CONTENTSTo open a chapter, click on the chapter name.To turn pages within a chapter, use the keyboard arrow keys.

To navigate within a chapter, use the navigation buttons on each page. ELECTRICITYChapter 1: Electronic Components and CircuitsChapter 2: Properties of Capacitors and InductorsChapter 3: Electrical Properties of SkinChapter 4: Electrical Stimulation of Nerve and MuscleChapter 5: Rectification and AmplificationFIELDSChapter 6: Electric and Magnetic FieldsChapter 7: Therapeutic Fields: Shortwave DiathermyChapter 8: Non-Diathermic FieldsWAVESChapter 9: SoundandElectromagneticWavesChapter 10: TherapeuticWaves:UltrasoundChapter 11: ElectromagneticWavesforTherapyGENERALChapter 12: DosageandSafetyConsiderationsChapter 13: ElectricalSafetyAppendices: Quantities, Units and Prefixes COMPONENTS AND CIRCUITS 1Although the central theme ofthis chapter is electroniccomponents and hardware,most of the concepts areneeded to understand therelevantphysiologyandelectrotherapy covered inlater chapters.1 Electronic Components and CircuitsThoughtheuseofelectricityinamedicalcontextisnorecentdevelopment.thelastseveraldecadeshaveseenanunparalleledgrowthinelectronictechnologyanditsapplicationinmedicine.Sophisticatedelectronicapparatusisnowascommonplaceinthehospitalandclinicalsettingasitpreviouslywasintheresearchlaboratory.Itisevidentthatallhealthprofessionalsmustbecomemoreandmoreaccustomedtotheuseofelectronicinstrumentationanditsproperplaceinthepracticeoftheirprofession.The physiotherapist makes use of such apparatus for a variety of diagnostic proceduresand treatments, collectively described by the terms electrodiagnosis and electrotherapy.The aim here is to provide a physical basis for electrodiagnosis and electrotherapy.Webeginbydiscussingthe'nutsandbolts'ofelectronicapparatus,thebasiccomponentswhich go to make up an electronic circuit.a third, smaller groupcomprisedofmaterialsdisplayingintermediateconduction properties: thisgroup includes thesemiconductors which formthe basis of modern electronics technology.Wewill talk more aboutsemiconductors in chaptertwo.Apart from conductorsand insulators there isElectroniccircuitsandelectroniccomponentsowetheirexistencetothefactthattheeverydaymaterialswithwhichwearefamiliarhavequitedissimilarelectricalproperties.Thevastmajorityofmaterialscanbereadilycategorizedasbelongingtooneoftwogroups:eitherconductorsorinsulators.Firstletusconsiderconductorsandinsulatorsand some additional concepts which are fundamental to an understanding of electricity.SOME FUNDAMENTALSConductors are familiar to us in everyday life: the filament of a light bulb, the copper wireconnectedtocarbatteriesandcopperwireinthecablesleadingtomostelectricalappliancesareafewexamples.Ifwepictureatomsasconsistingofapositivelychargednucleustogetherwithelectron'shells'surroundingthenucleusthentheatomsof conductors have a characteristic feature.Goodconductors(iron,copperandothermetalsareexamples)sharethecommonfeature that the outermost electrons of the atoms are only loosely bound to the nucleus.Forthisreasontheycanreadilytransmitelectrons.AgoodwaytopicturethisistoCOMPONENTS AND CIRCUITS 2imagineafairlyrigidlatticeofatomicnucleiwithaseaofelectronscontained within the boundaries of the piece of metal.An electron entering one end of a length of wire causes a 'ripple in the seaof electrons' and, a split second later, an electron is ejected from the otherend.Aflowofchargesthroughanymaterialiswhatwecallanelectriccurrentand materials which readily transmit charges are termed conductors.Thestandardunitofcurrentflowinthesystemeinternationale(Sl)istheampere(abbreviatedampandgiventhesymbolA).Thecurrentinamperesisameasureofthequantityofchargeflowingthroughaconductor each second.The unit is named after the noted French physicistAndr Marie Ampre.Insulatorshavetheirelectronstightlyboundintheirshellssothattheprocessofconductiondescribedabovecannottakeplaceundernormalconditions.Plasticsandceramicsaregoodinsulators;thuswecanusecopper wires coated in plastic for the power leads to electric toasters, jugs,TVsetsandsoon.Thetelegraphpolesinthestreethavethewiresfastened not to the wood but around small white ceramic insulators.Inordertocauseaflowofcurrentthroughaconductorweneedadevicewhichwillproduceanexcessofelectronsatoneendofawireandadeficiencyattheother:suchadeviceisasourceofelectricalpotentialenergyandissaidtoproduceapotentialdifference.Atypicalexampleofone of these devices is the battery found in a torch or transistor radio.Thetwo terminals of a battery are at a different electrical potential - that is, thereis a potential difference between the terminals.If a conductor is connectedbetweentheterminalselectronsmovefromtheterminalatahighpotentialto that at low potential.TheSlunitofpotentialdifferenceisthevolt(symbolV).Oftenthetermsa metal consists of a rigid and regular, crystalline array of metal ions ().A 'sea' or cloud of electrons ( ) fills the spaces between the ions.COMPONENTS AND CIRCUITS 3potential difference and voltage are used interchangeably.Just as we can have asubstantial pressure in thewater mains without any flowof water through a tap, so wecan have a substantialpotential difference withoutany flow of current.It is important to clearly understand the difference between voltage and current.Usingthe analogy with water in pipes we can liken voltage to the water pressure and currentto the volume rate of flow of water.Ifweconnectapieceofmaterialbetweentheterminalsofabatterytheamountofcurrent flow will depend on the battery voltage and the resistance to current flow whichthematerialoffers.Goodconductorsofferlittleresistancetocurrentflowwhileinsulatorsoffersubstantialresistance.TheSlunitofresistanceistheohm(abbreviated).Wesaysomethinghasaresistanceofoneohmwhenapotentialdifference of one volt produces a current of one amp through it.A good conductor suchasametrelengthofhouseholdmainswirewouldhavearesistancemeasuredinmilliohmswhileagoodinsulator,ablockofceramic,wouldhavearesistanceofseveralthousandmegohms.Resistancevaluesusedinelectroniccircuitsaretypically measured in ohms (), kilohms (k) or megohms (M).1 k = 103 = 1000 1 M = 106 =1 000 000 COMPONENTSAnelectronicdevicesuchasaradio,TVsetorCDplayercanbeverycomplexbutthecomplexityliesinthearrangementandtotalnumberofcomponents.Whenweexamineatypicalcircuitwefindonlyafewdifferent kinds of components.Resistorsarethemostcommoncircuitcomponents:theycomeinavarietyofshapesandsizesandusuallyhavetheirvaluescodedintheformofthreeorfourcolouredstripesonthebody.Highpowerresistorsarelargerandusuallyhavetheresistanceandpowerratingstampedonthe body.Rheostatsandpotentiometersarevariableresistorshavingtwo or three terminals respectively.COMPONENTS AND CIRCUITS 4Capacitorscomeinagreaterrangeofsizesandshapesthanresistorsbutcan readily be distinguished with a little practice.fixed value capacitorsTheyconsistoftwometalplates(usuallyaluminium)andaninsulator.Theinsulator may be air (in the case of variable capacitors), mica or a plastic film.Thedifferentkindsofcapacitorarenamedaftertheinsulatorused,thuswehavemicacapacitors,polyestercapacitors,ceramiccapacitorsandsoon.Typical examples of fixed value capacitors are shown, together with a variablecapacitorsuchasmightbeusedinthetuningsectionofaradio.Fortypicalvaluesofcapacitancelargeareasofmetalareinvolvedanditisconvenienttosavespacebymakingtheplates of thin aluminium foil and rolling them into a cylinder.Thusatubularcapacitorhastheinternalconstructionshown.Wewillreturntodiscusswhatacapacitordoesinthenextchapter.The unit of capacitance is the farad (F) but the values foundintypicalcircuitsaremeasuredinmicrofarads(F),COMPONENTS AND CIRCUITS 5nanofarads (nF) or picofarads (pF).1 F = 11 000 000 F = 10-6 F 1 nF = 11 000 F = 10-9 F1 pF = 11 000 000 F = 10-12 FElectrolyticcapacitorsarepolarized;thatis,theyhavedefinitepositiveandnegativeterminals and can only be connected one way around in a circuit.Ordinary capacitors(mica, polyester, etc.) are non-polarized and can be connected either way around.Inductorsandtransformersformourthirdcategoryofcomponents.Aninductorissimplyacoilofwire:thewiremaybewoundonvariouskindsofcore,dependingonthe specific role for which it is intended.An example is shown.Theunitofinductanceisthehenry(H).Youwillalsoseethetermsmillihenry(mH)and microhenry (H) used frequently.1 mH = 11000H = 10-3 H 1 H = 11000mH = 10-6 HWhentwoinductorsarewoundonthesamecoreincloseproximityoroverlapping we have a transformer.Valves,TransistorsandDiodesarethe'workhorses'ofanyelectroniccircuit,thedeviceswhichhavepermittedthedevelopmentoflongdistancevoicecommunication,radioandtelevision,computersandspaceexploration, guided missiles and the reproduction of music to name a few ofthemoreobviousapplications.Todayvalvesareseldomusedexceptforspecialapplications:theyarebulkyandinefficientandhavelargelybeensupersededbytransistors,theirsemiconductorequivalents.Thetransistorperformsmuchthesamejobasa valve but is physically much smaller.COMPONENTS AND CIRCUITS 6MostelectronicequipmenttodayusesIntegratedCircuits(IC's),smalldeviceshaving8ormorepinsandwhichcontaincompletecircuitshavingmanytransistors,diodes,resistorsandcapacitorsfabricateddirectlyinasinglepackage.Thepins are used for access to various points in the circuit so thatthedesignercantailorthecircuittospecificrequirementsbyconnectiontoexternalcomponents.Integratedcircuitshavepermittedafurtherreductioninthephysicalsizeofcomplexcircuitscomparabletothereductionthatwasachievedbythereplacementofvalvecircuitrywithtransistors.Thepocketcalculator,desktopcomputerandspacesatelliteshaveallbeenmadepossiblebytheminiaturizationpermittedwithintegrated circuits.ClRCUITSTheelectroniccomponentswehavemetsofararetypicallyfound with a complex maze of interconnections between them.Theparticularwayinwhichtheyareconnecteddefinesacircuit and the form that the circuit takes will depend on the jobitisdesignedtodo.Ifweconsider,forexample,aradio,thejobithastoperformisexceedinglycomplex.Itmustpickupradiotransmissions,converttheradiowavestoelectricalsignals,amplifythemandconvertthemtoaudiblesoundwaves.Not only that but it must also be capable of selecting aparticularfrequencyofradiowave(theonefromtheradiostationyouwishtotune-inon)andignoringtheremainder.Needlesstosay,thecircuitryrequiredtoperformthesetasksis very complex.In the next section we will consider extremely simple circuits inordertoexaminethecharacteristicbehaviourofsomeofthecomponentsdiscussedsofar.Butfirstafewwordsoncircuits themselves and their physical construction.COMPONENTS AND CIRCUITS 7The earliest circuits, those found for example in a valve radio receiver, were made upofalargenumberofbulkycomponents-thetransformer,valves,etc.,andthecircuitinterconnectionsreflectedthisbulkiness.Largecomponentssuchasthetransformerwereboltedtoametalframeorchassisalongwithsocketsforthevalves,andthecircuitwasbuilt-upbyinterconnectingwithsmallercomponents(resistorsandcapacitors) and lengths of wire.With the advent of the transistor and the consequent trend to miniaturization which thispermitted,itwasfoundtobemoreconvenientandsimplertoreplacemostofthewiringwithcopperstripsfirmlyattachedtoaninsulatingboardhavingholesdrilledinappropriateplaces.Alloftheminiaturecomponents(transistors,resistors,capacitors)couldthenbemountedontheboardandsolderedtothecopperconductors.Only