1.0 Unit Outline• apply the concepts of current, resistance, potential difference (voltage drop),power to the operation of electronic circuits comprising diodes, resistors,thermistors, and photonic transducers including light dependent resistors (LDR),photodiodes and light emitting diodes (LED); V = IR, P = VI• calculate the effective resistance of circuits comprising parallel and seriesresistance and unloaded voltage dividers;• describe energy transfers and transformations in opto-electronic devices• describe the transfer of information in analogue form (not including the technicalaspects of modulation and demodulation) using– Light intensity modulation i.e. changing the intensity of the carrier wave toreplicate the amplitude variation of the information signal so that the signalmay propagate more efficiently– Demodulation i.e. the separation of the information signal from the carrierwave• design, investigate and analyse circuits for particular purposes using technicalspecifications related to potential difference (voltage drop), current, resistance,power, temperature, and illumination for electronic components such as diodes,resistors, thermistors, light dependent resistors (LDR), photodiodes and lightemitting diodes (LED);• analyse voltage characteristics of amplifiers including linear voltage gain(ΔVOUT/ΔVIN) and clipping;• identify safe and responsible practices when conducting investigations involvingelectrical, electronic and photonic equipment
Chapter 1• Topics covered:• Electric Charge.• Electric Current.• Voltage.• Electromotive Force.• Electrical Energy.• Electric Power.
1.0 Electric Charge• The fundamental unit of electricalcharge is that carried by the electron(& the proton).• This is the smallest discrete chargeknown to exist independently and iscalled the ELEMENTARY CHARGE.• Electric Charge (symbol Q) ismeasured in units called COULOMBS(C).• The electron carries - 1.6 x 10-19C.• The proton carries +1.6 x 10-19C.If 1 electron carries 1.6 x 10-19CThen the number of electrons in 1 Coulomb of Charge= 1 C1.6 x 10-19= 6.25 x 1018electrons
1.1 Flowing Charges• When electric charges (in particularelectrons) are made to move or “flow”,an Electric Current (symbol I) is said toexist.• The SIZE of this current depends uponthe NUMBER OF COULOMBS ofcharge passing a given point in a givenTIME.Section of Current Carrying WireMathematically:I = Q/twhere:I = Current in Amperes (A)Q = Charge in Coulombs (C)t = Time in Seconds (s)If 1 Amp of current is flowingpast this point,then 6.25 x 1018electronspass here every second.
1.2 Electric Current• Electric CURRENTS usually flow alongwires made from some kind ofCONDUCTING MATERIAL, usually, butnot always, a METAL.• Currents can also flow through aLiquid (electrolysis), through aVacuum (old style radio “valves”), orthrough a Semiconductor (ModernDiodes or Transistors).• A Current can only flow around aCOMPLETE CIRCUIT.• A break ANYWHERE in the circuitmeans the current stops flowingEVERYWHERE, IMMEDIATLY.• The current does not get weaker as itflows around the circuit, BUTREMAINS CONSTANT.• It is the ENERGY possessed by theelectrons (obtained from the battery orpower supply) which gets used up asthe electrons move around the circuit.• In circuits, currents are measured withAMMETERS, which are connected inseries with the power supply.Typical Electric CircuitConnectingWiresResistor (consumesenergy)BatteryCurrentAMeasuresCurrentFlow
1.3 Conventional Current vsElectron CurrentPositive Terminal Negative TerminalConventional vs Electron CurrentResistorElectron Current:Never shown onCircuit DiagramsConventional Current:Always shown onCircuit DiagramsWell before the discovery ofthe electron, electric currentswere known to exist.It was thought that thesecurrents were made up of astream of positive particles andtheir direction of movementconstituted the direction ofcurrent flow around a circuit.This meant that in a Direct Current(D.C.) circuit, the current would flowout of the POSITIVE terminal of thepower supply and into the NEGATIVEterminal.Currents of this kind are calledConventional Currents, and ALLCURRENTS SHOWN ON ALLCIRCUIT DIAGRAMS EVERYWHEREare shown as Conventional Current,as opposed to the “real” orELECTRON CURRENT.
1.4 Voltage• To make a current flow around acircuit, a DRIVING FORCE is required.• This driving force is the DIFFERENCEin VOLTAGE (Voltage Drop orPotential Difference) between thestart and the end of the circuit.• The larger the current needed, thelarger the voltage required to drivethat current.• VOLTAGE is DEFINED as theENERGY SUPPLIED TO THE CHARGECARRIERS FOR THEM TO DO THEIRJOB ie.TRAVEL ONCE AROUND THECIRCUIT.• So, in passing through a Voltage of1 Volt, 1 Coulomb of Charge picksup 1 Joule of Electrical Energy.• OR• A 12 Volt battery will supply eachCoulomb of Charge passingthrough it with 12 J of Energy.Mathematically;V = W/Qwhere:V = Voltage (Volts)W = Electrical Energy (Joules)Q = Charge (Coulombs)Alessandro Volta
1.5 E.M.F.Voltage is measured with a VOLTMETER.The term EMF (ELECTROMOTIVE FORCE)describes a particular type of voltage.It is the VOLTAGE of a battery or powersupply when NO CURRENT is being drawn.This is called the “Open Circuit Voltage” ofthe battery or supplyVVoltmeterCircuit SymbolWith S closed, a current begins toflow and V drops and nowmeasures voltage available todrive the current through theexternal circuitResistorAVSV measures EMFVoltmeters are placed in PARALLEL withthe device whose voltage is beingmeasured.Voltmeters have a very high internalresistance, so they have little or no effectthe operation of the circuit to which they areattached.ResistorAV
10Electronics & Photonics RevisionQuestion Type:Q1: Which one of the following statements (A to D) concerning thevoltage across the resistor in Figure 1 is true?A. The potential at point A is higher than at point B.B. The potential at point A is the same as at point B.C. The potential at point A is lower than at point B.D. The potential at point A varies in sign with time compared to thatat point B.Potential Difference
1.6 Electrical EnergyThe conversion of ElectricalEnergy when a current passesthrough a circuit element (acomputer) is shown below.MathematicallyW = VQ ………1,where:W = Electrical energy (Joule)V = Voltage (Volts)Q = Charge (Coulomb)Current and Charge arerelated through:Q = It.substituting for Q, inequation 1 we get:W = VItVoltage= V voltsCharges (Q) enterwith high energyCharges (Q) leavewith low energyQ Coulombs ofElectricity entercomputerQ Coulombs ofElectricity leavecomputerIn time t, W units of energy are transformed to heat and lightElectrical Energy (W) isdefined as the product of theVoltage (V) across, times theCharge (Q), passing througha circuit element (eg. a lightglobe).
