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1. 1. 1. 1. 1. IntroductionIntroductionIntroductionIntroductionIntroduction 127127127127127

ElectronicsAtomic structureStructure ofelementsThe electronEnergy of an electronValence electronsFree electronsVoltagesourceConstant voltage sourceConstantcurrent sourceConversion of voltage source intocurrent sourceMaximum power transfertheoremThevenins theoremProcedure forfinding thevenin equivalent circuitNorton'stheoremProcedure for finding norton equivalentcircuit Chassis and ground.

C O N T E N T SC O N T E N T SC O N T E N T SC O N T E N T SC O N T E N T S

2.2.2.2.2. Electron EmissionElectron EmissionElectron EmissionElectron EmissionElectron Emission 28372837283728372837Electron emissionTypes of electron emissionTher-mionic emissionThermionic emitterCommonlyused thermionic emittersCathode constructionField emissionSecondary emissionPhoto electricemission.

3. 3. 3. 3. 3. Gas-Filled TGas-Filled TGas-Filled TGas-Filled TGas-Filled Tubesubesubesubesubes 38473847384738473847Gas-filled tubesConduction in a gasCold-cath-ode gas diodeCharacteristics of cold-cathodediodeApplications of glow tubesHot-cathodegas diodeThyratronApplications of Thyratron.

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4. 4. 4. 4. 4. Atomic StructureAtomic StructureAtomic StructureAtomic StructureAtomic Structure 48544854485448544854

Bohrs atomic modelEnergy levelsEnergy bandslmportant energy bands in solids Classification ofsolids and energy bands Silicon.

5.5.5.5.5. Semiconductor PhysicsSemiconductor PhysicsSemiconductor PhysicsSemiconductor PhysicsSemiconductor Physics 55755575557555755575SemiconductorBonds in semiconductorsCrystalsCommonly used semiconductors Energy banddescription of semiconductorsEffect of tempe-rature on semiconductorsHole currentIntrinsicsemiconductorExtrinsic semiconductorn-typesemiconductorp-type semiconductorCharge on n-type and p-type semiconductors Majority and

minority carrierspn-junctionProperties of pn-junctionApplying D.C.voltage acrosspn junctionVolt-ampere characteristics of pn-junctionImportant termsLimitations in the operating conditions of pn-junction.

6.6.6.6.6. Semiconductor DiodeSemiconductor DiodeSemiconductor DiodeSemiconductor DiodeSemiconductor Diode 7612476124761247612476124Semiconductor diodeCrystal diode as a rectifierResistance of crystal DiodeEquivalent circuit ofcrystal diodeCrystal diode equivalent circuitslmportant termsCrystal diode rectifiersHalf-waverectifierOutput frequency of Half-wave rectifierEfficiency of half-wave rectifierFull-wave rectifierCentre-tap full-wave rectifierFull-wave bridgerectifierOutput frequency of full-wave rectifierEfficiency of full-waverectifierFaults in centre-tap full-wave rectifierNature of rectifier outputRipple factorComparison of rectifiersFilter circuitsTypes of filter circuitsVoltage multipliersHalf-wave voltage doublerVoltage stabilisationZenerdiodeEquivalent circuit of zener diodeZener diode as voltage stabiliserSolving zener diode circuitsCrystal diodes versus vacuum-diodes.

7.7.7.7.7. Special-Purpose DiodesSpecial-Purpose DiodesSpecial-Purpose DiodesSpecial-Purpose DiodesSpecial-Purpose Diodes 125140 125140 125140 125140 125140Zener diodeLight-emitting diode (LED)LED voltage andcurrentAdvantages of LEDMulticolour LEDsApplica-tions of LEDsPhoto-diodePhoto-diode operationCharacteristics of photo-diodeApplications of Photo-diodesOptoisolatorTunnel diodeTunnel diode oscilla-torVaractor diodeApplication of varactor diodeShockley diode.

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8. 8. 8. 8. 8. TTTTTransistorsransistorsransistorsransistorsransistors 141191141191141191141191141191TransistorNaming the transistor terminalsSomefacts about the transistorTransistor actionTransistorsymbolsTransistor as an amplifierTransistorconnectionsCommon base connectionCharacteristics of common base connectionCommon emitter connectionMeasurement ofleakage currentCharacteristics of commonemitter connectionCommon collectorconnectionComparison of transistor connections

Commonly used transistor connectionTransistor as an amplifier in CEarrangementTransistor load line analysisOperating pointPractical wayof drawing CE circuitOutput from transistor amplifierPerformance oftransistor amplifierCut off and saturation pointsPower rating of transistorDetermination of transistor configurationSemiconductor devicesnumbering systemTransistor lead identificationTransistor testingApplications of common base amplifiersTransistors versus vacuum tubes.

9.9.9.9.9. TTTTTransistor Biasing ransistor Biasing ransistor Biasing ransistor Biasing ransistor Biasing 192239192239192239192239192239Faithful amplificationTransistor biasinglnherentvariations of transistor parametersStabilisationEs-sentials of a transistor biasing circuitStability fac-torMethods of transistor biasingBase resistormethodEmitter bias circuitCircuit analysis of emit-ter biasBiasing with collector feedback resistor-Voltage divider bias methodStability factor for po-tential divider biasDesign of transistor biasing cir-cuitsMid-point biasingWhich value of to beusedMiscellaneous bias circuits Silicon versus germaniumInstantaneouscurrent and voltage waveformsSummary of transistor bias circuits.

10.10.10.10.10. Single Stage TSingle Stage TSingle Stage TSingle Stage TSingle Stage Transistor Amplifiersransistor Amplifiersransistor Amplifiersransistor Amplifiersransistor Amplifiers 240279240279240279240279240279Single stage transistor amplifierHow transistor amplifies?Graphical demonstration of transistor amplifierPracticalcircuit of transistor amplifierPhase reversalInput/outputphase relationshipsD.C. and A.C. equivalent circuitsLoad line analysisVoltage gainA.C. emitter resistanceFormula for A.C. emitter resistanceVoltage gain of CEamplifierVoltage gain of unloaded CE amplifierVoltagegain of CE amplifier without CEInput impedance of CEamplifierVoltage gain stabilitySwamped AmplifierClassification ofamplifiersAmplifier equivalent circuitEquivalent circuit with signal sourceGain and transistor configurations.

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11.11.11.11.11. Multistage TMultistage TMultistage TMultistage TMultistage Transistor Amplifiersransistor Amplifiersransistor Amplifiersransistor Amplifiersransistor Amplifiers 280305280305280305280305280305Multistage transistor amplifierRole of capacitors intransistor amplifierslmportant termsProperties of dbgainR.C. coupled transistor amplifierTransformercoupled amplifiersDirect-coupled amplifierCom-parison of different types of couplingDifference be-tween transistor and tube amplifiers.

12.12.12.12.12. TTTTTransistor Audio Power Amplifiersransistor Audio Power Amplifiersransistor Audio Power Amplifiersransistor Audio Power Amplifiersransistor Audio Power Amplifiers 306334306334306334306334306334Transistor audio power amplifierSmall-signal and large-signal amplifiersOutput power of amplifierDifference between voltage and power

amplifiersPer formance quantities of poweramplifiersClassification of power amplifiersExpression for collector efficiencyMaximumcollector efficiency for series-fed class A amplifierMaximum collector efficiency of transformer coupledclass A power amplifierImportant points about classA power amplifierThermal runawayHeat sinkMathematical analysisStages of a practical power

amplifierDriver stageOutput stagePush pull amplifierMaximumefficiency for class B operationComplementary-symmetry amplifier.

13.13.13.13.13. Amplifiers With Negative FeedbackAmplifiers With Negative FeedbackAmplifiers With Negative FeedbackAmplifiers With Negative FeedbackAmplifiers With Negative Feedback 335363335363335363335363335363FeedbackPrinciples of negative voltagefeedback in amplifiersGain of negativevoltage feedback amplifierAdvantages ofnegative voltage feedbackFeedbackcircuitPrinciples of negative currentfeedbackCurrent gain with negative currentfeedbackEffects of negative currentfeedbackEmitter followerD.C. analysis of emitter followerVoltage gainof emitter followerInput impedance of emitter followerOutputimpedance of emitter follower Applications of emitter followerDarlingtonamplifier.

14.14.14.14.14. Sinusoidal OscillatorsSinusoidal OscillatorsSinusoidal OscillatorsSinusoidal OscillatorsSinusoidal Oscillators 364388364388364388364388364388Sinusoidal oscillatorTypes of sinusoidal oscillationsOscillatory circuitUndamped oscillations from tankcircuitPositive feedback amplifierOscillatorEssentials of transistor oscillatorExplanation ofbarkhausen criterionDifferent types of transistoroscillatorsTuned collector oscillatorColpitt's

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oscillatorsHartley oscillatorPrinciples of phase shift oscillatorsPhase shiftoscillatorWien bridge oscillator Limitations of LC and RC oscillatorsPiezoelectric crystalsWorking of quartz crystalEquivalent circuit of crystalFrequency response of crystalTransistor crystal oscillator.

15.15.15.15.15. TTTTTransistor Transistor Transistor Transistor Transistor Tuned Amplifiers uned Amplifiers uned Amplifiers uned Amplifiers uned Amplifiers 389410 389410 389410 389410 389410Tuned amplifiersDistinction between tunedamplifiers and other amplifiersAnalysis of paralledtuned circuitCharacteristics of parallel resonantcircuitAdvantages of tuned amplifiersWhy nottuned amplifiers for low frequency amplification?Frequency response of tuned amplifierRelationbetween Q and bandwidthSingle tunedamplifierAnalysis of tuned amplifierA.C.equivalent circuit of tuned amplifierDouble tunedamplifierBandwidth of double-tuned amplifierPractical application ofdouble tuned amplifierTuned class C amplifierClass C operationD.C.and A.C. loadsMaximum A.C. output power.