a limited number of bulky components (transformer, volume controlandspeakerinamains-poweredradio)needthentobemountedonachassisorcasing.Circuitsassembledontheseprintedcircuithoardscanberapidlymassproducedandeasilyputtogether.Consequentlytheyareusedalmostexclusivelybymanufacturers.Whenintegratedcircuitsareusedratherthantransistorsafurtherreductioninsizeisachieved.Theprintedcircuitboardscanbemadesmallerasthel.C.'s themselves are small and relatively few external components are required.SIMPLE CIRCUITS WITH RESISTORSPerhapsthesimplestkindofcircuitwhichcanbeconstructedcomprisesabattery(asourceofelectricalenergy)andasingleelectricalcomponentsuchasaresistororatorch globe.First consider the simple circuit arrangement shown in figure 1.1.Ontheleftisasmalllamp(torchglobe)connectedbymeansofwirestoadrycell(battery).Ontherightisthecircuitdiagramofthearrangement:notethecircuitsymbolsusedforthelampanddrycell.Whenconnectedthiswaycurrentwillflowfrom one terminal of the battery, through the lamp and back to the opposite terminal.Itiscommonconventiontosaythatcurrentflowsfromthepositiveterminalofthebatterytothenegativeterminal;althoughthisistheoppositedirectiontotheflowofCOMPONENTS AND CIRCUITS 8electrons.The difference between reality and convention is not important as it has noeffect on the magnitudes of currents and voltages in a circuit.Figure 1.1A very simple circuit and itsrepresentationusingacircuitdiagramUndernormalcircumstancestheflowofcurrentthroughthelampwillcauseittoglow-thenormalpurposeofsuchacircuit.If a break occurs in thewireforexample,orthewireisnotsecuredtothebattery,currentwillnotflowandwehave an open circuit.If the wireconnectedtooneterminalofthelampcomeslooseandtouchestheotherwire,currentcanflowdirectlythroughthewiresandbypassthelamp-thisconditionisknownasashortcircuit.Theswitchonatorchisjustameansofintroducinganopencircuitconditionatwill:preventingtheflowofcurrentandthusconservingtheelectricalenergy stored in the battery.Goingonestepfurtherthanthecircuitshowninfigure1.1wemayaddaresistortoproducethecircuitsshowninfigure1.2.Notethecircuitsymbol(zig-zagline)foraresistor.Figure 1.2(a) shows the resistor in series with the lamp, i.e. the resistor and lamp areconnectedendtoendsothatcurrentmustflowthroughboththelampandtheresistor.In1.2(b)theresistorisconnectedinparallelwiththelampsothatsomecurrent can flow through the lamp and some through the resistor.The resistor is thusallowing part of the current to bypass the lamp and return to the opposite terminal ofthe battery.COMPONENTS AND CIRCUITS 9Figure 1.2Two less simple circuits.(a) aseries and (b) a parallelarrangement.RESISTORS IN SERIESTwoormoreresistorsmaybeconnectedinseriesorinparallel.Whentheresistorsareconnectedinseriesasinfigure1.3thecurrentmustpassthrougheach resistance in turn and the total resistance offered by the circuit is equal to thesumoftheindividualresistances.Ingeneralforacircuithavingnresistorsconnectedinseriesthetotalresistanceof the circuit, R, is equal to the sum of the individual resistances i.e.R = R1 + R2 + R3 ........ + Rn.... (1.1)Asmentionedpreviously,thecurrentflowingisameasureoftherateofflowofcharges around the circuit and is measured in amperes, or amps for short.Sincetherecanbenoaccumulationofchargeswithinaresistorthecurrentflowingintoa certain resistor is equal to the current flowing out.Thus the same current mustflow through each resistor when they are connected in series.Althoughthecurrentineachresistoristhesame,thepotentialdifferenceacrosseachresistorinaseriescircuitis,ingeneral,different.Thesumofthepotentialdifferencesacrosstheresistorsisequaltothepotentialdifferencegeneratedbythe battery (the battery voltage).Figure 1.3ResistorsinseriesCOMPONENTS AND CIRCUITS 10RESISTORS IN PARALLELWhenanumberofresistorsareconnectedinparallelasinfigure1.4thecurrentflowingineachresistormaybedifferent.Inthiscasethetotalresistance of the circuit, R, is given by the equation1R=1R1 + 1R2 + 1R3or more generally, where there are n resistors in parallel in a circuit1R=1R1 + 1R2 + 1R3 ..... + 1Rn.... (1.2)Althoughthecurrentineachresistormaybequitedifferent,thevoltageorpotentialdifferenceacrosseachresistoristhesameandisequaltothebatteryvoltage.Thismustbesosincetheendsofeachresistorareconnecteddirectlytotheterminalsofthebattery.ThelargestamountofcurrentwillflowthroughthesmallestresistancewhenresistorsareFigure 1.4Resistorsinparallelconnected in parallel.OHM'S LAWTherelationshipbetweenpotentialdifference(V),resistance(R)andcurrent(I)througharesistorissummarizedbyOhm'slawwhichstatesthatthecurrentflowingthrough a resistor is proportional to the potential difference across the resistor and isinversely proportional to the resistance.The equivalent mathematical expression of Ohm's law isI=VR.... (1.3)COMPONENTS AND CIRCUITS 11whereIisthecurrentinamperes,VisthepotentialdifferenceinvoltsandRistheresistanceinohms.Example 1.If, in figure 1.3, R1 is 50 ohms, R2 is 100 ohms and R3 is 150 ohms thenthe total resistance R isR=50 + 100 + 150=300 If these are connected to a 10 volt battery the current flow will beI=VR= 10300=0.033 amps or 33 milliamps.Here we have calculated the total current drawn from the battery which is equal to thecurrent flowing through each resistor.The battery gives a total potential difference of10voltsacrossthecombinationofresistors.Ifwewishtocalculatethepotentialdifferenceacrosseachindividualresistorweagainuseequation1.3butapplytheequation to each resistor in turn.For example, for the 50 ohm resistor we know thatI=0.033 amps and R=50 soI=VRbecomesVR=0.033and henceV=50 x 0.033=1.7 volts.For the 100 ohm resistorV=100 x 0.033=3.3 voltsand for the 150 ohm resistorV = 150 x 0.033 = 5.0 volts.ThetotalpotentialdifferenceacrossthecombinationofresistorsisV=1.7+3.3+5.0 = 10 volts as expected.COMPONENTS AND CIRCUITS 12Example2.Nowconsiderthesamethreeresistorsasinthepreviousexampleconnectedinparalleltoa10voltbatteryasinfigure1.4.Thetotalresistance, R, is now given by1R=150 + 1100 + 1150=6300 + 3300 + 2300=11300so R=30011=27 ohmsThe total current is I=VR=1027= 0.37 ampsTocalculatethecurrentineachindividualresistorweusethefactthatthepotentialdifferenceacrosseachresistoristhefull10volts.Thenforthe50ohm resistor we haveI=VR=1050=0.20 amps.For the 100 ohm resistorI=10100=0.10 ampsand for the 150 ohm resistorI=VR=10150=0.07 ampsSo the total current drawn from the battery isI=0.20 + 0.10 + 0.07=0.37 amps as calculated previously.Thecalculationsaremaderelativelysimplebyrememberingthatwhenresistorsareconnectedinseriesthecurrentthrougheachisthesameandwhenresistorsareconnectedinparallelthepotentialdifferenceacrosseachisthesame.COMPONENTS AND CIRCUITS 13Starting with the expressionP = V.I and substituting V = I.Rwe haveP = (I.R).I orP = IR 2Starting with the expressionP = V.I and substituting I = V/Rwe haveP = V.(V/R) orP =2VRPOWER DISSIPATION IN RESISTORSWhenelectriccurrentflowsthroughaconductor,whethertheconductorisaresistor,copperwire,thefilamentofalightbulborwhatever,electricalenergyisconvertedtoheatenergyintheconductor.Anextremeexampleofthisisseenwithalamp-theheatgeneratedissogreatthatthefilamentwireglowsbrightly.Thesameheatingeffecttakesplace,thoughtoalesserextent,inaresistorandclearlytherewillbealimit to how much heat a resistor can generate per second before it starts glowing anddisintegratesormelts.Themaximumrateatwhicharesistorcansafelydissipateelectrical energy is called the power rating of the resistor and this depends on the sizeand physical construction.Theactualrateofdissipationofelectricalenergyatanyparticulartimeisthepowerdissipated.Thisdependsonboththecurrentflowingthroughtheresistorandthepotential difference across it.The power (in watts) is equal to the product of current (inamps) and potential difference (in volts).If we denote power by the symbol P the relationship may be writtenP=V.I.... (1.4)This formula can be usefully used with Ohm's law to calculate the maximum valuesfor voltage or current for a particular resistor of known resistance and power rating.Alternatively if we know any two of the values of V, I or R for a given resistor, we cancalculatethepowerdissipatedintheresistorusingequation1.4orthetwoexpressions obtained by solving equations 1.3 and 1.4.These are:P=V2R.... (1.5)where we have eliminated I in solving the two expressions andP=I2.R .... (1.6)where we have eliminated V.COMPONENTS AND CIRCUITS 14Withthesethreealternativeformulaeforcalculatingpowerwewouldchoosetheequation which gives the required answer with a minimum of calculations.Considerthethreeresistorsconnectedinseriesasinexample1previously.Ifwewishtocalculatethepowerdissipatedineachresistorwewouldfirstcalculatethecurrent (which is the same for each resistor) and then use equation 1.6.In example 2previouslytheresistorsareinparallelandthepotentialdifferenceisthesameforeach resistor.Equation 1.5 is thus the choice for power calculation.RESISTORS IN A SERIES/PARALLEL COMBINATIONTheideaswehaveconsideredpreviouslycanbeappliedtomorecomplexcircuitswhereresistorsarecombinedinaseries/parallelarrangement.Figure1.5showsonepossibleseries/parallelcombination.TocalculatethetotalresistanceinthiscircuitwewouldfirstneedtocalculatetheresistanceofR2 and R3inparallelusingequation1.2.ThetotalresistanceisobtainedbyaddingthistoR1.ThisisbecauseR1isinserieswiththeparallelcombination of R2 and R3 so equation 1.1 applies.To calculate voltages and currents in this circuit it is important tovisualize current flowing from the positive terminal of the battery,throughR1andthenthroughthecombinationofR2andR3.ThecurrentflowingthroughR1splits.SomegoesthroughR2andsomethroughR3.ThecurrentthoughR2plusthecurrentthroughR3mustequalthatthroughR1sincethemovingcharges cannot disappear nor can new ones appear.Figure 1.5Resistorsinaseries/parallelarrangementAnotherimportantideaisthatcurrent(charge)flowsaroundthecircuit,losingelectricalpotentialenergyintheprocess.SomeenergyislostaschargesflowthroughR1.TheremainderislostaschargesfloweitherthroughR2orR3.ThepotentialdifferenceacrossR2isone-and-the-sameasthepotentialdifferenceCOMPONENTS AND CIRCUITS 15acrossR3.ThepotentialdifferenceacrossR1 plus that across the R2/R3combinationmust equal the battery voltage.Example 3.If we have a resistor combination like that shown infigure1.5,withtheresistancevaluesandbatteryvoltageareasshownalongside:(i) what is the current through each of the resistors?(ii) do the currents calculated add-up i.e. is the current throughthe 50 resistor equal to the total current through the 100 and200 combination?(iii) what is the potential difference across the 50 resistor?(iv) what is the power dissipated in the 100 resistor?Answers:(i) First calculate the total resistance.For the parallel combination,1R= 1100+1200=3200so R=2003=67 The total resistance is thus 67+50=117 The battery voltage is 10 V so the total current is I=VR=10117= 0.085 AAll of this current flows through the 50 resistor so the current through it is 0.085 Aor 85 mA.The potential difference across the 50 resistor is V=I.R=0.085 x 50=4.3 V,whichanswersquestioniiiabove.The potential difference across the 100 /200 parallel combination is 10 - 4.3=5.7 V.COMPONENTS AND CIRCUITS 16The current through the 100 resistor is I=VR=5.7100= 0.057 AThe current through the 200 resistor is I=VR=5.7200= 0.029 A(ii) The total current through the 100 /200 parallelcombinationis0.057+0.029=0.086Awhich(allowingforroundingerrorsaswehaveusedtwo-figureaccuracy) agrees with the value calculated for the 50 resistor.