Electronics & Photonics RevisionQuestion Type:Q2: Determine the electrical energydissipated in the 100 Ω resistor of Figure 1 in1 second. In your answer provide the unit.Electrical EnergyA: Electrical energy W = VQ = VIt= (4.0)(40 x 10-3)(1)= 0.16 Joule
1.7 Electrical Power• Electrical Power is DEFINED as theTime Rate of Energy Transfer:P = W/twhere P = Power (Watts, W)W = Electrical Energy (Joule)t = Time (sec)• From W = VI t we get:P = VI• From Ohm’s Law (V = IR) [see nextchapter] we get:P = VI = I2R = V2/Rwhere: I = Current (Amps)R = Resistance (Ohms)V = Voltage (Volts)Electrical Power is sold toconsumers in units of Kilowatt-Hours. (kW.h)A 1000 W (1kW) fan heater operatingfor 1 Hour consumes 1kWh ofelectrical power.Since P = W/t or W = P x t, we can say:1 Joule = 1 Watt.secso1000 J = 1kW.secso3,600,000 J = 1 kW.houror3.6 MJ = 1 kW.h
1.8 A.C. Electricity• There are two basic types of currentelectricity:(a) D.C. (Direct Current) electricitywhere the current flows in onedirection only.(b) A.C. (Alternating Current) where thecurrent changes direction in aregular and periodic fashion.• The Electricity Grid supplies domesticand industrial users with A.C.electricity.• A.C. is favoured because:(a) it is cheap and easy to generate(b) it can be “transformed”; its voltagecan be raised or lowered at will bypassage through a transformer.• The only large scale use of highvoltage D.C. electricity is in publictransport, ie. trams and trains.VoltageTimeVP VPtoPTA.C. ELECTRICITY - PROPERTIESVPtoP = “Peak to Peak Voltage”for Domestic Supply VPtoP = 678 VT = “Period”for Domestic Supply T = 0.02 secVP = “Peak Voltage”for Domestic Supply VP = 339 V
1.9 R.M.S. Voltage and CurrentVt339-3390V2t1.15 x 1050 0Mean V25.8 x 104t0Mean V2240tGRAPHICAL DEVELOPMENT OF THE RMS VOLTAGE FROM AN A.C. VOLTAGEWith an A.C. supply, the average valuesfor both voltage and current = 0,so Vav and Iav cannot be used by thePower Companies to calculate theamount of electric power consumed byits customers.To get around this problem R.M.S. orRoot Mean Square values for ACvoltage and current were developed.RMS values are DEFINED as:The AC Voltage/Current whichdelivers the samevoltage/current to an electricaldevice as a numerically equalD.C. supply would deliver.An AC source operating at 240V RMS delivers the same powerto a device as a DC source of240 V.Yet, AC circuits do consume power,so a method of calculating it had tobe found.
