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"1.5V AA" 3 months continuously night lamp

Discussion in 'General Electronics Chat' started by TDA2030, Nov 5, 2009.

  1. TDA2030

    TDA2030 New Member

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    This circuit is usable as a Night Lamp when a wall mains socket is not available to plug-in an ever running small neon lamp device. In order to ensure minimum battery consumption, one 1.5V cell is used, and a simple voltage doubler drives a pulsating ultra-bright LED: current drawing is less than 500µA.
    An optional Photo resistor will switch-off the circuit in daylight or when room lamps illuminate, allowing further current economy.
    This device will run for about 3 months continuously on an ordinary AA sized cell or for around 6 months on an alkaline type cell but, adding the Photo resistor circuitry, running time will be doubled or, very likely, triplicated.

    circuit and component list link:-
    Battery-powered Night Lamp - RED - Page44

    i made it and its not working can any one help me plzzzzzzzzz
     
  2. kchriste

    kchriste New Member Forum Supporter

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    What is the EXACT part number of the timer chip you are using? An ordinary LM555 will NOT work. It needs to be the CMOS version and even then, it is being operated out of spec, since most of them, as far as I know, need a minimum supply voltage of 2V.
     
  3. TDA2030

    TDA2030 New Member

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    yeah i used 7555 number chip, and 1m ohms 10% tolerance resistors, i used polyester cap 100nf 100V instead of 63V, it is ok or its need exactly 63Vcap ????????
     
  4. dave

    Dave New Member

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  5. kchriste

    kchriste New Member Forum Supporter

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    The 100V part will be fine.
    Which manufacturer of the 7555 did you use? Some are only spec'd down to 3V.
    Also, try removing the CDS light sensor or going into a very dark room with the circuit to see if it works.
    If all else fails, post a clear picture of the circuit. (Top and bottom)
     
    Last edited: Nov 5, 2009
  6. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello there,


    The circuit, although interesting, is a bit superfluous. All the circuit is doing is
    increasing the voltage so that it can run an LED that has a forward voltage that
    is higher than a single battery. Why does the circuit have to do this? It's because
    the battery voltage isnt high enough to run the LED by itself. If the battery
    voltage WAS high enough, we wouldnt need a circuit anymore.
    Enter a *second* battery here, in series with the first.
    With TWO batteries in series we can easily drive a red LED and we dont need a
    circuit to do that, just a single resistor.

    What's the run time for two cells instead of one? Im glad you asked :)
    The run time will be longer because the current draw from the two cells
    will be approximately one half of what it would be with a voltage double
    circuit. This means the run time will be twice as long as with only one cell.

    So in the long run it is better to use two cells in series than to bother with
    a circuit unless there is some real need to run off of one single cell.
    It's true we would need two cells then, but they will last twice as long so
    that's about the same average battery usage meaning over time we will use
    the same number of batteries.

    Besides the circuit looking like it could use a lot of improvement, one drawback i
    see right away is that it doesnt look like it can run a white LED. That's a big
    drawback because we often dont want to have to navigate down a hallway
    with a dim red light, we would rather have white light. I've been there and done
    that so i have a good idea about the difference here. Try it if you want to see
    the difference.

    Using three cells and a resistor we can drive a small while LED now, and we can
    find white LEDs that run on very low current these days and are very efficient
    at low currents too. The run time will be about three times that of a single
    cell, so we wont be wasting batteries, and we get a nice white light to see by.
    Try it and you'll see the difference.
     
  7. TDA2030

    TDA2030 New Member

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  8. TDA2030

    TDA2030 New Member

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  9. TDA2030

    TDA2030 New Member

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    "7555", "IPA" company, in the bottom i think this is batch number or not I don't know
    "H 9340"
     
  10. audioguru

    audioguru Well-Known Member Most Helpful Member

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    The circuit blinks the red LED at 4Hz and saves battery power because the LED is dim and is turned off for most of the time.
    C2 is charged to the battery voltage less the voltage of D2 (0.3V) then the output of the Cmos timer goes low and applies the battery voltage plus the C2 voltage to the LED at a current of only about 2mA which is limited by the Cmos timer IC.

    The LED is dim with a current of only about 2mA.
     
  11. kchriste

    kchriste New Member Forum Supporter

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    What I meant was a real picture of the circuit you built (Top and bottom). Not the schematic. That way we can spot any mistakes you might have made.
    Sounds like it may be the 7555 from Intersil which should be OK as it is guaranteed to work down to 2V. Still out of spec, but less so than a 3V part.
     