16.16.16.16.16. Modulation And DemodulationModulation And DemodulationModulation And DemodulationModulation And DemodulationModulation And Demodulation 411441411441411441411441411441Radio broadcasting, transmission and receptionModulationTypes ofmodulationAmplitude modulationModulation factorAnalysis of ampli-

tude modulated waveSideband frequencies in AMwaveTransistor AM modulatorPower in AM wave-Limitations of amplitude modulationFrequencymodulationTheory of frequency modulationCom-parison of FM and AM DemodulationEssentials indemodulationA.M. diode detectorA.M. radio re-ceiversTypes of A.M. radio receiversStages ofsuperhetrodyne radio receiverAdvantages ofsuperhetrodyne circuitFM receiverDifference be-tween FM and AM receivers.

17.17.17.17.17. Regulated D.C. Power SupplyRegulated D.C. Power SupplyRegulated D.C. Power SupplyRegulated D.C. Power SupplyRegulated D.C. Power Supply 442467442467442467442467442467Ordinary D.C. power supplylmportant termsRegulated power supplyTypes of voltage regu-latorsZener diode voltage regulatorConditionfor proper operation of zener regulatorTransis-tor series voltage regulatorSeries feedbackvoltage regulatorShort-circuit protectionTran-sistor shunt voltage regulatorShunt feedbackvoltage regulatorGlow-tube voltage regulatorSeries triode voltage regu-

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latorSeries double triode voltage regulatorIC voltage regulatorsFixedpositive voltage regulatorsFixed negative voltage regulatorsAdjustablevoltage regulatorsDual-tracking voltage regulators.

18. Solid-State Switching Circuits18. Solid-State Switching Circuits18. Solid-State Switching Circuits18. Solid-State Switching Circuits18. Solid-State Switching Circuits 468505468505468505468505468505

Switching circuitSwitchMechanical switchElectromechanical switch orrelayElectronic switchesAdvantages of electronic switchesImportant

termsSwitching transistorsSwitching action of a tran-sistorMultivibratorsTypes of multivibratorsTransistorastable multivibratorTransistor monostablemultivibratorTransistor bistable multivibratorDifferen-tiating circuitIntegrating circuitImportant applica-tions of diodesClipping circuitsApplications of clip-persClamping circuitsBasic idea of a clamperPosi-tive clamperNegative clamper.

19.19.19.19.19. Field EfField EfField EfField EfField Effect Tfect Tfect Tfect Tfect Transistorsransistorsransistorsransistorsransistors 506553506553506553506553506553Types of field effect transistorsJunction field effect transistor (JFET)Prin-ciple and working of JFETSchematic symbol of JFETImportance of JFETDifference between JFET and bipolar transistorJFETas an amplifierOutput characteristics of JFETSa-lient features of of JFETImportant termsExpressionfor drain currentAdvantages of JFETParametersof JFETRelation among JFET parametersVariationof transconductance (gm or gfs) of JFETJFET bias-ingJFET biasing by bias batterySelf-bias for JFETJFET with voltage-divider biasJFET connectionsPractical JFET amplifierD.C. and A.C. equivalent circuits of JFETD.C. loadline analysisVoltage gain of JFET amplifier (with source resistance RS)JFETapplicationsMetal oxide semiconductor FET (MOSFET)Types of MOSFETSymbols for D-MOSFETCircuit operation of D-MOSFETD-MOSFET transfercharacteristicTransconductance and input impedance of D-MOSFETD-MOSFET biasingCommon source D-MOSFET amplifierD-MOSFETs versusJFETsE-MOSFETE-MOSFET biasing circuitsD-MOSFETs versus E-MOSFETs.

20.20.20.20.20. Silicon Controlled Rectifiers Silicon Controlled Rectifiers Silicon Controlled Rectifiers Silicon Controlled Rectifiers Silicon Controlled Rectifiers 554576554576554576554576554576Silicon Controlled rectifier (SCR)Working of SCREquivalent circuit of SCR-V-I characteristics of SCRSCRin normal operationSCR as a switchSCR switchingSCR half-wave rectifierSCR full-wave rectifierSingle-phase SCR inverter circuitApplications of SCR-Light-activated SCR.

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21.21.21.21.21. Power ElectronicsPower ElectronicsPower ElectronicsPower ElectronicsPower Electronics 577600577600577600577600577600Power electronicsThe triacTriac constructionSCRequivalent circuit of triacTriac operationTriac phasecontrol circuitApplications of triacThe DiacApplications of diacUnijunction transistor (UJT)Equivalent circuit of a UJTCharacteristics of UJTAdvantages of UJTApplications of UJT.

22.22.22.22.22. Electronic InstrumentsElectronic InstrumentsElectronic InstrumentsElectronic InstrumentsElectronic Instruments 601626601626601626601626601626Electronic instrumentsMultimeterApplications of multimeterSensitivity ofmultimeterMerits and demerits of multimeterMeter protectionElectronic

voltmetersVacuum tube voltmeter (VTVM)Applications of VTVMMerits and demerits of VTVMTransistor voltmeter circuitBridge rectifier voltmeterCathode ray oscilloscopeCathode ray tubeDeflection sensitivity of CRTApplying signal acrossvertical platesDisplay of signal waveform on CROSignal pattern on screenVarious controls of CROApplications of CRO.

23.23.23.23.23. Integrated CircuitsIntegrated CircuitsIntegrated CircuitsIntegrated CircuitsIntegrated Circuits 627641627641627641627641627641

Integrated circuitsAdvantages and disadvantagesof integrated circuitsInside an IC packageICclassificationsMaking monolithic ICFabrication ofcomponents on monolithic ICSimple monolithic ICsIC packingsIC symbolsScale of integrationSomecircuits using ICs.

24.24.24.24.24. Hybrid ParametersHybrid ParametersHybrid ParametersHybrid ParametersHybrid Parameters 642661642661642661642661642661

Hybrid parametersDetermination of hparametersh parameter equivalent circuitPerformance of a linear circuit in h parametersThe h parameters of a transistorNomenclature fortransistor h parametersTransistor circuitperformance in h parametersApproximate hybridformulas for transistor amplifierExperimentaldetermination of transistor h parametersLimitationsof h parameters.

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25.25.25.25.25. Operational AmplifiersOperational AmplifiersOperational AmplifiersOperational AmplifiersOperational Amplifiers 662728662728662728662728662728

Operational amplifierDifferential amplifier(DA)Basic circuit of differential amplifierOperation of differential amplifierCommon-mode and differential-mode signalsDouble-ended input operation of DAVoltage gains ofDACommon-mode rejection ratio (CMRR)D.C. analysis of differential amplifier (DA)Overview of differential amplifierParametersof DA (or OP-amp) due to mismatch of transistorsInput bias currentA.C.analysis of differential amplifierCommon-mode voltage gainOperationalamplifier (OP-amp)Schematic symbol of operational amplifierOutputvoltage from OP-ampA.C. analysis of OP-ampBandwidth of an OP-ampSlew rateFrequency response of an OP-amp OP-amp with negativefeedbackApplications of OP-ampInverting amplifierInput and outputimpedance of inverting amplifierNoninverting amplifierVoltage followerMultistage OP-amp circuitsEffect of negative feedback on OP-ampimpedancesFaults in feedback circuitsSummary of OP-ampconfigurationsSumming amplifiersApplications of summing amplifiersOP-amp integrators and differentiatorsOP-amp integratorCriticalfrequency of integratorsOP-amp differentiatorComparatorsComparator circuits.

26.26.26.26.26. Digital ElectronicsDigital ElectronicsDigital ElectronicsDigital ElectronicsDigital Electronics 729773729773729773729773729773

Analog and digital signalsDigital circuitBinarynumber systemPlace valueDecimal to binaryconversionBinary to decimal conversionOctalnumber systemHexadecimal number systemBinary-coded decimal code (BCD code)LogicgatesThree basic logic gatesOR gateANDgateNOT gate or inverterCombination of basiclogic gatesNAND gate as a universal gate-

Exclusive OR gateEncoders and decodersAdvantages anddisadvantages of digital electronicsBoolean algebraBoolean theoremsDeMorgans theoremsOperator precedenceCombinational logiccircuitsBoolean expressions for combinational logic circuitsAND and ORoperations in Boolean expressionTruth table from logic circuitDevelopinglogic circuit from its Boolean expressionSum-of-products formSimplification of Boolean expressionsBinary additionElectronic addersFlip-flops.

IIIIIndexndexndexndexndex 775778775778775778775778775778

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Introduction

1

1.1 Electronics1.2 Atomic Structure1.3 Structure of Elements1.4 The Electron1.5 Energy of an Electron1.6 Valence Electrons1.7 Free Electrons1.8 Voltage Source1.9 Constant Voltage Source1.10 Constant Current Source1.11 Conversion of Voltage Source into Current

Source1.12 Maximum Power Transfer Theorem1.13 Thevenins Theorem1.14 Procedure for Finding Thevenin

Equivalent Circuit1.15 Nortons Theorem1.16 Procedure for Finding Norton Equivalent

Circuit1.17 Chassis and Ground

GENERALGENERALGENERALGENERALGENERAL

In this fast developing society, electronics has come to stay as the most important branch ofengineering. Electronic devices are being used in almost all the industries for quality controland automation and they are fast replacing the present vast army of workers engaged in process-ing and assembling in the factories. Great strides taken in the industrial applications of electronicsduring the recent years have demonstrated that this versatile tool can be of great importance in in-creasing production, efficiency and control.

The rapid growth of electronic technology offers a formidable challenge to the beginner, whomay be almost paralysed by the mass of details. However, the mastery of fundamentals can simplifythe learning process to a great extent. The purpose of this chapter is to present the elementary knowl-edge in order to enable the readers to follow the subsequent chapters.

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2 Principles of Electronics1.1 ElectronicsThe branch of engineeringwhich deals with current con-duction through a vacuum orgas or semiconductor isknown as *electronics.

Electronics essentiallydeals with electronicdevices and their utilisation.An electronic device is thatin which current flows through a vacuum or gas or semiconductor. Such devices have valuableproperties which enable them to function and behave as the friend of man today.

Importance. Electronics has gained much importance due to its numerous applications in in-dustry. The electronic devices are capable of performing the following functions :

(i) Rectification. The conversion of a.c. into d.c. is called rectification. Electronic devicescan convert a.c. power into d.c. power (See Fig. 1.1) with very high efficiency. This d.c. supply can beused for charging storage batteries, field supply of d.c. generators, electroplating etc.