(iii)See above.(iv) Thepowerdissipatedinthe100resistorcanbecalculatedusingeitherequation1.4or1.5or1.6asweknowthepotentialdifference,currentandresistance.Using equation 1.4 we have P=V.I=5.7 x 0.057=0.33 W.AC AND THE MAINS SUPPLYDC and ACIf we connect a battery to a lamp, as in figure 1.1 previously, current will flowfrom one terminal of the battery, through the lamp and return to the oppositeterminal.Suchaflowofcurrentistermeddirectcurrent(DC)becausetheflowisalwaysinonedirection.Agraphofvoltageagainsttimeorcurrentagainst time would look like figure 1.6.DCisdirectlyavailablefromabatteryanditisDCthatweneedtopowermostelectroniccircuits.Thevoltageavailableatthepowerpointsinthelaboratoryorhomeisnot,however,constantordirectbutfluctuatesrapidlyandregularlybetweenabout+340and-340volts.Therepetitionrateorfrequencyofthealternating voltage is 50 times per second.If we plotted a graph of voltageagainst time it would look like figure 1.7.Figure 1.6Direct current as from a batteryCOMPONENTS AND CIRCUITS 17The alternating voltage supplied by the mains gives rise to an alternatingcurrent(AC)inanydevicewhichisconnectedtoapowerpoint.Duringthepositivehalf-cyclesofthewaveformshowninfigure1.7thecurrentflowsinonedirectionthroughtheapplianceandduringthenegativehalf-cyclesthecurrentflowsintheoppositedirection.Becausethewaveforminfigure1.7issymmetricalthecurrentflowinonedirectionduring the positive half-cycles will be exactly balanced by the current flowintheoppositedirectionduringthenegativehalf-cycles.Hencethereisnonettransferofchargeoverafullcycle;electronsaresimplymovedbackandforth.Thisisincontrasttodirectcurrentwhereelectronsaremoved in one direction only.ThefrequencyornumberoftimesanACwaveformrepeatsitselfpersecondisexpressedinunitsofhertz(abbreviatedHz).1Hzis1cycleFigure 1.7Alternating current as from the mainssupply persecond.ItiseasytocalculatethefrequencyofanACwaveformfromagraphlikefigure1.7.First we determine the period of the waveform.The period is simply the time taken foronecompletecyclewhichisreadfromthehorizontalaxis.Inthisexample,thealternating current is from the mains supply and the period, the time for one completecycleoroscillation,is0.02secondorI/50thofasecond.Thefrequency(f)isthereciprocal of the period (T).In symbols:f=1T.... (1.7)Hence for AC supplied by the mains the frequency is 1/0.02 = 50 Hz.During the first hundredth of a second the voltage in figure 1.7 has gone from zero to+340 volts and back to zero again.The effective (average) voltage will be some figurebetween these limits.The figure of 340 volts (the maximum value) is termed the peakvoltagebutwenormallyspecifyACbyanaverageratherthanpeakvalue.Theaverage during the first hundredth of a second is 340/or 240 volts in this example.COMPONENTS AND CIRCUITS 18Thus the appliances we plug into the mains must have a rating of 240 volts AC.Sincethevoltageisrapidlyoscillatingbetween+340and-340voltsthecurrentthrough a light globe connected to the mains will flow alternately in one direction andthentheotherandtheglobewillflickerinbrightnessaccordingtothevoltage.Itisonlybecausethefrequencyoffluctuationissufficientlyrapidthatthiseffectisnotnoticeable.TRANSFORMERS AND THE MAINS SUPPLYTransformersare the devices we use to step-up or reduce AC voltage.If we apply acertain AC voltage to one winding of a transformer (the primary winding) an AC signalofthesameshapebutdifferentamplitude(peakvoltage)willappearontheotherwinding (the secondary winding).PowerTransformersare specifically designed to convert 240 volts AC to some othervoltage: for example a stereo tuner/amplifier would require only 20 to 60 volts to powerthecircuitrycomprisingtheradioandamplifier,itwouldthusrequireapowertransformer to step down the mains voltage to an appropriate value.Anexampleofatransformerwasshowninsection1.1.Figure1.8illustratesthecircuit symbol and construction of a transformer.A transformer consists of two coils ofwirewhichareinsulatedfromeachotherbutincloseproximity.Inpracticethetwocoilsofwireareoftenwoundonacommoncore,madeofiron(seefigure1.8(b)).This allows the transformer to work more efficiently.When AC is applied to one coil ofwire(the'primary'coil),analternatingcurrentisinducedintheother(thesecondary)and the voltage will, in general, be different.Ifthetransformerisconstructedwithmoreturnsontheprimarywindingthanonthesecondarythenitwillactasastep-downtransformer.Forexampleifthereare200turns on the primary winding and 50 on the secondary the voltage from the secondary Figure 1.8(a) Circuit symbol and (b) construction of a transformerwill be one quarter of that applied to the primary.If 240 volts is applied to the primary60 volts will be produced at the secondary.COMPONENTS AND CIRCUITS 19The principle that 'you can'tget something for nothing'applies here.If the trans-former steps-up the voltage,the available current isreduced.If the transformersteps-down the voltage, the current is increased.Figure 1.9Isolation with a power transformerCurrent can only flow alongthe 'active' wire if it cansimultaneously return via the 'neutral' wire.Becausethe neutral wire is earthed, anything touching the earthis provided with a returnpathway via the neutral wire.Ifthetransformerisconstructedwithfewerturnsontheprimarywindingthanonthesecondary then it will act as a step-up transformer.For example if there are 20 turnson the primary winding and 100 on the secondary the voltage from the secondary willbefivetimesthatappliedtotheprimary.If240voltsisappliedtotheprimary1200volts will be produced at the secondary.Because of their physical construction - two inductors in close proximity but insulated-theyplayausefulroleinisolatingthevoltageappliedtoonewinding(theprimary)from that available at the other winding (the secondary).Most appliances are isolatedfromthemainsvoltageinthisway-bytheirtransformers.Theisolationgivenbyatransformerisimportantbecausethevoltagefrom the secondary is earth-free.The power from the mains comes from two cables, the active cable(colourcodedbrown)andtheneutralcable(colour-codedblue).The neutral cable is earthed at the power station and at distributionsubstations.Thatis,theneutralcableisphysicallyconnectedtothegroundthroughametalconductor(forexample,athicklengthofwireconnectedtoametalstakeorwatersupplypipeburiedinthe ground).The neutral cable is thus kept at zero volts potential.Theproblemwiththisarrangementisthatifaconnectionisinadvertently made between the active cable and earth, such as bya person accidentally touching the active cable, this will complete acircuit and current will flow from the active wire through the personto earth, giving an 'earth' shock.To obtain a shock from the 'earthfree'powersuppliedatthetransformersecondaryitwouldbenecessary for the person to touch both terminals simultaneously.Aslongasthereisnoconnectionbetweentheprimaryandsecondarywindings,touchingonlyonesecondaryterminalandthusmakingaconnectionbetweenthesecondary circuit and earth will not complete a circuit and no current can flow, henceno shock.Mains supply and the mechanisms of electric shock are described in moredetail in a later chapter.COMPONENTS AND CIRCUITS 20EXERCISES1The lamp shown in figure 1.1 has a resistance of 3.0 and is connected to a 1.5Vbattery.Calculatethecurrentflowingthroughthelampandthepowerdissipated.2Anelectricjugdrawsacurrentof1.2ampswhenconnectedtothe240voltsmains supply.Calculate the resistance and power rating of the heating element.3Thepowerdissipationofanelectricheateris1000wattswhenconnectedtoa240 volt supply.Calculate its resistance, and the current it will draw.4Threeresistorsareconnectedinseriesasshowninfigure1.3.Thebatteryvoltage is 12 V.The resistors have values R1 = 100 , R2 = 500 and R3 = 600.Draw a diagram of the circuit and label the values of the components.What is:(a)the current through each resistor(b)the potential difference across each resistor(c)thepowerdissipatedineachresistorDo the values calculated in (b) add to equal the battery voltage?5Threeresistorsareconnectedinparallelasshowninfigure1.4.Thebatteryvoltage is 5 V and the resistors have values R1 = 50 , R2 = 100 and R3 = 75.Calculate:(a) the current through each resistor(b)the potential difference across each resistor(c)thepowerdissipatedineachresistor(d)the total resistance of the circuit and hence(e)the total current drawn from the battery.COMPONENTS AND CIRCUITS 21(f) doesthevaluecalculatedinpart(e)agreewiththesum of the values found in part (a) above?6Threeresistorsareconnectedinthecircuitarrangementshown.The current through the 100 resistor is 50 mA.Determine:(a) the battery voltage and(b) the power dissipated in the 30 resistor.7In the course of an experiment a student sets-up the circuitshown below using a torch globe, 2 resistors and a battery.A voltmeter connected across the 200 resistor registers 8volts.(a)What is the resistance of the torch globe?(b)What is the current through the globe?COMPONENTS AND CIRCUITS 228Consider the circuit shown.The current through the 50 k resistor is measured as 67 A.Determinethe:(a)value of the unknown resistor, R.(b)total current drawn from the battery.(c) power dissipated in the 50 k resistor.(d) potential difference across the resistor, R.(e) current through the 100 k resistor.9Consider the circuit shown.The total current drawn from the battery is 0.2 A.What is the:(a)current through the 30 resistor.(b)value of the unknown resistor, R.(c) power dissipated in the 50 resistor.(d) potential difference across the 30 resistor.(e) power dissipated in the resistor, R.10(a)What are the two important functions of a transformer?(b)A transformer has 100 turns on the primary winding and40 turns on the secondary.If an AC signal of amplitude60Visappliedtotheprimarywindingwhatvoltagewillappear at the secondary?