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LED's on a vehicle

Discussion in 'Electronic Basics' started by Ken C, Jan 19, 2006.

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  1. Ken C

    Ken C Guest

    I am trying to determine the optimum arrangment for multiple LED's
    powered by a 12V battery/alternator system on a motor vehicle.

    I could mate each LED with a 500 ohm resistor, which is how they came
    from the HK eBay vendor. This will waste 2/3 of the energy consumed
    and might create a heat problem if I pot the electronics in epoxy.

    I could put three LED's in series with a 150 ohm resistor, but they
    would significantly brighten and dim as the vehicle's voltage ranged
    between 12.5 volts (engine off) and 15 volts (alternator charging).

    I am leaning toward using one 7808 regulator feeding multiple
    clusters, with each cluster containing two LED's in series and a 50
    ohm resistor. This would put most of the energy loss in the
    regulator, which I can put in a protected spot with a heat sink. The
    constant 8V would solve the variable brightness. The regulator, rated
    at 2.2 amps, could in this fashion illuminate over 200 LED's, which is
    more than enough.

    Any suggestions?

    Ken C
  2. Why not pulse the LEDs? Using power FETs, this should yield the best
    possible efficiency. By pulse, I mean full ON or full OFF. None of the
    losses associated with series resistors or linear voltage regulators.
  3. Bill Bowden

    Bill Bowden Guest

    Why not pulse the LEDs? Using power FETs, this should yield the best
    How many FETs would that require for 200 LEDs?

    I suppose you could put them all in series and hit them with a constant
    current pulse at 300 volts or so. But if one burned out, it would hard
    to find.

  4. Gareth

    Gareth Guest

    As I'm sure you know, you really want to limit the current rather than
    the voltage. You can do this with an adjustable voltage regulator and a



  5. Why not just wire them up, ten at a time, in series? There will be a 1.2
    volt drop across each of them which, as I recall but may be wrong about, is

  6. Pooh Bear

    Pooh Bear Guest

    Because a car's 12V supply isn't actually 12V. It varies widely.

  7. ehsjr

    ehsjr Guest

    Typical red LEDs drop ~ 1.8v. Put 6 of them in series with
    your 150 ohm resistor, and the string will draw ~28 ma at
    15 V, and ~11 mA at 12.5 volts. This will waste about 118 mW,
    worst case.

    You could run 4 in series with 68 ohms and a 7809
    regulator to get ~ 26 mA through the LEDs. It will
    waste 202 mW per string, worst case. Because the
    car is a hostile environment, you don't want to
    run the 7809 anywhere near its maximum rating.
    You need a heat sink and probably a pass transistor
    to run a decent number of LEDs. Plus you'll need
    to account for transients ( at a minimum a transorb
    and capacitors).

    All in all, I think the first approach is best.
    So what if it's dimmer when the engine is off?

  8. Ken C

    Ken C Guest

    Hadn't thought about that. I remember the LM3909.

    These are blue LED's with a 3.3V average drop at 20 mA. Could I put a
    50 ohm resistor in series with each LED and pulse them with the full
    12V on a 5% duty cycle at maybe 10kHz (or whatever looks continuous).

    What chip/circuit/kit would accomplish this?

    Each LED would be at the end of a 4 ft pair of wires. Would this
    arrangement disrupt radio reception?

    Ken C
  9. kell

    kell Guest

    Right now I am stuck using Google Groups, which doesn't have a
    nonproportional font for posting ASCII diagrams, so bear with me while
    I describe a simple constant current circuit. This uses a npn
    transistor at the low end of your led string. If you want the
    transistor at the top of the string, just use pnp and flip everything.

    Draw a npn transistor. Draw a common red led from transistor base to
    ground. Draw a resistor from transistor base to battery positive.
    Draw a resistor from emitter to ground. Draw your diode string between
    battery positive and the collector.
    The emitter resistor will have a fixed voltage impressed on it, which
    is the difference between the led voltage and the base-emitter voltage
    of the transistor. That will be about one volt, which makes calcs
    easy. Tempcos of the "base" led and b/e junction will cancel, or
    nearly so. Collector current, which is the led string current, will be
    equal to one volt divided by the value of the emitter resistor. Put
    enough leds in the string to minimize the voltage drop across the
    transistor. This is the most efficient way I can think of to run
    led's, if you are willing to wire one transistor for about every six or
    seven led's you run.
  10. 3 of the blues in series should work. Set the peak current with a current
    regulator (check the manufacturer's specs for recommended peak currents) and
    then set the duty cycle for the desired brightness (the maximum safe duty
    cycle will depend on the peak current that you choose). If you use a series
    resistor as your current regulator, the current will be temperature
  11. Rich Grise

    Rich Grise Guest

    Yes, it does, under "options" or something like that.

    And if you draft something in, say, notepad, you can paste it
    in any old font you want - it's just plain old ascii, so we
    can view it however we want to. :)

    Good Luck!
  12. kell

    kell Guest

    I used to post ASCII under Google. Can't do that any more.
  13. kell

    kell Guest

    I made another post before I figured out how to draw this in ASCII
    (thanks to Rich Grise for the suggestion about notepad).
    View in a fixed-width font.

    enough led's in series here to drop 10 or 11 volts
    | |
    | |/c
    B+---+--------R1--+-------| NPN
    | |\e
    | |
    red led |
    | R2
    | |
    | |
    | |

    Size R1 to illuminate the red led. Maybe 680 ohms. This red led just
    serves as a voltage reference. But it lights up too :)
    You can use R2, which has a fixed voltage of approximately 1 volt
    across it, to trim the current in the led string. If you want .02
    amps, R2 would be in the 50 ohm ballpark.
  14. Rich Grise

    Rich Grise Guest

    Thanks for the mention here - This is a very, very good way to start
    a Monday morning. :)
    Looks good to me!

    But then again, as has probably been noted, I'm easy. ;-)

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