Fig. 1.1

(ii) Amplification. The process of raising the strength of a weak signal is known as amplifica-tion. Electronic devices can accomplish the job of amplification and thus act as amplifiers (See Fig.1.2). The amplifiers are used in a wide variety of ways. For example, an amplifier is used in a radio-set where the weak signal is amplified so that it can be heard loudly. Similarly, amplifiers are used inpublic address system, television etc.

Fig. 1.2

(iii) Control. Electronic devices find wide applications in automatic control. For example,speed of a motor, voltage across a refrigerator etc. can be automatically controlled with the help ofsuch devices.

(iv) Generation. Electronic devices can convert d.c. power into a.c. power of any frequency(See Fig. 1.3). When performing this function, they are known as oscillators. The oscillators areused in a wide variety of ways. For example, electronic high frequency heating is used for annealingand hardening.

Current conduction through semiconductor

* The word electronics derives its name from electron present in all materials.

Introduction 3

Fig. 1.3

(v) Conversion of light into electricity. Electronic devices can convert light into electricity.This conversion of light into electricity is known as photo-electricity. Photo-electric devices are usedin Burglar alarms, sound recording on motion pictures etc.

(vi) Conversion of electricity into light. Electronic devices can convert electricity into light.This valuable property is utilised in television andradar.

1.2 Atomic StructureAccording to the modern theory, matter is electricalin nature. All the materials are composed of verysmall particles called atoms. The atoms are thebuilding bricks of all matter. An atom consists of acentral nucleus of positive charge around which smallnegatively charged particles, called electrons revolvein different paths or orbits.

(1) Nucleus. It is the central part of an atomand *contains protons and neutrons. A proton is apositively charged particle, while the neutron has thesame mass as the proton, but has no charge. There-fore, the nucleus of an atom is positively charged. The sum of protons and neutrons constitutes theentire weight of an atom and is called atomic weight. It is because the particles in the extra nucleus(i.e. electrons) have negligible weight as compared to protons or neutrons.

atomic weight = no. of protons + no. of neutrons(2) Extra nucleus. It is the outer part of an atom and contains electrons only. An electron is a

negatively charged particle having negligible mass. The charge on an electron is equal but opposite tothat on a proton. Also, the number of electrons is equal to the number of protons in an atom underordinary conditions. Therefore, an atom is neutral as a whole. The number of electrons or protons inan atom is called atomic number i.e.

atomic number = no. of protons or electrons in an atomThe electrons in an atom revolve around the nucleus in different orbits or paths. The number and

arrangement of electrons in any orbit is determined by the following rules :(i) The number of electrons in any orbit is given by 2n2 where n is the number of the orbit. For

example,First orbit contains 2 12 = 2 electronsSecond orbit contains 2 22 = 8 electronsThird orbit contains 2 32 = 18 electrons

* Although the nucleus of an atom is of complex structure, yet for the purpose of understanding electronics,this simplified picture of the nucleus is adequate.

Carbon Atom

Proton

Neutron

Nucleus

Electron

4 Principles of Electronicsand so on.

(ii) The last orbit cannot have more than 8 electrons.(iii) The last but one orbit cannot have more than 18 electrons.

1.3 Structure of ElementsWe have seen that all atoms are made up of protons, neutrons and electrons. The difference betweenvarious types of elements is due to the different number and arrangement of these particles withintheir atoms. For example, the structure* of copper atom is different from that of carbon atom andhence the two elements have different properties.

The atomic structure can be easily built up if we know theatomic weight and atomic number of the element. Thus takingthe case of copper atom,

Atomic weight = 64Atomic number = 29

No. of protons = No. of electrons = 29and No. of neutrons = 64 29 = 35Fig. 1.4 shows the structure of copper atom. It has 29

electrons which are arranged in different orbits as follows. Thefirst orbit will have 2 electrons, the second 8 electrons, thethird 18 electrons and the fourth orbit will have 1electron. The atomic structure of all known elements can beshown in this way and the reader is advised to try for a fewcommonly used elements.

1.4 The ElectronSince electronics deals with tiny particles called electrons, these small particles require detailed study.As discussed before, an electron is a negatively charged particle having negligible mass. Some of theimportant properties of an electron are :

(i) Charge on an electron, e = 1.602 1019 coulomb(ii) Mass of an electron, m = 9.0 1031 kg

(iii) Radius of an electron, r = 1.9 1015 metre

The ratio e/m of an electron is 1.77 1011 coulombs/kg.This means that mass of an electron is very small as comparedto its charge. It is due to this property of an electron that it isvery mobile and is greatly influenced by electric or magneticfields.

1.5 Energy of an ElectronAn electron moving around the nucleus possesses two typesof energies viz. kinetic energy due to its motion and potentialenergy due to the charge on the nucleus. The totalenergy of the electron is the sum of these two energies. Theenergy of an electron increases as its distance from the nucleusincreases. Thus, an electron in the second orbit possesses moreenergy than the electron in the first orbit; electron in the third

* The number and arrangement of protons, neutrons and electrons.

Fig. 1.4

This electron has thehighest energy.

Energy level

Thiselectronhas thelowestenergy

Energy levels increase asthe distance from the

nucleus increases

36

21

Introduction 5orbit has higher energy than in the second orbit.It is clear that electrons in the last orbit possess veryhigh energy as compared to the electrons in the inner orbits. These last orbit electrons play an impor-tant role in determining the physical, chemical and electrical properties of a material.

1.6 Valence ElectronsThe electrons in the outermost orbit of an atom are known as valence electrons.

The outermost orbit can have a maximum of 8 electrons i.e. the maximum number of valenceelectrons can be 8. The valence electrons determine the physical and chemical properties of a material.These electrons determine whether or not the material is chemically active; metal or non-metal or, agas or solid. These electrons also determine the electrical properties of a material.

On the basis of electrical conductivity, materials are generally classified into conductors, insula-tors and semi-conductors. As a rough rule, one can determine the electrical behaviour of amaterial from the number of valence electrons as under :

(i) When the number of valence electrons of an atom is less than 4 (i.e. half of the maximumeight electrons), the material is usually a metal and a conductor. Examples are sodium, magnesiumand aluminium which have 1, 2 and 3 valence electrons respectively (See Fig. 1.5).

Fig. 1.5

(ii) When the number of valence electrons of an atom is more than 4, the material is usually anon-metal and an insulator. Examples are nitrogen, sulphur and neon which have 5, 6 and 8 valenceelectrons respectively (See Fig. 1.6).

Fig. 1.6

(iii) When the number of valence electrons of an atom is 4 (i.e. exactly one-half of themaximum 8 electrons), the material has both metal and non-metal properties and is usually a semi-conductor. Examples are carbon, silicon and germanium (See Fig. 1.7).

6 Principles of Electronics

Fig. 1.71.7 Free ElectronsThe valence electrons of different materials possess different energies. The greater the energy of avalence electron, the lesser it is bound to the nucleus. In certain substances, particularly metals, thevalence electrons possess so much energy that they are very loosely attached to the nucleus. Theseloosely attached valence electrons move at random within the material and are called free electrons.

The valence electrons which are very loosely attached to the nucleus are known as freeelectrons.

The free electronscan be easily removed ordetached by applying asmall amount of externalenergy. As a matter offact, these are the freeelectrons which deter-mine the electrical conductivity of a material. On this basis, conductors, insulators and semiconduc-tors can be defined as under :

(i) A conductor is a substance which has a large number of free electrons. When potential differ-ence is applied across a conductor, the free electrons move towards the positive terminal of supply,constituting electric current.

(ii) An insulator is a substance which has practically no free electrons at ordinary temperature.Therefore, an insulator does not conduct current under the influence of potential difference.

(iii) A semiconductor is a substance which has very few free electrons at room temperature.Consequently, under the influence of potential difference, asemiconductor practically conducts no current.

1.8 Voltage SourceAny device that produces voltage output continuously isknown as a voltage source. There are two types of voltagesources, namely ; direct voltage source and alternating volt-age source.

(i) Direct voltage source. A device which producesdirect voltage output continuously is called a direct voltagesource. Common examples are cells and d.c. generators. Animportant characteristic of a direct voltage source is that it

No current flows Current flows

Electron Copper atom

Current moves through materials that conduct electricity.

Voltage source

Introduction 7maintains the same polarity of the output voltage i.e. positive and negative terminals remain the same.When load resistance RL is connected across such a source,*current flows from positive terminal tonegative terminal via the load [See Fig. 1.8 (i)]. This is called direct current because it has just onedirection. The current has one direction as the source maintains the same polarity of output voltage.The opposition to load current inside the d.c. source is known as internal resistance Ri. The equivalentcircuit of a d.c. source is the generated e.m.f. Eg in series with internal resistance Ri of the source as shownin Fig. 1.8 (ii). Referring to Fig. 1.8 (i), it is clear that:

Fig. 1.8

Load current, I = gL i

ER R+

Terminal voltage, V = (Eg I Ri) or I RL(ii) Alternating voltage source. A device which produces alternating voltage output continu-

ously is known as alternating voltage source e.g. a.c. generator. An important characteristic of alter-nating voltage source is that it periodically reverses the polarity of the output voltage. When loadimpedance ZL is connected across such a source, current flows through the circuit that periodicallyreverses in direction. This is called alternating current.

Fig. 1.9

The opposition to load current inside the a.c. source is called its internal impedance Zi. Theequivalent circuit of an a.c. source is the generated e.m.f. Eg (r.m.s.) in series with internal impedanceZi of the source as shown in Fig. 1.9 (ii). Referring to Fig. 1.9 (i), it is clear that :

Load current, I (r.m.s.) = gL i

EZ Z+

Terminal voltage, V = (Eg I Zi)** or I ZL

* This is the conventional current. However, the flow of electrons will be in the opposite direction.** Vector difference since a.c. quantities are vector quantities.

8 Principles of Electronics1.9 Constant Voltage Source

A voltage source which has very low internal *impedance as compared with external load im-pedance is known as a constant voltage source.

