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Infra Red Emitters

Discussion in 'Electronic Basics' started by Paul, Apr 19, 2004.

  1. Paul

    Paul Guest

    I'd like to produce a beam of IR, to trip a remote alarm when the beam
    is broken.

    Lots of IR Emitter LED's (and frequency matching IR Phototransistors)
    are available, but I have no idea how '90mW/Sr' translates to real
    life. (In fact, I don't know what 'Sr' stands for!)

    I would like to span about 3 meters outdoors - so rain, fog, etc must
    be catered for.

    Anyone any ideas about what I need, please?

    Thanks.

    Paul.
     
  2. Steradian is a measure of cone angular shape. A cone of 4*pi
    steradian is so wide that it folds back on itself and includes all
    directions. An angle of 1 radian is about 57 degrees. A steradian is
    sort of square radians, since it involves angle over one more
    dimension (the shape of a cone, instead of an angle) than an angle on
    a flat surface.
    http://www.physlink.com/Education/AskExperts/ae174.cfm
    Specifically, just as a radian is the angle you get if you form an
    angle from the center of a circle through the ends of a radius long
    section of the circumference of the circle. a steradian is the cone
    you get if you form it to pass from the center of the sphere through
    the circumference of a circle on the sphere that has an area of the
    radius squared.
    http://www.schorsch.com/kbase/glossary/solid_angle.html

    For long distances, you need a high energy in a small cone high watts
    per steradian). This doesn't mean that the total power out equals
    this number or that the light covers a steradian. But if the light
    did cover a cone this wide, and the intensity remained as bright as it
    is in the center of the actual beam, then the total power would equal
    the specified power.

    I like the TSAL6100 (940 nano meter wavelength, 10 degree beam) and
    TSFF5210 (870 nm wavelength and 10 degree beam) and similar devices
    from Vishay.
    http://www.vishay.com/docs/81009/81009.pdf
    http://www.vishay.com/docs/81090/81090.pdf
    http://www.vishay.com/ir-emitting-diodes/
     
  3. Paul

    Paul Guest

    Thanks for that John. Just a couple of questions:

    In order to cope with rain, fog etc, what kind of power would I need
    to span 3 metres?
    Am I better off with a shorter or longer wavelength for this
    application, please?

    Thanks again.

    Paul.
     
  4. I'm not sure which wavelength would be best. The shorter wavelength
    is closer to the peak response of most silicon detectors (at 940 nm,
    silicon is getting to be pretty transparent) and these are also happen
    to be slightly more powerful emitters. But from a signal to noise
    standpoint, you might get an advantage in using the longer wavelength
    ones if you use a silicon detector filtered to exclude all light that
    is much shorter wavelength than 900 nm, since the silicon has so
    little response left longer than about 1000 nm. In other words, the
    longer wavelength filtered silicon sees very little daylight between
    the filter cutting off the short wavelengths and the silicon cutting
    off the longer wavelengths. The longer wavelength should do a bit
    better in fog, also. But I doubt there is a 2:1 advantage to either.
    I would just run the LED at rated peak power, and possibly add a small
    plano convex lens (perhaps a plastic fresnel) to more nearly collimate
    the beam. If the whole thing operates in the dark, you may be able to
    get by with a DC excitation, but if there is any significant
    interfering light, you will need to modulate the emitter with some
    carrier frequency (40 kHz is common) and detect energy only in that
    band.

    Does you design allow a synchronous detection scheme (electrical
    signals shared by emitter and detector) or are you forced to a power
    detector following a bandpass filter, only? The nice thing about the
    latter is that they are available as an integrated optical long pass
    filter, detector, amplifier, bandpass filter, rectifier and comparator
    to produce a binary go/no go output from a modulated beam. (used for
    infrared data communication) But the synchronous detector has an
    advantage of higher ultimate signal to noise ratio capability if there
    is interference near the modulation frequency.

    Have you picked out a detector yet?
     
  5. Anand Dhuru

    Anand Dhuru Guest

    Thanks for that John. Just a couple of questions:

    Hi Paul,

    To achieve a long range, this is paramount; you need to *modulate* the
    IR beam, and then again modulate the modulated signal! This might
    sound complex but is really very easy to achieve.