1.10 Peak versus RMS Values• In AC supplies, the Peakand RMS values are relatedthrough simple formulae:• For Voltage:VRMS = VP/√2• For Current:IRMS = IP/√2• In Australia DomesticElectricity is supplied at240 V, 50 Hz• The Voltage quoted is theRMS value for the ACsupply.• Thus the Peak value forvoltage isVP = VRMS x √2= 240 x 1.414= 339 VVoltage (V)Time (s)VP+339 V- 339 VVP to P240 V
Chapter 2• Topics covered:• Resistance.• Ohm’s Law.• Resistors in Series and Parallel.• Voltage Dividers• Impedance Matching
2.0 Resistance• Electrical Resistance is a property ofALL materials, whether they beclassed as conductors, insulators orsomething in between. (ieSemiconductors)• The size of the resistance dependsupon a number of factors:(a) The nature of the material. This ismeasured by “resistivity” (ρ)(b) The length, L, of the material.(c) The cross sectional area, A, of thematerial.COMPARING RESISTANCELA 2A 1Wires 1 and 2 are made from thesame materialWire 1 has ½ the cross sectionalarea of Wire 2∴ Wire 1 has TWICE the resistanceof Wire 2Combining these mathematically:R = ρL/Awhere:R = Resistance (Ohms) Ωρ = Resistivity (Ohm.m) Ω.mL = Length (m)
2.1 Resistors in Series• Conductors which exhibit aresistance to current flow aregenerally called RESISTORS.• When connected “end to end” or in“SERIES”, the total resistance of thecombination = the sum of theindividual resistances of theresistors in the “network”.• Mathematically:RT = R1 + R2 + R3 + … …IN A SERIES CIRCUIT:(a) Since only ONE pathway around thecircuit exists, the current through eachresistor is the same.Thus: I = I1 = I2 = I3Resistors in SERIESThese three resistors can be replacedby a single resistor of valueRT = R1 + R2 + R3R1R2 R3VV1V2 V3Resistors in a Series Circuit(b) The sum of the voltage drops acrossthe resistors = the voltage of the powersupply,Thus: V = V1 + V2 + V3II1 I2I3The greater the number of resistors in a series network the greater thevalue of the equivalent resistance (RT)R1 R2 R3RT
2.2 Resistors in Parallel• Resistors connected “side by side”are said to be connected in“PARALLEL”.• The total resistance of a parallelnetwork is found from adding thereciprocals of the individualresistances.IN A PARALLEL CIRCUIT:(a) The current through each arm varies.Thus: I = I1 + I2 + I3R3R2R1These three Resistorscan be replaced by asingle Resistor ( RT )Resistors in ParallelResistors in a Parallel CircuitR3R2R1VI3I2I1IV1V3V2(b) The voltage drop across eacharm is the same.Thus: V = V1 = V2 = V3The greater the number of resistors in aparallel network the lower the value of theequivalent resistance (RT).Mathematically:1/RT = 1/R1 + 1/R2 + 1/R3RT
Electronics & Photonics RevisionQuestion Type:You wire up the circuit shown inFigure 1 but only have 10 kΩresistors to work with.Q3: Explain how you wouldconstruct the R1 = 5 kΩ resistorusing only 10 kΩ resistors.Include a sketch to show theconnections between theappropriate number of 10 kΩresistors.VINVOUTR2R1Connect two 10k resistors together in parallelVINVOUTR2R1Parallel Resistors
Electronics & Photonics RevisionQuestion Type:Q4: Which one of the following statements (A to D) concerning theRMS currents in the circuit of Figure 2 is true?A. The current in resistor A is identical to the current in resistor C.B. The current in resistor D is twice the current in resistor C.C. The current in resistor B is twice the current in resistor E.D. The current in resistor A is identical to the current in resistor D.20 VRMSVOUTA BCD ECurrent Flows
2.3 Ohm’s Law• OHM’S LAW relates the Voltageacross, the Current through and theResistance of a conductor.• Mathematically:V = IRwhere: V = Voltage(Volts) I = Current(Amps) R =Resistance (Ohms)• Any conductor which followsOhm’s Law is called an OHMICCONDUCTOR.Ohm’s Law - GraphicallyVIA graph of V versus I produces astraight line with Slope = R(Remember a straight linegraph has formula y = mx + c)The graph is a straight line,∴ the Resistance of Device 1 isCONSTANT (over the rangeof values studied).The slope indicates Device 2has a lower (but still constant)Resistance whencompared to Device 1.Slope = RDevice 1Slope = RDevice 2Georg Ohm
Electronics & Photonics RevisionQuestion Type:Figure 1 shows a resistor, alinear circuit component,with resistance R = 100 Ω.A DC current, I = 40 mA,passes through this resistorin the direction shown bythe arrows.Q5: What is the voltagedrop across thisresistor? Express youranswer in volts. A: V = IR= (40 x 10-3)(100)= 4.0 VOhm’s Law
2.4 Non Ohmic Devices• Electrical devices which followOhm’s Law (V = IR) are calledOhmic Devices.• Electrical devices which do notfollow Ohm’s Law are calledNon Ohmic Devices.• Non Ohmics show non linearbehaviour when a plot of V vs Iis produced, as can be seen inthe graphs for components Xand Y opposite.• Most of the individualcomponents covered in thissection of the course are NonOhmic Devices.Voltage (V)Current (A)Component Y0510152 4 6 8Current (A)Voltage (V)Component X0510151 2 43
Electronics & Photonics RevisionQuestion Type:A resistor is a linear device. Anexample of a non-linear deviceis a light-emitting diode (LED).Q6: On the axes provided, sketch atypical current-voltage characteristiccurve for each of the devices mentioned.In both cases label the axes and indicateappropriate units.I(Amp)V(volts)a linear device – a resistor a non linear device – an LEDV(volts)I(mA)Ohmic & Non Ohmic Devices
2.5 Voltage Dividers - 1For the circuit above:V = V1 + V2Since this is a series circuit ,the current ( I ) is the sameeverywhere:I = V1/R1 and I = V2/R2So V1/V2 = R1/R2R1V1R2V2VISuppose you have a 12 Vbattery, but you need only 4 Vto power your circuit. How doyou get around this problem ?You use a Voltage DividerCircuit.They are made by usingcombinations of fixed valueresistors or using variableresistors called rheostats.Voltage dividers are one of the mostimportant circuits types used inelectronics.Almost all sensor subsystems (egThermistors, LDR’s), use voltagedivider circuits, there is just no otherway to convert the sensor inputs intouseful “electrical” information.
