APPLIED ELECTRONICS Outcome 1Outcome 1 - Design and construct electronic systems to meet given specificationsWhen you have completed this unit you should be able to: State and carry out calculations using the current gain and voltage gain equations. Carry out calculations involving bipolar transistor switching circuits. Carry out calculations involving MOSFET transistor circuits. Identify and describe the uses of transistors in “push-pull” circuits. Carry out calculations involving Darlington pair circuits. Design transistor circuits for a given purpose.
APPLIED ELECTRONICS Outcome 1Before you start this unit you should have a basic understanding of: Input and Output transducers Voltage divider circuits Ohm’s Law - relationship between V and I in a d.c. circuit Kirchoff’s laws for current and voltage The operational characteristics of various electronic components Use of breadboards Use of circuit test equipment: multimeter and oscilloscope
APPLIED ELECTRONICS Outcome 1 Any electronic system can be broken down into three distinct parts INPUT PROCESS OUTPUT We are going to start by looking at INPUT TRANSDUCERSINPUT transducers convert a change in physical conditions (e.g. temperature)into a change in an electrical property (e.g. voltage) which can then beprocessed electronically to produce either a direct measurement of the physicalcondition (temperature in oC) or to allow something to happen at apredetermined level (e.g. switching ON the central heating at 20 °C).
APPLIED ELECTRONICS Outcome 1Changes in the resistance of an input transducer must be converted tochanges in voltage before the signal can be processed. This is normallydone by using a voltage divider circuit. Voltage divider circuits work on the basic electrical principle that if two resistors are connected in series across a supply, the voltage load across each of the resistors will be proportional to the value of the resistors. R2 Signal Voltage = × VCC Rtotal
APPLIED ELECTRONICS Outcome 1Common Input Transducers Physical condition to be Input Transducer Electrical property that monitored changes Temperature Thermistor Resistance Thermocouple Voltage Platinum Film Resistance Light LDR Resistance Selenium Cell Voltage/current Photo Diode Current/Resistance Distance Slide Potentiometer Resistance Variable Transformer Inductance Variable Capacitor Capacitance Force Strain Gauge Resistance Angle Rotary Potentiometer Resistance
APPLIED ELECTRONICS Outcome 1PUPIL ASSIGNMENT 1Calculate the signal voltages produced by the following voltage divider circuits:
APPLIED ELECTRONICS Outcome 1 AMPLIFICATION and BIPOLAR JUNCTION TRANSISTORS Input transducers rarely produce sufficient voltage to operate output transducers, (motors, bulbs, etc.) directly. To overcome this problem, we need to AMPLIFY theiroutput voltage or current. Amplifying devices are said to be active components as opposed to non-amplifying components such as resistors, capacitors etc. which are known as passive components. The extra energy required to operate the active component comes from an external power source such as a battery, transformer, etc.
