemiconductor ....................................................................... 84 2.13Einstein Relationship............................................................................................ 90 2. i 4Continuity Equation .............................................................................................. 90 2.15The Hall Effect ..................................................................................................... 92 2.16Semiconductor DiodeCharacteristics ................................................................. 96 2.17The p-n Junction Diode inReverse Bias............................................................. 98 2.18Thep-nJunction DiodeinForward Bias ............................................................ 98 2.19BandStructureof anOpen Circuit p-nJunction ................................................ 99 2.20The Current Componentsina p-n Junction Diode ...........................................102 2.21Lawof theJunction ........................................................................................... 103 2.22Diode Current equation ......................................................................................104 2.23Volt-AmpereCharacteristics of a p-n Junctiondiode....................................... 105 2.24Temperature Dependanceofp-n JunctionDiodeCharacteristics.................... 107 2.25SpaceChargeor Transition CapacitanceCT 108 2.26Diffusion Capacitance, CD111 2.27DiodeSwitching Times.....................................................................................113 2.28Break Down Mechanism...................................................................................118 2.29Zener Diode ........................................................................................................119 2.30The TunnelDiode........................................................................... :.. 0' 120 2.31Varactor Diode....................................................................... .' ...........................123 Summary ............................................................... :............................................ 129 Objective TypeQuestions ..................................................................................130 EssayTypeQuestions ........................................................................................13 1 Multiple Choice Questions.................................................................................132 Chapter 3 Rectifiers,FiltersandRegulators .............................................. 135-184 3.1Rectifiers ............................................................................................................136 3.2Half-Wave Rectifier ............................................................................................ 136 3.3FullWaveRectifier (FWR ) .............................................................................. 146 3.4Bridge Rectifiers................................................................................................ 150 3.5Comparison of Rectifier Circuits .......................................................................151 3.6Voltage Doubler Circuit ......................................................................................152 3.7InductorFilter Circuits......................................................................................152 3.8Capacitor Filter..................................................................................................157 3.9LCFilter .............................................................................................................161 3.10CLC or 1tFilter................................................................................................... 165 3.11Multiple LCFilters............................................................................................. 169 3.12IntroductiontoRegulators ................................................................................. 173 3.13Terminology ....................................................................................................... 182 Summary ............................................................................................................ 183 Objective Type Questions .................................................................................. 183 EssayTypeQuestions ........................................................................................ 184 Multiple Choice Questions .................................................................................184 Chapter 4 TransistorCharacteristics ...........................................................185-266 4.1BipolarJunctionTransistors(BJT's) ..............................................................186 4.2TransistorConstruction .....................................................................................190 4.3The Ebers-Moll Equation................................................................................... 191 4.4Typesof TransistorConfigurations.................................................................. 192 4.5ConventionforTransistorsandDiodes............................................................ 202 4.6FieldEffectTransistor(FET)........................................................................... 213 4.7FETStructure .................................................................................................... 215 4.8FET Operation................................................................................................... 219 4.9JFET Volt-AmpereCharacteristics .................................................................... 222 4.10TransferCharacteristicsof FET ....................................................................... 224 4.11FET Small Signal Model.................................................................................... 228 4.12FET Tree ............................................................................................................ 233 4.13The Depletion MOSFET .................................................................................... 240 4.14CMOSStructure(ComplementaryMOS) ......................................................... 243 4.15Silicon Controlled Rectifier ................................................................................ 246 4.16Unijunction Transistor (UJT) ............................................................................. 251 4.17LED's................................................................................................................. 255 4.18PhotoDiodes ...................................................................................................... 255 4.19PhotoTransistors.............................................................................................. 256 Summary ............................................................................................................ 257 Objective TypeQuestions .................................................................................. 258 EssayTypeQuestions ........................................................................................ 259 Multiple Choice Questions ................................................................................. 260 Chapter 5 Transistor Biasing andStabilization.......................................... 261-312 5.1Transistor Biasing .............................................................................................. 268 5.2Fixed Bias Circuit or (Base Bias Circuit) .......................................................... 270 5.3Bias Stability ....................................................................................................... 271 5.4Thermal Instability ............................................................................................. 271 5.5Stability Factor'S' forFixed Bias Circuit ......................................................... 272 5.6Collector to Base Bias Circuit ............................................................................ 273 5.7Self Bias or Emitter Bias Circuit ........................................................................ 276 5.8Stability Factor'S' forSelf Bias Circuit.. .......................................................... 277 5.9Stability Factor S I.................................................... 278 5.10Stability Factor S"forSelf Bias Circuit ........................................................... 280 5.11Practical Considerations ..................................................................................... 280 5.12Bias Compensation ............................................................................................. 281 5.13Biasing CircuitsFor Linear Integrated Circuits ................................................. 284 5.14Thermistor andSensistor Compensation.......................................................... 285 5.15ThermalRunaway .............................................................................................. 286 5.16Stability Factor S"forSelf Bias Circuit ........................................................... 292 5.17FETBiasing ....................................................................... _............................. 298 5.18BasicFET Circuits ............................................................................................. 302 Summary ............................................................................................................ 309 Objective TypeQuestions .................................................................................. 310 EssayTypeQuestions ........................................................................................ 310 Multiple Choice Questions ................................................................................. 311 Chapter 6 Amplifiers ..................................................................................... 313-380 6.1Introduction ....................................................................................................... 314 6.2Black Box Theory .............................................................................................. 314 6.3Transistor Hybrid Model .................................................................................... 318 6.4Transistor inCommonEmitter Configuration ................................................... 318 6.5Determinationof h-ParametersFromtheCharacteristicsof aTransistor ....... 319 6.6Common Collector Configuration ( CC ) .......................................................... 321 6.7Hybrid Parameter Variations ............................................................................... 