0200 - Find it - Dont lose it.pdf

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    PROJECTSANDCIRCUITS

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    A fuse blows and the lights goout. Everything goes dark. Youfumble for the torch you knowits there somewhere butwhere?

    Let this little gadget show youwhere!

    FIND ITWith the battery-powered

    Find It circuit, you will always beable to locate a torch (flashlight),bunch of keys, door lock justabout anything in darkness!

    While sufficient light reachesa sensor (light dependent resistor LDR) on the unit, nothinghappens. However, when it isdark enough, a light-emitting

    fitted to the box so that, say, abunch of keys or a torch couldbe hung from it (seephotograph).

    It would also be possible toattach the unit to a portableitem. In some cases, it mighteven be possible to build thecircuit panel inside a piece ofequipment but the reader willneed to make certain that he orshe is totally aware of anysafety implications and must becompetent at doing the jobcorrectly.

    If you are going to use thedevice to locate a door lock, itmay be convenient to have onlythe LED showing next to thelock and connect it to the unitmounted on the inside of thedoor.

    Locates almost anything in the dark!

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    diode (LED) begins to flashbriefly about once every fiveseconds. This helps to locatethe item.

    If preferred, you couldincrease or reduce the flashrate. However, any increasewould reduce the life of thebattery.

    LIGHTING THE WAYThere are many ways of

    using this circuit and readerswill, no doubt, have their ownideas. One method would be toattach the unit to a wall close tothe object to be found.Alternatively, a hook could be

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    Fig.1. Complete circuit diagram for Find it. Left, Find Itbeing used as an illuminated keyring hook.

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    As well as householdapplications, this circuit will befound handy in many outdoorpursuits. Campers and anglerswill certainly find uses for it. Notethat, in some cases, it will benecessary to waterproof the boxand this is left up to theconstructor.

    The circuit draws power fromtwo AA-size alkaline cells, whichshould give some one or twoyears of service. Since it requiresmore current while the LED isflashing, the actual life of thebatteries will depend on thenumber of hours of darkness in agiven 24 hour period. It alsodepends on what degree ofillumination is set for the unit tobegin operating.

    While sufficient light reachesthe sensor (so that the LED isoff), the current requirement ofthe prototype circuit is only 5uAwhich may be regarded asnegligible. While the LED isflashing, this rises to an average250uA approximately.

    This small operating currentis achieved by using a short dutycycle that is, the LED is off formuch longer than it is on about65 times longer. Thus, in eachfive second cycle the LED is onlyactually operating for some 0

    08s(80ms).

    Although while glowing theLED draws about 10mA, theaverage requirement is thereforeonly 150uA approximately. This isadded to the 100uAapproximately required by therest of the circuit, giving a total of250uA. If it is assumed that thereare eight hours of operation in a24 hour period, the averageoverall current requirement istherefore only 80uAapproximately.

    While the battery voltageexceeds about 2

    5V, the LED willflash brightly. It will become

    correspondingly dimmer downto about 2V which is thepractical end point.CIRCUITDESCRIPTION

    The complete circuitdiagram for the Find It project isshown in Fig.1. This may beconsidered to comprise twosections. The first is the lightsensing part based on IC1 andassociated components and thesecond, the LED flashercentered on IC2.

    Integrated circuit IC1 is anoperational amplifier (opamp).This has been specially selectedfor its ability to operate from alow supply voltage combinedwith an exceptionally smallstandby operating current.

    Looking at the light sensorstage first, the opamp invertinginput (pin 2) is maintained at avoltage equal to one-half that ofthe supply (nominally 1

    5V), dueto the effect of equal valueresistors R3 and R4 connectedas a potential divider across thepower supply. Since these have

    a very high resistance, thecontinuous current flowingthrough them is only a fraction ofa microamp.

    The opamps non-invertinginput, pin 3, is connected to afurther potential divider. The toparm of this comprises presetpotentiometer VR1 connected inseries with fixed resistor R1. Thelower one is simply light-dependent resistor (LDR) R2.

    As the intensity of lightreaching the LDRs sensitivesurface falls, its resistance risesand so does the voltage across itand hence at the non-invertinginput, pin 3. Depending on theadjustment of VR1, this voltagewill exceed that at the invertinginput, pin 2, at the operating lightlevel.

    A simple rule about opampsis this. When the voltage appliedto the non-inverting input exceedsthat at the inverting one (as willhappen here in dim light), theoutput (pin 6) will be high. Whenit is less (bright light), it will below.

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    pattern. Fig.3 (inset). How to make up R5 bywiring two resistors in series.

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  • Copyright 2000 Wimborne Publishing Ltd andMaxfield & Montrose Interactive Inc

    EPE Online, February 2000 - www.epemag.com - 111

    full output swing between thesupply voltage, and its outputwill therefore go from 0V to 3Vnominal as the light level falls tothe required operating point.

    LIGHT FLASHERNow let us look at the LED

    flasher based on IC2. Thisconsists of an astable (free-running pulse generator). Itsfrequency is related to the valueof resistor R5 (in the prototype,this consisted of two resistorsconnected in series to make up

    the required value), resistor R6and capacitor C1.

    The on times (during whichthe output, pin 3, is high) areprovided when C1 chargesthrough resistors R5 and R6 totwo-thirds of supply voltage (2Vapprox.). After that, the capacitordischarges via internal circuitrythrough resistor R6 alone to one-third of supply voltage (1Vapprox.) and this gives the offperiod during which pin 3 is low.

    This cycle repeats indefinitelyas long as a supply exists to pin 8and the reset input at pin 4 ishigh. With the values ofcomponents specified, each cycletakes about five seconds.

    Since resistor R5 has a muchhigher value than R6, capacitorC1 charging time is much longerthan the discharge time. Thus,the time during which output pin 3is high is much greater than whenit is l