APPLIED ELECTRONICS Outcome 1 AMPLIFICATION and BIPOLAR TRANSISTORSThe most common active device in an electronic system is the Bipolar JunctionTransistor (or simply transistor for short). Two types are available, NPN or PNP.The transistor has to beconnected into circuitscorrectly. The arrow head onthe emitter indicates thedirection of "conventional"current flow (positive-to-negative). NPN transistors operate when the base is made Positive PNP transistors operate when the base is made Negative
APPLIED ELECTRONICS Outcome 1TRANSISTOR NOTATIONSubscripts are normally used to indicate specific Voltages and Currentsassociated with transistor circuits,Ic - Collector currentIb - Base current V ccIe - Emitter currentVCC - Voltage of supply (relative to ground line)Vb - Voltage at the base junction (relative to ground line) R L V LVe - Voltage at the emitter junction (relative to ground line)Vce - Voltage between the collector and emitter junction IcVbe - Voltage between the base and emitter junctionVL - Voltage over the load resistor R b Ib V ce V be Ie V c V in V b R e V e V e 0V
APPLIED ELECTRONICS Outcome 1Common Emitter Mode b The transistor can be used in OUTPUT different modes, the most e common of which is the common INPUT emitter mode. (So called because the emitter COMMON LINE is common to both input and output signals.) c In the common emitter mode, a small current flowing between the base and emitter junction b will allow a large current to flow between the collector and emitter. b c e
APPLIED ELECTRONICS Outcome 1Common Emitter Mode c It can be seen that : b Ie = Ib + Ic b c Since Ib is usually much smaller than Ic, it follows that Ie is approximately = Ic e
APPLIED ELECTRONICS Outcome 1Common Emitter Mode Current Gain The bipolar transistor is a current-controlled amplifying device The current gain (or amplification) of the transistor is defined as the ratio of collector / base currents Collector current current gain = Base current Ic AI = Ib
APPLIED ELECTRONICS Outcome 1Common Emitter Mode Current Gain The accepted symbol for transistor current gain in dc mode is, hFE The maximum allowable currents will depend on the make of transistor used. These limits can be obtained from manufacturers data sheets. Forcing the transistor to carry currents greater than these maxima will cause the transistor to overheat and may damage it. If the transistor is used to amplify a.c. signals then the gain is defined as, ∆I c h fe = ∆I b
APPLIED ELECTRONICS Outcome 1Common Emitter Mode Current GainPupil Assignment 2 1. Calculate the gain of a transistor if the collector current is measured to be 10 mA when the base current is 0.25 mA. 2. Calculate the collector current through a transistor if the base current is 0.3 mA and hFE for the transistor is 250. 3. What collector current would be measured in a BC107 transistor if the base current is 0.2 mA and hFE is 100? 4. In questions 2 & 3, are the transistors ac or dc ? Explain why.
APPLIED ELECTRONICS Outcome 1 TRANSISTOR SWITCHING CIRCUITS In order to generate a c current in the base of the transistor, avoltage must be appliedbetween the base -emitter junction (Vbe). b It is found that no (or at least negligible) current b c flows in the base circuit unless Vbe is above 0.6 Volts. e
APPLIED ELECTRONICS Outcome 1 TRANSISTOR SWITCHING CIRCUITS c Increasing the base - emitter voltage further, increases the base current,producing a proportional increase in thecollector current. b When the base - emitter voltagereaches about 0.7 V, the resistance between the b cbase emitter junction starts to changesuch that the base - emitter voltageremains at about 0.7 V. e At this point the transistor is said to be saturated. Increasing the base current further has no effect on the collector current. Thetransistor is fully ON. It can be assumed that if the transistor is turned ON, Vbe = 0.7 V
APPLIED ELECTRONICS Outcome 1Pupil Assignment 3 For each of the circuits shown, calculate Vbe and state if thetransistor is ON or OFF.
APPLIED ELECTRONICS Outcome 1TRANSDUCER DRIVER CIRCUITS Output transducers can require large currents to operate them. Vcc Currents derived from input OUTPUT transducers, either directly, or TRANSDUCER from using a voltage divider circuit tend to be small. Ic A transistor circuit can be used Ib to drive the output transducer. A small current into the base of the transistor will cause a large current to flow in the collector/ 0V emitter circuit into which the output transducer is placed.