322 6.8Conversionof Parameters FromC.B.toC.E ................................................... 323 6.9Measurementof h-Parameters ........................................................................... 325 6.10General Amplifier Characteristics ...................................................................... 327 6.11Analysis of Transistor Amplifier Circuit Using h-Parameters ........................... 330 6.12Comparison of the CE,CB, CCConfigurations ................................................ 334 6.13SmallSignal Analysisof Junction Transistor .................................................... 337 6.14HighInput ResistanceTransistor Circuits........................................................ 354 6.15Boot Strapped Darlington Circuit...................................................................... 358 6.16The CascodeTransistor Configuration ............................................................. 361 6.17TheJFET LowfrequencyEquivalent Circuits .................................................. 365 6.18Comparisonof FET andBJT Characteristics ................................................... 369 6.19R.C. Coupled Amplifier ..................................................................................... 370 6.20Conceptof fa'fpandfT.....373 Summary ............................................................................................................ 375 Objective Type Questions .................................................................................. 376 EssayTypeQuestions ........................................................................................ 377 Multiple Choice Questions................................................................................. 378 Chapter 7 FeedbackAmplifiers .................................................................... 381-428 7.1FeedbackAmplifiers.......................................................................................... 382 7.2Classification ofAmplifiers ................................................................................ 382 7.3FeedbackConcept ............................................................................................. 385 7.4Typesof Feedback............................................................................................ 387 7.5Effect of Negative Feedback onTransfer Gain................................................ 387 7.6TransferGain withFeedback ............................................................................ 392 7.7Classifaction of Feedback Amplifiers ................................................................ 396 7.8Effectof FeedbackonInputResistance ........................................................... 397 7.9Effect of NegativeFeedback onR.....400 o 7.10Analysis of Feedback Amplifiers....................................................................... 406 Summary ............................................................................................................ 424 Objective Type Questions .................................................................................. 425 EssayTypeQuestions ........................................................................................ 426 Multiple Choice Questions ................................................................................. 427 Chapter 8 Oscillators .................................................................................... 429-453 8.1Oscillators .......................................................................................................... 430 8.2Sinusoidal Oscillators ......................................................................................... 433 8.3Barkhausen Criterion .... : ..................................................................................... 433 8.4R - C Phase-Shift Oscillator (Using JFET)...................................................... 434 8.5Transistor RCPhase-Shift Oscillator ................................................................ 437 8.6The Generalformof LCOscillator Circuit ....................................................... 440 8.7Loop Gain ........................................................................................................... 440 8.8Wien Bridge Oscillator ....................................................................................... 443 8.9Expressionfor f. ................................................................................................ 444 8.10Thermistor ......................................................................................................... 445 8.11Sensistor ............................................................................................................. 445 8.12Amplitude Stabilization ................................................... :................................... 445 8.13Applications ........................................................................................................ 446 8.14Resonant Circuit Oscillators .............................................................................. 446 8.15Crystal Oscillators .............................................................................................. 447 8.16Frequency Stability ............................................................................................ 448 8.17Frequency of Oscillations for Parallel Resonance Circuit.. ............................... 449 8.18I-MHz FET Crystals Oscillator Circuit ............................................................. 449 Summary ............................................................................................................ 450 Objective Type Questions .................................................................................. 451 EssayTypeQuestions ........................................................................................ 452 Multiple Choice Questions................................................................................. 452 Additional ObjectiveTypeQuestions(Chapter1-8)......................................................454 Answersto Additional ObjectiveTypeQuestions ........................................................... 455 Appendices .................................................................................................................... ...457 Appendix-IColour Codes for Electronic Components ....................................... 458 Appendix-IIResistor and CapacitorValues........................................................ 461 Appendix-IIICapacitors .... ................................................................................. 464 Appendix-IVInductors.......................................................................................... 470 Appendix-VMiscellaneous .................................................................................. 474 Appendix- VICircuit Symbols ............................................................... .................484 Appendix-VIIUnit Conversion Factors ................................................................. 486 Appendix- VIIIAmericanWireGauge Sizesand MetricEquivalents ...................... 489 Answers toObjectiveTypeand MultipleChoiceQuestions............. ............................. 491 Index ... ............................................................................................................ ................. 501 "This page is Intentionally Left Blank"a B C c d D D D f F h SYMBOLS Accelerationof electrons(m/secorcm/sec) Magnetic fieldIntensity (Wb/m2or Tesla) Chargeof electrons(Coulombs) Velocity of light=3x108 m/sec. Distance between theplatesina CRT Distancebetween thecentre of thedeflecting. plates andscreen. Diffusionconstant; Distortioninoutput waveform E =Electric fieldintensity (V/m or V/cm) frequency (Hzs/KHzs/MHzs) Force experiencedbyanelectroninNewtons Plank'sconstant=6.62x10-34 J-sec. D.C.current (rnAor IlA) A.C current (rnAor IlA) JCurrent density (A/m2 or mA/cm2) KBoltzman's constant=8.62x10-5 eV /OK KBoltzman's constant=1.38x10-23 J/ OK ILengthof deflectingplatesof CRT (cms) LDistancebetween thecentre of thefieldand screen(cm or rn) LDiffusion length mMassof electron(kgs) MMutualconductance nfreeelectronconcentration(No./m3or No./cm3) s S T V v W Y Y Z K.E PoE L C R a. (3* Acceptor AtomConcentration(No.lm3 or No.lcm3) Donor AtomConcentration(No/m3 or No/cm3) Holeconcentration(No.lcm3 or No./cm3) Q = Charge of anelectronincoulombs =1.6x10-19 C Spacing between the deflectingplates of CRT (in cms) Stability factor Periodof rotation(secsor 1.1.secs) Accelerating potential or voltage (volts) Velocity(m/secor cm/sec) WorkfunctionorEnergy(eV) Displacement of electrononthe CRT screen(cmsor mms) Admittance(inmhos U); (ohmsQ) Kinetic Energy (eV) Potential Energy (e V) Inductor Capacitor Resistor I DoClarge signalcurrent gainof BJT =f E Smallsignalcommonemitter forwardcurrent gain I DoClargesignalcurrent gain of BJT = f B oIpCInC TransportatIonfactorof BJT =-I- = -I-PEnE II Emitter efficiency of BJT =iE =InE EE Ripplefactor infilter circuits Conductivity of p-typesemiconductorin(U/cmor siemens) anConductivityof n-typesemiconductorin(U Icmor siemens) pResistivity(n - cm) eThermalresistance(inW/cm2) eAngle of deflection ~ Volt equivalent of work function(volts) LlIncremental value nResistance(ohms) UConductance(mhos) 11Efficiency(%) 03 Perrnitivityof freespace (F/m)=8.85x10-12 F/m IlMobility of electrons or holes (m2/V-sec) IloPermiabilityoffree space (Him) =1.25x10-6Him aWavelength (A 0) hIInputresistanceor inputimpedance(n) hrReverse voltage gain hoOutput admittance (U) ~ - Forwardshortcircuitcurrentgain = (xxi) "This page is Intentionally Left Blank".Brief History of Electronics Insciencewestudyaboutthelawsof natureanditsverificationandintechnology,we study the applicationsof theselawstohumanneeds. Electronicsis the science and technology of the passage of charged particles ina gas or vacuum orsemiconductor. Before electronic engineering came into existence, electrical engineering flourished.Electrical engineeringmainlydealswithmotionof electronsinmetalsonly,whereasElectronicengineering dealswithmotionof chargedparticles(electronsandholes)inmetals,semiconductors andalsoin vacuum.Anotherdifferenceis,inelectricalengineeringthevoltagesandcurrentsareof very high-kilovolts, and Amperes, whereas in electronic engineering one deals with few volts and rnA.Yet another difference is,inelectricalengineering, thefrequencies of operation are 50 Hertzs/60 Hertzs, whereasinelectronics,itisKHzs,MHz,GHzs,(highfrequency). Thebeginning forElectronics.wasmadein1895,whenH.A.Lorentz postulated the existence of discretechargescalledelectrons.Two yearslater,J.J.Thomsonprovedthesame