2.6 Voltage Dividers - 2If the main voltage supply (V) isconnected across the ends of therheostat, then the voltage requiredby RL is tapped between A and theposition of the slider.VARheostatRLSliderThe further from A the slider moves the larger thevoltage drop across the load resistor , RLUsing rheostats, the a voltage dividercan be set up as shown.Slider type rheostatVariousrotaryrheostats
2.7 Voltage Divider FormulaFor the VOUT circuit:VIN = I (R1 + R2)VIN CircuitVOUT CircuitR1R2VINVOUTIFor the VIN circuit:Applying Ohm’s LawThe Voltage divider circuit is a SERIES circuit.Thus, the SAME CURRENT flows EVERYWHEREIn other words, the SAME CURRENT flows through R1 AND R2∴ I = VIN(R1 + R2)…….(1)VOUT = IR2∴ I = VOUTR2……..(2)Combining 1 and 2 we get:VOUT = VINR2(R1 + R2)so, VOUT = VIN.R2(R1 + R2)This is the Voltage Divider Formula
Electronics & Photonics RevisionQuestion Type:20 VRMSVOUTA BCD EIn Figure 2, five identical 100 Ωresistors are used to construct avoltage divider. The voltage sourceacross this voltage divider is anAC supply with an RMS voltage of20 V. The resistors are labelled bythe letters A to E as shown.Q7: What is the RMS output voltage,VOUT?Voltage Divider NetworkA: Step 1 Determine the equivalentresistance for A and B and D and EFor A and B: 1/RE = 1/RA + 1/RB= 1/100 + 1/100= 2/100So RE = 100/2= 50 ΩReplace A and B with one 50Ω resistor, same with D and E.You can now redraw the circuit.
Electronics & Photonics RevisionQuestion Type:The series circuit in Figure 3 can befurther simplified as shown in Figure 420 VRMSVOUTA BCD E50 Ω50 Ω100 ΩFigure 3VOUT =VIN R2(R1 + R2)The original question (value ofVOUT) can now be calculated, usingthe Voltage Divider formula:=20(150)(50 + 150)= 15 VVoltage Divider Network20 VRMSVOUTA BCD E50 Ω50 Ω150 ΩFigure 4(R1)(R2)
Electronics & Photonics RevisionQuestion Type:An essential component in some ofthe practical circuits covered in thisexam paper is the voltage divider. ADC voltage divider circuit is shownin Figure 1.VINVOUTR2R1For the circuit of Figure 1, VIN = 30 V, R1 = 5 kΩand the output voltage VOUT = 6 V.Q8: What is the value of the resistance R2? Showyour working.Voltage Divider NetworkVOUTVIN(R1 + R2)R1= 65000=30(5000 + R2)R2 = 20,000 Ω = 20kΩ
Electronics & Photonics RevisionQuestion Type:In Figure 1 the 30 V DC input to thevoltage divider is replaced by a 100 mV(peak-to-peak) sinusoidal AC inputvoltage. The resistance values are nowR1 = 5 kΩ and R2 = 15 kΩ.VINVOUTR2R1100 mV5k15kA: Series circuit – add resistances so RT = 20kΩUse Ohm’s Law to find current V = IR so I =100 x 10-320 x 103= 5μAOhm’s LawQ9: What is the current through resistorR2? Show your working, and expressyour answer as a peak-to-peak currentin μA.
2.8 Impedance Matching 1IMPEDANCE is the TOTAL resistance to currentflow due to ALL the components in a circuit.In Voltage Divider circuits we only have resistors,so Total Impedance = Total Resistance.The current (I) in the circuit is:I = V/RT= 12/1200= 0.01 A.In the circuit shown a supply of 12 Vis connected across 2 resistors of500 Ω and 700 Ω in series.IR2V2VR1V17 V5 V 500 Ω700 Ω12The Voltage Drop across R1= I x R1= 0.01 x 700= 7.0 VThe Voltage Drop across R2= I x R2= 0.01 x 500= 5.0 V
CASE (b): Now RL = 5000 Ω,Then RT = (1/500 + 1/5000)-1= 454.5 Ω andI = V/RT= 0.011 A.This is only a 10 % increase incurrent.CASE (a):Suppose RL has a total impedance of50 ΩRL and R2 are in parallel,so Total Resistance RT for the parallelnetwork = (1/R2 + 1/RL)-1= (1/500 + 1/50)-1= 45.5 Ω∴I = V/RT= 5.0/45.5= 0.11 A.This is an 110% increase in thecurrent in the circuit.This will cause a dangerous heatingeffect in R1 and also decrease theVoltage across RL - both undesirableevents !Suppose a load (RL), requires5.0 V to operate.Conveniently, 5 V appearsacross R2.2.9 Impedance Matching 2IR2V2VR1V1500 Ω700 Ω127 V5 V RL50 Ω5 VIn other words it is important to “match”the impedance of the load RL to that ofresistor R2 such that: RL ≥ 10R25 V 500 Ω 5000 ΩLets look at 2 cases where the impedanceof RL varies.
3.0 Semiconductors• Most electronic devices, eg. diodes,thermistors, LED’s and transistors are“solid state semi conductor” devices.• “Solid State” because they are made upof solid materials and have no movingparts.• “Semiconductor” because thesematerials fall roughly in the middle ofthe range between Pure Conductor andPure Insulator.• Semiconductors are usually made fromSilicon or Germanium with impuritiesdeliberately added to their crystalstructures.• The impurities either add extra electronsto the lattice producing n typesemiconductor material.N - Type SemiconductorSi SiSi SiP SiSi SiextraelectronP - Type SemiconductorSi SiSi SiB SiSi Siholeor create a deficit of electrons (called“holes”) in the lattice producing ptype semiconductor material.Holes are regarded as positive (+)charge carriers, moving through thelattice by having electrons jump intothe hole leaving behind another hole.