APPLIED ELECTRONICS Outcome 1 VccTRANSDUCER DRIVER CIRCUITS The base current is derived from OUTPUT TRANSDUCER applying a voltage to the base of the transistor. Ic If the voltage between the base - Ib emitter junction (Vbe) is less than 0.6 V, the transistor will not operate, no current will flow in the emitter/collector circuit and the output transducer will be 0V OFF. If Vbe lies between 0.6 and 0.7, the transistor acts If Vbe is 0.7 V (or forced above in an analogue manner 0.7 V), the transistor will operate, and this may result in the a large current will flow in the output transducer emitter/collector circuit and the hovering around an on transducer will switch ON. and off state
APPLIED ELECTRONICS Outcome 1 VccWorked Example If the transistor is FULLY 150k 470RON, calculate the collectorcurrent and Vce , if hFE =200 andVCC = 9 VoltsStep 1The voltage between the base and emitter junction isalways about 0.7 V 0VSince the emitter is connected to the ground line (0V),Vb= 0.7 VStep 2The voltage dropped over the base resistor can then becalculated. Voltage drop = VCC - Vb = 9 - 0.7 = 8.3V
APPLIED ELECTRONICS Outcome 1 Vcc Worked Example continuedStep 3 150k 470RThe base current is calculated usingOhms law Vdropped 8.3Ib = = 0.0553 = 0.00553mA Rb 150k mA 0VStep 4Ic is calculated knowing hFEIc = hFE x Ib = 200 x 0.0553 = 11.06 mA
APPLIED ELECTRONICS Outcome 1 VccWorked Example continued 150k 470RStep 5VL is calculated using Ohms lawVL = Ic x RL = 11.06 mA x 470 = 5.2 V 0VStep 6Vce is calculatedVce = Vcc - VL = 9 - 5.2 = 3.8 V
APPLIED ELECTRONICS Outcome 1Pupil Assignment 4 A 6 V, 60 mA bulb is connected to the collector of a BFY50 transistor as shown below. 9V If the gain of the transistor is 30, determine the size of the Rb 6V, 60mA base resistor Rb required to ensure that the bulb operates at its normal brightness. BFY50 0V
APPLIED ELECTRONICS Outcome 1VOLTAGE AMPLIFICATION Although the transistor is a current amplifier, it can easily bemodified to amplify voltage by the inclusion of a load resistor, RL inthe collector and/or emitter line. Vcc If we apply a voltage Vin to the base of the transistor, the base current Ib VL will flow. RL This will causes a proportionalincrease (depending on the gain) of the Iccollector current Ic. Ib Since the current through the load BFY50resistor (Ic) has increased, the voltage Voutover RL has increased (VL = IcRL) and Vinhence Vout has decreased. (Vout =VCC - VL) 0V
APPLIED ELECTRONICS Outcome 1VOLTAGE AMPLIFICATION (continued) The Voltage gain of any amplifier is defined as Vcc voltage output Voltage gain = RL VL voltage input Ic Ib BFY50 Vo Vout AV = Vin Vi 0V
APPLIED ELECTRONICS Outcome 1 WORKED EXAMPLE Vcc 1k Calculate the voltage gain of this circuit if, Vin =1.7 Volt, hFE = 100 and VCC = 6V Vout Vin 2k 0V
APPLIED ELECTRONICS Outcome 1 WORKED EXAMPLE V ccStep 1The voltage between the base and R L V Lemitter junction (Vbe) is always about0.7 V hence:Ve = Vin - 0.7 = 1.0 V V cc V out VStep 2 beThe current through Re is calculated V inusing Ohms law R e V e 0V V 10 . Ie = e = = 0.5mA Re 2 k
APPLIED ELECTRONICS Outcome 1 WORKED EXAMPLE V ccStep 3For this value of hFE, Ib will be R L V Lsmall compared to Ic (onehundredth of the value), hence,Ic = Ie V cc V out V beStep 4 V inThe voltage over the load resistor (RL) is Rcalculated using Ohms law V e e 0VVL = Ic x RL = 0.5 mA x 1k = 0.5 V
APPLIED ELECTRONICS Outcome 1 Vcc WORKED EXAMPLE 1k 0.5VStep 5The output voltage can now be 0.5mAcalculated fromVout = VCC - VL 6.0V = 6 - 0.5 = 5.5 V 0.7V 5.5V 1.7VStep 6 2K 2k2The voltage gain is 1.0Vtherefore 0.5mA 0V VoAV = 5.5/1.7 = 3.2 Vi
APPLIED ELECTRONICS Outcome 1 Vcc Pupil Assignment 5 2kA transistor of very high current gain isconnected to a 9 Volt supply as shown. VoutDetermine the output voltage and Vinthe voltage gain when an input of 3 4k7Volts is applied. 0V
APPLIED ELECTRONICS Outcome 1Practical Considerations Care must be taken toensure that the maximum basecurrent of the transistor is notexceeded. When connecting the base of a transistor directly to asource, a base protection resistorshould be included. This willlimit the maximum current V ininto the base. Most data sheets will quote 0V the maximum collectorcurrent and hFE and so themaximum allowable base currentcan be calculated.
