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Right resistor for high power LED for sound reader?

Discussion in 'Electronic Basics' started by M, Aug 9, 2005.

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  1. M

    M Guest

    Hi, I'm a projectionist in a movie theater. I need to
    convert our old white light incandescent sound
    "exciter" lamps over to red LED.

    The entire movie industry is changing over to having
    the soundtracks on film based on cyan dye, instead of
    the traditional silver based soundtrack. The big
    difference is that the new way is cheaper for the
    studios to make. Another big difference is that almost
    every movie projector in the world will need the new
    red LED readers. Most have already changed over.

    The soundtrack runs along one side of the film, and the
    sound information is a tranparent wavy stripe which
    runs down the middle of the track. The old style
    soundtracks are opaque black, and the new tracks are
    semi-transparent cyan blue. There is an "exciter" lamp
    on one side of the film which shines through a barrel
    with lenses in it, which only lets a slit of light hit
    the film. On the other side of the film is a solar cell
    which picks up the light information.

    We have a movie starting Thursday which is our first
    cyan film. I need to rig up a red LED light source
    ASAP. I saw instructions on how to do this on
    www.film-tech.com (search using word "homebrew"), but I
    need more info. I need your help in choosing a proper
    resistor for the LED.

    I ordered and received 2 Luxeon Star red LEDs. Here are
    their specs:

    Part # LXHL-MD1D

    Max Current - 350ma (300ma or under may be preferable
    so I won't have to add a heat sink)

    Max Voltage - 3.5vdc

    Color - Red

    Wavelength - 625 NM

    Light Dispersal - LAMBERTIAN

    Typical Flux (Lumens) - 44

    The incandescent exciter lamp it is replacing is rated
    at 9v 4a 36w.

    I've checked the power going to the light, and it is
    only 7.5dc. That's OK because most film technicians set
    the power lower on exciter lamps to make them last
    longer. That's what I've heard, anyway. That power
    supply is adjustable, but I'd prefer to leave it at
    7.5v so I can still put in the old lamps whenever I
    need them. The power supply is rated at 5-10vdc and 5a.

    So, what kind of resistor would I use to get this LED
    to work? I wouldn't want to burn it out or run it too
    dim either. I've been wracking my brain trying to
    figure this out, and I'm not having much luck so far.
    Not sure which formulas in Ohm's Law to use.

    Thank you, Mitch

    spamsuckawell-wornhat at yeahwhoo dott caum
    Take off the well-wornhat to respond.
     
  2. Tom Biasi

    Tom Biasi Guest



    Hi,
    You have a device (LED) that will drop 3.5 volts at 300 mA.
    Your supply is 7.5 volts. You need a resistor that will drop 4.0 volts at
    300 mA. R=E/I
    P=EI, Closest standard values 10-15 Ohms 2 Watt.
    Tom
     
  3. M

    M Guest

    Thanks Tom! So your saying that:

    4V / 300mA = 13.33 Ohms, and
    4V * 300mA = 1.2W

    and the only resistor I can use is a standard 10-15
    Ohms 2 Watt. Can I use any kind of variable resistor
    and set it at 13.33 Ohms? If I can, what kind would be
    best? Would this be a waste of time, and would the
    fixed value resistor be fine?

    I know this is really basic stuff for you people, but
    this is the first time I've ever done this.

    Thanks, Mitch
     
  4. redbelly

    redbelly Guest

    Mitch,

    Best to keep your life simple. Just use 15 ohms, and don't worry about
    messing around with potentiometer settings. You'll be well below the
    max current (around 270 mA), so burning the thing out shouldn't be a
    concern, and you'll still get a reasonable amount of light output.

    Regards & good luck,

    Mark
     
  5. Tom Biasi

    Tom Biasi Guest

    I agree with Mark (AKA redbelly).
    Keep it simple. There is no need to set the resistance to 13.33 ohms. Any
    resistor between 10 and 15 ohms will be fine for this application. Two watt
    resistors are easy to come by so I would use one.
    Tom
     
  6. Assuming your power supply will be or can easily be set to your favored
    7.5 volts DC when drawing only 300-350 mA, then:

    The LED typical voltage drop is 2.95 volts according to the "DS23"
    datasheet available from the Lumileds website.

    Subtract this from 7.5 volts and this leaves 4.55 volts across the
    dropping resistor. Plus .64 volt, minus .56 volt tolerance worked out
    from subtracting the minimum and maximum voltage drops mentioned in this
    datasheet.