Fig. 1.10

In such a case, the output voltage nearly remains the same when load current changes.Fig. 1.10 (i) illustrates a constant voltage source. It is a d.c. source of 6 V with internalresistance Ri = 0.005 . If the load current varies over a wide range of 1 to 10 A, for anyof these values, the internal drop across Ri (= 0.005 ) is less than 0.05 volt. Therefore,the voltage output of the source is between 5.995 to 5.95 volts. This can be consideredconstant voltage compared with the wide variations in load current.

Fig. 1.10 (ii) shows the graph for a constant voltage source. It may be seen that theoutput voltage remains constant inspite of the changes in load current. Thus as the loadcurrent changes from 0 to 10 A, the output voltage essentially remains the same (i.e.V1 = V2). A constant voltage source is represented as shown in Fig. 1.11.

Example 1.1. A lead acid battery fitted in a truck develops 24V and has an internalresistance of 0.01 . It is used to supply current to head lights etc. If the total load is equal to100 watts, find :

(i) voltage drop in internal resistance(ii) terminal voltageSolution.

Generated voltage, Eg = 24 V

Internal resistance, Ri = 0.01

Power supplied, P = 100 watts(i) Let I be the load current.Now P = Eg I ( For an ideal source, V j Eg)

I = 10024gPE

= = 4.17 A

Voltage drop in Ri = I Ri = 4.17 0.01 = 0.0417 V

(ii) Terminal Voltage, V = Eg I Ri= 24 0.0417 = 23.96 V

* resistance in case of a d.c. source.

Fig. 1.11

Introduction 9Comments : It is clear from the above example that when internal resistance of the source is

quite small, the voltage drop in internal resistance is very low. Therefore, the terminal voltage sub-stantially remains constant and the source behaves as a constant voltage source irrespective of loadcurrent variations.

1.10 Constant Current SourceA voltage source that has a very high internal*impedance as compared with external load impedanceis considered as a constant current source.

In such a case, the load current nearly remainsthe same when the output voltage changes. Fig. 1.12(i) illustrates a constant current source. It is a d.c.source of 1000 V with internal resistance Ri = 900k. Here, load RL varies over 3 : 1 range from 50k to 150 k . Over this variation of load RL, thecircuit current I is essentially constant at 1.05 to 0.95mA or approximately 1 mA. It may be noted that output voltage V varies approximately in thesame 3 : 1 range as RL, although load current essentially remains **constant at 1mA. The beautifulexample of a constant current source is found in vacuum tube circuits where the tube acts as agenerator having internal resistance as high as 1 M.

Fig. 1.12 (ii) shows the graph of a constant current source. It is clear that current remains con-stant even when the output voltage changes substantially. The following points may be noted regard-ing the constant current source :

Fig. 1.12

(i) Due to high internal resistance of the source, the load current remainsessentially constant as the load RL is varied.

(ii) The output voltage varies approximately in the same range as RL, althoughcurrent remains constant.

(iii) The output voltage V is much less than the generated voltage Eg because ofhigh I Ri drop.

Fig. 1.13 shows the symbol of a constant current source.

* Resistance in case of a d.c. source

** Now I = gL i

ER R+

. Since Ri >> RL, I = g

i

ER

As both Eg and Ri are constants, I is constant.

Constant Current Source

Fig. 1.13

10 Principles of ElectronicsExample 1.2. A d.c. source generating 500 V has an internal resistance of 1000 . Find the

load current if load resistance is (i) 10 (ii) 50 and (iii) 100 .Solution.

Generated voltage, Eg = 500 VInternal resistance, Ri = 1000

(i) When RL = 10

Load current, I = 50010 1000g

L i

ER R

=+ + = 0.495 A

(ii) When RL = 50

Load current, I = 50050 1000+ = 0.476 A

(iii) When RL = 100

Load current, I = 500100 1000+ = 0.454 A

It is clear from the above example that load current is essentially constant since Ri >> RL.

1.11 Conversion of Voltage Source into Current SourceFig. 1.14 shows a constant voltage source with voltage V and internal resistance Ri. Fig. 1.15 showsits equivalent current source. It can be easily shown that the two circuits behave electrically the sameway under all conditions.

(i) If in Fig. 1.14, the load is open-circuited (i.e. RL ), then voltage across terminals A andB is V. If in Fig. 1.15, the load is open-circuited (i.e. RL ), then all current I (= V/Ri) flows throughRi, yielding voltage across terminals AB = I Ri = V. Note that open-circuited voltage across AB is V forboth the circuits and hence they are electrically equivalent.

Fig. 1.14 Fig. 1.15

(ii) If in Fig. 1.14, the load is short-circuited (i.e. RL = 0), the short circuit current is given by:

Ishort =i

VR

If in Fig. 1.15, the load is short-circuited (i.e. RL = 0), the current I (= V/Ri) bypasses Ri in favourof short-circuit. It is clear that current (= V/Ri) is the same for the two circuits and hence they areelectrically equivalent.

Introduction 11Thus to convert a constant voltage source into a constant current source, the following

procedure may be adopted :(a) Place a short-circuit across the two terminals in question (terminals AB in the present case)

and find the short-circuit current. Let it be I. Then I is the current supplied by the equivalent currentsource.

(b) Measure the resistance at the terminals with load removed and sources of e.m.f.s replaced bytheir internal resistances if any. Let this resistance be R.

(c) Then equivalent current source can be represented by a single current source of magnitudeI in parallel with resistance R.

Note. To convert a current source of magnitude I in parallel with resistance R into voltage source,Voltage of voltage source, V = I R

Resistance of voltage source, R = RThus voltage source will be represented as voltage V in series with resistance R.Example 1.3. Convert the constant voltage source shown in Fig. 1.16 into constant current

source.Solution. The solution involves the following steps :(i) Place a short across AB in Fig. 1.16 and find the short-circuit current I.

Clearly, I = 10/10 = 1 ATherefore, the equivalent current source has a magnitude of 1 A.(ii) Measure the resistance at terminals AB with load *removed and 10 V source replaced by its internal

resistance. The 10 V source has negligible resistance so that resistance at terminals AB is R = 10 .

Fig. 1.16 Fig. 1.17

(iii) The equivalent current source is a source of 1 A in parallel with a resistance of 10 asshown in Fig. 1.17.

Example 1.4. Convert the constant current source in Fig. 1.18 into equivalent voltage source.

Fig. 1.18 Fig. 1.19

Solution. The solution involves the following steps :

* Fortunately, no load is connected across AB. Had there been load across AB, it would have been removed.

12 Principles of Electronics(i) To get the voltage of the voltage source, multiply the current of the current source by the

internal resistance i.e.Voltage of voltage source = I R = 6 mA 2 k = 12V(ii) The internal resistance of voltage source is 2 k .The equivalent voltage source is a source of 12 V in series with a resistance of 2 k as shown in

Fig. 1.19.Note. The voltage source should be placed with +ve terminal in the direction of current flow.

1.12 Maximum Power Transfer TheoremWhen load is connected across a voltage source, power is transferred from the source to the load. Theamount of power transferred will depend upon the load resistance. If load resistance RL is made equalto the internal resistance Ri of the source, then maximum power is transferred to the load RL. This isknown as maximum power transfer theorem and can be stated as follows :

Maximum power is transferred from a source to a load when the load resistance is made equalto the internal resistance of the source.

This applies to d.c. as well as a.c. power.*To prove this theorem mathematically, consider a voltage source of generated voltage E and

internal resistance Ri and delivering power to a load resistance RL [See Fig. 1.20 (i)]. The current Iflowing through the circuit is given by :

I =L i

ER R+

Power delivered to the load, P = I2 RL = 2

LL i

E RR R

+

...(i)

Fig. 1.20

For a given source, generated voltage E and internal resistance Ri are constant. Therefore, powerdelivered to the load depends upon RL. In order to find the value of RL for which the value of P ismaximum, it is necessary to differentiate eq. (i) w.r.t. RL and set the result equal to zero.

Thus,L

dPdR

=2

24

( ) 2 ( )( )

L i L L i

L i

R R R R RE

R R

+ +

+ = 0

or (RL + Ri)2 2 RL (RL + Ri) = 0

or (RL + Ri) (RL + Ri 2 RL) = 0or (RL + Ri) (Ri RL) = 0

* As power is concerned with resistance only, therefore, this is true for both a.c. and d.c. power.

Introduction 13Since (RL + Ri) cannot be zero, Ri RL = 0or RL = Rii.e. Load resistance = Internal resistanceThus, for maximum power transfer, load resistance RL must be equal to the internal resistance Ri

of the source.Under such conditions, the load is said to be matched to the source. Fig. 1.20 (ii) shows a graph

of power delivered to RL as a function of RL. It may be mentioned that efficiency of maximum powertransfer is *50% as one-half of the total generated power is dissipated in the internal resistance Ri ofthe source.

Applications. Electric power systems never operate for maximum power transfer because oflow efficiency and high voltage drops between generated voltage and load. However, in the elec-tronic circuits, maximum power transfer is usually desirable. For instance, in a public address sys-tem, it is desirable to have load (i.e. speaker) matched to the amplifier so that there is maximumtransference of power from the amplifier to the speaker. In such situations, efficiency is **sacrificedat the cost of high power transfer.

Example 1.5. A generator develops 200 V and has an internal resistance of 100 . Find thepower delivered to a load of (i) 100 (ii) 300 . Comment on the result.

Solution.Generated voltage, E = 200 VInternal resistance, Ri = 100

(i) When load RL = 100

Load current, I = 200100 100L i

ER R

=+ +

= 1 A

Power delivered to load = I2 RL = (1)2 100 = 100 watts

Total power generated = I2 (RL + Ri) = 12 (100 + 100) = 200 watts

Thus, out of 200 W power developed by the generator, only 100W has reached the load i.e.efficiency is 50% only.

(ii) When load RL = 300

Load current, I = 200300 100L iE

R R=

+ + = 0.5 A

Power delivered to load = I2 RL = (0.5)2 300 = 75 watts

Total power generated = I2 (RL + Ri) = (0.5)2 (300 + 100) = 100 watts

Thus, out of 100 watts of power produced by the generator, 75 watts is transferred to the load i.e.efficiency is 75%.