(c) A transformer has 200 turns on the primary winding.AnACsignalofamplitude240Visappliedtotheprimarywindingandavoltageof12Visproducedbythesecondary.Howmanyturnsarethereonthesecondary?CAPACITORS AND INDUCTORS 23Figure 2.1Circuit symbol for (a) a capacitor and (b) an inductor.2 Properties of Capacitors and InductorsFigure 2.1 shows the circuit symbol for a capacitor and an inductor.Note the parallelbetweenthecircuitsymbolandthephysicalconstructionofthecomponent.Acapacitorconsistsoftwometalplates(oftensheetsofaluminiumfoil)separatedbyan insulator.An inductor is simply a coil of wire.Knowing the physical construction of each we can predict what resistance they offerwhen connected in a circuit to a battery.The capacitor, with an insulator between itsplates,offersanextremelyhigh(virtuallyinfinite)resistancetoDC.Currentcannotflowfromoneplatetotheother.Inductors,beingsimplycoilsofconductingwire,have an extremely low resistance to DC (almost zero).STORAGE CAPABILITY OF CAPACITORSAspecialpropertyofacapacitorisitsabilitytostorechargeandthuselectricalenergy.If we connect a battery to the capacitor current will flow for a very short timeand,sincethecurrentcannotflowcontinuouslythroughtheinsulator,thecapacitorwill accumulate charge on the plates.When the charge has built up until the voltageacross the capacitor is equal to the battery voltage, current will no longer flow.Thecapacitancetellsushowmuchchargeacapacitorwillaccumulateforagivenbattery voltage.Capacitance (C), charge (q) and voltage (V) are related by the formulaC=qV.... (2.1)Oncecharged,acapacitorcanstorethechargeindefinitely.Ifthebatteryisdisconnected,thechargeonthecapacitorandcanonlyleakawayifweconnectanexternalresistance.The circuit shown in figure 2.2 can be used to observe the discharge of a capacitor.CAPACITORS AND INDUCTORS 24Figure 2.2A circuit for measuring the discharge of a capacitorSupposethatthecapacitorisfirstfullychargedto12V,thenconnectedintothecircuit.Uponclosingtheswitch,currentflowsthrough the 500 k resistor, allowing the capacitor to discharge.Thereadingonthevoltmeterprogressivelyfalls.Table2.1showsmeasuredvaluesofthevoltagetakenat2secondintervalsfor20secondsusinga25Fcapacitor.The25Fcapacitorwasthenreplacedwitha50Fcapacitorandthesecondsetofresultswasobtained.Table 2.1Measured voltages for the discharge of capacitors through a resistance of 500 kTimeVoltage on the Voltage on the(s) 25 F capacitor 50 F capacitor0 12 122 10.2 11.14 8.7 10.26 7.4 9.48 6.3 8.710 5.4 8.012 4.6 7.414 3.9 6.816 3.3 6.318 2.8 5.820 2.4 5.4Agraphofthesevoltagesagainsttimeisshowninfigure2.3.Notethat it takes about 8.5 seconds for the 25 F capacitor to discharge tohalfoftheoriginalvoltageand17secondsforthe50Fcapacitor.On this basis we predict that it would take about 85 seconds for a 250Fcapacitortodischargetohalftheoriginalvoltagethroughthesameresistance.CAPACITORS AND INDUCTORS 25Figure 2.3Discharge of capacitors througharesistanceWhat we observe is that the time to half discharge (1/2)isdirectlyproportionaltothecapacitance(C)ofthecapacitor.In symbols this is written1/2C Other experiments, where the resistance through whichthecapacitordischargesisvaried,showthatthetimeforhalf-dischargeisalsoproportionaltotheresistance(R) through which the capacitor discharges.That is,1/2R Combiningtheseequationswehavethatthetimeforhalfdischargeisdirectlyproportionaltotheproductofthe resistance and the capacitance.In symbols1/2R.C.... (2.2)Thisisageneralrelationshipbetweenresistance,capacitanceanddischargetimeforaresistor/capacitorcombination.Sometimestheterm'half-dischargetime'isused,sometimes'RCtime-constant'.Theyarerelated,butarenotthesamething.TheRCtime-constantissimplytheproductR.C.Theconstantofproportionalityinequation2.2is0.693(thenaturallogarithm of 2) so if we know the RC time constant, 1/2is easily calculated.Notice that for a short period of time, such as 1/50th of a second, the 25 F capacitorin table 2.1 will not have discharged appreciably.If it discharges by 0.9 volt in the firstsecond, in l/50th of a second it will discharge by less than 0.02 volt.The importanceof this and the previous ideas will be considered later.CAPACITORS AND INDUCTORS 26Most of you would have madean electromagnet at school.Insulated copper wire iswound around, say, a nail tomake an inductor.When the ends of the wire are connect-ed to a battery, the nailbecomesmagnetized.Infants can hear soundfrequencies of 20 kHz or more.With age, this maximumfrequency drops so that mostpeople over 35 can not hearfrequenciesaboveabout14kHz.INDUCTORS AND MAGNETIC FIELDSWhencurrentflowsthoughaninductor,amagneticfieldisproduced.Infact,whenevercurrentflows,thereisanassociatedmagneticfieldsoastraightpieceofwirewhichiscarryingcurrentwillhaveamagneticfieldaroundit.Thereasoninductorsarespecialisthatthewireiswoundincloselyspacedloopsandthisresults in a much more intense magnetic field.The field can be further intensified bywrappingthecoilaroundapieceofironormildsteel.Boththemagneticfieldintensity and the inductance are increased if the inductor is wound around a core ofiron or other so-called ferromagnetic material.We will have more to say about magnetic fields (and their effects on tissues) in laterchapters.IMPEDANCE OF CAPACITORS AND INDUCTORSAmajorpartofthefieldofelectronicsisconcernedwithprocessingalternatingcurrentofvariousfrequenciesanditisforthisreasonthatcapacitorsandinductorsaresoimportant.Resistorsareimpartialasfarasdirectcurrentandalternatingcurrent is concerned: they offer the same resistance to AC and DC.Capacitors andinductors display no such impartiality: they both react quite differently to AC and DC.Beforedescribingthepropertiesofcapacitorsitisusefultohavesomeideaofthekind of frequencies we are dealing with.Soundswhichareaudibletohumanslieintherangefrom20Hztouptoamaximum of about 20 kHz (1 kHz = 1000 Hz).Normal (AM) radio transmissions aremadeinthefrequencyrangefrom500to1500kHz.FMradiotransmissionsaremade in the range 88 to 108 MHz.(l MHz = 1000 kHz or 1000 000 Hz).Check-outthenumbersonyourtunerorpocketradiodisplaypanel.Televisiontransmissionsare made in the frequency range on either side of the FM range, i.e.between 40 and80 MHz and above 108 MHz up to about 130 MHz.In each case the sound, radio ortelevisionfrequencywavecanbeconvertedtoalternatingcurrentofthesamefrequencywhichcanbeamplified,filteredandprocessedbyanelectroniccircuittoproduce the resulting sound or image.CAPACITORS AND INDUCTORS 27Figure 2.4A circuit for measuring the impedance of a capacitorCapacitorsandinductorsareusefulbecausetheirresistancetoalternatingcurrentdepends on the frequency of the AC.Next we examine this frequency dependence.Knowingthephysicalconstructionofcapacitorsandinductorswecanpredictwhatresistance they offer when connected in a circuit to a battery.The capacitor, with aninsulatorbetweenitsplates,offersanextremelyhigh(virtuallyinfinite)resistancetoDC.Current can not flow from one plate to the other.Inductors, being simply coils ofconducting wire, have an extremely low resistance to DC (almost zero).Foralternatingcurrenttheresistanceofthesecomponentsisnotpredictedquitesosimply.Insteadofattemptingapredictionletuslookatsomeexperimentalobservations with AC.When we talk about resistance to the flow of alternating current we no longer use theterm'resistance'butimpedance.Theimpedanceisstillmeasuredinohmsbutwereserve 'resistance' for use when talking about DC.The symbol Z is usually used forimpedance.Impedancecanbecalculatedfrommeasurementsusingacircuitsuchas that shown in figure 2.4.The potential difference across R and that across C arefoundtodependontheACfrequency.Thetwovaluesaddedequalthepotentialdifference produced by the AC source.Thevoltageacrosstheresistorcanbeusedtocalculatethecurrentflowingaroundthecircuit(usingOhm'sLaw,I = V/R).Since C and R are in series, this equals thecurrentflowingtothecapacitor.TheimpedanceofthecapacitoristhencalculatedusingZ=VI.... (2.3)Which is Ohm's law in its more general form.Ifthecapacitorinfigure2.4isreplacedbyaninductor,impedancevaluesfortheinductorcanbecalculatedinasimilarmanner.Table2.2givesthemeasuredimpedanceofa20millihenry(mH)inductoranda1microfarad(F)capacitoratvariousfrequencies.CAPACITORS AND INDUCTORS 28Frequency Impedance of 1 F Impedance of 20 mHf (kHz) capacitor(ohms) inductor(ohms)0.2 795 25.10.3 530 37.70.4 398 50.20.6 265 75.41.0 159 1262.0 79.5 2513.0 53.0 3774.0 39.8 5025.0 31.8 628Table 2.2Variation of impedance withfrequency for a 1 Fcapacitor and a 20 mHinductorAgraphofthesevaluesofimpedanceateachfrequencyisshowninfigure2.5.Noticethatfortheinductorthegraphisastraightlinepassingthroughtheorigin.ThistellsusthattheimpedanceoftheinductorisdirectlyproportionaltothefrequencyoftheAC.Forthecapacitorthecurvesuggestsaninverserelationshipbetweenimpedanceandfrequency:thisis,infact,thecase.Tocheckthattherelationshipisaninverseoneratherthaninversesquareorinversecubewewouldneedtoplotimpedanceversus1/frequencyandestablishthatthisgivesastraightline graph.This is left as an exercise for the interested reader.ThemathematicalexpressionforthedependenceofimpedanceonfrequencyforeachofthesecomponentsisZ f for the inductorand Z 1/f for the capacitor.Bysubstitutingdifferentvaluesofinductanceandcapacitance,itisestablishedthatfor inductors, Z L and for capacitors, Z 1/C.CAPACITORS AND INDUCTORS 29Sinceforinductors,impedanceisproportionaltothefrequencyandisalsoproportionaltotheinductance,wecancombinetheseexpressionsandwriteZ f.L.WhentheimpedanceisspecifiedinOhms,theinductanceinHenrysandthefrequencyinHz,theconstantofproportionalityis2.Therelationshipbetween Z and f for an inductor is thusZ=2fL .... (2.4)Similarlyforthecapacitor:sinceimpedanceisproportionaltothereciprocaloffrequencyandthereciprocalofthecapacitancewecancombinetheexpressions and write Z a 1/f.C.WhentheimpedanceisspecifiedinOhms,thecapacitanceinFaradsandthefrequencyinHz,theconstantofproportionalityis1/2p.Therelationshipbetween Z and f for a capacitor is thusZ=12fC.... (2.5)Figure 2.5Impedance versus frequency graph for (a) a 1 F capacitor and (b) a 20 mH inductorFromequations2.4and2.5wecanseethatitispredictedthatdoublingthevalueofinductancewoulddoubletheimpedanceatallfrequencies.Doublingthevalueofcapacitancewouldhalvetheimpedanceatallfrequencies.Thesepredictions are easily confirmed experimentally.Twoimportantpointstonotefromtheequations(orthegraphs)arethatcapacitors have an infinite impedance to direct current (zero frequency) whereasinductorshavezeroimpedancetoDC.Thisisinagreementwithourpreviouspredictions based on the physical construction of each.CAPACITORS AND INDUCTORS 30RESONANT CIRCUITSAtraditionalandsimplewayofproducingalternatingcurrentreliesonthepropertiesofcapacitorsandinductorsandtheirbehaviourwhenconnectedtogetherinacircuit.Aninductorandcapacitorcombinedinparallelasinfigure2.6formswhatiscalled a parallel resonant circuit.Theimpedanceoftheinductorincreaseswithfrequencywhiletheimpedanceofthecapacitordecreases with increasing frequency (figure 2.5).Theparallelcombinationwillhavealowimpedanceatlow frequencies (because of the inductor) and at highfrequencies(becauseofthecapacitor).Atanintermediatefrequencythecombinationhasamaximumimpedance.Thisiswhentheimpedancesofthecomponentsareequal.Thepeakinthegraphshown in figure 2.6 occurs at this frequency.Aformulaforcalculatingthefrequencyatwhichtheimpedancesofthecapacitorandinductorareequalisobtainedasfollows.Theimpedanceofthecapacitor is given by equation 2.5, and the impedanceoftheinductorbyequation2.4.WhentheimpedancesofthecomponentsareequalFigure 2.6A parallel resonant circuit and how its impedance varies with AC frequency12fC=2fLwhich rearranges to givef2=142LCCAPACITORS AND INDUCTORS 31or f=12LC.... (2.6)Thisfrequencyistermedtheresonantfrequencyofthecombination, and for good reason.Imaginethatthecapacitorinfigure2.6isdisconnectedfromtheinductorandconnectedtoabatterytochargeit.Thebatteryisthenremovedandthecapacitorconnectedtotheinductor again.What subsequently happens?Thecapacitorimmediatelybeginstodischargeandcurrentflowsthroughtheinductor.Insodoingthecurrentflowsetsupamagneticfieldaroundtheinductor.Immediatelythechargedisappearsthecurrentceases,causingthemagneticfield to collapse.The collapsing magnetic field then induces acurrentintheinductorandthecapacitorrecharges,butwithoppositepolarity.Thesequenceofeventsisillustratedinfigure 2.7 where the arrows show the direction of current flow.Thecapacitornextbeginstodischargeintheoppositedirection(notshowninfigure2.7-thiswouldfollow(d))andthe whole cycle is repeated again and again.Whatishappeningisthattheelectricalenergystoredinthecapacitorisconvertedtomagneticfieldenergyaroundtheinductor.Because a magnetic field can only exist if current isflowing,themagneticfieldmustcollapseasthecapacitorbecomesdischarged.Themagneticfieldenergycannotdisappear,butisconvertedbackintoelectricalenergywhichmoveschargearoundthecircuitandreverse-chargesthecapacitor.The reverse-charged capacitor (figure 2.7(d)) wouldthen discharge through the inductor.Current would flow in theFigure 2.7Initial current flow in a resonantcircuitCAPACITORS AND INDUCTORS 32reverse direction to that in figure 2.7, producing a magnetic field which would collapseandre-chargethecapacitortoitsoriginalstate(figure2.7(a).Theprocesswouldrepeat indefinitely.Ifthiswerethewholestorythecircuitwouldresonate,generatingacontinuousalternating flow of current through the inductor.A graph of current versus time wouldbe an undamped sine wave as shown in figure 2.8(a).A graph of voltage across thecapacitor against time would also resemble figure 2.8(a).Inpractice,capacitorsandinductorsarenotideal.TheinductorwillhavesomeDCresistanceandthecapacitorwillallowsomeleakageofcurrentsothatenergyislostduringeachoscillation.Theoscillationswillbedampedandmusteventuallycometoanend.Thenetresultwillbeadampedoscillation as shown in figure 2.8(b).Thenaturalfrequencyofoscillationisthesamefrequencyindicatedbythepeakinthegraphshowninfigure2.6:theresonantfrequency.Thisisalsothefrequencyatwhichtheimpedancesofthecapacitorandinductorareequal(wherethegraphsFigure 2.8(a)undampedand(b)dampedsinewavesintersect in figure 2.5).