3.1 p-n junctionsp nJoining together p type and n typematerial produces a so called “p-njunction”When brought together, electronsfrom the n type migrate to fill holesin the p type material.npnpAs a result, a “depletion layer”, (aninsulating region containing very fewcurrent carriers), is set up betweenthe two materials.depletion layerThe “majority” current carriers areholes in p type material and electrons inn type material.However, each also has some“minority” carriers (electrons in p, holesin n) due to impurities in thesemiconductor and their dopeantsNote: undoped semiconductormaterial, pure silicon orgermanium, is called “intrinsicsemiconductor material”.
3.2 Forward and Reverse Biasp ndepletion layerit draws the charge carriers awayfrom the junction and makes thedepletion layer bigger meaningcurrent is even less likely to flowand the junction is now “reversebiased”p ndepletion layerit draws thecharge carriers towardthe junction and makesthe depletion layersmaller.If an external supply isnow connected asshownThe current carriers nowhave enough energy tocross the junction whichnow becomes“conducting” or “forwardbiased”If the externalsupply is nowreversed,
3.3 The Diode• Diodes are electronic devices made bysandwiching together n type and ptype semiconductor materials.• This produces a device that has a lowresistance to current flow in onedirection, but a high resistance in theother direction.Cathode (-)Anode (+)ConventionalCurrent FlowCurrent (mA)Voltage (V)0.7 VThe “Characteristic Curve”(the I vs V graph) for atypical silicon diode isshown.This diode will not fully conductuntil a forward bias voltage of 0.7V exists across it.Notice that when thediode is reverse biased itdoes still conduct - butthe current is in the pA orμA range.This current is due tominority carriers crossingwhat is for them a forwardbiased junction.V (μA)Circuit Symbol
3.4 The TransistorThere are two general groups oftransistors:•BJT (Bipolar Junction Transistors)•FET (Field Effect Transistors)There are two basic types of BJT’s:•NPN Transistors•PNP TransistorsLets look at theConstruction of a BJTnpn type transistorEmitterCollectorBaseBaseCollectorEmitterCircuit symbolNPNNote: npn transistors havethe arrow:Not Pointing iNThe arrow points in thedirection of conventionalcurrent flowAn npn type transistor
3.5 Transistor UsesTransistors are used to perform three basic functions.They can operate as either(a) a switch; or(b) an amplifier;There are over 50million transistorson a singlemicroprocessorchip.(The Intel®Pentium 4 has 55million transistors)This is first ever solid state amplifier(transistor) and was created in 1947at Bell Labs in the USor (c) an oscillator•The term transistor comes from the phrasetransfer-resistor because of the way its inputcurrent controls its output resistance.
4 . 0 Ph ot on ic sPhotonics is the technology of usinglight to transmit “information” fromone place to another.The light source used is almost alwaysthe laser and the means oftransmission is the optical fibre.Light has the ability to transmit“information” at a much faster ratethan electrons in copper wires.Photonics main use today is intelecommunications.With optical fibres costing only afraction of previously used copperwires and having the ability to carryfar more information,telecommunications has beenrevolutionised by the use ofPhotonics.Photonic devices fall into 2 generalcategories:Photovoltaics they generate theirown voltage and do not require anexternal power supply, examplesolar cells.Photoconductive require anexternal supply and operate bymodifying the current, exampleLight Dependent Resistor (LDR) orPhotodiode
4.1 PhotodiodesThe photovoltaic detectormay operate withoutexternal bias voltage.A good example is thesolar cell used onspacecraft and satellites toconvert the sun’s light intouseful electrical power.Photodiodes are detectorscontaining a p-nsemiconductor junction.Photodiodes arecommonly used incircuits in which there isa load resistance inseries with the detector.The output is read as achange in the voltagedrop across the resistor.The magnitude of thephotocurrent generated by aphotodiode is dependent uponthe wavelength of the incidentlight.Silicon photodiodes respondto radiation from the ultravioletthrough the visible and into thenear infrared part of the E-Mspectrum.RL VOUT+V0 VThey are unique in that theyare the only device that cantake an external stimulusand convert it directly toelectricity.
4.2 PhototransistorsLike diodes, all transistors arelight-sensitive.Phototransistors are designedspecifically to take advantage ofthis fact.The most-common variant is anNPN bipolar transistor with anexposed base region.Here, light striking the basereplaces what would ordinarily bevoltage applied to the base -- so, aphototransistor amplifiesvariations in the light striking it.Phototransistors may or may nothave a base lead (if they do, thebase lead allows you to bias thephototransistors light response.Note that photodiodes alsocan provide a similarfunction, although withmuch lower gain (i.e.,photodiodes allow muchless current to flow thando phototransistors).Phototransistors are usedextensively to detect lightpulses and convert theminto digital electricalsignals.In an optical fibre networkthese signals can be useddirectly by computers orconverted into analoguevoice signals in atelephone.
4.3 Phototransistor ApplicationsRL+V0VVOUTRL+V0VVOUTWhen light is onVOUT is HighWhen light is onVOUT is LowPhototransistors can be used as light activated switches.Further applications1. Optoisolator- the opticalequivalent of an electricaltransformer. There is nophysical connectionbetween input and output.2. Optical Switch – anobject is detected when itenters the space betweensource and detector.