Comments. Although in case of RL = Ri, a large power (100 W) is transferred to the load, butthere is a big wastage of power in the generator. On the other hand, when RL is not equal to Ri, the

* Efficiency =2

2output powerinput power ( )

L

L i

I RI R R

=+

= RL2 RL = 12 = 50% ( RL = Ri)** Electronic devices develop small power. Therefore, if too much efficiency is sought, a large number of

such devices will have to be connected in series to get the desired output. This will distort the output aswell as increase the cost and size of equipment.

14 Principles of Electronicspower transfer is less (75 W) but smaller part is wasted in the generator i.e. efficiency is high. Thus,it depends upon a particular situation as to what the load should be. If we want to transfer maximumpower (e.g. in amplifiers) irrespective of efficiency, we should make RL = Ri. However, if efficiencyis more important (e.g. in power systems), then internal resistance of the source should be consider-ably smaller than the load resistance.

Example 1.6. An audio amplifier producesan alternating output of 12 V before the connec-tion to a load. The amplifier has an equivalentresistance of 15 at the output. Whatresistance the load need to have to produce maxi-mum power ? Also calculate the power outputunder this condition.

Solution. In order to produce maximumpower, the load (e.g. a speaker) should have a re-sistance of 15 to match the amplifier. The equiva-lent circuit is shown in Fig. 1.21.

Load required, RL = 15

Circuit current, I = 1215 15TVR

=+ = 0.4 A

Power delivered to load, P = I2 RL = (0.4)2 15 = 2.4 W

Example 1.7. For the a.c. generator shown in Fig. 1.22 (i), find (i) the value of load so thatmaximum power is transferred to the load (ii) the value of maximum power.

Fig. 1.22Solution.(i) In a.c. system, maximum power is delivered to the load impedance (ZL) when load imped-

ance is conjugate of the internal impedance (Zi) of the source. Now in the problem, Zi = (100 + j50).For maximum power transfer, the load impedance should be conjugate of internal impedance i.e. ZLshould be (100 j50) . This is shown in dotted line in Fig. 1.22 (ii).

ZL = (100 j50)

(ii) Total impedance, ZT = Zi + ZL = (100 + j50) + (100 j50) = 200 *

Circuit current, I = 50200TVZ

= = 0.25 A

Maximum power transferred to the load = I2 RL = (0.25)2 100 = 6.25 W

Fig. 1.21

* Note that by making internal impedance and load impedance conjugate, the reactive terms cancel. Thecircuit then consists of internal and external resistances only. This is quite logical because power is onlyconsumed in resistances as reactances (XL or XC) consume no power.

Introduction 151.13 Thevenins Theorem

Sometimes it is desirable to find a particular branch current in a circuit as the resistance of thatbranch is varied while all other resistances and voltage sources remain constant. For instance, in thecircuit shown in Fig. 1.23, it may be desired to find the current through RL for five values of RL,assuming that R1, R2, R3 and E remain constant. In such situations, the *solution can be obtainedreadily by applying Thevenins theorem stated below :

Any two-terminal network containing a number of e.m.f. sources and resistances can be re-placed by an equivalent series circuit having a voltage source E0 in series with a resistance R0 where,

E0 = open circuited voltage between the two terminals.R0 = the resistance between two terminals of the circuit obtained by looking in at

the terminals with load removed and voltage sources replaced by their internalresistances, if any.

To understand the use of this theorem, consider the two-terminal circuit shown in Fig. 1.23. Thecircuit enclosed in the dotted box can be replaced by one voltage E0 in series with resistance R0 asshown in Fig. 1.24. The behaviour at the terminals AB and AB is the same for the two circuits,independent of the values of RL connected across the terminals.

Fig. 1.23 Fig. 1.24

(i) Finding E0. This is the voltage between terminals A and B of the circuit when load RL isremoved. Fig. 1.25 shows the circuit with load removed. The voltage drop across R2 is the desiredvoltage E0.

Current through R2 =1 2

ER R+

Voltage across R2, E0 = 21 2

E RR R

+

Thus, voltage E0 is determined.

Fig. 1.25 Fig. 1.26

* Solution can also be obtained by applying Kirchhoffs laws but it requires a lot of labour.

16 Principles of Electronics(ii) Finding R0. This is the resistance between terminals A and B with load removed and e.m.f.

reduced to zero (See Fig. 1.26). Resistance between terminals A and B is

R0 = parallel combination of R1 and R2 in series with R3

= 1 2 31 2

R RR

R R+

+

Thus, the value of R0 is determined. Once the values of E0 and R0 are determined, then thecurrent through the load resistance RL can be found out easily (Refer to Fig. 1.24).

1.14 Procedure for Finding Thevenin Equivalent Circuit(i) Open the two terminals (i.e. remove any load) between which you want to find Thevenin

equivalent circuit.

(ii) Find the open-circuit voltage between the two open terminals. It is called Thevenin voltageE0.

(iii) Determine the resistance between the two open terminals with all ideal voltage sourcesshorted and all ideal current sources opened (a non-ideal source is replaced by its internalresistance). It is called Thevenin resistance R0.

(iv) Connect E0 and R0 in series to produce Thevenin equivalent circuit between the two termi-nals under consideration.

(v) Place the load resistor removed in step (i) across the terminals of the Thevenin equivalentcircuit. The load current can now be calculated using only Ohms law and it has the samevalue as the load current in the original circuit.

Example 1.8. Using Thevenins theorem, find the current through 100 resistance connectedacross terminals A and B in the circuit of Fig. 1.27.

Fig. 1.27

Solution.(i) Finding E0. It is the voltage across terminals A and B with 100 resistance removed as

shown in Fig. 1.28.E0 = (Current through 8 ) 8 = 2.5* 8 = 20 V

* By solving this series-parallel circuit.

Introduction 17

Fig. 1.28 Fig. 1.29

(ii) Finding R0. It is the resistance between terminalsA and B with 100 removed and voltage source short cir-cuited as shown in Fig. 1.29.

R0 = Resistance looking in at terminals A and B in Fig. 1.29

=

10 20 12 810 20

10 20 12 810 20

+ + + + +

= 5.6 Therefore, Thevenins equivalent circuit will be as shown in Fig. 1.30. Now, current through 100

resistance connected across terminals A and B can be found by applying Ohms law.

Current through 100 resistor = 00

205.6 100L

ER R

=+ +

= 0.19 A

Example 1.9. Find the Thevenins equivalent circuit for Fig. 1.31.Solution. The Thevenins voltage E0 is the voltage across terminals A and B. This voltage is

equal to the voltage across R3. It is because terminals A and B are open circuited and there is nocurrent flowing through R2 and hence no voltage drop across it.

Fig. 1.31 Fig. 1.32

E0 = Voltage across R3= 3

1 3

1 201 1

RV

R R =

+ + = 10 V

The Thevenins resistance R0 is the resistance measured between terminals A and B with no load(i.e. open at terminals A and B) and voltage source replaced by a short circuit.

R0 =1 3

21 3

1 111 1

R RR

R R+ = +

+ + = 1.5 k

Therefore, Thevenins equivalent circuit will be as shown in Fig. 1.32.

Fig. 1.30

18 Principles of ElectronicsExample 1.10. Calculate the value of load resistance RL to which maximum power may be

transferred from the circuit shown in Fig. 1.33 (i). Also find the maximum power.

Fig. 1.33

Solution. We shall first find Thevenins equivalent circuit to the left of terminals AB inFig. 1.33 (i).

E0 = Voltage across terminals AB with RL removed

= 120 2040 20

+ = 40 V

R0 = Resistance between terminals A and B with RL removed and 120 V sourcereplaced by a short

= 60 + (40 || 20 ) = 60 + (40 20)/60 = 73.33 The Thevenins equivalent circuit to the left of terminals AB in Fig. 1.33 (i) is E0 (= 40 V) in

series with R0 (= 73.33 ). When RL is connected between terminals A and B, the circuit becomes asshown in Fig. 1.33 (ii). It is clear that maximum power will be transferred when

RL = R0 = 73.33

Maximum power to load =2 20 (40)

4 4 73.33L

ER

= = 5.45 W

Comments. This shows another advantage of Thevenins equivalent circuit of a network. OnceThevenins equivalent resistance R0 is calculated, it shows at a glance the condition for maximumpower transfer. Yet Thevenins equivalent circuit conveys another information. Thus referring toFig. 1.33 (ii), the maximum voltage that can appear across terminals A and B is 40 V. This is not soobvious from the original circuit shown in Fig. 1.33 (i).

Example 1.11. Calculate the current in the 50 resistor in the network shown in Fig. 1.34.

Fig. 1.34

Introduction 19Solution. We shall simplify the circuit shown in Fig. 1.34 by the repeated use of Thevenins

theorem. We first find Thevenins equivalent circuit to the left of *XX.

Fig. 1.35

E0 =80 100

100 100

+ = 40V

R0 = 100 || 100 = 100 100100 100

+ = 50

Therefore, we can replace the circuit to the left of XX in Fig. 1.34 by its Thevenins equivalentcircuit viz. E0 (= 40V) in series with R0 (= 50). The original circuit of Fig. 1.34 thenreduces to the one shown in Fig. 1.35.

We shall now find Thevenins equivalentcircuit to left of YY in Fig. 1.35.

0E = 40 80

50 30 80

+ + = 20 V

0R = (50 + 30) || 80 = 80 8080 80

+ = 40

We can again replace the circuit to the left of YY in Fig. 1.35 by its Thevenins equivalentcircuit. Therefore, the original circuit reduces to that shown in Fig. 1.36.

(a) (b)

E0 = Current in 100 100 = 80 100

100 100

+ = 40V [See Fig. (a)]

R0 = Resistance looking in the open terminals in Fig. (b)

= 100 || 100 = 100 100100 100+ = 50

Fig. 1.36

*

20 Principles of ElectronicsUsing the same procedure to the left of ZZ, we have,

0E =20 60

40 20 60

+ + = 10V

0R = (40 + 20) || 60 = 60 6060 60+ = 30

The original circuit then reduces to that shown in Fig. 1.37.By Ohms law, current I in 50 resistor is

I =10

30 20 50+ + = 0.1 A

1.15 Nortons TheoremFig. 1.38 (i) shows a network enclosed in a box with two terminals A and B brought out. The networkin the box may contain any number of resistors and e.m.f. sources connected in any manner. Butaccording to Norton, the entire circuit behind terminals A and B can be replaced by a current sourceof output IN in parallel with a single resistance RN as shown in Fig. 1.38 (ii). The value of IN isdetermined as mentioned in Nortons theorem. The resistance RN is the same as Thevenins resistanceR0. Once Nortons equivalent circuit is determined [See Fig. 1.38 (ii)], then current through any loadRL connected across terminals AB can be readily obtained.