[A graph similar to that in figure 2.6 could be arrived at simply by adding impedancesgraphically,saybyusingfigure2.5.Themethodgivesacorrectqualitativeresult.Itdoesnotgiveacorrectquantitativepredictionasnoconsiderationisgiventothephase relationship between the current in each component.The 'resonant frequency'obtainediscorrectbuttheheightoftheimpedancepeakisnot.Arigorousmathematical treatment shows that, for ideal components, the impedance is infinite atthe resonant frequency.For real components the impedance will be finite because ofthe resistance of the inductor and leakage of the capacitor].CAPACITORS AND INDUCTORS 33Althoughresonantcircuitsaremostcommonlyencounteredintheparallelcombination shown in figure 2.6, they can occur as a series combination (figure 2.9).Againthereisadefiniteresonantfrequencyatwhichtheimpedancesofthetwocomponentsareequal.Figure 2.9A series resonant circuit and how its impedance varies with AC frequencyFor a series resonant circuit the impedance has a minimumvalueattheresonantfrequency.Becauseofthenon-idealnatureofthecomponentstheimpedanceisnotzeroattheresonant frequency but when component losses are low theimpedancecansuddenlydroptoaverylowvalueasthisfrequency is approached.COUPLING AND RESONANT CIRCUITSIfanoscillatorisdesignedtooperateatafixedfrequencytheusualwayofcouplingitssignaltoanamplifieror,moregenerally,betweencircuitsisbyuseofaresonantcircuit.Thismethodismostcommonlyusedathighfrequencies(above 100 kHz).Figure 2.10 shows how a resonant circuit is used to transferelectrical energy from one circuit to another most efficiently.In other words to 'couple' the circuits most efficiently.Thesignalfromthefirstcircuitsuppliesenergyfortheresonantcircuittooscillate.Thesignalisinducedinthesecond circuit by the transformer action of the two inductors.In order for transformer action to operate most efficiently the signal produced by circuit1 should have the same frequency as the resonant circuit.This means either that thefrequency of circuit 1 can be adjusted to match that of the resonant circuit or that thefrequency of the resonant circuit can be adjusted to match that of circuit 1.The latterwould require the inductor or capacitor in the resonant circuit to be variable so that theresonant frequency could be adjusted to suit.CAPACITORS AND INDUCTORS 34Figure 2.10Coupling with a resonant circuit.This is analogous to pushinga swing.The swing has itsown, natural or resonantfrequency.If you applypushes at the samefrequency, the swingoscillations are reinforced.Pushing at a differentfrequency would becounterproductive.ThecombinationofLandCwillhavearesonantfrequencywhichisdeterminedby their inductance and capacitance values according to equation 2.6.The currentthroughtheinductorconnectedtocircuit1willinduceacurrentofthesamefrequencyinL.IfthecurrentinducedinLdoesnothavethesamefrequencyasthe resonant circuit, little energy will be transferred.If the resonant frequency of LandCincombinationisthesameasthefrequencyofthesignalsuppliedbycircuit1,thenthecircuitwillresonateandmaximumelectricalenergywillbetransferred.SUSTAINED OSCILLATIONByappropriatechoiceofthecapacitorandinductoraresonantcircuitcanbemadetogenerateanyfrequencyofsinewave.Aresonantcircuitaloneisnotsufficient, however, to generate a sustained oscillation.To produce a continuous,steadysupplyofalternatingcurrent(asinfigure2.8a)wemustarrangefortheresonantcircuittobecontinuouslysuppliedwithenergytoovercomethelossesin the components and keep it oscillating.By use of an amplifying circuit, we can provide the energy to overcome the circuitlossesandpreventtheoscillationsfromdying.Anamplifierprovidespositivefeedback.Amplifiers and positive feedback are considered in chapter 5.CAPACITORS AND INDUCTORS 35Figure 2.11A piezo-electric crystal compressedandstretchedPIEZOELECTRIC CRYSTALSSofarourdiscussionofoscillatorshasbeenrestrictedtoresonantcircuitsintheformofinductorandcapacitorcombinations.Unfortunately,suchcombinationstendtodriftinfrequencyoveraperiodoftimeandwithchangesinambienttemperature.Theeffectisonly slight and not very important at low frequencies unless high accuracy is required.Theeffect is much more significant at radio frequencies (greater than 500 kHz) where it is veryimportanttohavegoodfrequencystability-imagineswitchingonyourradioandneverknowing quite where to find your favourite radio station!TheeffectisreallyquiteseriousinTVandradiotransmissionandcommunications.Thefrequenciesallocatedtousersarequitecloselyspacedandanydriftintransmissionfrequencycouldresultinoverlapwithadjacenttransmitters.Similarlytheshortwavediathermyequipmentusedinphysiotherapyoperatesatradiofrequenciesandradiatesacertainamountofenergyasradiowaves-forthisreasononlycertainfrequenciesarepermitted for their operation and little deviation or drift is permitted.Good frequency stability can be achieved by use of a piezoelectric crystal.These crystalshavethespecialpropertythatwhensqueezedorstretched,apotentialdifferenceisproduced between each surface.This piezo-electric effect is illustrated in figure 2.11.Theothersideofthecoinwithpiezoelectriccrystalsisthatifapotentialdifferenceisappliedtotheiroppositesides,theychangeinthickness.Thusintheexampleshowninfigure2.11,ifapotentialdifferenceisappliedtoanunstressedcrystalandthevoltageispositiveontop,thecrystalwillshrinkinthickness.Ifthepotentialdifferenceisappliednegative on top in this example, the crystal will expand.Aninterestingthinghappensifapotentialdifferenceissuddenlyandbrieflyapplied.Thepiezoelectric crystal reacts like a bell which is struck by a hammer and starts ringing.Thethicknessofthecrystalchangeswhenthevoltageisappliedbutthemoleculeshaveamomentumwhichcausesthemtoovershoottheirequilibriumpositions.Theyarepulledback and overshoot in the opposite direction.The cycle continues and the result is that thecrystalvibratescontinuously.Figure2.12illustratestheprocessandwhatwouldbeobserved in practice.CAPACITORS AND INDUCTORS 36Thecrystalresonatesmechanicallyataparticularfrequency.Itsthicknessoscillatesaboutameanwhichisitsnormal,unstressedthickness(thehorizontal axis in figure 2.12a).Asthecircuitresonatesmechanically,thesurfacesbecomechargedaccordingtowhetherthecrystalisstretchedorcompressed(figure2.11).Hence,asthethickness oscillates, the charge on each surface will vary and a graph of chargeversustimewouldresemblefigure2.12(a).Agraphofpotentialdifferenceversus time would look the same.Crystalresonatorsconsistofaquartzwaferbetweentwoelectrodes.Thephysicaldimensionsofthecrystaldeterminetheresonantfrequencyandifthecrystalismaintainedataconstanttemperatureaveryhighorderoffrequencystabilitycanbeobtained.Whenincludedinacircuitthecrystalonlypermitscurrent to flow when the frequency of the current is equal to the natural frequencyof oscillation of the crystal.Quartzcrystalscanthusreplacetheresonantcircuitsoffigure2.6and2.9andsetthefrequencyoftheoscillatormoreprecisely.Thefrequencystabilityofquartzcrystalsisexploitedinapplicationswhereprecisetimekeepingisrequired.Everyday applications include wristwatches and the clock which is theheart of every computer.We will meet quartz crystals again in a later chapter describing the production ofFigure 2.12(a) The change in thickness of a piezoelectric crystal in response to (b) a briefly applied voltage ultrasoundwaves.CAPACITORS AND DEPOLARISATIONEarlierinthischapterwediscussedthechargestoringcapabilityofcapacitorsandintroducedthehalf-dischargetime,1/2whichcharacterizestherateofdischarge of a capacitor through a resistor (equation 2.2).1/2R.C.... (2.2)CAPACITORS AND INDUCTORS 37Whenacapacitorischargedthrougharesistor,equation2.2alsoappliestothechargingbehaviour.Henceinacircuitsuchasthatshowninfigure2.13,whentheswitch is closed the voltage across the capacitor will increase as shown.Thevoltageacrossthecapacitorincreasesandapproachesthebatteryvoltageasymptotically.For the graph shown the battery voltage is 12 V.From the graph, 1/2is approximately 6 seconds.Thereasonthatthegraphhasthisshapeisthatthecurrentflowingtochargethecapacitor decreases as the capacitor charges.Initially, the capacitor is uncharged sowhen the switch is closed, the potential applied to the left side of the resistor is 12 Vand the potential on the right side is zero.The potential difference across the resistoris maximum (in this circuit, 12 V) so the current flow is maximum.When the capacitorhaschargedto,say,3V,thepotentialdifferenceacrossRis9volts,sothecurrentCAPACITORS AND INDUCTORS 38Figure 2.14A pulsed voltage applied to a resistor-capacitorcombination. (a) potential difference across the capacitor and (b) current through the resistor versus time.flow (calculated using Ohm's law or measured directly) is less.When the capacitor ischargedto9V,thepotentialdifferenceacrossRisonly3Vandthecurrentisone-quarterofitsoriginalvalue.Thecloserthecapacitorcomestofullycharged,thesmalleristhepotentialdifferenceacrossRandthesmalleristhechargingcurrent.Consequently,therateofcharging,determinedbythecurrentflowthroughtheresistor,becomessmallerandsmallerasthecapacitorapproachesfullychargedand the capacitor voltage approaches the battery voltage but never quite gets there.Nextconsiderwhathappensif a pulsed voltage is appliedtotheresistor-capacitorcombinationshowninfigure2.14.Atthestartofthepulse,thecurrentflowthroughtheresistorishighandthecapacitorchargesrapidly.Asthecapacitorcharges,the rate of charge decreasesandthepotentialdifferenceacrossthecapacitor(VC)approachesaplateau(figure2.14a).AgraphofcurrentflowthroughtheresistororpotentialdifferenceacrosstheresistorversustimemirrorsthegraphofVC(figure2.14b).Theinitialchargingcurrentishighbutreducesrapidlyasthe capacitor charges.NotethatthecurrentflowthroughtheresistorisAC.Currentflowsthroughtheresistor as the capacitor charges and an equal amount flows in the reverse (negative)direction as the capacitor discharges.CAPACITORS AND INDUCTORS 39Figure 2.15Response of the circuit shown in figure 2.14 when the pulse duration is (a) much greater than R.C, (b) comparable to R.C and (c) less than R.C.Thepulsedurationinthisexampleissignificantlygreaterthanthehalf-discharge/half-chargetime,1/2sothepulseisableto(almost)completelychargethecapacitorbeforecomingtoanendandallowingthecapacitor to discharge.If the RC time constant(and hence 1/2) of theresistor-capacitorco-mbinationwascom-parabletothepulsewidth,thecapacitorwouldnotfullycharge(figure2.15b)andthecurrentflowthroughtheresistorwouldbemoresustained.