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Small low-power light sensor

Discussion in 'LEDs and Optoelectronics' started by seanspotatobusiness, Sep 7, 2018.

  1. seanspotatobusiness

    seanspotatobusiness

    190
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    Sep 11, 2012
    I would like to make a very small battery (~100 mAh) powered light that turns on automatically when the light level is low. The device will spend most of it's time in light (i.e. stand-by) so it's important that the sensor consumes low power. The LEDs will probably be one or two red SMDs. I think a conventional solution using discrete components would be too bulky and consume too much power but this integrated circuit called OPT101 consumes 120 µA or an NPN phototransistor called TEMT6000 which on its own can consume more than 120 µA, depending on how bright the ambient light gets. I think both would need another transistor to power the LED on in the dark.

    The datasheet for the OPT101 is confusing http://www.ti.com/lit/ds/symlink/opt101.pdf - is "responsivity" the responsiveness to light? It talks about DC gain and bandwidth. I'm not certain how the light threshold is set but this chip seems it would use less power than the TEMT6000, at least in daylight when the TEMT6000 less more current though (potentially more than 1 mA?).

    Does anyone know of an alternative light sensor, ideally using SMD components and of course not requiring a microcontroller.

    Edit: I don't want the light to turn on very gradually either; I'd like it to be practically on vs off with as little room in between as reasonably achievable.
     
    Last edited: Sep 7, 2018
  2. BobK

    BobK

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    Jan 5, 2010
    The output of that chip is analog, proportional to the light falling on the sensor. You would need a comparator in addition to it to get a digital output.

    What type of battery are you talking about? 100 mAH is about the output of an LR44 coin cell, which is not going to power an LED for long.

    Bob
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I'd use a circuit like this:

    upload_2018-9-8_11-46-52.png

    This has minimal current drain when the LED is off, and I'd use an ultra-bright LED at a low current (say 1mA).

    The IC could either be a 40106 (which will operate at 3V) or 74HC14 (which should operate down to 2V) or similar. Note that these chips have 6 schmitt trigger inverters, and the unused ones need the inputs tied to ground.

    There are also single gate versions of this in surface mount form. the part numbers are something like 74LVC1G14.

    Note that the power supply connections to the IC are not shown. A red LED will operate down to around 1.7V.

    If the LED fails to come on at the desired level of darkness, you can reduce R1. However for better (lower) current drain, place two or more LDRs in series.

    If the LED illuminates the LDR the LED light can turn the curcuit off. This will make the LED flash. This does have the advantage of making the LED more noticable AND reducing the average current. In most cases it's a fault, but in this case it could be a feature!
     
    seanspotatobusiness likes this.
  4. kellys_eye

    kellys_eye

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    Jun 25, 2010
    Copyright Apple Corp.....
     
  5. seanspotatobusiness

    seanspotatobusiness

    190
    4
    Sep 11, 2012
    Thanks very much. I ordered some 74LVC1G14. I may use a larger battery but I only expect it to be used for minutes per day so it will last long enough.
     
  6. Audioguru

    Audioguru

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    553
    Sep 24, 2016
    Why not use a solar garden light that costs only $1.00?
    It has a solar panel, a 300mAh AAA Ni-MH battery cell, a circuit that switches from charging in the light or boosting the battery voltage to power the LED in darkness. It uses the solar panel to detect light or darkness.

    Get one with a glass face on the solar panel. Ones with a plastic face get sunburned. Replace the cheap Chinese battery cell and LED with ones that last.
     
  7. seanspotatobusiness

    seanspotatobusiness

    190
    4
    Sep 11, 2012
    I ended up implementing this in a slightly different arrangement than I first intended. It worked as expected for a day or two, turning on only in low light, but now when I turn on the circuit, the light comes on immediately and stays on regardless of the intensity of ambient light. If I try to measure the voltage across the LDR, the light turns off immediately. It turns back on after ~20 seconds if I take the probes away. I guess the impedance of the DMM isn't high enough to not interfere? I don't understand why it takes so long to turn back on though. Has anyone got any ideas what might have gone wrong and what to check?

    The IC is just a constant-current regulator called CN7511, set to about 30 mA.

    Thanks.
    [​IMG]
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,160
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    Jan 21, 2010
    Try replacing the LDR and see if that fixes it.

    You would get that fault if the LDR is open.
     
  9. Audioguru

    Audioguru

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    Sep 24, 2016
    A few years ago, LDRs were used in solar garden lights. Most of mine failed and needed replacement, but the replacements also soon failed. Now solar garden lights do not use an LDR, instead they use the reliable glass solar panel instead. I think the cheap LDRs and cheap plastic solar panels get sunburned.
     
  10. seanspotatobusiness

    seanspotatobusiness

    190
    4
    Sep 11, 2012
    I can test the LDR resistance in-circuit and it seems to be working fine (20 K in light, 1 M in dark) so I don't think it's the LDR at fault.
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,160
    2,676
    Jan 21, 2010
    A problem with your circuit is that all the current is sourced from the output of the inverter.

    The two Vcc connections should go to the power supply and the inverter should simply control the enable pin.

    That doesn't explain the fault though.

    I would start by seeing what is happening at the input and output of the inverter.
     
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