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Series connection of 10 LEDs?

Discussion in 'Electronic Design' started by [email protected], Jan 17, 2006.

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  1. Guest

    Hello everyone!

    I need to connect in series about 10 LED and supply them with a
    constant current for a life test, but would not like to interrupt the
    circuit if one breaks down...

    Could you please give a hint if there could be a simple circuit to do
    this, I was thinking of connection an appropriate Zener diode in
    parallel of each LED?

    Any help is greatly appreciated!

  2. If it's driven with a constant current source why not just a diode
    in parallel with each LED?
  3. Mochuelo

    Mochuelo Guest

    A normal diode would absorb most of the current, leaving the LED
    almost off. A zener diode is better, as the OP proposed.
  4. Tim Williams

    Tim Williams Guest

    Well, blue diode maybe.. protector Vf > LED Vf. Otherwise you're testing
    the protection diode instead of the LED ;)

  5. Sure, wrongly written in haste. <smacks self> A couple of diodes
    works, depending on the LED though. I'm not a huge fan of zeners
    for such things.
  6. DJ Delorie

    DJ Delorie Guest

    Isn't this what antifuses are for?

    I'd go with a zener too. Remember that the zener will dissipate E*I
    so go with the smallest voltage that's still above the LED's working

    Compare the cost of a zener with the cost of a current limiting
    resistor though. Wiring a pile of LEDs in parallel, each with a
    resistor, to a standard 5Vdc power supply, might be less expensive.
  7. That would work. You just need to select a zener that draws
    essentially zero current at the normal LED drop but comes on without
    using up all the voltage compliance of your current source. Of
    course, it has to handle the power of the current times its voltage
    drop, but if these are ordinary LEDs that should be no problem. I
    think a 4.7 volt, .25 or .5 watt zener would work with most LEDs.
  8. Derek Potter

    Derek Potter Guest

    Or even three :)
    Agreed, I suppose you could go down to 2.7V but generally zeners add
    so much voltage to the chain that if more than one LED goes o/c you
    start to run out of drive voltage.

    I have to admit I'm curious about the point of this, though, as LEDs
    are very reliable and you are unlikely to learn anything about MTBF in
    a normal human lifetime :)
  9. Rich Grise

    Rich Grise Guest

    [crossposted to,sci.electronics.basics,
    followups-to sci.electronics.basics]

    Put an appropriate resistor in series with each. The resistor should
    be (Vsupply - Vled) / Iled so, for example, with a 12V supply, a
    forward voltage of 1.2V, and 10 mA target current, you'd use a
    ((12 - 1.2)/.01) ohm resistor.

    Doing the actual arithmetic, of course, is left as an exercise for
    the reader. ;-)

  10. Mochuelo

    Mochuelo Guest

    If he pushes the limits (current, temperature, etc), he can easily see
    degradation in short time periods, but it is not trivial how to then
    extrapolate those results to nominal conditions. There's a lot of
    theory behind that.
  11. Hmmm, If you accelerate the human to near C... ;-)

    Or accelerate the failure modes in a T&H oven, BIAS, etc. We used
    to do 30:1 sort of acceleration factors[*] for silicon devices
    (microprocessors and memory) that needed rating for 100kPOH.

    [*] Don't ask me, I wasn't in Quality Assurance.
  12. John Fields

    John Fields Guest

  13. Sjouke Burry

    Sjouke Burry Guest

    Google for a LED switching supply.Those can
    convert up from the powersupply, and sense the
    resulting current at ground connection.
    If a Led fails "shorted",the others just keep
    on working.
    They work by switching a coil current into
    the stack of LEDS,and adjust switching freq.
    to get the wanted current.
    Note that it works without series resistor,
    so you use your energy very efficiently.
  14. Guest

    If you're short on constant-current power supplies, you can give each
    diode its own LM317 regulator, run as a current regulator.
  15. Put two more leds across each of the ones on test - or use a single one
    of a different colour which operates at a higher voltage.

    Thus if a red leds are on test, wire a yellow one across each.
  16. Mochuelo

    Mochuelo Guest

    Just curious. What do you mean by "sense the resulting current at
    ground connection" and "without series resistor"?

    Also, varying the switching frequency usually does not modify the
    average of any current or voltage, but only the amplitude of the
    ripple. This assumes that the converter stays in continuous conduction
    mode. An easy way of changing average voltages and currents is by
    modifying the duty ratio.

  17. Sjouke Burry

    Sjouke Burry Guest

    There is a small resistor from the bottom LED
    to ground.This is filtered,and connected to the
    current sense input of the switch IC.
    The switch IC connects a coil to the power supply,
    until a set current is flowing.
    The IC is then swithes the coil accross the
    pile of LEDS,the coil will dump its stored power
    into the LEDS,just like it does into any parallel
    protection diode.
    Thats one currentpulse. If the sense input senses
    average LED current as to low yet, another dump
    cycle is performed.
    This results in a high enough dump rate to get
    the required average LED current.
    To decrease the peak LED current, you can also
    put a filtercap across the LED stack,so only
    the filtercap sees the current pulses.

    Al this does not require a current limiting series
    resistor,but only a sense resistor.
    Googling for LED powersupply,one of the first entry's:
  18. Mac

    Mac Guest

    On Tue, 17 Jan 2006 18:21:19 +0000, Derek Potter wrote:
    I have a 20+ year-old LED clock. It is so dim I can only read it at night.
    But your point is taken. ;-)

  19. Derek Potter

    Derek Potter Guest

    Well, let's not get bogged down in this, but you're going to have to
    overstress them a lot to get a significant failure rate with a one-off
    simple test. Arrhenius equation is already out of its depth when
    applied to failure rates (AFAIK it is basically about chemical
    reactions) and becomes quite meaningless once you exceed the
    manufacturer's ratings to force a failure. New failure mechanisms
    spring into existence at higher temperatures. Once the die melts, for
    example, the MTBF is zero :)

    I'm always intrigued by manufacturers' current ratings. If anything is
    hurt by current as such, rather than heat, it should show up if you
    pulse the LED at low duty cycle keeping the same mean current. AFAIK,
    this effect does not take place - at least, not until the RMS current
    gets to silly values that damage the chip bonds. I suppose there could
    be a kind of second breakdown effect but I don't think this can
    generally be produced in diodes. Other than that it's hard to imagine
    a mechanism for current as such to cause failure - carrier flow does
    not disrupt the device structure.

    Really there's not much else apart from temperature that you *can*
    push. Voltage is determined by the diode and the current, you'd need
    enormous pulses to increase the filed strength sufficiently to cause
    migration of doping agents in the chip. Hardly in the same league as a
    string of LEDs running off a DC PSU.

    Hmm. Did I say "let's not get bogged down in this"? i seem to have
    rambled on a bit.
  20. Derek Potter

    Derek Potter Guest

    Was it always that dim and have you measured the LED current?
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