4.4 Optoisolator CircuitHow does VOUT respond tochanges to VIN ?As the input signal changes,IF changes and the light levelof the LED changes.This causes the base currentin the phototransistor tochange causing a change inboth IC and hence VOUTThe response of the phototransistor is notinstantaneous, there is a lag between achange in VIN showing up as a change in VOUTIFtICtAssume VIN varies such that the LEDswitches between saturation (full on) andcut off (full off), producing a square wavevariation in IFIC will respond showing a slight time lagevery time IF changes state
4.5 Opto-electronic DevicesAn op amp (operational amplifier)is a high gain, linear, DC amplifierThe inputs marked as (+) and (-)do not refer to power supplyconnections but instead refer toinverting and non invertingcapabilities of the amplifier.
Electronics & Photonics RevisionQuestion Type:You are asked to investigate theproperties of an optical coupler,sometimes called an opto-isolator. This comprises a light-emitting diode (LED) thatconverts an electrical signal intolight output, and aphototransistor (PT) that convertsincident light into an electricaloutput. Before using an opto-isolator chip you consider typicalLED and PT circuits separately.A simple LED circuit is shown inFigure 4 along with the LEDcurrent-voltage characteristics.The light output increases as theforward current, IF , through theLED increases.Q10: Using the information in Figure 4,what is the value of the resistance, RD, inseries with the LED that will ensure theforward current through the LED is IF = 10mA?For 10 mA to flow through LED requires avoltage of 1.5 V (read from graph)Because LED and R are in seriesVRD = 10 – 1.5 = 8.5 VVRD = IRD so RD = VRD/I = 8.5/(10 x 10-3)= 850 ΩOhm’s Law
Electronics & Photonics RevisionQuestion Type:Q11: Will the light output of theLED increase or decrease if thevalue of RD is a little lower thanthe value you have calculated inthe last question? Justify youranswer.Justification: Reducing the valueof RD will not affect the voltagedrop across it.The Voltage across RD iscontrolled by the LED which willremain at 1.5 V thus VRD will stillequal 8.5 V.So if V remains the same and R goesdown I must go up. So a largercurrent flows through the LEDmeaning an increased light outputIncreased outputOhm’s Law
Electronics & Photonics RevisionQuestion Type:You now consider the phototransistor (PT)circuit of Figure 5 with RC = 2.2 kΩ. The lightis incident upon the base region of the PTand produces a collector current, IC.Q12: As the light intensity incident on the PT increases, which one of thefollowing statements concerning the PT-circuit of Figure 5 is correct?A. The collector current remains constant, but VOUT increases.B. The collector current remains constant, but VOUT decreases.C. The collector current increases, but VOUT decreases.D. The collector current decreases and VOUT decreases.Phototransistor
4.6 CD ReadersCD pitsdigitalsignalanaloguesignallaserphotodiodeDACdigital toanalogueconverteramplifier speakerCompact discs store information in Digital form.This information is extracted by a laser andphotodiode combination.The data is passed through a series of electronicprocesses to emerge from the speaker as sound
Electronics & Photonics RevisionQuestion Type:The information on an audioCD is represented by a seriesof pits (small depressions) inthe surface that are scannedby laser light. When there isno pit the reflected light givesa maximum light intensity, I1,detected by a photodiodecircuit. When the laser lightstrikes a pit, the light intensityis reduced to I0. A plot of atypical light intensity incidenton the photodiode is shown inFigure 4.
Electronics & Photonics RevisionQuestion Type:Q13: With no light incidentupon the photodiode, thecurrent in the photodiodecircuit, the “dark current”,is 5 µA.What is the output voltage,VOUT, across the 100 Ωresistor in the circuit ofFigure 5b?Ohm’s LawA: The photodiode and resistor are in series.The same current flows through each.VOUT = V100Ω = IR= (5 x 10-6)(100)= 5 x 10-4VThe variation in current as a function of lightintensity for the photodiode is shown in Figure5a, together with the circuit used to determinethis, which is shown in Figure 5b.
5.0 Analogue DataThe world is divided into twostreams:Analogue and DigitalHumans perceive the world as ananalogue place i.e. we receive ourinput is a continuous stream, thiscontinuous stream is what definesanalogue data.On the other hand digital data (astream of 1’s and 0’s) estimatesanalogue data by “sampling” itat various time intervalsAnalogue data is usually moreaccurate than digital data.However digital data is easierto store and manipulate and ofcourse computers can onlycope with digital dataDigital systems arenot just a moderninvention.Examples of ancientdigital systemsinclude:The AbacusMorse CodeBrailleSemaphore
5.1 ModulationModulation is a a way of changing ananalogue signal so data or informationcan be transmitted over a communicationnetwork.“Modulated” signals consist of2 components(a) A carrier signal(b) An information or datasignalThe carrier is usually of one frequencyand the wave (usually a sine wave) isy(t) = A sin (ft + φ)WhereA = Amplitudef = Frequencyφ = PhaseChanging (modulating) this wave canonly occur by changing one of A, f or φChanging A leads to Amplitude ModulationChanging f leads to Frequency ModulationChanging φ leads to Phase Modulation
5.2 DemodulationDemodulation is the inverse process ofmodulation. The modulated wavesignal is transmitted to a receiver at thereceiving station.Then information components areextracted from the carrier signal(recovering information). Theprocess is called demodulation.