[Thedashedlinesinfigure 2.15 indicate thevoltageacrossthecapacitorwhenitisfullycharged(inthiscase,itisthepulsevoltage).]If the RC time constantwasgreaterthanthepulsewidth,thecapacitorwouldchargeminimally(figure2.15c)andthecurrent flow through the resistor would be well sustained.TheimportanceoftheRCtimeconstantwillbeapparentwhenweconsidertranscutaneouselectricalnervestimulationinsubsequentchapters.CAPACITORS AND INDUCTORS 40A final point which should be noted is that in each example in figure 2.15, the currentthrough R is AC.The capacitor charges by a certain amount then discharges, so thenet movement of charge is zero.Whatever charge flows through R during the pulsemust flow back afterwards.EXERCISES1 Thetablebelowshowsreadingsofvoltageobtainedat2secondintervalsasacapacitordischargedthrougharesistance.Thecircuitarrangementusedisshown in figure 2.2.In this case the value of the capacitor was known to be 50 F but the resistancewasunknown._______________________________________________________TimeVoltage on the 50 F TimeVoltage on the 50 F(sec) capacitor (sec) capacitor_______________________________________________________0 12.0 12 1.82 8.7 14 1.34 6.3 16 0.96 4.6 18 0.78 3.3 20 0.510 2.4_______________________________________________________(a)Plot a graph of voltage against time for comparison with figure 2.3.(b)Howlongdoesittakeforthe50Fcapacitortodischargetohalftheoriginalvoltage?(c)Whatwouldtheunknownresistanceneedtobetogivetheresultsshownabove?CAPACITORS AND INDUCTORS 412 A capacitor is charged to a potential of 15 V then connected to a 250 k resistorandallowedtodischarge.Thetimetakentohalfdischargeis2.4seconds.How long would the capacitor take to half discharge if:(a) the resistor was replaced by one of value 500 k(b) the resistance was halved in value(c) the capacitance was doubled in value(d) the resistance was halved in value and the capacitance doubled(e) the resistance was doubled in value and the capacitance also doubled.3What is the impedance of:(a) a 10 F capacitor(b)a 0.047 F capacitor(c)a 10 H inductor(d)a 2 mH inductorat a frequency of 1 kHz?How would the impedance change if the frequency was doubled?4A capacitor is connected directly to a 12 volt AC supply (of frequency 50 Hz).Thecurrentflowingthroughthecapacitorisfoundtobe15mA.Whatisthecapacitance?5A capacitor has an impedance of 600 ohms at a frequency of 2 kHz.What is itsimpedance at a frequency of:(a)100 Hz(b)1 MHz.6 Figure2.5includesagraphofimpedanceversusfrequencyfora20mHinductor.Thegraphisobtainedusingthevaluesofimpedanceandfrequencylistedintable2.2.UsethevalueslistedtoplotyourowngraphofimpedanceCAPACITORS AND INDUCTORS 42versusfrequencyfora20mHinductor.Onthisgraphdrawlines showingtherelationship between impedance and frequency for:(a)a 10 mH inductor(b)a 50 mH inductor.7Use the values in table 2.2 to plot a graph of impedance versus frequency for a 1Fcapacitor.Onthesamegraphdrawcurvesrelatingimpedanceandfrequency for:(a)a 0.5 F capacitor(b)a 2 F capacitor.8 Figures2.6and2.9showthevariationofimpedancewithfrequencyforseriesandparallelresonantcircuits.Explaintheshapeofeachgraphintermsofthevariation of impedance with frequency of the individual inductors and capacitors.9 Aresonantcircuitismadebycombininga0.1Fcapacitorwitha3.0mHinductor.What is the resonant frequency of such a combination?10A50mHinductorincombinationwithacapacitorisfoundtohavearesonantfrequency of 1 kHz.What is the value of the capacitor?11Whenapulseofcurrentisappliedtothecircuitshowninfigure2.22it'resonates' generating an alternating voltage as shown in figure 2.24(b).(a)explain what is meant by resonance and why the circuit resonates(b) explain why the AC is damped.12Twocircuitsarecoupledtogetherasshowninfigure2.10.Thevaluesofthecomponents are C = 100 pF and L = 1 mH.(a)Underwhatcircumstanceswillpowerbetransferredmostefficientlybetween the circuits?CAPACITORS AND INDUCTORS 43(b)At what frequency will maximum powertransfer occur?(c) IfCwasincreasedto500pF,whatwouldbethenewfrequencyformaximumpowertransfer?13 An AC signal of frequency 100 kHz is generated by circuit 1 in figure 2.10.Thevalue of C is 0.01 F.(a) What value of L is needed for maximum power transfer to circuit 2?(b) If L is 3 mH, at what frequency will maximum power be transferred?14ADCpulseisappliedtoaresistor/capacitorcombination.Theresultisapotential difference and flow of current through the resistor as shown.(a) why does the graph of current flow through the resistor droop so rapidly?(b) why must the current flow through R be purely AC?(c) ifthecapacitancewasincreased,howwouldthisaffecttheshapeofthegraph of current through R?Why?(d) underwhatcircumstanceswillthegraphofcurrentflowthroughtheresistor resemble that of the applied voltage?CAPACITORS AND INDUCTORS 4415DCpulsesasshownbelowareappliedtoaresistor/capacitorcombination.(a)what can you conclude about the RC time constant for this circuit?(b)iftheRCtimeconstantwasreduced(bydecreasingCorR)howwouldthisaffecttheshapeofthegraphofpotentialdifferenceorcurrentflowthrough the resistor? Give an explanation of your reasoning.ELECTRICAL PROPERTIES OF SKIN 45In most areas of thebody, the subcutis ispredominantlyadipose (fat storing) tissue.It is, nonetheless,quiteconductive because of its water (and ion) contentand the presence of an extensive network ofblood vessels. 3 Electrical Properties of SkinPhysiologistsstudythebioelectricpropertiesofnerveandmuscleusinginvasivesurgicaltechniqueswherethenervetrunkisexposedandsorenderedaccessibletodirectelectricalstimulation.Sometimesnervefibresareteased-outandstimulatedindividually.Inthiswaymuchhasbeenlearnedabouttheelectricalcharacteristicsofnerve.Thisknowledgeisusedbythephysiotherapisttochooseelectricalstimuluscharacteristics(suchastheintensity,treatmenttime,pulsewidthandfrequency)toinvokethedesiredphysiologicalresponseduringpatienttreatment.Acomplicationisthatelectricalstimulationfortherapyisalmostinvariablyappliednon-invasively.Itisnormallyachievedbyapplyinganelectricalstimulusthroughtheskinusingsurfacemountedelectrodes.Theskinhascomplexelectricalcharacteristicsandthesemustbe taken into account when transcutaneous electrical nerve stimulation (TENS) is used.SKIN AND APPENDAGESThebiologicalmakeupofskinanditsphysiologicalstatedetermineitselectricalpropertiessoitisworthwhiletoconsidersomeaspectsofitsstructurewhicharerelevanttoelectricalstimulation.Examinedmicroscopically,theskinisfoundtohavetwo distinct layers, the dermis and epidermis, as illustrated in figure 3.1.Thedermisandepidermistogetherconstitutetheskin.Theepidermisispuncturedbytheskinappendages:thesweatglandductsandhairfollicles.Beneaththeskinisthesubcutis,alsoreferredtoasthesuperficialfasciaorsimplysubcutaneoustissue.Bloodvessels,lymphvesselsandnervesinfiltratethesubcutisanddermisbutnottheepidermis.Thereareotherfundamentaldifferencesbetweenthedermisandepidermis.Thedermis is well hydrated.It consists of a matrix of collagen and elastin fibres embeddedina'groundsubstance'richinproteoglycansandhyaluronicacid.Fibroblastsarethepredominantcellsinthislayer.Thehighdegreeofhydrationmakesthedermiselectricallyconductive.Theepidermisislesshydratedandconsistsofamatrixofkeratinfibres.ThecellspresentinthislayerarepredominantlykeratinocyteswhichELECTRICAL PROPERTIES OF SKIN 46Figure 3.1Some important features of skin.receivenutrientsfromcapillariesintheunderlyingdermis.Thebasallayer of the epidermis is metabolicallyveryactive,withthecellsregularlyundergoingmitosis.Keratinocytes,formedandpushedupwardsfromthislayer,synthesisekeratinandstoreitwithinthecytoplasm.Intheirlifecycle,thekeratinocytesmovetowardstheskinsurface,becominglessmetabolicallyactiveasdiffusionlimits the rate of nutrient supply.Nearthesurfacethecellsdieandshrivel,turningintolittlesacsof(mostly)keratin.Thedeadcellsformascalyshellcalledthestratumcorneum.Thestratumcorneumisthusthewithered,dehydratedremainsofkeratinocytespackagedfullofkeratin.Fromformationinthebasallayertodesquamation(flakingoffastinyscalesfromthesurface)takesthekeratinocyteapproximately40days.Thebalancebetweendesquamationandmitosisinthebasallayerkeeps the thickness of the epidermis constant.Thestratumcorneumisadry,insulatingbutverythinshellwhichseparatesand isolates the highly hydrated soft tissues of the human body from the drierandfarmorechangeableexternalenvironment.Ifthestratumcorneumisremoved, the body loses water and if too great an area of the stratum corneumis lost or damaged, the resulting water loss can be fatal.As noted above, thefibrous protein, keratin forms the bulk of this dead, insulating layer which is soessentialforwaterhomeostasis.Fromthepointofviewoftranscutaneouselectrical nerve stimulation, the presence of the stratum corneum is significantELECTRICAL PROPERTIES OF SKIN 47Figure 3.2An electrical model fortranscutaneousstimulation.because it is an electrical insulator.Theskinappendages,thehairfolliclesandsweatglandducts,areimportantbothphysiologicallyandelectrically.Thekeratinousstructureswhichwecallhairsservetwo important physiological roles; one sensory, one thermal.Hairs are relatively rigidstructureswhichareanchoredintheskin.Sensoryreceptorsclosetothehairrootdetectandrespondtomovementofthehair,soprovidingsensitivetouchreceptors.Asufficientdropintemperaturewillsignalareflexresponsefromreceptorswhichtriggers contraction of smooth muscle fibres connected to the hair root.The myofibrilcontraction causes the hairs to 'stand on end', trapping a volume of warmer air whichseparatesthelivingtissuefromtheexternalenvironment.Thiseffectispronouncedinhumansandisalsoseeninotherspecieswhichhavegreaterhaircovering.Anelevatedtemperaturetriggersanincreasedbloodflowtothedermis,thusincreasingtherateofheattransferfromdeeperstructuresandalsoactivatesthesweatglandsto release water on the skin surface and provide evaporative cooling.Electrically, thehairfolliclesandsweatglandductsareconductivepathwaysthroughtheinsulatinglayer, the stratum corneum.AN ELECTRICAL MODEL FOR TRANSCUTANEOUS STIMULATIONWhenanelectrodeisplacedontheskinsurfacewecreateasituationwheretwoconductorsareseparatedbyaninsulator,inotherwords,acapacitor.The electrode and the tissue beneath thestratumcorneumarethetwoconductors.Thestratumcorneumistheinsulatoralthough,aswehave seen, the stratum corneum is a 'leaky' insulatoras it is punctured by conductive channels created bytheskinappendages.Thecombinationofelectrodesandtissuecanthusbemodelledbyaresistor capacitor combination.Figure 3.2 shows anelectricalmodelfortwoelectrodesplacedontheskinsurface.ELECTRICAL PROPERTIES OF SKIN 48In this model, the resistanceof the