APPLIED ELECTRONICS Outcome 1 Practical Considerations CURRENT FLOWINGIf the transistor is to be connected to a FROM Vcc INTO BASEpotential divider circuit then themaximum possible current into the base Vccwill depend on R1 R1 RLThe maximum possible currentthrough R1 (and hence into the base)would be = VCC R1 R2 Rehence if R1 is large, the base current 0Vwill be small and therefore nodamage should occur.
APPLIED ELECTRONICS Outcome 1 Practical Considerations V ccIf R1 is small (or has the capability ofgoing small e.g. using a variable resistoras R1), a protection resistor must be R Lincluded in the base. R 1 R bIf R1 = 0, the maximum possible currentinto the base = VCC R R e Rb 2 0Vhence Rb can be calculated if VCC and themaximum allowable base current is known.
APPLIED ELECTRONICS Outcome 1 Pupil Assignment 6 9VAssume Ic(max) for thetransistor shown is 100 mA and RbhFE is 200. Vin = 5V 0VCalculate: The maximum allowable base current. The size of protection base resistor required (remembering Vbe = 0.7V, and R = V/I)
APPLIED ELECTRONICS Outcome 1 CIRCUIT SIMULATION It is possible to use circuit simulation software such as ‘Crocodile Clips’ to investigate electric and electronic circuits. Circuit simulation is widely used in industry as a means of investigating complex and costly circuits as well as basic circuits. Circuit simulators make the modelling and testing of complex circuits very simple. The simulators make use of libraries of standard components along with common test equipment such as voltmeters, ammeters and oscilloscopes.Question: What do you think the main advantage of simulation of circuits is?
APPLIED ELECTRONICS Outcome 1CIRCUIT SIMULATION (Base Protection) Using the simulation software, construct the circuit shown, using a 5 V supply. Switch on and see what 5V happens. 5V Now insert a 10k base protection resistor and see what happens when you switch on now. Use the simulation to determine the smallest value of resistor required to protect this transistor.
APPLIED ELECTRONICS Outcome 1 CIRCUIT SIMULATION (Base Protection)Construct the circuit shown. 100R See what happens when you reduce the size of the variable 5V resistor. 10k Now re-design the circuit to include a base protectionresistor.
APPLIED ELECTRONICS Outcome 1 Pupil assignment 7An NTC thermistor is used in the circuit shown below to indicate if thetemperature falls too low. When the bulb is on the current through it is 60 mA. 6V RESISTANCE(Ω ) 10k 2000 1500 GRAPH OF R/T FOR 1000 NTC THERMISTER -t 500 TEMP(° C) 10 20 30 40 0V
APPLIED ELECTRONICS Outcome 1Pupil assignment 7 If hFE for the transistor is 500, determine the base current required to switch on the bulb. What voltage is required at the base of the transistor to ensure that the bulb indicator switches ON? Calculate the voltage dropped over, and hence the current through the 10 k resistor. Calculate the current through the thermistor and the resistance of the thermistor when the bulb is ON? Using the information on the graph, determine at what temperature the bulb would come ON. How could the circuit be altered so that the bulb would come on at a different temperature? How could the circuit be altered so that the bulb would come when the temperature is too high?