experimentally in1897. Inthesameyear,Braunbuiltthefirsttube,basedonthemotionof electrons,andcalledit Cathoderay tube(CRT). In1904,Fleming invented the Vacuumdiode called'valve'. In1906,asemiconductor diodewasfabricatedbut they couldnotsucceed,inmaking itwork. So,semiconductor technologymetwithprematuredeathandvacuumtubesflourished. In1906 itself.DeForest put a third electrode into Fleming's diode and he called it Triode.A small changeingrid voltageproduceslarge changeinplate voltageinthisdevice. In1912Instituteof RadioEngineering (IRE)wassetupinUSAtotakecareof thetechnical interestsof electronicengineers.Beforethat,in1884Instituteof ElectricalEngineerswas formed and in1963both institutes merged into one association called IEEE (Institute of Electrical and ElectronicEngineers). The firstradio broadcasting stationwasbuiltin1920inUSA. In1930,black andwhite television transmissionstartedinUSA. In1950,Colour televisionbroadcasting wasstarted. The electronicsIndustry canbedividedinto4 categories: Components Communications Control Computation Transistors,ICs,R,L,Ccomponents Radio, Television, Telephone - wireless, landline communications Industrialelectronics,controlsystems Computers VacuumTubesruledtheelectronicfieldtilltheinventionof transistors.Thedifficultywith vacuumtubeswas,itgeneratedlotof heat.Thefilamentsgetheated to>2000 k,sothat electron emission takes place.The filamentsget burnt and tubesoccupy large space.Soin1945, Solid State Physicsgroupwasformedtoinvent semiconductor devicesinBellLabs,USA. Majormilestonesindevelopmentof Electronics: 1895:H.A.Lorentz - Postulatedexistance of Electrons 1897:J.J.Thomson- Provedthesame 1904:Fleming invented Vacuum Diode 1906:DeForest developed Triode 1920:Radio Broadcasting inUSA 1930:Black and White Television Transmission inUSA. 1947:Shockley- invented the junction transistor.(BJT) 1950:Colour Television Transmission startedinUSA. 1959:Integratedcircuit concept wasannouncedbyKilbyat anIREconvention. 1969:LSI,IC- LargeScaleIntegration,withmorethan1000but10,000components per chip.ICsweremade. 1975:CHMOS- ComplimentaryHighMetalOxideSemiconductorICswereannouncedby INTEL group. 1975:MSI(Multiplenum, Address)100- 1000components/chipwasdeveloped. 1978 : 1980 : 1981: 1982: 1984 : 1985: 1986 : 1987 : 1989: 1990s: 1992: 1998 : 2001: 2002 : 2003: 2004 : 2010 : LSI8bitmicroprocessors(flP),ROM,RAM1000- 10,000components/chip VLSI>1,00,000components/chip,Ex:16bitand32bitflPS 16bitflP>1,00,000components/chip,Ex:16bitand32bitflPS 100,000Transistors,(80286)wasdeveloped CHMOS> 2,00,000components/chipEx:16bit and32bitflPS 32bitflP> 4,50,000components/chipEx:16bit and32bitflPS 64bitflP>10,00,000components/chip Ex:16bitand32bit flPS MMICSMonolithicMicrowaveIntegrated Circuits i860 Intel's 64 bit CPU developed ULSI> 500,000Transistors;UltraLargeScaleIntegration GSI>1,000,000Transistors;GiantScaleIntegration 3millionTransistors, (Pentiumseries) 2 Million Gates/Die 5 Million Gates / Die 1 Gigabit Memory Chips 10nanometer patterns,linewidth CommercialSuper CompterlOT.FlipFlopsdeveloped. Neuro - Computer Using Logic Structure Based onHuman Brain likely StillNatureissuperior.There are 10' cells/cm3 inhumanbrain Development ofVLSI Technology: 3flTechnology J, 0.5flTechnology ,J... 0.12flTechnology ASICS (ApplicationSpecificIntegrated Circuits) HYBRIDICs BICMOS MCMs (Multi Chip Modules) 3-Dpackages Tableshowing predictionsmade in1995 onVLSITechnology 19951998200120042007 Lithography(/-l)0.350.25O.IS0.120.1 No.Gates/Die:SOOK2M5M10M20M No. Bits/Die Dram64M256IG4G16G Sram16M64 N256 MIG4G WaferDia(mm)200200-400-400-400-400 Power(/-lW/Die)15304040-12040-200 PowerSupply.V.3.32.22.21.51.5 Frequency MHz100175250350500 (xxvi) ,namlCS InthisChapter, The path or trajectories of electrons under the influence of Electric Fields, Magnetic Fields and combined Electric and Magnetic Fields are given, The Mathematical Equations describing the Motion are derived, ThePracticalApplicationof thisstudyinaCathodeRayOscilloscopeis also given, 2Electronic Devices and Circuits 1.1ELECTRONDYNAMICS The term Electron Dynamics refers to the analogy between an electron under electric and magnetic fields, and a body falling under gravity.If a shell is fired from a cannon, it traverses a path and falls under gravity.The motion of an electron is similar to the trajectory of a shell.Inthis chapter, we study the motion of electrons in electric fields and magnetic fields.First we consider only uniform electric fieldsandthenuniformmagneticfields,parallelelectricandmagneticfieldsandthen perpendicular electric and magnetic fields. Theradiusof anelectronisestimated as10-15 metresandthatof anatomas10-10 metre.Theseare verysmallandhenceallchargesareconsidered asPoints of Mass. Thechargeof anelectronis1.6x10-19 Coulombs.Themassof anElectronis 9.11x10-31 Kgs. There aretwodifferent types of Electron Models. 1.ClassicalModel 2.Wave-MechanicalModel. Theassumptionthatelectronisatinyparticle possessingdefinitemassand charge, istheClassicalModel,whiletheassumptionthatelectronstravelinthe formof wavesis called theWave-MechanicalModel.ClassicalModelsatisfactorilyexplainsthebehaviorof electronsinelectricandmagneticfields.Forlargescalephenomena,suchas,electron transactioninavacuumtubeClassical Model givessatisfactory results.But,inthesubatomic systems,suchas,electronbehaviorinacrystalorinanatom,classicaltheoryresultsdonot agreewithexperimentalresults.Wave-MechanicalModelsatisfactorlyexplainsthose phenomena. Weshallnow consider the trajectories of electrons under different conditions. 1.2MOTIONOFCHARGEDPARTICLESINELETRICANDMAGNETICFIELDS 1.2.1THE FORCE ONCHARGED PARTICLES INANELECTRIC FIELD The force experienced by a unit positive charge at any point inan electric fieldis the electric field intensity'E'at trat point.Itsunits are V 1m For unit pusitive charge,force=1 xENewtons. :.For apositive charge'q', theforce,F =qxENewtons where F isinNewton's, qisincoulombs, and EinV1m But by Newton's Second Law of Motion, F=mxaandF =qxE dvdv m x - =qXE..a=-dtdt By solving this equation, the trajectory of the electron in the electric field can be found out. For accelerating potential, considering electron charge as e, F=-eXE .......... (1.1) Inthis case negative signindic.1tes that forceis opposite to the direction of E. ElectronDynamicsand CRO 3 Let A andB are two horizontal plates, separatedby distance 'd' as shown inFig 1.1. LetVbetheappliedpotential.Thedirectionof electric field isalways from positive tonegative.Soin thiscaseitisactingdownwardsandisE=V /d.The electric field willbe uniformif'V' is the same.Suppose an electron is present in the electric field and it is desired to A;:+ B _______---l:'- ! investigate its trajectory : Fig 1.1Direction of electric field. Let the initial velocity = voxand displacement = Xoi.e.,at t = 0,Vx= vax'x = xo' According to Newton's law, F=m xaxandF =exE exE=m x ax. p. Hole concentration 'p' is greater than free electron concentration.p >D. InstrinsicSemiconductorD= p. 2.4CONDUCTIVITYOF ANINTRINSICSEMICONDUCTOR WhenvalenceelectronsareexchangingtheIrposItions,wesayholesaremoving.Currentis contributedby theseholescurrentisnothingbutrateof flowof charge.Holesarepositively charged.Soholemovementcontributesforflowof current.Becauseof thepositivecharge movement, the direction of hole current issame as that of conventional current.Suppose to start with, there are many freeelectrons, and these wiIlbe movinginrandom directions.A current is constitutedby theseelectrons.Soatanyinstant,thetotalcurrent densityissummationof the currentdensitiesduetoholesandelectrons.Thechargeof freeelectronsisnegativeandits mobility isJ..ln.The hole ispositive, andits mobility isJ..lp.The charge of both holes and electrons are same'e'.A hole can move fromoneion to thenearest where asanelectronisfree to move anywhere tillit collides with another ionor free electron.Electrons and holes move inopposite directions in an electric field E.Though they are of opposite sign, the current due to each ionisin the same direction. Current Density,J =(JE J = (nJ..ln+ pJ..lp)xex E = (JxE n = Magnitude of Free Electron Concentration p = Magnitude of Hole Concentration (J=Conductivity n / cmor Seimens (J= (nJ..ln+ P J..lp)e = neJ..ln+ peJ..lp A pure or intrinsic semiconductor is one in which n =p =nj where nl is intrinsic concentration. Ina pure Germanium at room temperature, thereisabout one hole-electronpair for every 2 x109 Germanium atoms.As the temperature increases, covalent bonds are broken and so more free electrons and holes are created.Sonl increases, as the temperatureincreases,inaccordance with the relationship, Junction Diode Characteristics 3 n.'=AxT2 e-EGo/2kT I EG=Energy Gapinev. e= Charge of anElectron or hole A = Constant for a semiconductor; for GeA= 9.64x1021; for GeEG=0.785eVat 0 oK forSiEcJ=1.21eVat0 oK EGat room temperature, for Ge =0.72eV forS i=1.1e V T = Temperaturein0 oK k = Boltzman 's Constant. 67 ..........(2.12) As nl increases with T, the conductivity also increases, with temperature for semiconductor. Inotherwords,theresistivitydecreaseswithtemperature forsemiconductor.Ontheother handresistanceincreaseswithtemperature formetals.Thisisbecause,anincreasein temperature for metals resultsingreater thermalmotion of ions and hence decrease inthe mean freepath of thefreeelectrons.Thisresultsindecrease of themobility of freeelectrons andso decrease in conductivity or increase inresistivity for metals. 