  12. colin55

    colin55 Well-Known Member

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    The circuit you have posted will not work because the electrolytic will not charge when the output of the chip is HIGH as the output will be about 1v lower than rail and the diode removes another 0.6v.
    The output will be 1v lower than rail due to the electrolytic being a DEAD SHORT and the chip will not deliver a charging current.
    What you need is the circuit I designed using two transistors. This circuit will even flash a white LED even though a white LED requires about 3.2v to 3.6v under normal conditions.
    [​IMG]
     
  13. madhippiescientist

    madhippiescientist New Member

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    How about using one of those "funky"
    little (I believe) 6V batteries, that are
    about the diameter of a AA and 2/3
    it's length ?
     
  14. audioguru

    audioguru Well-Known Member Most Helpful Member

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    No.
    The 7555 is a Cmos 555 that has an output that goes rail to rail (as high as the power supply voltage).

    No.
    The 1N5819 Schottky diode has a voltage drop of only 0.25V or less at the low current used for charging the capacitor.

    Therefore the capacitor charges to 2.75V which is plenty to light a 1.8V red LED.
    When the 1.5V battery drops to only 1.0V then the LED will still be driven with 1.75V which will make it blink dimly.
     
  15. audioguru

    audioguru Well-Known Member Most Helpful Member

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    A little battery does not have much capacity so it will not last long.
     
  16. colin55

    colin55 Well-Known Member

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    This is entirely wrong:
    The 7555 will only source 1mA when the supply is 5v and the output voltage is 1v lower than rail. The capacitor provides a Dead-Short across the output and thus the capacitor will never charge. Secondly, any slight voltage produced by the capacitor will be lost when the output goes low as the output is about 0.4v above 0v rail.
    How did you get the impression that the charge-rate will be low?
    How did you get the idea that the capacitor will charge to 2.75v?
    You should come and work for me. You will make me a fortune with your technical specifications.
     
    Last edited: Nov 7, 2009
  17. audioguru

    audioguru Well-Known Member Most Helpful Member

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    No.
    Intersil's datasheet shows a sourcing current typically about 7mA.

    No.
    The capacitor never discharges to zero volts. It discharges to about 0.3V when the LED stops conducting.

    No.
    The output of the ICM7555 sinks about 2mA into the LED and discharges the capacitor from 1.25V to 0.3V.

    The sourcing current with a 2V supply is shown as typically 1mA. It is probably 0.8mA with a 1.5V supply. The capacitor needs to charge from 0.3V to 1.25V in 167ms. It cannot do it so the capacitor will charge to only about 0.6V.


    I meant that the voltage available to the LED is the 1.5V battery plus the 1.25V of the capacitor's charge.

    I would make one to prove that it works but I already have an LM3909 LED flasher working from a 1.0V to 1.5V battery. Its flash is much brighter than this weak circuit.
     

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    Last edited: Nov 7, 2009
  18. colin55

    colin55 Well-Known Member

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    This is incorrect. You are assuming the voltage on the capacitor is 1v or more. But the electro never charges.
    From the graph you provide it is clear that the current at 2v Vdd is only about 1 mA. The other 7555 chips state 1mA for 5v not 7mA so the current capability at 1.5v will be even less than 1mA.
    The whole concept of the circuit is a failure. If one person could not get it to work, then it's a failure. I have been shown these circuits many times and none of them work at 1.5v.
    The circuit I designed puts a high current into the LED to produce a bright flash. It's not the most efficient design but it's cheap and works well. The 47R is wasteful but cannot be avoided.
     
    Last edited: Nov 7, 2009
  19. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Colin,
    Which manufacturer of a 7555 has an output high current of only 1mA?
    Intersil, Harris, Philips and Maxim are all the same with about 5mA to 8mA. National and Texas Instruments do not show typical current graphs.
     
  20. colin55

    colin55 Well-Known Member

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  21. audioguru

    audioguru Well-Known Member Most Helpful Member

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    That is correct.
    The capacitor charges with a little less than 1mA for a duration that is twice as long as its discharge a little into the LED.

    It is a horrible weak circuit and the 7555 oscillator might need to make many cycles for the capacitor to become charged enough to dimly light the LED.
    It probably won't work if the battery voltage is less than 1.4V.
     

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