Fig. 1.38

Hence Nortons theorem as applied to d.c. circuits may be stated as under :Any network having two terminals A and B can be replaced by a current source of output IN in

parallel with a resistance RN.(i) The output IN of the current source is equal to the current that would flow through AB when

terminals A and B are short circuited.(ii) The resistance RN is the resistance of the network measured between terminals A and B with

load (RL ) removed and sources of e.m.f. replaced by their internal resistances, if any.Nortons theorem is converse of Thevenins theorem in that Norton equivalent circuit uses a

current generator instead of voltage generator and resistance RN (which is the same as R0) in parallelwith the generator instead of being in series with it.

Illustration. Fig. 1.39 illustrates the application of Nortons theorem. As far as circuit behindterminals AB is concerned [See Fig. 1.39 (i)], it can be replaced by a current source of output IN inparallel with a resistance RN as shown in Fig. 1.39 (iv). The output IN of the current generator is equalto the current that would flow through AB when terminals A and B are short-circuited as shown inFig. 1.39 (ii). The load R on the source when terminals AB are short-circuited is given by :

R = 2 3 1 2 1 3 2 312 3 2 3

R R R R R R R RR

R R R R+ +

+ =+ +

Fig. 1.37

Introduction 21Source current, I = 2 3

1 2 1 3 2 3

( )V R RVR R R R R R R

+= + +

Short-circuit current, IN = Current in R2 in Fig. 1.39 (ii)

= 3 32 3 1 2 1 3 2 3

R V RI

R R R R R R R R =

+ + +

Fig. 1.39

To find RN, remove the load RL and replace the voltage source by a short circuit because itsresistance is assumed zero [See Fig. 1.39 (iii)].

RN = Resistance at terminals AB in Fig. 1.39 (iii).

= 1 321 3

R RR

R R+

+

Thus the values of IN and RN are known. The Norton equivalent circuit will be as shown inFig. 1.39 (iv).

1.16 Procedure for Finding Norton Equivalent Circuit(i) Open the two terminals (i.e. remove any load) between which we want to find Norton equiva-

lent circuit.(ii) Put a short-circuit across the terminals under consideration. Find the short-circuit current

flowing in the short circuit. It is called Norton current IN.(iii) Determine the resistance between the two open terminals with all ideal voltage sources

shorted and all ideal current sources opened (a non-ideal source is replaced by its internalresistance). It is called Nortons resistance RN. It is easy to see that RN = R0.

(iv) Connect IN and RN in parallel to produce Norton equivalent circuit between the twoterminals under consideration.

22 Principles of Electronics(v) Place the load resistor removed in step (i) across the terminals of the Norton equivalent

circuit. The load current can now be calculated by using current-divider rule. This load current will bethe same as the load current in the original circuit.

Example 1.12. Using Nortons theorem, find the current in 8 resistor in the network shown inFig. 1.40 (i).

Solution. We shall reduce the network to the left of AB in Fig. 1.40 (i) to Nortons equivalentcircuit. For this purpose, we are required to find IN and RN.

(i) With load (i.e., 8 ) removed and terminals AB short circuited [See Fig. 1.40 (ii)], thecurrent that flows through AB is equal to IN. Referring to Fig. 1.40 (ii),

Load on the source = 4 + 5 || 6

= 5 645 6

++

= 6.727

Source current, I = 40/6.727 = 5.94 A

Fig. 1.40

Short-circuit current in AB, IN =6

6 5I

+ = 5.94 6/11 = 3.24 A

(ii) With load (i.e., 8 ) removed and battery replaced by a short (since its internal resistance isassumed zero), the resistance at terminals AB is equal to RN as shown in Fig. 1.41 (i).

Fig. 1.41

RN = 5 + 4 || 6 = 5 + 4 64 6

+ = 7.4

The Nortons equivalent circuit behind terminals AB is IN (= 3.24 A) in parallel with RN (= 7.4 ).When load (i.e., 8 ) is connected across terminals AB, the circuit becomes as shown inFig. 1.41 (ii). The current source is supplying current to two resistors 7.4 and 8 in parallel.

Current in 8 resistor = 7.43.24 8 7.4

+ = 1.55 A

Introduction 23Example 1.13. Find the Norton equivalent circuit at terminals X Y in Fig. 1.42.Solution. We shall first find the Thevenin equivalent

circuit and then convert it to an equivalent current source. Thiswill then be Norton equivalent circuit.

Finding Thevenin Equivalent circuit. To find E0,refer to Fig. 1.43 (i). Since 30 V and 18 V sources are in opposi-tion, the circuit current I is given by :

I =30 18 1220 10 30

=+ = 0.4 A

Applying Kirchhoffs voltage law to loop ABCDA, we have,

30 20 0.4 E0 = 0 E0 = 30 8 = 22 V

(i) (ii)Fig. 1.43

To find R0, we short both voltage sources as shown in Fig. 1.43 (ii). Notice that 10 and 20 resistors are then in parallel.

R0 = 10 || 20 = 10 2010 20

+ = 6.67

Therefore, Thevenin equivalent circuit will be as shown in Fig. 1.44 (i). Now it is quite easy toconvert it into equivalent current source.

IN =0

0

226.67

ER

= = 3.3A [See Fig. 1.44 (ii)]

RN = R0 = 6.67

(i) (ii) (iii)

Fig. 1.44

Fig. 1.44 (iii) shows Norton equivalent circuit. Observe that the Norton equivalent resistance hasthe same value as the Thevenin equivalent resistance. Therefore, RN is found exactly the same way.

Fig. 1.42

24 Principles of ElectronicsExample 1.14. Show that when Thevenins equivalent circuit of a network is converted into

Nortons equivalent circuit, IN = E0 /R0 and RN = R0. Here E0 and R0 are Thevenin voltage andThevenin resistance respectively.

Solution. Fig. 1.45 (i) shows a network enclosed in a box with two terminals A and B broughtout. Thevenins equivalent circuit of this network will be as shown in Fig. 1.45 (ii). To find Nortonsequivalent circuit, we are to find IN and RN. Referring to Fig. 1.45 (ii),

IN = Current flowing through short-circuited AB in Fig. 1.45 (ii)= E0/R0

RN = Resistance at terminals AB in Fig. 1.45 (ii)= R0

Fig. 1.45 (iii) shows Nortons equivalent circuit. Hence we arrive at the following two importantconclusions :

(i) To convert Thevenins equivalent circuit into Nortons equivalent circuit,IN = E0/R0 ; RN = R0

Fig. 1.45

(ii) To convert Nortons equivalent circuit into Thevenins equivalent circuit,E0 = IN RN ; R0 = RN

1.17 Chassis and GroundIt is the usual practice to mount the electronic components on a metal base called chassis. For example,in Fig. 1.46, the voltage source and resistors are connected to the chassis. As the resistance of chassis isvery low, therefore, it provides a conducting path and may be considered as a piece of wire.

Fig. 1.46 Fig. 1.47

Introduction 25It is customary to refer to the chassis as ground. Fig. 1.47 shows the symbol for chassis. It may

be seen that all points connected to chassis are shown as grounded and represent the same potential.The adoption of this scheme (i.e. showing points of same potential as grounded) often simplifies theelectronic circuits. In our further discussion, we shall frequently use this scheme.

MULTIPLE-CHOICE QUESTIONS1. The outermost orbit of an atom can have a

maximum of .............. electrons.(i) 8 (ii) 6

(iii) 4 (iv) 32. When the outermost orbit of an atom has

less than 4 electrons, the material is gener-ally a ..............(i) non-metal (ii) metal

(iii) semiconductor (iv) none of above3. The valence electrons have ..............

(i) very small energy(ii) least energy

(iii) maximum energy(iv) none of the above

4. A large number of free electrons exist in..............(i) semiconductors (ii) metals

(iii) insulators (iv) non-metals5. An ideal voltage source has .............. inter-

nal resistance.(i) small (ii) large

(iii) infinite (iv) zero6. An ideal current source has .............. inter-

nal resistance.(i) infinite (ii) zero

(iii) small (iv) none of the above7. Maximum power is transferred if load

resistance is equal to .......... of the source.(i) half the internal resistance

(ii) internal resistance(iii) twice the internal resistance(iv) none of the above

8. Efficiency at maximum power transfer is..............(i) 75% (ii) 25%

(iii) 90% (iv) 50%9. When the outermost orbit of an atom has

exactly 4 valence electrons, the material isgenerally ..............(i) a metal (ii) a non-metal

(iii) a semiconductor(iv) an insulator

10. Thevenins theorem replaces a complicatedcircuit facing a load by an ..............

(i) ideal voltage source and parallel resistor(ii) ideal current source and parallel resistor(iii) ideal current source and series resistor(iv) ideal voltage source and series resistor

11. The output voltage of an ideal voltagesource is ..............(i) zero (ii) constant

(iii) dependent on load resistance(iv) dependent on internal resistance

12. The current output of an ideal current sourceis ..............(i) zero (ii) constant

(iii) dependent on load resistance(iv) dependent on internal resistance

13. Nortons theorem replaces a complicatedcircuit facing a load by an ..............(i) ideal voltage source and parallel resistor

(ii) ideal current source and parallel resistor(iii) ideal voltage source and series resistor(iv) ideal current source and series resistor

14. The practical example of ideal voltagesource is ..............(i) lead-acid cell (ii) dry cell

(iii) Daniel cell (iv) none of the above15. The speed of electrons in vacuum is

.............. than in a conductor.(i) less (ii) much more

(iii) much less (iv) none of the above16. Maximum power will be transferred from a

source of 10 resistance to a load of..............(i) 5 (ii) 20

(iii) 10 (iv) 40 17. When the outermost orbit of an atom has

more than 4 electrons, the material is gen-erally a ..............(i) metal (ii) non-metal