2.5DONORTYPEORn-TYPESEMICONDUCTORS Intrinsic or pure semiconductor is of no use since its conductivity isless and it can not be charged much.If apUiesemiconductorisdopedwithimpurityitbecomesextrinsic.Dependingupon impurity doped, the semiconductor may become n-type, where electrons are the majority carriers or donor type,since it donates an electron.On the other hand ifthe majority carriers are holes, it isp-typeoracceptortypesemiconductor,becauseitacceptsanelectrontocompletethe broken covalent bond. Germanium atom with its electrons arranged inshells willhave configuration as Is22s22p23s2 3p63d1o 4s2 4p2 Geistetravalent(4).'Ge'becomesn-typeif apentavalent(5),impurityatomssuchas Phosphorus (P),or Arsenic are addedtoit. The impurity atoms have size of the same order as that ofGe atoms.Because of the energy suppliedwhiledoping,theimpurity atomdislodgesonefromitsnormalpositioninthecrystal lattices takes up that position.But since the concentration of impurity atoms is very small (about I atom per million ofGe atoms), the impurity atom is surrounded by Ge atoms.The impurity atom ispentavalent.That is,it has5 electrons inthe outermost orbit (5valence electrons).Now 4 of these are sharedby Ge atoms,surrounding the impurity atom and they form covalent bonds.So one electron of the impurity atomisleft free.The energy required to dislodge this fifthelectron from its parent impurity atom is very little of the order of 0.01eV to 0.05 eY.This free electron is inexcess to the free electrons that will be generated because of breaking of covalent bonds due to thermalagitation.Since anexcess electronis available for eachimpurity atom, or it can denote anelectronit iscalled n-type,or donortypesemiconductor. \.J 68Electronic Devices and Circuits 2.6ACCEPTOR TYPEOR P- TYPESEMICONDUCTORS Anintrinsic semiconductor when doped with trivalent (3) impurity atoms like Boron, Gallium Indium, Aluminium etc., becomes p-type or acceptor type. Because of the energy suppliedwhile doping, the impurity atom dislodges ane Geatom from the crystal lattice.The doping level is low,i.e., there is one impurity atom for one million Ge atoms, theimpurity atomissurrounded by Ge atom.Now the three valence electrons of impurity atom are sharedby 3 atoms.The fourthGe atomhasnoelectron toshare with the impurity atom.So the covalent bondisnot filled or a hole exists.The impurity atom tries to steal one electron from the neighboring Ge atoms anditdoes so when sufficient energy issupplied to it.So hole moves. There willbe a natural tendency inthe crystal to form4 covalent bonds. The impurity atom (and not just 3) since all the other Ge atoms have 4 covalent bonds and the structure ofGe semiconductor is crystalline and symmetrical.The energy required for the impurity atom to steal one Ge electron is0.0 I e V to0.08 e Y.This holeisinexcess to thehole created by thermalagitation. 2.7IONIZATIONENERGY Ifintrinsic semiconductor is doped with phosphorus, it becomes n-type as Phosphorus is pentavalent. The 4electronsinthe outer orbit of Phosphorus aresharedby the 4Germanium atomsandthe fifthelectron of Phosphorusinthe outer orbit isa freeelectron.But inorder that this electron is completelydetachedfromtheparentPhosphorusatom,someenergyistobesupplied.This energy required to separate the fifth electron is called Ionization Energy.The value of ionization energy for Germanium is 0.012 eV, andinSilicon, it is 0.044 eY.For different impurity materials, these values will be different, inSilicon and Gennanium.As this energy is small, at room temperature, we assume that all the impurity atoms are ionized. 2.8HOLESANDELECTRONS In intrinsic semiconductors,n=p=ni.Or theproductnxp =ny.Inextrinsic semiconductor say n-type semiconductor practically the electron concentration, 0OJ.As a result holes, minority carrier inn type, encounter with free is much larger since np.So when a hole encounter a freeelectrc)\:;,both electrons and holes recombine and the place of hole is occupied by the free electrons and this probability is muchlager since 0p. The result is that bothfreeelectronandholearelost.So theholedensity'p'decreases andalso that of electron density' n'butstill0OJ.Thisisalsotrueinthecaseof p-typesemiconductorpOJ,n decreasesinacceptor typesemiconductor,asaresultof recombination,'p' alsodecreasesbut p>>OJ.Ithasbeenobservedpractically that thenetconcentrationof theelectronsandholes follows the realtion n x p =n7.This is an approximate formula but still valid.Though 'p' decreases inn-typesemiconductorwithrecombination,'0'alsodecreases,but0OJandbecauseof breaking of covalent bonds,more freeelectrons may becr.eated :and'0'increases. In the case ofp-type semiconductor the concentration of acceptor atoms Nanl.Assuming thatalltheacceptoratomsareionized,eachacceptoratomcontributeatleastonehole.So pOJ.Holesarethemajoritycarriersandtheelectronsminoritycarriers.AspOJ,the current is contributed almost all, by holes only and the current due to electrons is negligibly small. If impuritiesof bothdonor type andacceptor typearesimultaneouslydopedinintrinsic semiconductor, the net result will be, it can be either,p-rype or n-rype depending upon their individual concentration. To give a specific example, suppose donor atoms concentration is100 nl,and acceptor atoms concentration in10 nl"Then No =0.1NI.The number of electrons combine contributed by Junction Diode Characteristics69 ND combine with number of holes contributed by Na.So the net free electrons will be equal to 0.9 No =90Ni.Suchasemiconductor,canberegardedasntypesemiconductor.If NA =NDthe semiconductor remains intrinsic. 2.S.1INTERSTITIALATOMS Intrinsic or pure semiconductor ispractically not available.While doping a semiconductor with impurities, pure Phosphorous, Arsenic, Boron or Aluminium may not be available.These impurities themselves will contain some impurities.Commonly found such undesirable impurities are Lithium, Zinc, Copper, Nickel etc.Sometimes they also act as donor or ac:::eptor atoms.Such atoms are called as interstitial atoms,except Copper andNickle other impurities donot affect much. 2.S.2EFFECTIVEMASS Whenquantummechanismisusedtospecify themotionof electrons or holeswithinacrystal, holes and electrons are treated as imaginary particles with effective masses mp and mn respectively. This is valid when the external applied fieldis smaller than the internal periodic fields produced by the lattice structure. Mostmetalsandsemiconductorsarecrystallineinstructure,i.e.,theyconsistof space arrayof atomsinaregulartetrahedraloranyotherfashion.rileregular patternof atom arrangement is called lattice.Inthecase of metals,ineachcrystal,the atoms are very close to each other.So the valency electron of one atomare asmuchassociatedwith the other atoms as with theparent atom.Inother words, the valance electrons areloosely bound to the parent atom andthe electronsof oneatomaresharedbyanotheratom.Soeverysuchvalence electron has almost zero affinity with any individual atom.Such electrons are free to move within the body of themetalunder theinfluence of applied electric field.So conductivity of metals is large. On the other hand, for a semiconductor also, the valence electrons of one atomaresharedby theother atoms.But these binding forces are very strong.So thevalence electrons are very muchlessmobile. Hence conductivity is less;As the temperature isincreased, the covalent bondsbindingthevalanceelectronsarebrokenandelectronsmade free to move, resulting in electrical conduction. Valence Electrons are the outer most electrons orbiting around the nucleus. . . \J f\ . .+4 . . Free electrons are those valence electrons whichare separated fromthe parent atom. Since the covalent bonds are broken. Germanium has4valenceelectrons.Numberof electronsisequaltonumberofFig2.6 Covalentbonds protons.The atomisneutralwhenno electric fieldisapplied.Inthe adjacent figure,theionishaving acharge(circles) with4electrons aroundit.The covalent bonds are shown by Iines Iinking one electron of one atom to the nucleus of other atom (Fig. 2.6). 2.S.3HOLESANDEXCESSELECTRONS Whenacovalent bondisbrokenduetothermalagitation,anelectronisreleasedandaholeis created in the structure of that particular atom.The electron so released is calledfree electron or excesselectronsinceitisnotrequired tocompleteanycovalentbond initsimmediate neighborhood.Now theionwhichhaslost electronwillseekanother new electrontofillthe 70Electronic Devices and Circuits vacancy.So because of the thermal agitation of the crystallattice, an electron of another ion may come very close to theionwhich has lost the electron.The ionwhichhaslost the electron will immediately stealanelectron fromthe closest ion,to fillitsvacancy.The holes move from the firstiontothesecondion.Whennoelectric fieldisapplied,themotionoffreeelectronis random in nature.But when electric fieldis applied, allthe free electrons are lined up and they movetowardsthepositiveelectrode.Thelife period of a freeelectronmaybelp-sec to1 millisecond after which itisabsorbed by another ion. 2.9MASSACTIONLAW Inanintrinsic Semiconductor number of freeelectronsn =nl = No. of holes p = PI Since the crystalis electrically neutral, nlPI= n ~ .Regardless of individual magnitudes ofn and p,the product is always constant, np=~3-EGO nj =AT2e2KT ..........(2.13) This iscalled Mass Action Law. 2.10LAWOFELECTRICAL NEUTRALITY Let No isequaltotheconcentration of donor atomsinadopedsemiconductor.Sowhenthese donor atoms donate an electron, it becomes positively charged ion, since it has lost an electron.So positive charge densitycontributedby themisNo.If 'p'isthe hole density then totalpositive charge density is No + p.Similarly if NAis the concentration of acceptor ions, (say Boron which istrivalent,ion,accepts anelectron,sothat 4electronsinthe outer shellaresharedby theGe atoms),itbecomesnegativelycharged.Sotheacceptorionscontributecharge=(NA+ n). Since the Semiconductor is electrically neutral, when no voltage is applied, the magnitude of positive charge density must equal that of negative charge density. Totalpositive charge, No + P = Totalnegative charge (NA + n ) No+ p=NA + n Thisisknown as Law of Electrical Neutrality. Consider n-typematerialwithacceptor iondensity NA=O.Sinceitisn-type,number of electrons isnumber ofholes. So 'p' canbe neglected in comparison withn. nn:::No.(Since every Donor Atom contributes one freeelectron. ) Inn-type material, the free electron concentration is approximately equal to the density of donor atoms. Inn-type semiconductor the electron density nil = No.Subscript n indicates that it is n-type semiconductor. Butnxp=n2 nnI n2 Pn= The hole densityinn-type semiconductor =N ~n2 I PI1=No ..........(2.14) Junction Diode Characteristics Similarly Hole concentration inp-type semiconductor, Pp:::::NA _2 npx Pp- nl n2 71 I n=-pNA ..........(2.15) Problem2.14 A specimen of intrinsic Germanium at 300 oK having a concentratiun of carriers of2.5 x1013/cm3 isdoped withimpurity atoms of one for every milliongermanium atoms.Assuming that allthe impurity atoms areionisedandthat the concentrationof Ge atomsis4.4x1022 I cm3,findthe resistivity of doped material. (Ilnfor Ge =3600 cm2/volt-sec). Solution The effect of minority carriers (impurity atoms) is negligible.So the conductivity will not appreciably change. Conductivity, an =nelln ( neglecting Hole Concentration) 4.4 x 1022 n=4.4x1022/cm3 , No =Icm3 Resistivity, Problem2.15 106 Iln=3600 cm2 I Volt - sec an=4.4x1016 x3600x1.6x10-19 =25.37mhos I cm I Pn=25.37=0.039 O-cm Determine theconductivity (a) andresistivitd:(p.) of puresilicon,at 3000Kassuming that the concentration of carriers at 300 oKis1.6x101 Icm3 for Siand mobilities as Iln =1500 cm2N-sec; /lp=500cm2N-sec. Solution Problem2.16 0'1=(Iln + /lp) enl =1.6x10-19 x1.6x1010 x(1500 + 500) =5.12x10-6 u/cm Pi=5.12xlO-6=195,3000-cm Determine the concentration of free electrons and holesinasample ofGe at 300 oKwhich has a concentration of donor atoms equal to 2x1014 atoms Icm3 and a concentration of acceptor atoms =3x1014 atom lem3 Is this p-type or n-type Germanium? Solution n xp = nr -Ego n=AT3/2e2kT I 72 A =9.64x1014 EG= 0.25ev n ~ = 6.25x1026/cm3 NA+ n = No + P Totalnegative charge = Total positive charge Electronic Devices and Circuits orp - n = NA- No =(3- 2)x1014 =1014 orp =n + 1014 Thenn(n + 1014)= 6.25x1026 orn =5.8x1012 electrons./cm3 andp =n + 1014 =1.06x1014 holes/cm3 As p> n,this is p-type semiconductor. Problem2.17 Find the concentration of holes and electrons ina p type germanium at 300oK, if the conductivity is100 n -cm.IIpMobility of holesinGermanium =1800 cm2 I V - sec. Solution Problem2.18 Itisp-type p n. O"p= P e IIp 0"100 Pp=ellp =1.6 x 10-19 x1800= 3.47x1017 holes/cm3 n xp =n7 ni = 2.5x10 13 /cm3 (2.Sx 1013 f n =...l..