5.3 Fibre OpticsAll forms of modern communication--radio and television signals, telephoneconversation, computer data--rely on acarrier signal.By modulating the carrier, we canencode the information to betransmitted; the higher the carrierfrequency, the more information asignal can hold.In 1960, an idea first introducedby Albert Einstein more than 40years earlier bore practical fruitwith the invention of the laser.The idea of using visible light as amedium for communication hadoccurred to Alexander Graham Bellback in the late 1870s, but he didnot have a way to generate a usefulcarrier frequency or to transmit thelight from point to point.This achievement promptedresearchers to find a way to makevisible light a communicationmedium--and a few years later fibreoptics arrived.A.G. Bell
Figure 9 is a sketch of an electro-optical system that allowssound to be transmitted over a distance via a fibre opticcable, using light.Electronics & Photonics RevisionQuestion Type:Q14. Explain the terms modulation and demodulation as they applyto the transmission of sound by this system.In the context of this question, modulation referred to variation of thelight intensity. Demodulation meant that the variation in the light intensitycreated an electrical signal.Modulation and Demodulation
Figure 8a, below, shows a schematicdiagram for an intensity-modulatedfibre-optic link that is used to transmitan audio signal.Electronics & Photonics RevisionQuestion Type: Modulation and DemodulationTo test the device an audio signalis fed into the microphone. Thesignal at point W is shown inFigure 8b.Q15. Which of the diagrams (A–D)below best represents the signalobserved at point X in Figure 8a?At X there is light of varyingintensity, so the answer was D.The light intensity cannot gobelow zero, so B was not anoption. Even when there is nosignal at W to modulate the outputof the laser diode, there is auniform brightness emitted.
Figure 8a, below, shows a schematicdiagram for an intensity-modulatedfibre-optic link that is used to transmitan audio signal.Electronics & Photonics RevisionQuestion Type: Modulation and DemodulationTo test the device an audio signalis fed into the microphone. Thesignal at point W is shown inFigure 8b.Q 16. Which of the diagrams (A–D)above could represent the signalthat would be observed at point Yin Figure 8a?The varying brightness incident onthe photodiode would cause avoltage like C to be produced.
6.0 Input TransducersTransducers are devices which convert nonelectrical signals into electrical signals.Input Transducers convert mechanical andother forms of energy eg. Heat, Light orSound into Electrical Energy.Light Emitting Diode (LED)Light is emitted when the diodeis forward biasedLight Dependent Resistor (LDR)The resistance changes aslight intensity variesSymbolExamples of a few such devicesare shown here.PhotodiodesCurrent flows when light of aparticular frequency illuminatesthe diodeThermistorThe resistancechanges as thetemperaturechanges
6.1 Light Emitting Diodesanode (+)cathode (-)flat edgeLEDs emit lightwhen an electriccurrent passesthrough them.LEDs must be connected the correct wayround.The diagram may be labelled a or + foranode and k or - for cathode (yes, it reallyis k, not c, for cathode!).The cathode is the short lead and theremay be a slight flat region on the body ofround LEDs.Circuit Symbola kLEDs must have aresistor in seriesto limit the currentto a safe valueNotice this is a voltagedivider circuitMost LEDs are limited to a maximumcurrent of 30 mA, with typical VL valuesvarying from 1.7 V for red to 4.5 V for blue
Electronics & Photonics RevisionQuestion Type:The LED in Figure 4 is an electro-optical converter.Q17. Which one of the followingstatements (A to D) regardingenergy conversion for the LED iscorrect?All the electrical energy supplied from the DC power supply is convertedA. only to heat energy in both the resistor, RD, and the LED.B. partly to heat energy in the resistor, RD, the remainder to light-energy outputfrom the LED.C. partly to heat energy in both the resistor, RD, and the LED, with the remainderto light-energy output from the LED.D. to heat energy in the LED, with the remainder to light-energy output from theLED.Energy Conversion
Electronics & Photonics RevisionQuestion Type:Q18: Describe the basic purpose of eachof the following electronic transducers.i. Light-Emitting Diode (LED)ii. Photodiode(i) Emits visible light when a current flows through it.Converts electrical energy to light.(ii) Switches on (allows a current to flow through it) when exposed to light.Converts light to electrical energy.Transducer Properties
6.2 Light Dependent Resistors (1)The light-sensitive partof the LDR is a wavytrack of cadmiumsulphide.Light energy triggersthe release of extracharge carriers in thismaterial,so that its resistancefalls as the level ofillumination increases.A light sensor uses an LDR aspart of a voltage divider.Suppose the LDR has a resistanceof 500Ω , (0.5 kΩ), in bright light,and 200 kΩ in the shade (thesevalues are reasonable).When the LDR is inthe light, Vout will be:When the LDR is inthe dark, Vout will be:In other words, this circuit gives a LOW voltagewhen the LDR is in the light,and a HIGH voltage when the LDR is in the shade.A sensor subsystemwhich functions like thiscould be thought of as adark sensor and couldbe used to controllighting circuits whichare switched onautomatically in theevening.
