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110V powered LEDs

Discussion in 'Electronic Design' started by [email protected], Jan 24, 2006.

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  1. Guest

    I saw a LED assembly some time ago that consisted of two red LEDs in
    parallel, a 3/8" diameter "bump" inline with the red & black wire with
    a black & white wire out of that. The bump was about 1" long and
    covered in heat shrink. The point was to run the two LEDs off 110VAC. I
    don't know who made it, or even what country the assembly come from.
    The point to all this, is to fire two red LEDs at an absolute minimum
    cost from 110VAC. Wall-warts cost a lot.

    Any ideas?

    Thanks in advance!
     
  2. If the LED's were wired in opposite directions, the bump could be a
    capacitor. A resistor would be much cheaper, but have to dissipate
    about a watt of heat.
     
  3. Rich Grise

    Rich Grise Guest

    The dead simplest way to do this is to simply use a dropping resistor,
    and put the LEDs in antiparallel. Size the resistor for 110V * .02A,
    and I'll leave the arithmetic as an exercise for the student. :)

    Good Luck!
    Rich
     
  4. Mmm... Capacitors tend to be big and expensive. Besides, the inrush current
    may be high enough to blow the LEDs. To stay on the safe side you will need
    a capacitor and a resistor. Assuming the LEDs will produce light enough at
    1mA (and a lot of them will do) you only need a 100k or so resistor which
    dissipates only about 100mW. You can even go downto 68k or 56k for more
    light (and more heat) but still use a common 250mW resistor.

    petrus bitbyter
     
  5. I hope you meant "size" in the sense of wattage and not in the sense
    of resistance.

    Also, might try a .47uF capacitor. I see a metallized polyprop film
    from EPCOS at Digikey for USD 0.67/ea. I believe they are designed
    for AC lamp applications. Part number is 495-1315-ND. (The rating on
    Digikey's site shows 400V, but that is for DC.)

    Jon
     
  6. John Fields

    John Fields Guest

    ---
    "Absolute Minimum Cost", IMO, would include not only the cost of
    ownership of the device, but also its operating costs. Such being
    the case, a device with a reactive ballast might cost more to buy
    but less to run, in the long term, than a device with a resistive
    ballast.

    Consider a system with two white 3.5V 20mA LEDs connected in
    antiparallel, in series with resistive, capacitive, or inductive
    ballasts across 120V mains.

    Which would you recommend, and why?
     
  7. By comparison with 1/4 watt resistors, very much so. But I did find a
    67 cent cap at Digikey. Not horrible and almost the same price as a 2
    watt resistor, bought at Digikey in 1s. (All this assumes you have
    enough order quantity on other parts, of course, and discounts any
    shipping.)
    Is this true? The maximum slope at 60Hz is d/dt (160*sin(wt)), which
    is 2*PI*60*160. I = dV/dt * C, which means the actual worst case is
    2*PI*60*160*.47e-6 or about 28mA peak current, roughly.

    I don't see how this can be a problem for most common LEDs.

    Jon
     
  8. Clark

    Clark Guest

    I agree with Jonathan, the last time I saw Leds powered from 120vAC was
    using a 100 ohm on one lead and a .47uF cap 250 Vac cap on the other. Hardly
    no heat was generated. This only problem with this is a spike on the line
    will pass right through to the leds.
     
  9. How do you *guarantee* that it doesn't EVER, even once, get switched
    on right at the peak of the AC line? Might take more than one or two
    components to do that.



    Best regards,
    Spehro Pefhany
     
  10. That would be a problem. We have the effective dV/dI resistance slope
    of the diode, itself. But that would not be enough. I could easily
    see a few amps for periods of a few microseconds. A smaller resistor
    value of about 1k would nip that to something reasonable, but it would
    still need to be about 1/2 watt in size, I think.

    Thanks for pointing that out.

    Jon
     
  11. Zak

    Zak Guest

    I think you can with an extra cap and two diodes:

    o-----||---- +->|--+----o
    | |
    line +->|-+ ===== LED
    | -----
    | |
    o-------+----------+----o


    The output cap needs to be fairly large, but it is low voltage (voltage
    limited by the LED).


    Thomas
     
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