    Design 2 oscillators, one at 38KHz, the other at around 500Hz; you
    could use a pair of 555s for this. Connect your transmitter LED to the
    outputs of these; anode to the 500Hz, cathode to the 38KHz.

    At the receiver end, use one of the 3 pin IR receiver/demodulator
    modules, at the output of which you will get your 500Hz signal. You
    could then use a 555 based missing-pulse-detector to detect the
    absense of this pulse train.

    This arrangement is essentially what is used in consumer IR (TVs, DVDs
    etc.); it will give you ample range, and is quite immune to ambient
    conditions.

    Of course, in place of 2 555s you could use a 556, or a PIC
    microcontroller to reduce the component count.

    Hope this helps!

    Regards,

    Anand Dhuru
     
  6. Paul

    Paul Guest

    Thanks again for your help, John.
    That's a good idea - thanks.
    This will be operating in daylight too. Anand also suggests this
    approach.
    No, it won't be synchronous - it would merely detect any absence of a
    signal greater than 100 ms.
    No, not yet John - I'm still trying to understand as much as possible.
    I started out on this by thinking it would be quite simple - I now see
    that if this is going to work with any degree of reliability, a lot
    more knowledge is required.

    Paul.
     
  7. Paul

    Paul Guest

    (Anand Dhuru) wrote in message
    Thanks for that Anand. Alas, I fear your assumption regarding my level
    of knowledge is somewhat more than reality :). Sorry to ask what are
    probably very basic questions:
    Why is such a signal desirable? Does it make it less prone to
    interference?
    So - the transmitted signal is a pair of square waves, one about 70
    times the frequency of the other, superimposed?
    Do you, by any chance, have an example mfctr no. of such a device?
    An ordinary 500Hz square wave signal? Because the demodulator
    'removes' the 38kHz element? (I'm assuming a 555 only produces a
    square wave, BTW - this assumption may be incorrect of course!)
    OK, I understand that.
    Yes, thank you.
     
  8. Søren

    Søren Guest

    Hi Paul,

    The integrated IR-receivers demodulate the signal (removes the carrier
    wave) but they need it to differentiate between the ambient light and the
    IR signal.

    If you used only the carrier wave to modulate the IR-LED, the AGC
    (Automatic Gain Control) circuit in the receiver would assume it was some
    kind of static "noise" and turn down its gain.

    By making the carrier intermittent (in pulses), the receiver keep the gain
    high.

    The signal you need looks something like:

    __||||____||||____||||__

    If you want to save on the power, you can use shorter bursts of the
    carrier wave, with longer delays in between, making it look something
    like:

    __||______||______||__


    Most IR-detectors needs at least 6..12 periods of the carrier for AGC
    adjustment, så with eg. a 38kHz carrier, so use bursts of at least 0.5ms.

    A person breaking the beam will do that for at least 60ms (and they have
    to be speedy to accomplish this), as IR tends to "creep" around objects a
    slight bit, perhaps 50ms is a better number to be on the safe side.
    That means, that you would have to use at least 2 bursts each 50ms and the
    lowest frequency is as such 40 Hz.

    Your average power consumption is then very low, as illustrated by this:
    ______|__________________________________________________|_______

    Only the pulses use any real power.

    It is better to let one oscillator gate the other, like this:

    +-----------------------------+---O +12V
    _ | _ |
    +-| \ _ Vdd +-| \ [120]
    | | )O--------+-----| \ | | )O--+ _|_
    +-|_/ | | )O---------+---+-|_/ | _\_/_ // IR-LED
    | | +-|_/ | | | |
    +----[680k]----+ | | | _ | |
    | | +--[36k]--[4k7]-+ +-| \ | b|/c
    +--[18k]--|<|--+ | ^ | | | )O--+--[3k3]--| BC337
    _|_ 1N914 _|_ |___| +-|_/ |\e
    --- 47nF or equ. --- 1nF |
    | | IC=4093 |
    +------------------+----------------------------------------+---O Gnd

    Adjust the 4k7 trimmer for 38kHz (and get 38kHz receivers ;)
     
  9. Paul

    Paul Guest

    Hi Soren,

    Thanks very much for that, and the excellent ASCII circuit diagram.

    I think I'm going to have to do a lot more background study on this
    subject, and then have a play.

    Anyway, the answers have given me a lot of useful information.

    Many thanks to you and others who have helped with my understanding.

    Regards,

    Paul.
     
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