(iii) semiconductor (iv) none of the above

26 Principles of Electronics18. An ideal source consists of 5 V in series with

10 k resistance. The current magnitudeof equivalent current source is ..............(i) 2 mA (ii) 3.5 mA

(iii) 0.5 mA (iv) none of the above19. To get Thevenin voltage, you have to

..............(i) short the load resistor

(ii) open the load resistor(iii) short the voltage source(iv) open the voltage source

20. To get the Norton current, you have to..............(i) short the load resistor

(ii) open the load resistor(iii) short the voltage source(iv) open the voltage source

21. The open-circuited voltage at the terminalsof load RL in a network is 30 V. Under theconditions of maximum power transfer, theload voltage will be ..............(i) 30 V (ii) 10 V

(iii) 5 V (iv) 15 V22. Under the conditions of maximum power

transfer, a voltage source is delivering apower of 30 W to the load. The powerproduced by the source is ..............(i) 45 W (ii) 60 W

(iii) 30 W (iv) 90 W23. The maximum power transfer theorem is

used in ..............(i) electronic circuits

(ii) power system(iii) home lighting circuits(iv) none of the above

24. The Norton resistance of a network is 20 and the shorted-load current is 2 A. If thenetwork is loaded by a resistance equal to20 , the current through the load will be..............(i) 2 A (ii) 0.5 A

(iii) 4 A (iv) 1 A25. The Norton current is sometimes called the

..............(i) shorted-load current

(ii) open-load current(iii) Thevenin current(iv) Thevenin voltage

Answers to Multiple-Choice Questions1. (i) 2. (ii) 3. (iii) 4. (ii) 5. (iv)6. (i) 7. (ii) 8. (iv) 9. (iii) 10. (iv)

11. (ii) 12. (ii) 13. (ii) 14. (i) 15. (ii)16. (iii) 17. (ii) 18. (iii) 19. (ii) 20. (i)21. (iv) 22. (ii) 23. (i) 24. (iv) 25. (i)

Chapter Review Topics1. What is electronics ? Mention some important applications of electronics.2. Describe briefly the structure of atom.3. Explain how valence electrons determine the electrical properties of a material.4. Explain constant voltage and current sources. What is their utility ?5. Derive the condition for transfer of maximum power from a source to a load.6. State and explain Thevenins theorem.7. Write short notes on the following :

(i) Atomic structure (ii) Valence electrons (iii) Free electrons

Problems1. A dry battery developing 12 V has an internal resistance of 10 . Find the output current if load is

(i) 100 (ii) 10 (iii) 2 and (iv) 1 . [(i) 0.1A (ii) 0.6A (iii) 1A (iv) 1.1A]2. Convert the current source in Fig. 1.48 into the equivalent voltage source.

[36 V in series with 900 ]

Introduction 27

Fig. 1.48 Fig. 1.49

3. Convert the voltage source in Fig. 1.49 into equivalent current source. [3 mA in parallel with 8 k]4. Using Nortons Theorem, find the current in branch AB containing 6 resistor of the network shown

in Fig. 1.50. [0.466 A]

Fig. 1.50 Fig. 1.51

5. Fig. 1.51 shows Nortons equivalent circuit of a network behind terminals A and B. Convert it intoThevenins equivalent circuit. [2.56 V in series with 1.71 ]

Fig. 1.52 Fig. 1.53

6. A power amplifier has an internal resistance of 5 and develops open circuited voltage of 12 V. Findthe efficiency and power transferred to a load of (i) 20 (ii) 5 . [(i) 80%, 4.6 W (ii) 50%, 7.2 W]

7. Using Thevenins theorem, find the current through the galvanometer in the Wheatstone bridge shownin Fig. 1.52. [38.6 A]

8. Using Thevenins theorem, find the current through 4 resistor in the circuit of Fig. 1.53. [0.305A]

Discussion Questions1. Why are free electrons most important for electronics ?2. Why do insulators not have any free electrons ?3. Where do you apply Thevenins theorem ?4. Why is maximum power transfer theorem important in electronic circuits ?5. What are the practical applications of a constant current source ?

AdministratorStamp

28 Principles of Electronics

2.1 Electron Emission2.2 Types of Electron Emission2.3 Thermionic Emission2.4 Thermionic Emitter2.5 Commonly Used Thermionic Emitters2.6 Cathode Construction2.7 Field Emission2.8 Secondary Emission

2.9 Photo Electric Emission

Electron Emission

2

INTRINTRINTRINTRINTRODUCTIONODUCTIONODUCTIONODUCTIONODUCTION

The reader is familiar with the current conduction (i.e. flow of electrons) through a conduc-tor. The valence electrons of the conductor atoms are loosely bound to the atomic nuclei. Atroom temperature, the thermal energy in the conductor is adequate to break the bonds of thevalence electrons and leave them unattached to any one nucleus. These unbound electrons move atrandom within the conductor and are known asfree electrons. If an electric field is applied acrossthe conductor, these free electrons move throughthe conductor in an orderly manner, thus consti-tuting electric current. This is how these free elec-trons move through the conductor or electric cur-rent flows through a wire.

Many electronic devices depend for their op-eration on the movement of electrons in an evacu-ated space. For this purpose, the free electronsmust be ejected from the surface of metallic con- Electron Emission

AdministratorStamp

Electron Emission 29ductor by supplying sufficient energy from some external source. This is known as electron emission.The emitted electrons can be made to move in vacuum under the influence of an electric field, thusconstituting electric current in vacuum. In this chapter, we shall confine our attention to the variousaspects of electron emission.

2.1 Electron EmissionThe liberation of electrons from the surface of a substance is known as electron emission.

For electron emission, metals are used because theyhave many free electrons. If a piece of metal isinvestigated at room temperature, the random motionof free electrons is as shown in Fig. 2.1. However, theseelectrons are free only to the extent that they may transferfrom one atom to another within the metal but theycannot leave the metal surface to provide electronemission. It is because the free electrons that start atthe surface of metal find behind them positive nucleipulling them back and none pulling forward. Thus atthe surface of a metal, a free electron encounters forcesthat prevent it to leave the metal. In other words, themetallic surface offers a barrier to free electrons and isknown as surface barrier.

However, if sufficient external energy is given to the free electron, its kinetic energy is increasedand thus electron will cross over the surface barrier to leave the metal. This additional energy requiredby an electron to overcome the surface barrier of the metal is called work function of the metal.

The amount of additional energy required to emit an electron from a metallic surface is knownas work function of that metal.

Thus, if the total energy required to liberate an electron from a metal is 4 eV* and the energyalready possessed by the electron is 0.5 eV, then additional energy required (i.e., work function) is4.0 0.5 = 3.5 eV. The work function of pure metals varies roughly from 2 to 6 eV. It depends uponthe nature of metal, its purity and the conditions of its surface. It may be noted that it is desirable thatmetal used for electron emission should have low work function so that a small amount of energy isrequired to cause emission of electrons.

2.2 Types of Electron EmissionThe electron emission from the surface of a metal is possible only if sufficient additional energy(equal to the work function of the metal) is supplied from some external source. This external energymay come from a variety of sources such as heat energy, energy stored in electric field, light energy orkinetic energy of the electric charges bombarding the metal surface. Accordingly, there are followingfour principal methods of obtaining electron emission from the surface of a metal :

* Work function is the additional energy required for the liberation of electrons. Therefore, it should havethe conventional unit of energy i.e. joules. But this unit is very large for computing electronics work.Therefore, in practice, a smaller unit called electron volt (abbreviated as eV) is used.One electron-volt is the amount of energy acquired by an electron when it is accelerated through a poten-tial difference of 1 V.Thus, if an electron moves from a point of 0 potential to a point of +10V, then amount of energy acquiredby the electron is 10 eV.

Since charge on an electron = 1.602 1019 C and voltage = 1 V, 1 electron-volt = Q V = (1.602 1019) 1 Jor 1 eV = 1.602 1019 J

Fig. 2.1

30 Principles of Electronics(i) Thermionic emission. In this method, the metal is heated to sufficient temperature (about

2500C) to enable the free electrons to leave the metal surface. The number of electrons emitteddepends upon the temperature. The higher the temperature, the greater is the emission of electrons.This type of emission is employed in vacuum tubes.

(ii) Field emission. In this method, a strong electric field (i.e. a high positive voltage) isapplied at the metal surface which pulls the free electrons out of metal because of the attraction ofpositive field. The stronger the electric field, the greater is the electron emission.

(iii) Photo-electric emission. In this method, the energy of light falling upon the metal surfaceis transferred to the free electrons within the metal to enable them to leave the surface. The greater theintensity (i.e. brightness) of light beam falling on the metal surface, the greater is the photo-electricemission.

(iv) Secondary emission. In this method, a high velocity beam of electrons strikes the metalsurface and causes the free electrons of the metal to be knocked out from the surface.

2.3 Thermionic EmissionThe process of electron emission from a metal surface by supplying thermal energy to it is known asthermionic emission.

At ordinary temperatures, the energy possessed by free electrons in the metal is inadequate tocause them to escape from the surface. When heat is applied to the metal, some of heat energy isconverted into kinetic energy, causing accelerated motion of free electrons. When the temperaturerises sufficiently, these electrons acquire additional energy equal to the work function of the metal.Consequently, they overcome the restraining surface barrier and leave the metal surface.

Metals with lower work function will require less additional energy and, therefore, will emitelectrons at lower temperatures. The commonly used materials for electron emission are tungsten,thoriated tungsten and metallic oxides of barium and strontium. It may be added here that hightemperatures are necessary to cause thermionic emission. For example, pure tungsten must be heatedto about 2300C to get electron emission. However, oxide coated emitters need only 750C to causethermionic emission.