APPLIED ELECTRONICS Outcome 1 Pupil assignment 8For each of the circuits, calculate the base current, the emitter voltage and current
APPLIED ELECTRONICS Outcome 1 Pupil assignment 8For each of the circuits, calculate the base current, the emitter voltage and current
APPLIED ELECTRONICS Outcome 1 The Darlington Pair In order to obtain higher gains, more than one transistor can be used, the output from each transistor being amplified by the next (known as cascading). RL Increasing the gain of the circuit means: The switching action of the circuit is more immediate; Tr1 A very small base current is required in switching; The input resistance is very high. Tr2 A popular way of cascading two transistors is to use a Darlington pair (Named after the person that first designed the circuit)
APPLIED ELECTRONICS Outcome 1 The Darlington Pair The current gain of the "pair" isequal to the product of the two individualhFEs. RL If two transistors, each of gain 50 are used, the overall gain of the pair will be 50 x 50 = 2500 Tr1 AI = hFE 1 × hFE 2 Tr2 Because of the popularity of this circuit design, it is possible to buy a single device already containing two transistors
APPLIED ELECTRONICS Outcome 1 The Darlington Pair In a Darlington pair, bothtransistors have to be switched onsince the collector-emitter current of 0.7VTr1 provides the base current for Tr2. 1.4V In order to switch on the pair, each base-emitter voltage 0.7Vwould have to be 0.7V 0V The base-emitter voltagerequired to switch on the pairwould therefore have to be1.4V.
APPLIED ELECTRONICS Outcome 1Worked ExampleFor the Darlington pair shown, calculate: h FE1 = 2 0 0 The gain of the pair; The emitter current; h FE2 = 5 0 The base current 8V 27R 0V
APPLIED ELECTRONICS Outcome 1 Worked Example Step 1 The overall gain = product of the individual gains h FE1 = 2 0 0AI = hFE 1 × hFE 2 = 200 × 50 = 10000 h FE2 = 5 0 Step 2 8V The voltage over the load resistor must be the input voltage to the 27R base minus the base-emitter voltage required to switch on the 0V pair VL = Vin - Vbe = 8 - 1.4 = 6.6 V
` Worked ExampleStep 3The emitter current in the loadresistor can be obtained fromOhm’s law h FE1 = 2 0 0 V L 6.6 Ie = = = 0.244 A R L 27 h FE2 = 5 0 8VStep 4Since the gain is very high, Ic = Ieand the gain for any transistor 27Rcircuit = Ic/Ibhence knowing Ic and AI, Ib can be 0Vcalculated Ic Ic 0.244 Ai = ⇒ Ib = = = 24.4 × 10 −6 A Ib Ai 10000
APPLIED ELECTRONICS Outcome 1 Pupil Assignment 9For the circuit shown, the gain of RL Tr1 is 150, the gain of Tr2 is 30.Calculate: Tr1 The overall gain of the Darlington pair; The base current required Tr2 to give a current of 100 mA through the load resistor.
APPLIED ELECTRONICS Outcome 1MOSFETS Although the base current in a transistor is usually small (< 0.1 mA), some input devices (e.g. a crystal microphone) have very small output currents. In many cases, this may not be enough to operate a bipolar transistor. In order to overcome this, a Field Effect Transistor (FET) can be used. COLLECTOR DRAIN BASE GATE EMITTER SOURCE BIPOLAR TRANSISTOR FIELD EFFECT TRANSISTOR
APPLIED ELECTRONICS Outcome 1 MOSFETS Applying a voltage to the Gate connection allows current to flow between the Drain and Source connections. DRAIN This is a Voltage operated device. GATE It has a very high input resistance (unlike the transistor) and therefore requires very little current to operate it (typically 10-12 mA). SOURCE Since it operates using very little current, it is easy FIELD EFFECT to destroy a FET just by the static electricity built up TRANSISTOR in your body. FET’s also have the advantage that they can be designed to drive large currents, they are therefore often used in transducer driver circuits
APPLIED ELECTRONICS Outcome 1 D D MOSFETS Two different types of FET’s are G Gavailable: S S N-CHANNEL JFET P-CHANNEL JFET JFET (Junction Field Effect Transistor) D D MOSFET (Metal Oxide Semiconductor Field Effect Transistor) G G All FET’s can be N-channel or P- S Schannel. N-CHANNEL ENHANCEMENT P-CHANNEL ENHANCEMENT MOSFET MOSFETEnhancement-type MOSFETs can be used in a similar way to bipolar transistors.N-channel enhancement MOSFET’s allow a current to flow between Drain andSource when the Gate is made Positive (similar to an NPN transistor).P-channel enhancement MOSFET’s allow a current to flow between Drain andSource when the Gate is made Positive (similar to an PNP transistor
APPLIED ELECTRONICS Outcome 1 MOSFETS The simplicity in construction of the MOSFET means that it occupies very little space. Because of its small size, many thousands of MOSFET’s can easily beincorporated into a single integrated circuit. The high input resistance means extremely low power consumptioncompared with bipolar transistors. All these factors mean that MOS technology is widely used within theelectronics industry today.