electrodes isassumedtobenegligible,but could be included putting a resistor at eachend of the model or includingthe electrode resistance In R.sCisthecapacitanceofthestratumcorneum.Rp,theresistorinparallelwithC,representstheconductivepathwaycreatedbytheskinappendages.Rs,theseriesresistance,representstheresistanceofthetissuevolumeunderthestratumcorneumbetweenthetwoelectrodes.Theresistanceoftheelectrodesisassumedtobenegligible.TheactualvaluesofRpandCdependonfactorssuchastheelectrode size, ambient temperature and skin condition.The quantity Rs varies little.An important point which we will return to is that nerve fibres, whether sensory motororpain,arelocatedbeneaththestratumcorneum.Intermsoftheelectricalmodel,thismeansthatthepotentialdifferenceacrossRsdeterminesthestimulationintensityexperiencedbynervefibres.ThepotentialdifferenceacrossCandRprepresents lost electrical energy as far as nerve fibres are concerned.RESPONSE OF THE MODEL TO STEADY DIRECT CURRENTIfasourceofdirectcurrent,suchasabatteryisconnectedtothecircuitshowninfigure3.2,somecurrentwillflowthroughthethreeresistorsinseriesandsomewillflowontothecapacitors,sochargingthem.Theinitialflowofcurrentislargeasitincludeschargingcurrentflowingtothecapacitors(figure3.3a)whichislimitedonlybyRs.Afterthecapacitorshavecharged,thecurrentflowislowerasthereisonlycurrent flow through the resistors (figure 3.3b).IfRp,theparallelresistanceofthestratumcorneum,ishigh(asitis,undernormalcircumstances),the'steadystate'currentflow(figure3.3b)willbelow-appreciablylowerthantheinitialcurrentflowwhichincludeschargingcurrentforthecapacitors.At the extreme, if the stratum corneum was not 'leaky' i.e. if the parallel resistance, Rp,infigure2wasinfinite,nosteadydirectcurrentcouldflow.Therewouldbeabriefflow of current as the capacitors charged, then no further current flow.In reality, Rp allows current flow, so after C is charged the only current flow is throughRs and the two resistances Rp which are connected in series.ELECTRICAL PROPERTIES OF SKIN 49Figure 3.3Response of the electrical model to direct current (a) initially and (b) after the capacitors have charged.Question 1:Howmuchsteadydirectcurrentwouldflowinatypical'reallife'situationiftheappliedDCvoltage is 50 V and electrodes with an area of 10cm2areused?Information:Skincapacitanceisproportionaltotheelectrodearea.A typical value for the capacitance per unitareais0.05F.cm-2, so for a 10 cm2electrodearea,thecapacitance,C,is0.5F.Skinresistanceisinverselyproportionaltotheelectrodearea.Thelargerthearea,thegreaterthenumberofconductivechannelsthroughthestratumcorneumandthelowerthetotalresistance.Theparallelresistance x unitareamighttypicallybe10k.cm2,sofora10cm2electrodearea,theparallelresistance,Rp,isabout 10 k.cm2/10 cm2 = 1 k.The resistanceofthetissuevolumebeneaththestratumcorneum is typically about 200 .Answer:Understeady-stateconditions,thecapacitorsare charged so no current flows to the capacitorsand the value of C is irrelevant.The combinationofelectrodesandtissuebehavesasthreeresistorsinseries.Thetotalresistanceis(1k+ 200 + 1 k) = 2200 .The applied potentialdifference is 50 V so the resulting current flow is,from Ohm's Law, I = V/R = 50/2200 = 0.0227 A =22.7 mA.ELECTRICAL PROPERTIES OF SKIN 50Question 2:How big is the initial current flow for the 'real life' situation described in question 1?Information:Initially,thecapacitorsactasa'shortcircuit'.Theyareunchargedandoffernoresistance to current flow.As each capacitor charges, the potential difference acrossthe capacitor increases and this opposes further flow of current.When fully charged,an ideal capacitor has infinite resistance.Uncharged, the 'resistance' is zero.Answer:The initial flow of current is resisted only by Rs.Thecapacitorsofferzeroresistancewhen they are uncharged so the current flow isI = V/R = 50/200 = 0.25 A = 250 mA.This is more than ten times larger than the steady-state current (question 1).Question 3:What is the steady-state potential difference across Rp and Rs for an applied potentialdifference of 50 V?Answer:Sincethethreeresistorsareinseries,thesamecurrent(22.7mA)flowsthrougheach.ThepotentialdifferenceacrosseachresistorcanbecalculatedusingOhm'sLaw.TheresistanceRs is 200 sothepotentialdifferenceacrossitisV=I.Rs =0.0227x200=4.5Volts.Foreachparallelresistance,Rp, V = I.Rp=0.0227x1000=22.7 Volts.THE USE OF STEADY DCThepreviouscalculationsdemonstrateanimportantideaconcerningtheuseofsteady direct current for patient treatment.Namely, that steady DC is of little practicaluseforstimulationofnervefibres.Forsteady,continuousflowofdirectcurrentELECTRICAL PROPERTIES OF SKIN 51Notice that in the answers toquestion 3, for an appliedpotential difference of 50 V,only 4.5 V is producedacross the tissue beneaththe stratum corneum.through the skin, the tissue impedance is high (Rs + 2Rp) and the potential differenceacrossRsissmall.Nervefibresarelocatedinthetissuesunderlyingthestratumcorneum(representedbyRs)sothepotentialdifferenceacrossthemissmallandtheirstimulationintensityislow.Mostoftheappliedvoltagedropoccursacrossthestratumcorneum(Rp).Thisisonereasonwhysteadydirectcurrentisnotusedfortranscutaneousnervestimulation.A second, more important, reason is that nerves are relatively insensitivetosteadyDCbecausetheyaccommodate.ThefiringthresholdprogressivelyincreaseswithaconstantlyappliedDCstimulus,sothenervewillceasefiringoncethe threshold has risen above the applied DC stimulus.Accommodation is describedin more detail in chapter 4.IONTOPHORESISItisforthereasonsoutlinedabovethatsteadydirectcurrentisnotusedfortranscutaneousnervestimulation.SteadyDCdoes,however,haveanimportantclinicalroleiniontophoresis,wheredrugsorotherchemicalagentsaredriventhroughtheskintotheunderlyingtissue.Iontophoresisonlyworksifthechemicaltobedrivenischarged.ItusessteadyDCsothatthechemicalionsaredrivencontinuallyinonedirectionwhilenonervestimulation(inparticular,nostimulationofpain or motor fibres) occurs.Anexampleisacetateioniontophoresisfordecalcificationofconnectivetissue.Asolubleacetate,suchassodiumacetateisdissolvedinconductivegel.Thisisappliedtotheskinunderthepositiveelectrode(anode).SteadyDCisappliedatacurrentlevelofabout10mAoveratreatmentperiodof10to20minutes.Thenegativelychargedacetateionsaredriventhroughtheskinintodeepertissue.Thecalciumincalcifiedtissueisintheformofcalciumphosphatecrystals,whichreactwiththeacetateionstoformthesolublesubstance,calciumacetate.Inthiswaythecalcium phosphate crystals are removed.Iftheiontobedriventhroughtheskinwaspositivelycharged,itwouldbeappliedunder the negative electrode (the cathode).ELECTRICAL PROPERTIES OF SKIN 52RESPONSE OF THE MODEL TO PULSED DIRECT CURRENTIf a pulse of direct current is applied to the circuit shown in figure 3.2, the initial currentflowthroughRs will be large, but will quickly drop as the capacitors charge.The timeconstant for the charging process is proportional to Rs.C.Figure 3.4 shows the effectof stimulus pulse width and time constant on the current flow.In(a)thepulsedurationislongcomparedtoRs.Csothecapacitorschargeduringthepulseandthecurrentquicklydropstoresistivecurrentflowonly.ThismeansthatifRpislarge,thecurrentwilldroptoaverylowvalue.Attheendoftheappliedvoltagepulsethecapacitorsdischarge,producinganegative-goingcurrentspike.Nervefibres,locatedindeepertissuebeneaththestratumcorneum(representedbyRs),wouldthusexperiencetwoshort-durationpulsesofstimulatingcurrent,oneatthebeginningandoneattheend of the applied voltage pulse.MostoftheappliedvoltageisineffectiveasfarasstimulationofnervefibresisFigure 3.4Response of the electrical model to pulsed direct current (a) when the pulse duration is long compared to R.C and (b) when the pulse duration is short compared to R.C.ssconcerned.In(b)thepulsedurationisshortcomparedtoRs.C.Thecurrentdecreasesonlyfractionallyduringthepulseandmostoftheresultingcurrent is due to the capacitors charging.The resulting current throughRsishighanddoesnotdecreasemuchduringthepulse.Mostoftheapplied voltage pulse is effective.ELECTRICAL PROPERTIES OF SKIN 53Figure3.4showsexamplesclosetotherangeofextremesforstimulationwithrectangularpulsedcurrent.Notethatthehorizontal(time)axisscaleisdifferentineach example. In figure 3.4(a) the time scale is relatively long compared to Rs.C andsharpspikesareseeninthegraphsofcurrentthroughRsversustime.Infigure3.4(b) the time scale is relatively short compared to Rs.C and the current through Rsshowsonlyasmalldecreaseduringthestimuluspulse.Duringthespikesin3.4(a),appreciablecurrentflowsthroughRs,meaningthatelectricalenergyisdissipatedinthe tissue volume beneath the stratum corneum.Between spikes, little current flowsthroughthedeepertissue(Rs), and most of the electrical energy is dissipated in thestratumcorneum.Theshortdurationpulsein3.4(b)resultsinappreciableandsustainedcurrentflowsthroughRs,somostoftheelectricalenergyisavailablefornervestimulation.Theaboveisoneofthereasonswhymodernstimulatorsuseshortdurationpulsesfortranscutaneousnervestimulation.Asecond,perhapsmoreimportantreasonisthatbetterdiscriminationbetweensensory,motorandpainfibrestimulationisachieved with narrower stimulus pulses.This will be considered further in chapter 4.RESPONSE OF THE MODEL TO ALTERNATING CURRENTThe term 'low frequency' currents refer to those in the 'biological'frequencyrangefrom 0 to approximately 100 Hz.'Medium frequencies' arein the kHz to tens of kHz range.'High frequencies' are in the 100 kHz and upward range.Theresponseofthemodeltoalternatingcurrentisparticularlyinterestingasalternatingcurrent(atkilohertzfrequencies)iscommonlyusedinphysiotherapyforpatient treatment, either in the form of 'interferential currents' or 'Russian stimulation'.Interferentialcurrentsare'mediumfrequency'ACcurrentsnormallyappliedatafrequency of 4 kHz.Russian current has a frequency of 2.5 kHz.Part of the reasonthat these frequencies are used is that at kHz frequencies, the stratum corneum hasa low electrical impedance.TheimpedanceofthestratumcorneumisthetotalimpedanceofRpandCintheelectricalmodel(figure3.2).Itisstronglydependentonthealternatingcurrentfrequency because of the capacitance, C.The impedance of C is given by the formula:ELECTRICAL PROPERTIES OF SKIN 54In most descriptions ofinterferential or Russiancurrents, the choice of akilohertz frequency is justified as being due to the low skinimpedanceatkHz frequencies.The low skinimpedance is due to the lowimpedance of both thestratum corneum and theunderlying skin layers. Zc=12fC.... (2.5)where Zc is the impedance and f is the alternating current frequency.Question 3:What is the capacitative impedance of the stratum corneum, Zc, at frequencies of (a)50 Hz, (b) 500 Hz and (c) 5 kHz?Information:Atypicalvaluefortheskincapacitanceperunitareais0.05F.cm-2.Atypicalelectrode area is 10 cm2 so the capacitance, C, in the model is about 0.5 F.Answer:(a) Using equation 2.5, the impedance at 50 Hz isZc =12fC=12 x 3.142 x 50 x 0.5 x 10-6=10650 x 3.142=6400 (b) Using equation 2.5, the impedance at 500 Hz isZc =12fC=12 x 3.142 x 500 x 0.5 x 10-6=106500 x 3.142=640 (c) Using equation 2.5, the impedance at 5 kHz isZc =12fC=12 x 3.142 x 5000 x 0.5 x 10-6=1035 x 3.142=64 Ifthestratumcorneumwaspurelycapacitative,itsimpedancewouldvaryascalculatedabove.Infacttheskinappendagesprovideaparallelresistivepathforcurrent