Richardson-Dushman equation. The amount of thermionic emission increases rapidly as theemitter temperature is raised. The emission current density is given by Richardson-Dushman equa-tion given below :

Js = A T2

bTe

amp/m2 ...(i)

where Js = emission current density i.e. current per square metre of theemitting surface

T = absolute temperature of emitter in KA = constant, depending upon the type of emitter and is measured

in amp/m2/K2

b = a constant for the emittere = natural logarithmic base

The value of b is constant for a metal and is given by :

b =e

k

where = work function of emittere = electron charge = 1.602 1019 coulombk = Boltzmanns constant = 1.38 1023 J/K

b =19

231.602 10

1.38 10

= 11600 K

Electron Emission 31Putting the value of b in exp. (i), we get,

Js =11600

2 TAT e ...(ii)

The following points may be noted from eqn. (ii) :(i) The emission is markedly affected by temperature changes. Doubling the temperature of an

emitter may increase electron emission by more than 107 times. For instance, emission from puretungsten metal is about 10 6 ampere per sq. cm. at 1300C but rises to enormous value of about 100amperes when temperature is raised to 2900C.

(ii) Small changes in the work function of the emitter can produce enormous effects on emis-sion. Halving the work function has exactly the same effect as doubling the temperature.

Example 2.1. A tungsten filament consists of a cylindrical cathode 5 cm long and 0.01 cm indiameter. If the operating temperature is 2500 K, find the emission current. Given thatA = 60.2 104 A/m2 / K2, = 4.517 eV.

Solution. A = 60.2 104 amp/m2/K2, T = 2500 K, = 4.517 eV b = 11600 K = 11600 4.517 K = 52400 KUsing Richardson-Dushman equation, emission current density is given by :

Js = 2bTAT e

amp/m2 = 60.2 104 (2500)2 524002500(2.718)

= 0.3 104 amp/m2

Surface area of cathode, a = d l = 3.146 0.01 5 = 0.157 cm2 = 0.157 104 m2

Emission current = Js a = (0.3 104) (0.157 104) = 0.047 A

Example 2.2. A tungsten wire of unknown composition emits 0.1 amp/cm2 at a temperature of1900 K. Find the work function of tungsten filament. Determine whether the tungsten is pure orcontaminated with substance of lower work function. Given that A = 60.2 amp/cm2/K2.

Solution. Js = 0.1 amp/cm2 ; A = 60.2 amp/cm2/K2 ; T = 1900 K

Let electron-volt be the work function of the filament. b = 11600 KUsing Richardson-Dushman equation, emission current density is given by :

Js = 2bTA T e

amp/cm2

or 0.1 = 60.2 (1900)2 11600

1900e

or11600

1900e = 2

0.160.2 (1900) = 4.6 10

10

or e6.1 = 4.6 1010

or 6.1 loge e = loge 4.6 10 loge 10or 6.1 = 1.526 23.02

=1.526 23.02

6.1

= 3.56 eV

Since the work function of pure tungsten is 4.52 eV, the sample must be contaminated. Thoriatedtungsten has a work function ranging from 2.63 eV to 4.52 eV, depending upon the percentage ofmetallic thorium. Therefore, the sample is most likely to be thoriated tungsten.

32 Principles of Electronics

Thermionic Emitter

2.4 Thermionic EmitterThe substance used for electron emission is known as an emitter or cathode.The cathode is heated in an evacuated space to emit electrons. If the cathodewere heated to the required temperature in open air, it would burn up because ofthe presence of oxygen in the air. A cathode should have the following properties:

(i) Low work function. The substance selected as cathode should havelow work function so that electron emission takes place by applying small amountof heat energy i.e. at low temperatures.

(ii) High melting point. As electron emission takes place at very hightemperatures (>1500C), therefore, the substance used as a cathode should havehigh melting point. For a material such as copper, which has the advantage of alow work function, it is seen that it cannot be used as a cathode because it meltsat 810C. Consequently, it will vaporise before it begins to emit electrons.

(iii) High mechanical strength. The emitter should have high mechanicalstrength to withstand the bombardment of positive ions. In any vacuum tube, no matter how carefulthe evacuation, there are always present some gas molecules which may form ions by impact withelectrons when current flows. Under the influence of electric field, the positive ions strike the cath-ode. If high voltages are used, the cathode is subjected to considerable bombardment and may be dam-aged.

2.5 Commonly Used Thermionic EmittersThe high temperatures needed for satisfactory thermionic emission in vacuum tubes limit thenumber of suitable emitters to such substances as tungsten, thoriated tungsten and certain oxidecoated metals.

(i) Tungsten. It was the earliest material used as acathode and has a slightly higher work function (4.52 eV).The important factors in its favour are : high melting point(3650 K), greater mechanical strength and longer life. Thedisadvantages are : high operating temperature (2500 K),high work function and low emission efficiency. Therefore,it is used in applications involving voltages exceeding 5 kVe.g. in X-ray tubes.

(ii) Thoriated tungsten. A mixture of two metals mayhave a lower work function than either of the pure metalsalone. Thus, a tungsten emitter with a small quantity ofthorium has a work function of 2.63 eV, compared with 3.4eV for thorium and 4.52 eV for tungsten. At the same time, thoriated tungsten provides thermionicemission at lower temperature (1700C) with consequent reduction in the heating power required.

In the manufacture of this type of cathode, tungsten filament is impregnated with thorium oxideand heated to a very high temperature (1850C to 2500C). The thorium oxide is reduced to metallicthorium and coats the filament surface with a thin layer of thorium. Thoriated tungsten cathodes areused for intermediate power tubes at voltages between 500 to 5000 volts.

(iii) Oxide-coated cathode. The cathode of this *type consists of a nickel ribbon coated with

Thoriated Tungsten

* Oxides of any alkaline-earth metal (e.g. calcium, strontium, barium etc.) have very good emissioncharacteristics. In the manufacture of this type of emitter, the base metal (e.g. nickel) is first coated with amixture of strontium and barium carbonates. It is then heated to a high temperature in vacuum glass tubeuntil the carbonates decompose into oxides. By proper heating, a layer of oxides of barium and strontiumis coated over the cathode surface to give oxide-coated emitter.

Electron Emission 33barium and strontium oxides. The oxide-coated cathode has low work function (1.1 eV), operates atcomparatively low temperature (750C) and has high emission efficiency. However, the principallimitation of oxide-coated cathode is that it cannot withstand high voltages. Therefore, it is mostlyused in receiving tubes or where voltages involved do not exceed 1000 volts.

S.No. Emitter Work Function Operating Emissiontemperature efficiency

1 Tungsten 4.52 eV 2327C 4 mA/watt2 Thoriated tungsten 2.63 eV 1700C 60 mA/watt3 Oxide-coated 1.1 eV 750C 200 mA/watt

2.6 Cathode ConstructionAs cathode is sealed in vacuum, therefore, the most convenient way to heat it is electrically. On thisbasis, the thermionic cathodes are divided into two types viz directly heated cathode and indirectlyheated cathode.

(i) Directly heated cathode. In this type, the cathode consists of oxide-coatednickel ribbon, called the *filament. The heating current is directly passed through this ribbon whichemits the electrons. Fig. 2.2 (i) shows the structure of directly heated cathode whereas Fig. 2.2 (ii)shows its symbol.

Fig. 2.2

The directly heated cathode is more efficient in converting heating power into thermionic emission.Therefore, it is generally used in power tubes that need large amounts of emission and in small tubesoperated from batteries where efficiency and quick heating are important. The principal limitation ofthis type of cathode is that any variation in heater voltage affects the electron emission and thusproduces hum in the circuit.

(ii) Indirectly heated cathode. In this type, the cathode consists of a thin metal sleeve coatedwith barium and strontium oxides. A filament or heater is enclosed within the sleeve and insulatedfrom it. There is no electrical connection between the heater and the cathode. The heating current ispassed through the heater and the cathode is heated indirectly through heat transfer from the heaterelement. Fig. 2.3 (i) shows the structure of indirectly heated cathode whereas Fig. 2.3 (ii) shows itssymbol.

* Filament. The term filament (literally means a thin wire) denotes the element through which the cathodeheating current flows. In case of directly heated, cathode is itself the filament. If indirectly heated, heateris the filament.

34 Principles of Electronics

Fig. 2.3

Indirectly heated cathode has many advantages. Ascathode is completely separated from the heating circuit,therefore, it can be readily connected to any desiredpotential as needed, independent of the heater potential.Furthermore, because of relatively large mass of cylindricalcathode, it takes time to heat or cool and as such does notintroduce hum due to heater voltage fluctuations. Finally,a.c. can be used in the heater circuit to simplify the powerrequirements. Almost all modern receiving tubes use thistype of cathode.

2.7 Field EmissionThe process of electron emission by the application ofstrong electric field at the surface of a metal is known asfield emission.

When a metal surface is placed close to a high voltageconductor which is positive w.r.t. the metal surface, theelectric field exerts attractive force on the free electronsin the metal. If the positive potential is great enough, itsucceeds in overcoming the restraining forces of the metalsurface and the free electrons will be emitted from themetal surface as shown in Fig. 2.4.

Very intense electric field is required to produce field emission. Usually, a voltage of the orderof a million volts per centimetre distance between the emitting surface and the positive conductor isnecessary to cause field emission. Field emission can be obtained at temperatures much lower (e.g.room temperature) than required for thermionic emission and, therefore, it is also sometimes calledcold cathode emission or auto- electronic emission.

2.8 Secondary EmissionElectron emission from a metallic surface by the bombardment of high-speed electrons or otherparticles is known as secondary emission.

Fig. 2.4

Electron Emission 35

Fig. 2.6

* An interesting aspect of secondary emission is that a high-speed bombarding electron may liberate as manyas 10 secondary electrons. This amounts to a multiplication of electron flow by a ratio as great as 10 andis utilised in current multiplier devices.

When high-speed electronssuddenly strike a metallic surface,they may give some or all of theirkinetic energy to the free electronsin the metal. If the energy of thestriking electrons is sufficient, itmay cause free electrons to escapefrom the metal surface. Thisphenomenon is called secondaryemission. The electrons that strikethe metal are called primaryelectrons while the emittedelectrons are known as secondaryelectrons. The intensity ofsecondary emission depends uponthe emitter material, mass andenergy of the bombardingparticles.

The principle of secondaryemission is illustrated in Fig. 2.5.An evacuated glass envelope con-tains an emitting surface E, the collecting anodeA and a source of primary electro