___---,-:_=_=1.8x109 electrons/cm3 3.47x 1017 ( a)Findtheconcentrationof holesandelectronsinp-typeGermaniumat300oK,if 0"=100u/cm. (b)Repeat part ( a) for n-type Si,if 0"= 0.1u/cm. Solution Asitis p-type semiconductor, pn. 0"=peup cr100 p=ellp =1.6xlO-19x1800 = 3.47x1017 holes/cm3 nxp=n; 3EGO ni =AT2e--2kT = 2.5x1013/cm3 Junction Diode Chara'cteristics73 p =3.47x1017/cm3 n =n ~ = (2.5 x 1013 L =1.8x109 electron/m3 p3.47xl017 (b)0"=nef..I11; n - 0.1=4.81x1014/cm3 - 1300x 1.6 x 10-19 =4.81x1014/cm3 n,=1.5x1010 ; P=nf=(1.5x10lOL =4.68x1011hole/m3 n4.81 x 1014 Pwblem2.19 A sample of Geisdopedtotheextent of 1014 donor atoms/cm3 and7x1013 acceptoratoms/ cm3.Atroom temperature, the resistivity of pureGeis 60 O-cm.If the applied electric fieldis 2 V/cm, findtotal conduction current density. Solution For intrinsic Semiconductor, n =p =nj O"j=n,e (J..lp+ J..ln) 1 =60u/cm. J..lpforGeis1800 cm21V-sec; J..ln= 3800 cm21V-sec. cr II nj=e(llp+lln)60{1.6xIO-19)(3800+1800) = 1.86x1013electron/cm3 p xn= nr= ( 1.86x1013)2 NA+ n =No+ P NA=7 x1013/cm3 No =1014/cm3 (p - n ) =NA- No =-3x1013 Solving (I) and (2) simultaneously to get p and n, p =0.88x1013 n = 3.88x1013 J= ( nJ..ln+ PJ..lp).eE ={( 3.88)( 3800) + (0.88)(1800)}x1013.eE =52.3mA/cm3 ........ ( 1) ........ (2; 74Electronic Devices and Circuits Problem2.20 Determinetheconcentrationof freeelectronsandholesinasampleof Germaniumat3000K which has a concentration of donor atoms =2x1014 atoms /cm3 and a concentration of acceptor atoms =3x1014 atoms/cm3.Is this p-type or n-type Ge?Inother words,is the conductivity due primarily to holes or electrons? Solution n x p = NA+ n = P + No Solving (a) and ( b ) to get nand p, p - n =NA- No n2 p=_. n n2 _n2 =(NA - No) or n n.2 - n2 =nx(NA- No ) orn2 + n ( NA- No) - n? = 0 This is inthe form ax2 + bx + c = 0 x-4ac 22a n =_ (NA- No) +i(NA- ND? = 0 2V2 Negative sign is not taken into consideration since electron or hole concentration cannot be negative. n>0andp> O. NA=3x1014/cm30 No = 2x1014/cm3 njat 3000K=2.5x1013 /cm3 l-EG n.=AT2e2KT E=O.72eV n? =6.25x1026/cm3 n can be calculated.Similarly p is also calculated. n= 5.8xIOIS/cm3 p =10.58x1019/m3 as p> n,it is p-type semiconductor. Junction Diode Characteristics75 Problem2.21 Calculate the intrinic concentration of Germanium incarries/m3 at a temperature of320oK given that ionization energy is 0.75 eV and Boltzman's Constant K =1.374 xI 0-23J 10K.Also calculate the intrinsic conductivity given that the mobilities of electrons andholesinpure germanium are 0.36 and O. I 7 m21 volt-sec respectively. Solution ~ -e EGO nl=AT2e2KT K =Boltzman's Constant inJ I oK K =Boltzman's Constant ineV I oK _16)(10-19 xO75 -21 = 9.64x1021 x (320)3/2.e2xl 37xl0. x320 nl =6.85x1019 electrons (or holes)/m3. Inintrinsic semiconductor, n = p = nl. 0"1=enj(f.ln+ f.lp) =1.6x10-19 x 6.85x1019 (0.36 + O. 17) = 5.797 u/m Problem2.22 Determine the resistivity of intrinsic Germanium at room temperature Solution T =300 oK A=9.64xI021 E =0.75eY. 3-EGO nl =AT2e= 2.5x1019 electrons (or holes) 1m3. f.ln=0.36m2 I V- sec f.lp=0.17m2 I V- sec O"j= enj(f.ln+ f.lp) =1.6x10-19 x 2.5x1019 (0.35+ 0.17) =2.13u/m 11 p =- =-- =0.47O-m (J2.13 2.11THEFERMIDIRACFUNCTION N(E ) =Density of states. i.e.,The number of statesper ev per cubic meter (number of statesle V 1m3 ) The expression for N(E) =Y EY'where y isaconstant. 76Electronic Devices and Circuits 47t Y =h3(2m)3/2(1.6x10-19)3/2= 6.82x1027 m = Mass of ElectroninKgs h =Planck's Constant isJoule-sees. Theequationfor./{E)iscalledthe Fermi Dirac ProbabilityFunction.It specifies the fractionof allstates at energyE (eV) occupiedunder conditions of thermalequilibrium.From Quantum Statistics, it is found that, where ./(E) =(E-EF) l+ekT k =Boltzmann Constant, eVfK T= Temp oK EF=FermiLevelor Characteristic Energy ........(2.16) The momentum ofthe electron can be uncertain.Heisenberg postulated that there is always uncertainty in the position and momentum of a particle, and the product of these two uncertainities isof the order of magnitude of Planck's constant 'h'. Ifp is the Uncertainty in the Momentum of a particle,~ n is the uncertainty inthe position of a particle ~ p x~ x ~ h. 2.11.1EFFECTIVEMASS When an external fieldis applied to a crystal, the free electron or hole in the crystal responds, as if itsmassis different from the true mass.This mass iscalled the Effective Mass of the electron or the hole. By considering this effective mass, it will be possible to remove the quantum features of the problem. This allows us to use Newton's law of motion to determine the effect of external forces onthe electrons andholes within the crystal. 2.11.2FERMILEVEL Named after Fermi,it is the Energy State,with50% probability of being filled if no forbidden band exists.Inother words,itisthe mass energy levelof the electrons, atOaK. IfE =E6 ./(E) = ~ FromEq.(2.17) . If a graphisplottedbetween (E - EF) and./{E), it isshowninFig.2.7 At T = OaK,if E >EF then,./{E) = o. That is, there is no probability of finding an electron having energy> EF at T = 0 K.Since fermilevelis the max.energy possessed by the electrons at Oak . ./(E) varies with temperature as shown inFig. 2.7. Junction Diode Characteristics77 f(E) + _____--1._....;;:=_-...;__ eV -1.0 ( E- E,) +1.0 Fig.2.7 Fermi levelvariationwithtemperature. 2.11.3UNCERTAINTY PRINCIPLE This was proposed by Heisenberg. The measurement of a physical quantity is characterized in an essential way bylack of precision. 2.11.4THEINTRINSICCONCENTRATION I j( E ) =1+ e(E-EF )/KT e(E-EF)/KT I-j(E)= l+e(e-EF)/KT:::e-(Ef-E)/KT e(E-EF)/KT[l +e(E-Er)/KT l' :::e-(E,-E)/KT Fermi functionfor ahole =I - j( E ). (EF- E ) KT forE .::::Ev the number of holes per m3 in the Valence Band is, Ev p =f Y{Ev-xdE where Similarly -00 =Nv xe-(EF-Ey)/KT 3

n = Ne-(Ec-EF)fKT e nxp =Ny xNee-E"/KT Substituting the values ofNe and Nv, nxp =n; =AT3 e-E"/KT ..........(2.17) 78Electronic Devices and Circuits 2.11.5CARRIERCONCENTRATIONSINASEMICONDUCTOR dllN(E )x J(E )xdE dnnumber of conduction electrons per cubic meter whose energy lies betweenE andE + dE J( E)=The probability that a quantum state with energy E is occupied by the electron. N(E)=Density of States. 1 J(E)1+ e(E-EF )/KT The concentration of electrons inthe conduction band is 00 n =fN{E)x j{E)x dE Ec for E ~ E c(E - Es)isKT. J(E) = e-(E-E, )/KT e(E-Er )/KTI 00 n =fY(E- Edl/2 x e-{E-EF)KT . dE Er Simplifying this integral, we get, where N.isconstant. n =Nc x e-(EcEF)/KT 3 N=2(21tlnnKTJ2 Ch 2 K= Boltzman's Constant inJ/oK wheremn= effective massof theelectron. Similarly the number of holes / m3 in the Valence Band p = Nye-(EF-Ey)/KT 3 _[21tmp KxTJ2 Ny- 22 h where Fermi Level isthemaximum energy levelthat canbeoccupiedbytheelectrons at 0 oK. Fermi Level or characteristicenergyrepresents the energy state with50%probability of being filledifno forbiddenbond exists.If E = EF, thenJ( E) = Y2for any value of temperature. J( E) is the probability that a quantum state with energy E is occupiedby the electron. Junction Diode Characteristics 2.11.6FERMILEVEL ININTRINSICSEMICONDUCTOR or n = p = nj - Ne-(ECEF )/KT n- c -Ne-(EF-Ev)/KT p- V n=p Nce-(EcEF)/KT=Nv e-(EF-Ev)/KT Electrons in the valence bond occupy energy levels upto'EF'.EFisdefinedthatway.Then theadditional energythathastobosuppliedsothatfreeelectronwill move from valence band to the conduction band isEc N -(EF-EV)(Ec+EF) C+ -= eKTKT Ny = e Taking logarithms on both sides, NcEc+Ev-2EF InNy=KT KTInNc 2Nv 3 N=2(21tmnKTJ2 Ch 2 3 _[21tm p .KT)2 Ny- 2 h2 79 Conduction Band r EF --------------Eo ~e Valence Band oo.s1.0 ~ f ( E )Fig.2.8Energyband diagram. ..........(2.18) ..........(2.19) wheremnandmpareeffectivemassesof holesandelectrons.If weassumethatmn=mp' ( though not valid ), Nc=Ny InNc=0 Ny Ec +Ev . .EF =2..........(2.20) The graphical representation is as shown in Fig. 2.8.Fermi Level in Intrinsic Semiconductor liesinthe middle of Energy gapEG 80Electronic Devices and Circuits Problem2.23 Inp-type Ge at room temperature of 300 oK,for what doping concentration will the fermilevel coincide with the edge of the valence bond?AssumeJ.!p= 0.4m. Solution when 3 Nv = 4.82x101S( ':Yx T3/2 = 4.82x101\0.4)3/2(300)3/2 = 6.33x1018. Doping concentration N A= 6.33x1018 atoms/cm3. Problem2.24 Ifthe effective mass of an electron is equalto twice the effective mass of a hole, find the distance in electron volts (ev) of fermilevel in as intrunsic semiconductor from the centre of the forbidden bond at room temperature. Solution For Intrunsic Semiconductor, If then mp=mn Nc=Nv Hence EF will be at the centre of the forbiddenband.But if mp =I'mn- EFwillbe away from the centre of the forbiddenband by KTNckTmn 2In.Nv =%T.lnmp ( - )3/2 21tmnKT Nc=22 n ( - J3/2 21tmpkT Nv=22 n 3 = 4"x0.026 In (2) = 13.5m.eV Junction Diode Characteristics 2.11.7FERMILEVEL INADOPEDSEMICONDUCTOR cr= ()lnn+ )lpp )e, 81 Sotheelectricalcharacteristicsof asemiconductordependsupon'n'and'p',the concentration of holes and electrons. -{E-E)/KT-(E-E)/KT The expression for n = NceCFand the expression for p = NveFY These are valid for both intrinsic and extrinsic materials. The electrons and holes, respond to an external field as if their mass is m*( m* = O.6m) and not om'.So thism*isknownas Effective Mass. With impurity concentration, only EF will change. In the case of intrinsic semiconductors, EF isin the middle of the energy gap, indicating equal concentration of holes and electrons. If donor type impurity is added to the intrinsic semiconductor it becomes n-type.So assuming that allthe atoms areionized, eachimpurity atom contributes at least one freeelectron.So the first ND states in the conduction band will be filled.Then it will be more difficult for the electrons to reachConduction Band, bridging the gap between Covalent Bond and Valence Bond.So tbe number of electron hole pairs, thermally generated at that temperature willbe decreased.Fermi level isanindictionof the probability of occupancyof theenergy states.SinceBecauseof doping, more energy states in the ConductionBand are filled, the fermilevel will move towards the Conduction Band. EXPRESSIONFOREG But and Taking logarithms, or or -(ECEF) n = Nc xeKT -(EF-Ey) p=Nv.eKT -{EcEy} nxp = Nc xNv xeKT (Ec- Ev) =EG nxp = nr 2- ~nj=NcxNv xeKT n2 - ~_--,-1_= eKT NcxNv ..........