flow (Rp in figure 3.2) so the total impedance of the stratum corneum is lower,ELECTRICAL PROPERTIES OF SKIN 55Strictly speaking, capacitative andresistiveimpedances must be added using complex number algebra.If a resistor, R and a capacitor, C, are connected in parallel, the total impedance, Z, is given by the formula:= ZRp(1+(2fRC) p21/2)psignificantlysoatlowfrequencies.Theparallelresistance x unitareaofthestratumcorneumistypicallyabout10k.cm2,sofora10cm2electrodearea,theparallelresistance, Rp, is about 10 k.cm2/10 cm2 = 1 k.Atafrequencyof5kHz,thecapacitativeimpedanceoftheskinisabout64andplacinga1kresistanceinparallelmakesvirtuallynodifferencetothetotalimpedance.At a frequency of 500 Hz, the capacitative impedance of the skin is about640andplacinga1kresistanceinparallelreducesthetotalimpedancetoappreciably less than 640 .At a frequency of 50 Hz, the capacitative impedance oftheskinisabout6400 and placing a 1 kresistanceinparallelreducesthetotalimpedance to less than 1 k.Despitetheeffectofskinappendagesofferingaconductivepathway(Rp)atlowfrequencies,thetotalimpedanceofthestratumcorneum(CandRpinparallel)stillshowsadramaticdecreasewithincreasingfrequency.Thus,foragivenstimulusvoltage,greatercurrentflowsathigherfrequencies.ThismeansthatthecurrentthroughRs,representingtheunderlyingtissues,ishigherathigherfrequenciesandthe potential difference across Rs is correspondingly higher.Question 4:What is the impedance of the stratum corneum at frequencies of (a) 50 Hz, (b) 500 Hzand (c) 5 kHz?Usethemodelinfigure3.2andvaluesofCandRpusedpreviouslyassuminganelectrode area of 10 cm2.Use the formula for two resistances in parallel to add the two impedances.Information:As indicated previously, a typical value for the skin capacitance for an electrode areaof 10 cm2 is about 0.5 F.The capacitative impedance is thus 6400 , 640 and 64 at frequencies of 50 Hz, 500 Hz and 5 kHz respectively.The parallel resistance, Rp,is about 1 k.ELECTRICAL PROPERTIES OF SKIN 56Answer:The impedance of the stratum corneum at each frequency is the total impedance of Cand Rp in parallel.Two resistances in parallel can be added using the formula1RT= 1R1+1R2Adding the impedances of C and Rp in this way we use1Z= 1Rp+1Zcwhere Zc is the impedance of C at the frequency concerned.(a) Using the above formula, the impedance at 50 Hz is calculated as follows:1Z=16400+11000=1000 + 64001000 x 6400=65001000 x 6400soZ=6400 x 10006500=985 (b) The impedance at 500 Hz is given by1Z= 1640+11000=1000 + 6401000 x 640=16401000 x 640soZ=640 x 10001640=390 (c) The impedance at 5 kHz is given by1Z=164+11000=1000 + 641000 x 64=10641000 x 64soZ=64 x 10001064=60 The impedance of the stratum corneum (C and Rp in parallel) thus shows a dramaticdecrease with increasing frequency: fromabout 1 k at 50 Hz to about 60 at 5 kHz.ELECTRICAL PROPERTIES OF SKIN 57ItisthepotentialdifferenceacrossRswhichdeterminesthestimulationintensityexperiencedbynervefibres,whicharelocatedunderthestratumcorneum.ThisdependsonbothZandRs. Rs is about 200 anddoesnotvaryappreciablywithfrequency.SinceZdecreasesappreciablywithincreasingfrequency,thepotentialdifferenceacrossRswillcorrespondinglyincrease.Question 5:What is the potential difference across Rs at frequencies of (a) 50 Hz, (b) 500 Hz and(c)5kHzforanappliedpotentialdifferenceof50V?UsethevaluesofZandRscalculated previously assuming an electrode area of 10 cm2.Information:AccuratecalculationofthepotentialdifferenceacrossRsrequirescomplexnumberalgebra,butagoodfirst-approximationcanbemadebytreatingtheimpedancesassimpleresistancesinseries.Answer:(a) Theimpedanceofthestratumcorneumat50Hzis985andRsis200.Thusthetotalimpedanceis985+200+985=2170.Foranappliedpotentialdifferenceof50V,thecurrentthroughtheseriescombinationisI=V/Z=50/2170=0.023A=23mA.ThisisalsothecurrentthroughRs,sothepotentialdifferenceacross Rs is V = I.Rs = 0.023 x 200 = 4.6 volts.(b) Theimpedanceofthestratumcorneumat500Hzis390andRs is 200 .Thus the totalimpedance is 980 .For an applied potential difference of 50 V,thecurrentthroughtheseriescombinationisI=V/Z=50/980=0.051A=51mA.This is also the current through Rs, so the potential difference across Rs isV = I.Rs = 0.051 x 200 = 10.2 volts.ELECTRICAL PROPERTIES OF SKIN 58(c) The impedance of the stratum corneum at 5 kHz is 60 and Rs is 200 .Thusthetotalimpedanceis320.Foranappliedpotentialdifferenceof50V,thecurrent through the series combination is I = V/Z = 50/320 = 0.156 A = 156 mA.Thus the potential difference across Rs is V = I.Rs = 0.156 x 200 = 31.2 volts.Noticethatat5kHz,thepotentialdifferenceacrossRsis31.2volts,whichismorethan 60% of the applied (50 V) stimulus.At 500 Hz, the potential difference across Rsisabout20%oftheappliedstimulusandat50Hz,about9%.Iftheobjectiveistostimulate nerve fibres which are located beneath the stratum corneum, the higher thealternating current frequency, the less the energy wasted in the stratum corneum andthe greater the current through, and voltage across, the deeper tissues.Ifskinimpedancewastheonlyfactor,thenthehigherthealternatingcurrentfrequency,thelowertheskinimpedanceandthemoreefficientlynervescouldbestimulated.Followingthislogic,if5kHzstimulationismore efficient than 500 Hz, then 50 kHz should be even more so.In factthis is not the case.The reason is that the impedance of the nerve fibrealso varies with frequency.NERVE FIBRE IMPEDANCETheelectricalpropertiesofnervefibresareanalogoustothoseofskinandunderlyingtissue.Thenervefibremembrane,likethemembraneofallcells,isaphospholipidbilayerwithanembeddedpatchworkormosaicofproteinmolecules.Thisfluid-mosaicstructure,illustratedalongside,explainsthemeasuredelectricalproperties.Orangeshapesarethepolar,high-water-content,proteinmolecules.Bluecircleswithtwodanglingchainsrepresentthephospholipidmoleculeswhichhavetwo long, non-polar 'tails'.Thephospholipidbilayerisaninsulator.Thehydrocarbontailsofthephospholipidmoleculesseparatetheconductiveintracellularfluidfromthe equally conductive extracellular fluid.Thus the nerve-fibre membraneactsasacapacitor.ProteinmoleculeswhichspanthewidthoftheELECTRICAL PROPERTIES OF SKIN 59Figure 3.5An electrical model for a nerve fibremembraneprovideconductivechannels,allowingsomeionstocrossthemembranecontinuously.Forthisreasonthemembranebehaveselectricallylikeacapacitor, C, with a parallel resistance Rp.Electrical stimulation of the nerve fibre involves currentflowacrossthemembrane,throughtheintracellularfluid and out across the membrane at some other point(usuallytheadjacentnodeofRanvier).Theinternalcontentsofthenervefibre,theintracellularfluid,ispurelyresistivesothenervefibrecanbeelectricallymodelled as shown in figure 3.5.Cinfigure3.5isthecapacitanceofthenerve-fibremembrane:aninsulatingphospholipid bilayer.Rp is the resistance of conductive channels through the bilayer(theproteinmoleculeswhichspanthefullwidthofthemembrane).Rsistheresistance of the intracellular fluid.Toexciteanervefibre,thatis,tocausethenervetofireandproduceanactionpotential,themembranepotentialmustbechangedfromitsrestingvaluetotheexcitationthreshold.Fornerveexcitationitisnotthepotentialdifferenceacrossthefibre which determines this but the potential difference across the fibre membrane (Cand Rp in figure 3.5).Sincethemembraneactsasacapacitor,ifalternatingcurrentisusedfornervestimulation,thepotentialdifferenceacrossthefibremembranewillbelessathigheralternatingcurrentfrequencies.ThedecreaseinmembraneimpedancewithincreasingfrequencymeansthatagreaterpotentialdifferencewillbeproducedacrossRsandacorrespondinglylowerpotentialdifferencewillbeproducedacrossC and Rp.At high frequencies, nerve fibres are less excitable because the potentialdifferenceproducedacrossthefibremembraneisreduced.Inpracticaltermsthismeansthatelectricalstimulationwithalternatingcurrentdoesnotbecomemoreandmoreefficientathigheralternatingcurrentfrequencies.thedecreaseinimpedanceofthestratumcorneum,whichresultsinahigherpotentialdifferenceacrosstheELECTRICAL PROPERTIES OF SKIN 60Figure 3.6Change in potential difference across the nerve fibre membrane (voltage across C and R in the electrical model shown in figure 3.5) when the pulse duration is (a) much longer than R.C and (b) shorter than R.C. s spunderlyingtissuesatkilohertzfrequenciesiscounteredbythedecreaseinimpedanceofthenervefibremembrane.Intheory,thedecreasednervefibresensitivitycouldbecompensated-forbyincreasingthestimulusintensitybutinpracticethiswouldresultinahighcurrentflowthroughtheskinandconsequentlyahighelectricalpowerdissipationandtheriskoftissuedamage.Theresultisthatfrequencies above about 10 kHz are of little use for eliciting a nerve response.Asi mi l arargumentappl i estostimulationwithrectangularpulses.Asindicatedpreviously,longdurationpulsesarerelativelyineffectivefortranscutaneousnervestimulationasmostoftheelectricalenergyisdissipatedinthestratumcorneumandthecurrentthroughthedeepertissuesisintheformoftwospikes,attheonsetandcessationofthestimulus(figure3.4(a)).Withshortdurationpulses(figure3.4(b)),littleenergyisdissipatedinthestratumcorneumandthecurrentthroughthedeepertissuesi ssustai nedduri ngthepul se.However,ifthepulsedurationisext r emel yshor t ( t ensofmi croseconds), t herewi l l beinsufficient time during the pulse for the nerve membrane capacitance tocharge.Figure3.6illustratestheresponseoftheelectricalmodelshown in figure 3.5 to short and longer duration current pulses.Withasufficientlylongdurationpulse,thereistimeforthemembranecapacitancetocharge(figure3.6(a))andthechangeinmembranepotentialismaximum.Withaveryshortdurationpulse,thereisinsufficienttimeforthemembranecapacitancetofullychargeandthechangeinmembranepotentialissmall(figure3.6(b)).TheuseofaELECTRICAL PROPERTIES OF SKIN 61higherstimulusintensitycancompensatefortheshortpulseduration,butiftheintensityrequiredistoolarge,thereistheriskofskindamage.Specifically,atstimulusintensitiesofseveralhundredvolts,thereistheriskofskinelectricalbreakdownwheretinyregionsofskinundertheelectrodessuddenlybecomehighlyconductive,allowinganextremelyhighcurrenttoflowandcausingtissuedamage.The net effect is that for transcutaneous stimulation using rectangular pulses, there isa'window'ofpulsewidthsbetweentensofmicrosecondsandabout1ms,outsidewhich stimuli are either ineffective or inefficient and potentially dangerous.EXERCISES1 Explain,intermsofthestructureofskin,whytheelectricalmodelshowninfigure3.2isanappropriatemodelforadescriptionoftranscutaneouselectricalnervestimulation.2 WhenaconstantDCstimulusisapplied,viaelectrodes,totheskinsurface,thereisalargetransientflowofcurrentthroughthetissue.Explainwhytheinitial flow of current is high and why the steady-state current is lower.3 (a) Whatisiontophoresis?(b) Why must the active agent be charged (i.e. in the form of an ion)?(c)What is the advantage of iontophoresis for application of medication?4 Figure3.4,page52,showsgraphsofthecurrentflowthroughtissueinresponsetostimuluspulsesofdifferentduration.(a) explainwhylongdurationpulsesproduceonlyshort-durationcurrentflows, as in figure 3.4(a).(b) Whyisthecurrentflowsustainedwhenshortdurationpulses,asinfigure3.4(b), are applied transcutaneously?5 Fortranscutaneousnervestimulation,bodytissuecanbemodelledbytheresistor/capacitorcombinationshownbelow.SkincapacitanceperunitareaisELECTRICAL PROPERTIES OF SKIN 62approximately0.05F.cm