APPLIED ELECTRONICS Outcome 1 MOSFETS ID Like a bipolar transistor, if the Gatevoltage is below a certain level (the thresholdvalue, VT), no current will flow between theDrain and Source (the MOSFET will be DRAINswitched off). GATE VDS If the Gate voltage is above SOURCE VT, the MOSFET will start to switch on. VGS Increasing the Gate voltage 0V will increase ID
APPLIED ELECTRONICS Outcome 1 MOSFETS ID For a given value of VGS(above VT), increasing VGSincreases the current until saturationoccurs. DRAIN Any further increase will cause no further increase in ID. The GATE VDS MOSFET is fully ON and can SOURCE therefore be used as a switch. VGS 0V Saturation occurs when VDS = VGS - VT.
APPLIED ELECTRONICS Outcome 1 Worked ExampleThe threshold gate voltage for the MOSFETshown is 2 V. Calculate the gate voltagerequired to ensure that a saturation currentof 10 mA flows through the load resistor.Step 1The Drain - Source channel acts as aseries resistor with the 100R, since thecurrent is the same in a series circuit,the voltage over the 100R can becalculated.Using Ohm’s lawV = IR = 10 mA x 100 = 1 Volt
APPLIED ELECTRONICS Outcome 1Worked ExampleStep 2Using Kirchoff’s 2nd law,the voltage over the channel + thevoltage over the load resistor = supplyvoltagehence VDS = 5 - 1 = 4 VoltsStep 3For saturation to occur,VDS = VGS-VTVGS = VDS + VTVGS = 4 + 2 = 6 Volts.
APPLIED ELECTRONICS Outcome 1 MOSFETSMOSFET’s can be designed to handle very high drain currents , this meansthat they can be used to drive high current output transducers drivers withoutthe need for relay switching circuits (unlike the bipolar transistor). V cc R L The load resistor could be any output transducer, bulb, motor, relay etc. Since MOSFET’s are particularly sensitive to high voltages, care must be taken to include a reverse biased diode over transducers that may cause a back emf when V in switched off. 0V
APPLIED ELECTRONICS Outcome 1Possible application of a mosfet VccA variable resistor can beused in a voltage divider RLcircuit and adjusted to ensurethat the input voltage to thegate = VTThe load resistor could be abulb, motor, relay coil, etc. 0V
APPLIED ELECTRONICS Outcome 1The Push-Pull Amplifier NPN bipolar transistors and n-type enhancement +MOSFETs operate when thebase or gate is made positivewith respect to the zero volt N PNline. V in O V PNP and p-type MOSFETs R L operate off negative signals. PN P A push-pull amplifier _consists of one of each typeof bipolar transistor (or MOSFET)connected in series with a + and- supply rail.
APPLIED ELECTRONICS Outcome 1 The Push-Pull Amplifier + If Vin is Positive with respect to 0V, the NPN N PN transistor will switch on, current will flow from the V in O V + supply line through the R L collector-emitter junction, PN P through the load resistor down to the 0Volt line _ If Vin is Negative with respect to 0V, the PNP transistor will switch on,current will flow from the 0Volt linethrough load resistor, through the emitter-collector junction, to the - supply line.