(2.21) 82Electronic Devices and Circuits The position of Fermi Levelis as showninFig. 2.9 ConductionConduction BandBand t 1!diO! oR- dv 0..!diO!1 But ItISgIVenthatdvmax= 10 mhos Therefore,R shouldbe asleast100, so that thereisnonegativeresistanceregioninthe characteristic. Problem2.39 TheZenerDiodecanbeusedtopreventoverloadingof sensitive meter movements without affecting meter linearity. The circuit showninFig. 2.38 represents a D.C volt meter which reads 20V fullscale.The meter resistanceisS60 0 andR)+ R2=99.SkO.If the diode is a16V Zener, findR) andR2so that. whenVi> 20V,the Zener Diodeconducts andthe overload currentisshunted away fromthemeter. Solution When VI=20V, the Zener should not conduct. R5600 m V.: 20V I Fig2.38For Problem2.39. 20V=(R)+ R2+ Rm)xI= (99.S+0.S6)x103 xI 20V I100KO=200 ).lA on fullscale. When VI > 20V, the voltage across R)and Rmmust be equal to the Zener voltage V z =16V. or(R\+~ ) xI =16V 16V R)+Rm=200x 10-0=80KO R\=80KO - soon ::::79.SKO R2=99.SKO - 79.SKO =20KO Problem2.40 What is the ratio of the current for a forwardbias ofO.OSV to the current for the same magnitude of reverse bias for aGermaniumDiode? Solution For Forward Bjas,V ForReverseBias,V VT TJ v e VT-I v eVT-1 =+O.OSV=+SOm V. =- O.OSV=-SOm V. = + 0.026V = +26mV =1 for Germanium Diode. e50126 -I =---e-50/26 _1 6.82-1 0.147-1=-6.83. Negative sign ISoecause, the direction of current is opposite when the diode isreverse biased. 128Electronic Devices and Circuits Problem2.41 ItispredictedforGermaniumthereversesaturationcurrentshouldincreaseby0.11 C-I.Itis foundexperimentally ina particular diode that at a reverse voltage of 1OV,the reverse current is 5rnAandthetemperaturedependenceisonly0.07C-I.Whatistheresistanceshunting the diode? Solution 1= 10+ IR ~ =dl o dTdT IRthe reverse saturation current, willnot change with temperature. 10 1= 10+ R ~ dlo For Germanium,10xdT= 0.11. Taking the ratio of 1dl IxdT= 0.07. 1dlo -x-10dT0.11 =- =--1dl100.07 - x ~ -IdT I + IOV 10= 0.636 I.Fig2.39ForProblem2.41. IR= 1-0.636 or - ~ I -I = 0.364I- 0.6360- 0.57210 10=5mA IR= 0.572(5mA) = 2.68rnA. V/10 R =fiR=2.86mA= 3.5KO Problem2.42 Over whatrangeof inputvoltagewillthezenerregulatorcircuitmaintains30Vacross2KO resistor, assuming Rs= 200n andmax.zener current is 25rnA Solution 30V IL=2KO=15mA Max.Zener Current =25mA . .Total Current = 25 + 15= 40mA VImax= 30 V +Rsx1= 30V + 200(15 +25) = 38V Junction Diode Characteristics SUMMARY Energy possessed by an electron rotating in an orbit with radius r; e2 E=----81tEor n2h2 E Expression for the radius of orbit, r =2 1tme E .fiIhf.dd~ 12,400 xpresslOnor wave engt0emttteratatlOn, I\. =() E2-E) Types of Electronic Emissions from the surface. 1.Thermionic Emission2.Secondary Emission 3.Photo electric Emission4.High Field Emission Expression for Threshold FrequencyiT in photoelectric emission, e ~iT= h Energy Gap EGdecreases with temperature in semiconductors. Mobility J..ldecreases with temperature.Units are cm2/v-sec. 23/2eEG n p = n.; n.= ATe- 2KT II Law of Electrical Neutrality N A+ P = ND + n; Fermi Level lies close to Ec inn-type semiconductor Fermi Level lies close to Ev in p-type semiconductor cr=neJ..l+ peJ..l np BI Hall Voltage, VH = -pro Expressionfor current througha p-n junction. Diodeis 1 10[ e " ~ ' -I] 129 Forward Resistance Rfof a diode will be small of the order offew O.Reverse Resistance Ry will be very high, of the order ofMO Cut involtage Vvfor Germanium diode is 0.1 V and for Silicon diode it is 0.5V at room temperature.It decreases withincreaseintemperature. 10 the reverse saturation current of a diode wi II be of the order of a few J..lA.It increases with temperature.10 gets doubledfor every10Criseintemperature. When the diodeisreversebiased,transitioncapacitance CT results when the diodeis forward biased, diffusion capacitance Co is exhibited. 130Electronic Devices and Circuits The three types of breakdown mechanisms insemiconductor diodes are 1.AvalancheBreakdgwn 2.ZenerBreakdown 3.Thermal Breakdown. Tunnel diode exhibits negative resistance characteristics Varactor diodes are operated inreverse bias and their j unction capacitance varies with the voltage. Zener diode is also operated inreverse bias for voltage regulation. OBJECTIVETYPEQUESTIONS 1.Free electronconcentrationinsemiconductorsisof theorder of ............................. . 2.Insulators willhaveresistivity of theorder of ............................ .. 3.Expressionfor 'Jintermsof Eandcris............................ .. 4.ExpressionforJinthecaseof asemiconductorwithconcentrationsnandpis 5.Valueof Eo at room temperature forSiliconis............................ .. 6.AccordingtoLawof Mass ActionInsemiconductors,............................ .. 7.IntrinsicconcentrationdependsontemperatureTas............................. . 8.The equation governing Lawof Electrical Neutrality,is............................. . 9.CutinVoltages..............................withtemperature 10.Depletionregionwidthvaries withreversebiasvoltagesas............................. . 11.In p-type semiconductor,FermiLevelliescloseto............................. . 12.Inn-typesemiconductor,FermiLevelliescloseto............................. . 13.The rate at withIchanges with temperatureinSilicondiodeis............................. . 14.Valueof Voltequivalent of Temperatureat-25Cis............................. . 15.Einstein's relationshipinsemiconductors-is ............................. . 16.A verypoor conductor of electricityiscalled............................. . 17.Whendonorimpuritiesareaddedallowableenergylevelsareintroducedalittle ..............................the..............................band. 18.Under thermal equilibrium, the product ofthe free negative and positive concentration is a constant independent of the amount of .............................. doping and this relationship, calledthe mass actionlowandisgivenbynp=.............................. . 19.Theunneutralizedionsintheneighbourhoodof thejunctionarereferredtoas ...............................chargesandthisregiondepletedof mobilechargesiscalled ...............................chargeregion. Junction Diode Characteristics131 20.Generalexpression forEo'the contact difference of potentialinanopen circuited p-n junction' intermsof Nc, N Aandnjis...................... ,...... . 21.Typicalvalueof Eo= .............................. eV 22.The equationgoverning thelaw of the junctionis............................. . 23.The expressionforthecurrentinaforwardbiased diodeis............................ .. 24.Thevalueof cutinvoltageinthecaseof Germaniumdiode..............................and SiliconDiode..............................atroomtemperature. 25.ExpressionforVBthebarrierpotentialintermsof depletionregionwidthWis VB=............................. . 26.Expressionfor10intermsof temperatureTandV Tis............................. . 27.Zener breakdown mechanism needs relatively .............................. electric field compared to AvalancheBreakdown. 28.Expression for the diffusion capacitance Co in terms ofLp and Dp is............................. . 29.ExpressionforVoltequivalentof temperatureV Tintermsof temperatureTis VT =............................. . 30.The salient feature of a Tunneldiodeis,it exhibits ..............................characteristics. ESSAY TYPEQUESTIONS 1.Explaintheconceptof 'hole'.Hown-typeandp-typesemiconductorsareformed? Explain. 2.Derive the expression for Eo in the case of intrinsic semiconductor. 3.Derive the expression forEOinthe case of of p-type and n-type semicunductors. 4.Withthehelpof necessary equations,ExplainthetermsDriftCurrent andDiffusion Current. 5.Explain' aboutHallEffect.DerivetheexpressionforHallVoltage.Whatarethe applications of Hall Effect? 6.Distinguish between Thermistors and Sensistors. 7.Derivetheexpressionfor contact differenceof potentialVinanopencircuitedp-n o junction. 8.Draw the forward and reverse characteristics of a p-n junction diode and explain them qualitatively. 9.Derive the expression for Transistor Capacitance CT i ~ the case of an abrupt p-n junction. 10.Compare 1 So the ripple voltage exceeds D.C voltage.Hence HWR is a poor circuit for convertingAC to DC. 3.2.10POWERSUPPLYSPECIFICATIONS The input characteristics whichmust be specified for a power supply are: 1.Therequired outputD.Cvoltage2.Regulation 3.Averageand peak currentsineachdiode4.Peakinversevoltage( PIV) 5.Ripple factor. 3.3FULL WAVERECTIFIER (FWR ) Since halfwave rectifier circuit has poor ripple factor, for ripple voltage is greater than DC voltage, it cannot be used.So now analyze a fullwave rectifier circuit. The circuit is as showninFig. 3.9.

.. ::i II !U l'0:1t:21t II o A.C. \.r---IA II .. WaveformsFWRCircuit Fig 3.9 Rectifiers, Filters and Regulators During the +ve half cycle, 0] conducts and the current through O2 is zero During the - vehalf cycle, O2 conducts andthe current through0] is zero A centre tapped transformerisused. 147 The totalcurrenti flowsthroughtheloadresistorRLandtheoutput voltageV 0istaken acrossRL. I2lt.I For half wave rectifier, circuit, IDe = -2 flmslOada= --'1l 1t01t For fullwave rectifier, circuit, IDe =Twice that of half wave rectifier circuit IDe= 2Im/1t 1It For half wave rectifier circuit, Inns=- f I ~ sin ada= .!.m. 21t02 1It For full wave rectifier circuit, Inns=2 x - f I ~ sin2 ada 21t0 Ym 1=---'-'-'-mRf + RL Rf is theforwardresistance of eachdiode. A centre tapped transformer is essential to get fullwave rectification.So there is a phase shift of 1800,because of centre tapping.So 0]is forwardbiased during the input cycle of 0 to 1t,O2 is forwardbiased during the period 1t to 21t since the input to O2 has a phase shift of 1800 compared to theinput to0]. Sopositivehalf cycleforO2 starts at a fullwaverectifier circuit, while the D.C. current starts through the load resistance RLis twice that of the Half Wave Rectifier Circuit. Therefore, for Half Wave Rectifier Circuit,V 1m I=-DC1t For Full Wave Rectifier Circuit, 21m IDe =---;-Hence, Ripple Factor isimproved. 3.3.1RIPPLEFACTOR 1t 2J2=1.11 o i o Fig 3.10FWRvoltageoutput. I22.