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LED Lighting PSU

Discussion in 'Electronic Basics' started by Bigus, Jul 9, 2009.

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  1. Bigus

    Bigus Guest

    Hi

    I want to experiment with powering LEDs from a 220V mains supply. I note you
    can get GU10 style LED bulbs these days but was wondering what circuitry
    they have in them. Has anyone seen/got a circuit diagram for such a thing?

    Presumably, if using standard LEDs then the circuit could consist of a
    bridge rectifier, capacitor & then a suitable resistor?

    I want to also play with Luxeon LEDs aswell. Trouble is a single Luxeon 3W
    white LED connected to the mains and driven at 800mA would mean a power
    dissipation of 175W, so that would be transformer territory wouldn't it?

    Regards
     
  2. T

    T Guest

    You could use a step down transformer before you hit the rectifier that
    way you can get the voltage down towards the 5V or a little over using a
    voltage regulator.

    The thing with LED's is current. You need to supply enough for the
    junction to work.
     
  3. Bigus

    Bigus Guest

    Hi. Thanks for the link to that circuit. My knowledge isn't brill - I can
    create a circuit from a diagram but things get hazy when it comes to
    modifying a circuit to vary things, like the number and types of LEDs.

    That circuit looks like it has a current limit of 100mA, so it sounds like
    20 LEDs would be the limit there (some LEDs are 20mA so might squeeze a few
    more in a I suppose). I guess that's the kind of circuit that'd be used in a
    GU10 LED array as it probably can be made pretty small.

    For currents like the Luxeon III though, with current demands of 800mA for
    just one 3W LED (and I was ganting to try 3 at a time in that department!),
    do you think a transformer would be the only way?

    Regards
    SPencer
     
  4. Bigus

    Bigus Guest

    Thanks, that's a good (easy to follow!) article. I've bookmarked your site
    too :)

    One other question. I just bought a switchable mains adapter that I was
    going to use for initial testing of the LEDs because it said it can handle
    up to 1.2 Amps. Looking at the label though it says the following:

    Input: 230V / 0.28A MAX
    Output: 3/4.5/5V = 1.2A, 6/9V = 1A, 12V = 0.9A

    This is my patchy electronics knowledge showing but I don't understand why
    the input says 0.28A. I thought there was just one current in an electronic
    device, that which the load creates - i.e: in the case of the Luxeon 3W LED
    around 800mA (it can actually take up to 1A). Where does the 0.28A come into
    it and will it be OK for driving the LED?

    Thanks
    Bigus
     
  5. Bigus

    Bigus Guest

    Yes! Awesome, thank you. I love it when someone explains something v.clearly
    the penny finally drops :)

    When I first looked at the supply current of the Luxeon and saw it could
    take a max current of 1A, I was thinking "blimey, so I could only run 13 of
    them on one mains plug!!!" (given a 13A fuse). In actual fact one could run
    more like 1000 of them :)
    I see. Theoretically then, the current on the 230V input side when supplying
    12V at 0.9A, should be about 47mA. So, you think the fact it says 0.28A max
    on the input side is a clue to the inefficiency of the adaptor rather than
    some kind of surge rating?

    If I did build a power supply myself, with a fixed supply and using the
    transformer/rectifier/resistor approach, would that have an inherent level
    of inefficiency as well?

    Regards
    Bigus
     
  6. greg

    greg Guest

    There will be some inefficiency, particularly if you use
    a very small transformer, since the efficiency of a
    transformer gets better the bigger you make it.

    I'd be surprised if it were *that* inefficient, though.
    17% sounds rather apallingly bad!

    In any case, you'll get better efficiency if you choose
    a transformer with a somewhat higher power rating than
    you intend to use. (Not *too* much higher, though, since
    the no-load loss goes up with the transformer size as
    well.)
     
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