6.3 Light Dependent Resistors (2)The position of the LDR and the fixedresistor are now swapped.Remember the LDR has a resistance of500Ω , (0.5 kΩ), in bright light, and 200kΩ in the shade.In the light:In the dark:This sub system could bethought of as a “lightsensor” and could be usedto automatically switch offsecurity lighting at sunrise.How does this change affect thecircuit’s operation ?Vout 1010 + 0.5= x 9 = 8.57 VVout 1010 + 200= x 9 = 0.43 V
The graph opposite shows thevariation in resistance of a lightdependent resistor (LDR) withchanges in light intensity i.e. anillumination of 105lux produces aresistance of 102ohms.Electronics & Photonics RevisionQuestion Type:Q19. What is the resistanceof the LDR when the lightlevel is 103lux?Read from graph R = 104Ω= 10,000 ohmsLDR
6.4 ThermistorsA temperature-sensitive resistor iscalled a thermistor.There are severaldifferent types:The resistance ofmost commontypes ofthermistordecreases as thetemperature rises.They are callednegativetemperaturecoefficient, or ntc,thermistors.Note the -t° nextto the circuitsymbol.Different types ofthermistor aremanufactured and eachhas its owncharacteristic pattern ofresistance change withtemperature.Resistance (Ω)Temp (oC)20 40 60 80100100010000100000Note the log scale for resistanceThe diagram showscharacteristic curvefor one particularthermistor:
6.5 Thermistor CircuitsR = 10 kHow could you make asensor circuit for useas a fire alarm?At 80oRThermistor = 250 Ω (0.25 kΩ)1010 + 0.25= x 9 = 8.78 VVoutR = 10 kYou want a circuit whichwill deliver a HIGHvoltage when hotconditions are detected.You need avoltage dividerwith the ntcthermistor in theposition shown:How could you makea sensor circuit todetect temperaturesless than 4°C to warnmotorists that theremay be ice on theroad?You want a circuitwhich will give aHIGH voltage incold conditions.You need a voltagedivider with thethermistor in theposition shown:At 4oRThermistor = 40 kΩ4010 + 40= x 9 = 7.2 VVout
Electronics & Photonics RevisionQuestion Type:A thermistor is a device theresistance of which varies withtemperature. The resistance-temperature characteristicfor a thermistor is shown in Figure 7.Q20. What is the value of the resistanceof the thermistor at 20°C?From the graph read off the resistance for a temp of 200C.1000 ΩThermistors
Electronics & Photonics RevisionQuestion Type:The thermistor is incorporated intothe control circuit for therefrigeration unit of a cool room.The circuit is shown in Figure 8The relay switches the refrigerationunit ON when voltage, V, acrossvariable resistor R ≥ 4V andswitches OFF when V < 4V.The refrigeration unit must turn onwhen the temperature of the coolroom rises to, or exceeds, 5°C.Q21. At what value should theresistor R be set so that therefrigeration unit turns on at thistemperature?You must show your working.At 50C, the thermistor resistance =4000 ohms.Use the voltage divider relationship todetermine the value of R.4 = 12R / (4000 + R)R = 2000 ΩThermistors
TransducersElectronics & Photonics RevisionQuestion Type:Q22. From the list of components below (A–D) selectthe one that would be most suitable for use in thecircuit shownin Figure 9 at position P and the one most suitable foruse at position Q.A. LDR (light dependent resistor)B. LED (light emitting diode)C. transistorD. diode The LED was the best option for P and the LDR wasthe best for Q.Figure 9 is a sketch of an electro-optical system that allows sound to betransmitted over a distance via a fibre optic cable, using light.
7.0 Transistor AmplifiersShown opposite is asingle stage commonemitter amplifier.+V0 VR1R2VINVOUTSingle stage becauseit has only 1 transistorCommon emitterbecause the emitter iscommon to both inputand output.The voltage divider consisting of R1and R2provides the forward bias so the base will bepositive with respect to the emitter.Resistors R1 & R2 are sized to set the quiescent(Q) or steady state operating point at the middleof the load line (shown by the green dot on loadline, see next slide).RLRE C2RL is chosen to limit the collector current tothe maximum allowed value.RE is chosen to set VCE at the voltage whichwill allow the biggest “swing” in the outputsignal to occur.The device can be regardedas a black box with an inputand an outputC1So this amplifier is nowcorrectly biased and canoperate to produce anenlarged (amplified), invertedoutput.
7.1 Gain+V0 VR1R2VINVOUTC1C2RERLSingle stage NPN TransistorCommon Emitter AmplifierThe gain of theamplifier can becalculated from:Gain = VOUT/VIN
Electronics & Photonics RevisionQuestion Type:The graph of vOUT versus vIN forthe transistor amplifier is shownin Figure 4.Q23. What is the voltageamplification of the transistoramplifier?You must show your working.Voltage amplification (gain) is themagnitude of gradient of the graph.Gain = (3 – 0) /(0- 60 x 10-3) = - 50.+ 50 is also acceptable. Q24. Explain the shape of the graph inFigure 4.Negative slope: An increase in VINleads to a corresponding decrease inVout. This is an inverting amplifierAmplifiersYour explanation should include why thegraph shown has a negative slope, andwhy it has horizontal sections at vIN > +60mV and vIN < –60 mV.Horizontal section for VIN > +60 mV: the amplifier is saturated, i.e. maximumcurrent flows through the transistor.Horizontal section for VIN < −60 mV: the amplifier is at cut-off, i.e. minimum
7.2 Clipping+V0 VR1R2VINVOUTC1C2RERLSetting the Q point of theamplifier at an incorrect levelcan lead to the output signalbeing distorted, cut off or“clipped”VCE (V)VBE (V)QVVOUTQ set too low– bottom ofsignal clippedQVINVOUTQ set correctly –no clippingQVINVOUTQ set too high –top of signalclippedSingle stage NPN TransistorCommon Emitter AmplifierThe load line for an amplifier is aplot of the collector emitter voltageagainst the base emitter voltageTrying to drive the amplifier too hard, byhaving too large an input signal will alsolead to clipping of the output signal
Electronics & Photonics RevisionQuestion Type:The input signal, vIN, she is usingis shown in Figure 5.Q25. On the graph below, sketch theoutput signal, VOUT.AmplifiersA student is studying theperformance of the invertingamplifier in question 23 . It has again of 50Clipping occurs because the input signal can only vary between ± 60 mV (seeFig 4 previous question)