=~ I r m / -Ioc2 I'rmsV1rms-IDe'YIoe Therefore, y =1.21for HWR, andit isis0.482 for FWR. 148Electronic Devices and Circuits 3.3.2REGULATION FOR FWR The regulation indicates how D.C voltage varies as a function of load current. The percentage of regulation is defined as VNL - VFL % Regulation =Vx100 FL The variation of DC voltage as a function ofload current is as follows VDe = IDeRL 2:rn. RL[.: Irn=Rf:rn RL] _2Vrn R[ .. I=Vrn] - 1t(Rf +RL)'L.rnRf +RL _2Vrn [1- Rf ] - 1tRf + RL 2Vrn 2Vrn =--1t1t(Rf + Rd IVDe=~ - IDe' RrI Vrn V=-NL1t Vrn VFL= - - IDe' Rr 1t 2Vrn_2Vrn+ Ioc. Rf % Regulation ~ 1 t ~__1 t ~_______x100 2Vrn -IDC .Rf 1t IDC Rf =x100 2Vrn -IR dcf 1t (1 l2Vrn x(Rf +Rd -lJ 1t2Vrn.Rf 1t R % Regulation =x100 L x100 Rectifiers, Filters and Regulators 3.3.3RATIO OF RECTIFICATION D.C. powerdeliveredto Load A.C. input power from transformer secondary PDe= 16e x RLfor FWR, =21m=4.1m2 R IDe1t1[2V 149 PAC is what a Wattmeter would indicate if placed in the FWRCircuitwith its voltage terminals connectedacross thetransformer secondary. 1m Pac = (Irmsf. (RF +RL);IrmsforFWR =.fi ()'.(RF+ Rd 2 ~ ( R +R). 2FL If we assume that Rf is the forward resistance of the diode is very small, compared to RL. Rr+RL=RL e PAC = -I!L(Rd 2 Poe4Im2RLx28 Ratio of Rectification =--=22= -2 =0.812 PAC1tI mx R L1t For HWR it is 0.406.Therefore, for FWR the ratio of rectification is twice that of HWR. 3.3.4TRANSFORMER UTILIZATIONFACTOR: TUF Infullwave rectifier using center-tapped transformer, the secondary current flows through each half separately in every half cycle. While the primary of transformer carries current continuously. HenceTUFiscalculatedforprimary andsecondary windingsseparately andthen the average TUF is determined. DCpower del,vered to load Secondary TUF =---...:.----'------AC rating of transformer secondary = if RfRL'then secondary TUF = 0.812. ( ~ r R LVm1m -x-J2J2 150Electronic Devices and Circuits The primary of the transformer in feeding two HWR's separately. These two HWR's work independently of each other but feed a commonload. Therefore, 0.287 TUF for primary = 2xTUF of HWR = 2x--I Rf +-RL ifRfRL, then TUF for primary = 0.574 The average TUF for FWR using center tapped transformer TUF of primary + TUF of secondary0.812 + 0.574 22 =0.693 Therefore, the transformer is utilized 69.3% in FWR using center tapped transformer. 3.3.5PEAK INVERSE VOLTAGE (PIV) FOR FULL WAVE RECTIFIER WithreferencetotheFWRcircuit,duringthe-vehalf cycle,D,isnotconductingandD2is conducting.Hence maximumvoltage acrossRLisV m'since voltageisalsopresent between A and0of the transformer, the total voltage acrossD,= V m + V m = 2Vm(Fig.3.9 ). 3.3.6D.C. SATURATION InaFWR,theD.C.currentsI,and12flowingthroughthediodesD,andD2areinopposite direction and hence cancel each other. So there is no problem of D.C. current flowing through the core of the transformer and causing saturation of the magnetic fluxinthe core. PIV is 2 V m for each diode in FWR circuit because, when D, is conducting, the drop across itis zero.Voltage delivered to RLis V m.D ~ isacross RL. During the same half cycle,D2is not conducting.Therefore, peak voltage across It is V mfrom the second half of the transformer.The totalvoltage acrossD2isV m + V m = 2Vm 3.4BRIDGERECTIFIERS The circuit isshowninFig.3.11.During the positive half cycle, DJ and D2are forward biased. D3and D4are open.So current will flow through DJ first and then through RLand then through D2 back to the ground. During the -ve half cycle D 4 and D ~ are forward biased and they conduct. The current flowsfromD3through RLto D4. Hence the dIrection of current is the same.So we get full wave rectified output. InBridge rectifier circuit, there is no need for centre tapped transformer.So the transformer secondaryline toline voltageshouldbe onehalf of that,usedfor theFWR circuit,employing two diodes. II Fig3.11Bridgerectifiercircuit. 21m 1t Rectifiers, Filters and Regulators where Rs+2Rf+RL 2Vrn =---IDC(RS + 2RF) 7t. Resistanceof Transformer Secondary. =ForwardResistance of theDiode. =LoadResistance. 151 2Rfshould be used since two diodes inseries are conducting at the same time.The ripple factor and ratio of recti ficationare the same asforFu IIWaveRectifier. 3.4.1ADVANTAGES OF BRIDGE RECTIFIER 1.Thepeak inversevoltage(PIV)acrosseachdiodeisVmand not2Vm asinthe caseof FWR.HencetheVoltageratingof thediodescanbeless. 2.Centretappedtransformerisnotrequired. 3.ThereisnoD.C.Current flowingthroughthetransformersincethereisno centre tapping and the return path istotheground.Sothetransformer utilization factorishigh. 3.4.2DISADVANTAGES 1.Fourdiodesaretobeused. 2.Thereissomevoltagedropacrosseachdiodeand sooutputvoltagewillbe slightlylesscomparedtoFWR.Butthese factorsareminorcomparedtothe advantages. Bridge rectifiers are available ina package with all the 4 diodes incorporated in one unit. It willhave two terminals for A.C.Input and two terminals for DCoutput.Selenium rectifiers are also available as apackage. 3.5COMPARISONOF RECTIFIERCIRCUITS HWRFWRBridgeRectifier Circuit PeakInverseVo ltage Vrn 2Vrn Vrn Vm2Vm2Vrn No Load Voltage -- --1t1t1t Ripple factor1.210.4820.482 Number of Diodes required124 Ratio of RectificationPdc 0.4060.8120.812 PAC DC Power delivered to the load TUF= AC ratings of transformer secondary 0.2870.6930.812 152 I Electronic Devices and Circuits 3.6VOLTAGEDOUBLERCIRCUIT 3.6.1HALF WAVE'VOLTAGE DOUBLEwEIRCUIT The circuit is shown in Fig. 3.12 and 'e wave forms are shown inFig. 3.13.During the +ve half cycle, 01 isforwardbiased.So C1 gets charged toV m.During the -ve half cycle,Dt jsreverse biased and the PIV ~ r o s s it V m.Therefore, the total voltage during -ve half cycle V m across C1 +Vm of -ve half cycle = 2Vm.So,O2 isforwardbiasedduring the -ve half cycle andC2 gets charged to 2Vm.Thisisthe no-load voltage.The actualvoltage dependsuponthe valueof RL connected. InA.C. Voltage measurements, the A.C.input isgiven to a voltage doubler and filter circuits.The output of D.C. meter (2Vm)isproportionalto A.C.Input.The meter calibrated in terms of the rms value of the A.c. Input. D2 Fig 3.12Hal/wavevoltagedoublercircuit. 3.6.2FULL WAVE VOLTAGE DOUBLER CIRCUIT ~ tFig 3.13Outputwaveform. The circuit is shown inFig. 3.14.Wave forms are shown in Fig. 3.15.During +ve half cycle, 01 conducts C1 gets charged to V m.During -ve half cycle, C2 charges throughO2 and to V m.RL is across the series combination of C1 and C2Therefore, the total output voltage V 0=2Vm.Here the PIV across each diode isonly V m and the capacitors are charged directly fromtheinput and notC2 willnotgetchargedthroughC1 Duringthetimeperiod0to21twehavetwocycles. Current flowsfrom0 to 1tand 1tto21talso.So itisFW Voltage Doubler Circuit. Fig 3.14Fullwavevoltagedoublercircuit. 3.7INDUCTORFILTERCIRCUITS FILTERS i V o I i --+t Fig3.15Outputwaveforms. A power supply must provide ripple free source of power fromanA.C.line But the output of a rectifier circuit contains ripple components in addition to a D.C. term.It is necessary to include a Rectifiers, Filters and Regulators153 filterbetween therectifier and theloadinorder toeliminatetheseripple components.Ripple componentsarehighfrequency A.C.SignalsintheD.Coutputof therectifier.Thesearenot desirable, so they must be filtered.So filter circuits are used. Flux linkages per ampere of current L =. The ability of a component to develop induced voltage when alternating current isflowing through the element isthe property of theinductor. Types of Inductors are: 1.IronCored2.Air Cored An inductor opposes any change of current in the circuit. ,So any sudden change that might occur in a circuit without an inductor are smoothed out with the presence of aninductor. In the case of AC, there is change in the magnitude of current with time. Inductor is a short circuit for DC and offers some impedance for A.C. (XL = jmL).So it can be used as a filter.ACvoltage is dropped across the inductors,where as D.C. passes through it. Therefore, A.C is minimized in the output. Inductor filter is used with FWR circuit.Therefore, HWR gives 121 % ripple and using filter circuit for such high ripple factor has no meaning. FWR gives 48% ripple and by using filter circuit we can improve it. According to Fourier Analysis, current, I in HWR Circuit is .1m1m.21mCos4rot I=-+-Smmt-- ---1t21t3 21m4Im4Im cos4rot Current in the case of FWR isi:::::--- --cos 2 rot - -----"=----- 1t31t151t The reactance offered by inductor to higher order frequencies like 4mt can be neglected. So the output current 1S i:::::21m- 41mcos2rot - 1t31t the fundamental harmonic m is eliminated. The circuit for FWR with inductor filter is as shown in Fig. 3.16. An II '""'-"""" ......... IDU= Fig 3.16Inductor filter circuit One winding oftransformer can be used as 'L'.For simplification, diode and choke (inductor) resistances can be neglected, compared with RL.TheD.C.component of the current is Vm RL Impedance due to L and RLin Series IZI=+ (2mL),2 For second Harmonic, the frequency is 2m. 154Electronic Devices and Circuits Therefore, A.C.Component of current 1m=Vm IR L 2 + 4oo2L2.Sosubstitutingthese values inthe expression, for current, 2141 I =---1!!... - ---1!!... Cos 200 t n3n by inductor to higher order frequencies like 400t etc., we get, .2V4V I =_m_ - mcos (2oot - e ) nRL wheree is the angle by which the load current lags behind the voltage and isgiven by 9=Tan-12roL RL. 3.7.1RIPPLEFACTOR If Y =Irnn/IO.C.=I' nn/IOCloc =2VminRL I- 1m_4Vm rnns- .J2 - + 4oo2L2 4vmxnR Ripple factor =L +4oo2L2x2Vm 21 - --x -;======:= - 3.J24oo2L 2 1+2 RL 4oo2L2 --2- 1, then RL RLx2 Ripple Factor r =3-v2 x200L

Fig 3.17RippleVoltageV V Ripple factor =Ji,L 3200L ..........(3.5) Therefore, for higher values ofL, the ripple factor is low.IfRL is large, then also y is high. Hence inductor filter should be used where the value ofRL is low.Suppose the output wave form from FWR supply is as shown in Fig. 3.17, then, V yp _ pis the peak to peak value of the ripple voltage.Suppose Voc = 300V, and V yp_pislOY. 10Vymax5 then VR maximum ='2= 5Y.Therefore, V/ms =.J2=.J2=3.54V .Vrrms_3.54_00118 RIpple Factor y =--- ---.. Vx300 %ripple = Ripple factorxl 00 % =1.18 Rectifiers, Filters and Regulators 3.7.2REGULATION v=2ImRL=2Vm DCnn ImRL = Vm Therefore, V DCis constant irrespective of RL. But this is true if L isideal.In practice 2Vm V DC=---IDc-Rf 1t where Rr is the resistance of diode. 155 Aninductor stores magnetic energy when the current flowing through it is greater than the average value andreleases this energy when the current isless than the average value. Another formula that is used for Inductor Filter for Rc +RL Ripple Factor = 0.236ro..........(3.6) L where RLisloadresistanceRcis the series resistance of theinductors.Thisis the same as the above formula~32roL Problem3.1 1 ~ =0.236. 3",2 AFWRisusedtosupplypower toa20000 Load.ChokeInductorsof 20Hinductance and capacitors of l6,...f areavailable.Compute the ripple factorusing1.OneInductor filter 2.One capacitor filter 3. Single L type filter. Solution 1.One Inductor Filter: 21m41m1 (2roL) I =~ - 31tcos (200t- ~ ) where~ =Tan- R 2Vm IDC=nRL I rippleRL100 V=loe= 3 ~ 2 0 0 L =3x1.414x2=0.074 2.Capacitor filter: The ripple voltage for a capacitor filter is of Triangular waveform approximately. The rms EdVnns value for Triangular wave isr::;orr::; 2",32",3 Ed Erms=2J3 156Electronic Devices and Circuits Suppose the discharging of the capacitor will cut one for one full half cycle .then the change last by the capacitor 1 Q=rot = IDeX-. 