Connect with us

1 watt luxeon of 12 volts dc

Discussion in 'Electronic Basics' started by mark krawczuk, Sep 18, 2004.

Scroll to continue with content
  1. hi , i am using 2 1 watt luxeon red leds for brake lights , can i run
    these of 12 volts via a resistor, or do i need a driver circuit ?
    what value resister would i have to use , (if it can be done this way ./.)

    thanks, mark k

    "Are you still wasting your time with spam?...
    There is a solution!"

    Protected by GIANT Company's Spam Inspector
    The most powerful anti-spam software available.
  2. Ian Stirling

    Ian Stirling Guest

    What colour of luxeon?

    You need to find the forward voltage of them.
    If it's for example 3.5 volts, and you are putting them in series (a
    good idea) then the total voltage across them is some 7V.
    Take this number away from 12V, and you get the voltage that
    you need on the resistor.
    This is some 5V.

    If the current is .3A, then you need 5V/.3A = 15 ohms.
    And the power needed for that resistor is .3A*5V = 1.5W.
  3. According to the datasheet at

    luxeon reds have a forward voltage of between 2.25 and 3 volts,
    depending on the current you want to put through them.

    If you want 300mA, then they will be at about 2.8V each, according to
    the current to forward voltage graph. So, put them in series, that'll
    give you 5.6V. Then, you need a resistor to drop the voltage
    sufficiently. That is 300mA over 6.4V, which works out to about 22 ohms.
    However, the power dissipated in the resistor will then be

    P = 6.4^2 / 22 = almost 2 W. You'll need a big resistor.

    You can use a network of resistors if you can't get 2W ones; if you use
    10 220 ohm resistors in parallel, you can use standard quarter watt
    resistors. The amount of heat generated will be the same, but it'll be
    spread over the whole set. If you mount them so there is a bit of space
    between them, it should be OK.

    Another option would be to use a switching power supply, like the 'roman
    black' step-down supply, shown here:
    Its simple to build, and you can easily make it put out 5.6V (which will
    cover your two LEDs) by increasing the zener to a 6.2V one. The nice
    thing about this is that its much more efficient; this may not be a
    concern for you, though.

    Bob Monsen
  4. Do not apply fixed/regulated voltage to an LED - the current will be
    unpredictable. The LED has a tolerance in its voltage drop (assuming the
    amount of current specified in the datasheet for characterizing voltage
    drop). The actual voltage drop *at specified current* has some
    significant tolerance, and varies with temperature. Since the voltage
    drop does not vary much with current, a tolerance in voltage drop
    characteristics can mean current being far from what you want if you apply
    a fixed voltage.
    If you apply a fixed voltage, the amount of current usually varies
    greatly with temperature. Such current variation with temperature can
    reinforce the temperature change, leading to the current changing even
    more. Sometimes this even results in "thermal runaway", which is the
    situation where increasing temperature and increasing current reinforce
    themselves - sometimes in a spectacular "runaway spiral", sometimes to a
    lesser extent but to an extent great enough to seriously overheat the LED.
    The LED may fail in minutes or may merely have a seriously shortened life.

    What to do: Use a switching current regulator (or any current
    regulator), or use a voltage source significantly more than the voltage of
    the LED (or series string thereof) and use a "dropping resistor".
    Maybe set the "roman black" power supply to 6.9 volts, and drop the 1.4
    volt difference with a 4.7 ohm resistor.

    The necessity of limiting LED current has been discussed here before
    enough times. I mention a bit more of this (and calculating dropping
    resistors) in a web file of mine,

    - Don Klipstein (,
  5. Ban

    Ban Guest

    snip resistor method>>
    Much better would be a fixed current output switching regulator such as the
    LT1618. For a bike it is important not to waste any energy in form of heat.
    You can even have the options of current adjustment and automatic on/off
    switch. Only a few external components needed, no power resistor. See design
    notes in last EDN.
  6. Ban

    Ban Guest

    I copied the schematic which is a bit difficult to explain, because a boost
    converter is used as a buck. Also that original is faulty, as the SW is not
    connected, I guessed it goes to the other end of the coil. Efficiency is
    claimed as being 70% with a 3.6V/700mA Lumiled. Hope I got it right.
    | |
    | ___ 700mA ___ |
    | | 0.07R | LED 47uH | PMEG2010
    | | | | schottky
    | | .-------------. |
    | | | ISN | |
    | | | | |
    | +--|ISP LT1618 | |
    | | | | |
    | +--|Vin SW|----+
    | | |
    | | |
    | o---|SHDN\ FB|-----------+
    | | | |
    | | | |
    | |Iadj GND Vc | |
    | '-------------' |
    | | | | ___ |
    --- | | +-|___|---+ |
    --- | | | 10k | |
    |4u7 | | --- --- |
    | | | --- --- |
    | | | |220p |2n2 |
    created by Andy´s ASCII-Circuit v1.24.140803 Beta
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day