APPLIED ELECTRONICS Outcome 1The Push-Pull Amplifier + N PN The direction of current V inflow through the load resistor O Vwill therefore depend on whether R Lthe input voltage is positive or PN Pnegative. _ If the load resistor isreplaced by a motor, the directionof rotation of the motor can bealtered dependent on the inputvoltage, Vin.
APPLIED ELECTRONICS Outcome 1Circuit simulationUsing Crocodile Clips construct the following circuit. 5VInvestigate whathappens when 1kthe potentiometer 10kslider is altered 5V
APPLIED ELECTRONICS Outcome 1Circuit simulationUsing Crocodile Clips construct the following circuit.Set the resistance +5Vof each LDR tothe same value. 1k 1k 0VSet the variableresistor to itsmiddle position. -5VAlter the value ofone LDR and Alter the value of the otherobserve the LDR and observe themotor. motor.
APPLIED ELECTRONICS Outcome 1 SEB & SQA Past Paper exam Questions1995, Paper 1, question 2 12VThe following electronic systemis set up for a test with variousammeters and voltmetersconnected as shown. 1kIn the condition shown, thetransistor is fully saturated witha base current of 5mA.Write down the readings which 300Ryou would expect to see oneach of the four voltmeters (V1- V4) and the two ammeters(A1 - A2). 0V
APPLIED ELECTRONICS Outcome 1 SEB & SQA Past Paper exam Questions1994, Paper 1, question 1 Vcc +9VA designer is asked to construct RELAYan electronic circuit which will (RESISTANCE 180R)energise a relay at a set lightlevel. Having investigated thecharacteristics of the lighttransducer, she finds that the 200k Rresistance of the transducer at“switch on” level is 2.1 M . The 0Vproposed design is shown Determine, assuming the transistor is in aopposite. fully saturated condition:The transistor saturates when (a) The value of the unknownVbe = 0.6V. resistor R required to make the transistor operate correctly; (b) The power dissipated in the relay coil.
APPLIED ELECTRONICS Outcome 1 SEB & SQA Past Paper exam Questions1993, Paper 1, question 2The control circuit for a cooling fan is based on a thermistor.The graph shows the operating characteristics of the thermistor and theproposed circuit diagram is also shown. +6V 300 RELAY -t ANCE (kΩ) 200 FAN MOTOR RESIST 100 10k 0V 100 200 300 TEMPERATURE (°C)
APPLIED ELECTRONICS Outcome 1 SEB & SQA Past Paper exam Questions 1993, Paper 1, question 2 Continued(a) The motor should switch on when Vbe = 0.6V. For this condition, calculate the value of Rt. From the graph, determine the temperature at which the fan should switch on.(b) When the circuit is built and tested, it is found that the relay does not operate at the switch - on temperature. Suggest one reason why the transistor fails to operate the relay. Redraw the circuit diagram to show how a Darlington pair could be used to overcome this problem.
APPLIED ELECTRONICS Outcome 1 SEB & SQA Past Paper exam Questions1998, Paper 2, question 4(c)(amended) +12VAn instant electric shower isdesigned to deliver water at a -tfixed temperature from a cold RELAY OPERATINGwater supply. CURRENT 250mAAn additional safety feature is to 12kbe added which will switch off the hFE = 100power to the shower if the water Ib R hFE 100temperature produced by theheating element becomesdangerously high (greater than 0V50 oC).The relay requires an operatingcurrent of 250 mA. Theresistance of the thermistor at
APPLIED ELECTRONICS Outcome 1 SEB & SQA Past Paper exam Questions1998, Paper 2, question 4 (c)Continued, +12VName the transistor configuration used in this circuit. -t RELAY OPERATINGState one advantage of using this CURRENT 250mA configuration. 12k hFE = 100For the relay to operate: Ib R hFE 100calculate the base current, Ib;calculate the potential differenceacross the 12kresistor; determine 0Vthe voltage across the fixedresistor R; calculate the value ofR.