2f Q voltage = -C i --+t Fig 3.18Voof Capacitor filter. 1 ==0.09 4xJj x50xl6xl0-6 x 2000 3.LType factor: Problem 3.2 A diode whose internal resistance is 20 0is to supply power to a10000 load from allOV rms) source of supply.Calculate(a) The peak loadcurrent.(b) The DCloadcurrent (e) ACLoad Current(d) The DC diode voltage.(e) The total input power to the circuit. (1) % regulation from no load to the given load. Solution Since only one diode is being used, it is for HWR Circuit. I=Vm. =110.Ji = 152.5mA. (a)mRf + RL1020 (b) (e) (d) (e) (I) Problem3.3 - ~ =152.5= 48 5A IDe - .m. 1t1t I= ~ =152.5 = 76.2mA. nns22 V De= -ImRL =--48.5 x 1 =--48.5V. 1t Pi=Innsx(Rf+RL)=5.920. Show that themaximumDCoutputpower PDe = VDe IDeinahalf wavesinglephase circuit occurs when the loadresistance equals the dioderesistanceRf. Rectifiers, Filters and Regulators157 when RLis very large, or Problem 3.4 A 1rnA meter whose resistance is Ion is calibrated to read rms volts when used in a bridge circuit withsemiconductor diodes.The effectiveresistanceof each element maybeconsidered tobe zerointheforwarddirectionand00in thereversedirection.Thesinusoidalinputvoltageis applied in series with a 5 - KO resistance. What is the fullscale reading of this meter? Solution 21 IDe = ---.!!L for bridge rectifier circuit. 1t V I- --1!!... m- RL' Vm=.J2vrms 2..[i x Vrms IDe =1tR; L RL = 5KO + 100 = (5000 + 10)0 = 5010 0 lmA =2..[i x Vrms 1txSOIO. 1 x 10-3 hx 5010= S.S6V. Vrms=22 3.8CAPACITORFILTER HereXf should be smaller thanRL.Because,current should pass throughCand Cshouldget chargeo.If Cvalueisverysmall,Xewillbelargeandhencecurrent flowsthroughRonly (Fig. 3.19) and no filtering action takes place.During +ve half cycle for a H W R circuit, with C filter,C getscharged when the diodeisconducting andgetsdischarged(when the diodeisnot conducting)throughRL.Whentheinputvoltagee= EmSinrotisgreater thanthecapacitor voltage, C gets charged.When the input voltage isless than that of the capacitor voltage,C will discharged through RL.Thestored energyin the capacitor maintains theload voltage at a high value for a long period. The diode conducts only for a short interval ofhigh current. The waveforms areasshowninFig.3.20.Capacitor opposessuddenfluctuationsinvoltage acrossit.So the ripple voltage is minimised. 158 AC Input Electronic Devices and Circuits Fig 3.19FWRCircuit withC filter. v o t ;, .1 :' .. .. w : 9 192 rot==---. Fig3.20Output of capacitor filtercircuit. So the voltage across RLis more or less constant and hence ripple is reduced, as shown in the graph,the voltage V 0is closer to DCwave form.At 9)'C' getschargedand at 92 Cstarts discharging. The point at which diode starts conducting and C gets charged is known as cut in point 9 .. Thepointatwhichthediodestopsconductingor thecapacitorgetsdischargedisknownas Cut Out Point 92, Cwill not start charging at 1titself.Therefore,even though thesecond diode is forward biased,ec isalmost=::.Ym'Sothefirstdiodeconductsfrome)toe2andtheseconddiode conductsfrom 1t + 8) to 1t + 82,Diode conducts only for a short of time when input current is high and charges Cto the peak voltage. Considering one diode, iB = ie + iR ( Fig. 3.19 ) ._dV =C dec Ie - C.dt.d. .ec 1=-RR t where ec is the voltage to which cgetscharged. Rectifiers, Filters and Regulators159 Butec =V m Sin cot. Therefore, at that point, diode conducts andCgets charged. 81=cotl dec dt = coY mCos cot. General expression for .C VVIn S IS=comcos cot + Rm cot By some mathematical manipulation, ibcan be written as is=-; .jl + co2R2C2Sin(cot + 4 where~ =Tan-l(coRC) At an coinstant 82,when the capacitor fully discharges,is =0 EmS80 coCVcos 82 +- In2=. mR or82 =tan-i (-nRC) Thus the value of conduction angle depends upon the values of Rand C, and also the value of is. Ouring the non conducting interval, the capacitor discharges into the load R supplying load current. -ic= iR dee The circuit equation is -C x~ =~t orec =;Ae-t/RL Atcot=82 ec =V m sin82. The final expression for EDC seems alittle complicated. So a simplified expression which is widely usedisV DC=(V m - VR.), where ERis the peak to peak ripple voltage. The expression 2 for Ripple Factor, V'Ae_1t+(81 -82) y =V;;; - 2JjcoRC. where(81 - 82)is the angle of conduction 8c.For large coRC(typicalvalue100),8 c=14.6. The larger the value of R,thesmaller the y. 3.8.1RIPPLE FACTOR FOR CFILTER Initially, the capacitor charges to thepeak value Vwhen the diode 0\ is conducting. When the capacitor voltageis equal to the input voltage, the diode stops conductmg or the current through the diode 01is zero. Now the capacitor starts discharging throughRLdependingupon the time constant C.RL.Therefore, RL value is large.Rate of charging is different from rate of discharging. Whenthevoltageacrossthecapacitor fallsbelow theinputvoltageandwhenthe diode 02 is forward biased, the capacitor willagain charge to the peak value Vm and the current through D2 becomes zero when Vc=Vm.Thus the diodes 1 and 02 conduct for a very short period 81to 82 and 1t+ 8 Ito 1t+ 82 respectively (Fig.3.21). 160Electronic Devices and Circuits V,) i \ \ I\ I\ I\ I\ III i\/D,_II I\I - - __J - - - __- -__________ /_Conducting \ I\IIt!"I I\II IIII I\II I\I\ I\II I\/I I\/\ [.... 11 D, Conducting

)(l)t Fig3.21Outputwaveformsof acapacitor filter. T2=Time period ofthe discharging of the capacitor T]=Timeperiod of the charging of the capacitor The amount of chargelost by the capacitor whenit isdischarging =Because, IDe istheaverage valueof thecapacitor dischargecurrent.This chargeisreplacedduring ashort intervalT]during whichthe voltage across the capacitor changesbyanamount =peak topeak voltage of the rippleV' rp_p' Q=Vxc. Qcharge=V p-px C Qcharge=QDischarge. V'p_pxC=IDCxT2' or V'=IDex T2 p-pC TheoutputwaveformcanbeassumedtobeatriangularwaveT2T](T]+T2)=T/2 when the diodeisconducting (0 ton).(T]+ T2corresponds to one half cycle). TIl T - "T-2 ="2 - 2f"2- 2f . ,_'ocx1"T__1_1 Vp-p - C2f.2- 2f'for the triangular wave, form factor = Jj' V' V' Ims2Jj Voe IDe=RL V' rms =4JjfCRL' Rectifiers, Filters and Regulators Ripple Factor v'nns y =VDC =4.J3fCRL Therefore,Ripple factor maybedecreasedbyincreasing Cor RLor both. Expression forVDC: or 3.9LCFILTER V'p-p VOC=Vrn- 2 VDC Voc=Vrn- 4fC.RL Voc(I+-l-)=Vm 4fCRL '. ..V' P P - IDeI_VDe - - 'De-2fCRL 161 Inaninductor filter,ripple decreases withincreaseinRL.Ina capacitor filter,rippleincreases withincreaseinRL.A combinationof these twointoa choke input or L-Section Filter should then make the ripple independent of load resistance. If L is small, the capacitor will be charged to V rn'the diodes will be cut off allowing a short pulseof current.AsLisincreases,thepulsesof current aresmoothed andmade toflowfora larger period, but at reduced amplitude.But for a critical value of inductance Lc'either one diode or the other will be always conducting, with the result that the input voltage Vand I to the filter are fullwave rectified sine waves. The graph of DC output voltage for LCFilter is as showninFig. 3.22. 2Vrn VDC =-1t-for Full Wave Rectifier Circuit.considering ideal elements, conductionangleisincreasedwheninductorisplacedV R because there is some drop across L.So Ccharge i toV rn'Therefore, the diode willbe forwardbiased fora longer period. Theripplecircuit whichthroughL,isnotIe --. a.Howed to develop much ripple voltage across RL, if Xc atFig3.22LC filtercharacteristic. rIpplefrequencyISsmallcomparedtoRL,Because the current will pass through C only.Since Xcissmalland Capacitor will get charged to a constant voltage.So V 0across RL will not vary or ripple will not be there.Since for a properly designed LC Filter, XcRL andXLXcat (0= 21t/ XL should be greater than Xc because, all the Af3twukl be or-opped across XL itself so that AC Voltage across C is niland hence ripple islow. 162Electronic Devices and Circuits 3.9.1RIPPLE FACTOR INLC FILTER AC Component of current through L is determined by XL' XL=2coL RMS value of ripple current for Full Wave Rectifier with L Filter is, I4Vm Irms ==3refiXL =RMS Value of Ripple Current for L Filter I2 Irms =~3v2XL 2Vm =--re I2.J2vDC ..Irms =3.J2XL. VDe =3XL The ripple voltage in the output is developed by the ripple current flowing through Xe' II.J2XcV ..Vrms = Irms. Xe= )'X' DC V'rms Ripple factor = -V--DC .J2.xc y=3XL 1 Xe = ~ ,XL= coL .J211 Y=)X2fficX2;L L XL = 2coL since ripple is being considered at a frequency, twice the line frequency. IfIf=60HI=O.83xl0--6 ....____z......, _YLC 3.9.2BLEEDER RESISTANCE For the LCfilter,the graphbetween IDeand VDe is asshown inFig.3.23.For lightloads,i.e., when theload current issmall,the capacitor gets charged to the peak value and so the noload voltageis2VIre.Astheloadresistanceisdecreased,IDeincreasessothedropacrossother elements V iz~ i o d e s and choke increases and so the average voltage across the capacitor will be less than the peak value of2VmIre.The out put voltage remains constant beyond a certain point IB andsotheregulationwillbegood.ThevoltageVoeremainsconstantevenif IDeincreases, because, the capacitor gets charged every time to a value just below the peak voltage, even though the drop across diode andchokeincreases.So for currents above IB, the filter actsmorelike an inductor filter than C filter and so the regulation is good.Therefore, for LC filters, the load is chosen such that,theI ~ e ~ lB'The corresponding resistanceisknownasBleeder Resistance ( RB).So Bleeder Resistance is the value for which loc~ IBand good regulation is obtained, and conduction angle is180. When RL =RB, the conduction angle = 180, and the current is continuous. So just aswehavedeterminedthecriticalinductanceLe>whenRL= RB,theBleederResistance, IDC= the negative peak of the second harmonic term. Rectifiers, Filters and Rellulators i Fig 3.23LC filtercharacteristic. I=2Vm DC1tRs 4V1 If peak = ---1!L._. 31tXL 2Vm 4Vm 1 1tRs=J;-'XL 3.xL RB=-2-;RBisBleeder Resistance. 163 IDC shouldbe equalto or greater thanI peak,since,XLdetermines the peak valueof t.. ripple component. If XL is large I' peak can be~ IDC and so the ripple is negligible or pure D.C. is obtained, where XL = 2mL corresponding to the second harmonic.Therefore, R should be at least equal to this value ofRB, the Bleeder Resistor. 3.9.3SWINGINGCHOKE The value ofthe inductance in the LC circuits should be > Lc the minimum value so that conduction angle of thediodes is 1800 and ripple is reduced. But if the current IDC is large, then the inductance for air cored inductors as L =~ I I (flux linkages per ampe