    To get 350 mA with 4.55 volts across the dropping resistor: Divide 4.55
    by .35, and you get 13 ohms. The next common value up is 15 ohms,
    although 13 ohms is a semi-standard value. Power dissipation will
    typically be 4.55 volts times .35 amp, or about 1.59 watts. I would use a
    5 watt "sandstone" style 15 ohm resistor.

    To be extra conservative, let's use 300 mA (.3 amp) and 2.25 volts LED
    voltage (minimum at .35 amp is 2.31 volts), leaving a worst case of 5.25
    volts across the droping resistor assuming a 7.5V supply.

    5.25 volts divided by .3 amps is 17.5 ohms. The next higher "standard"
    resistor value is 18 ohms. Again, I recommend the 5 watt "sandstone"
    type.

    For "bad worst case" designing, use 22 ohms 5 watt rectangular sandstone
    style. Expect typical LED current then of about 230 mA with a 7.5V
    supply. And the resistor will typically be dissipating 1.15 watts, which
    is enough to get a 5 watt rectangular sandstone style resistor very warm.

    Another note: Performance of Lumileds LEDs is with the "junction
    temperature" at 25 degrees C, and the red one has quite a high sensitivity
    of light output to temperature. The DS23 datasheet gives a thermal
    resistance of 23 degrees C per watt for this model, and at 2.95 volts .35
    amp (1.03 watt) expect to require a heatsink temperature of 1-2 degrees C
    to achieve this! And this red model has output slightly less than 70% of
    "full" when the "junction temperature" is 25 degrees C warmer than this!

    Not that I don't think it will work, but I advise to drive these things
    conservatively, heatsink them adequately to excessively, and make your
    expectations of light output realistic - I would say 30 lumens at most,
    plan on 25 lumens even with a heatsink, and less if you have a
    worse-than-average LED - worst case is about 30.4% below "typical"
    according to the datasheet. This may well be plenty good for your
    application however!

    - Don Klipstein ()
     
  7. I would use a fixed resistor with value at the high end of the range for
    possible answers - at least 15 ohms.

    Two reasons:

    1. The chips in these LEDs have a nonlinearity, with efficiency being
    maximized at currents in the general ballpark of 50-60% of "full current"

    2. These LEDs (red Luxeons) have light output very sensitive to
    temperature, with output doubling by having the junction 45 degrees C
    cooler than the 25 C "characterizing temperature", and halved by having
    the junction 45 C warmer than this. This means a 1 degree C temperature
    change causes light output to change about 1.5%, with higher temperature
    being unfavorable. So I consider it good to operate these LEDs
    conservatively, and expect little to gain in light output from pushing
    them with current past about 300 mA.
    In addition, I recommend heatsinking them to an extent many
    would call excessive.

    - Don Klipstein ()
     
  8. Tom LeMense

    Tom LeMense Guest

    ....and let's hope that that's a fairly well regulated supply, at that: I
    would imagine that any variations in the intensity of the LED (due to line
    regulation) would show up as audible noise!
    I've worked with the Luxeon Star LEDs and I heartily recommend a heatsink.
    The LED life goes down very quickly as Tj rises, so keeping it cool will
    improve the reliability of your projector retrofits.

    I have a red Luxeon Star on an old i486 (passive) heatsink and that seems to
    work pretty well. I also didn't want to worry about a regulated supply, so
    I built a "constant current regulator" onto the same heatsink using an LM317
    adjustable regulator IC. The circuit is right out of the LM317 datasheet
    (see figure 26 of the On Semi datasheet located at
    http://www.onsemi.com/pub/Collateral/LM317-D.PDF ) and I used a 1W 3.6 ohm
    resistor to set the current at approx 300mA. The benefit of this circuit is
    that the LED will be fed with a constant 300mA current without depending on
    the input voltage. Not bad for a three component circuit (counting the
    LED)!

    TJL
     
  9. M

    M Guest

    Oh man, thanks guys! That's what I was hoping to hear!
    I'm going to have at least 300 people counting on me to
    put on a good show, so it's a big deal. I've been
    trying to figure this out for myself and am making some
    small progress. I'm just really stuck for time and
    appreciate your help a lot. This is my very first foray
    into the world of electronics.

    I live in a small town and went to the local Radio
    Shack. They don't have any 2 watt resistors. They have
    15 ohm 1/2 watt, 10 ohm 1/2 watt, and 10 ohm 1 watt.
    What combinations of common resistors can I put
    together to get a 10 ohm 2 watt?

    Someone told me it has to do with "Kirchoff's Law." I'm
    trying to learn about that from some badly written web
    pages. It's not making a dent yet.

    -Mitch
     
  10. Mitch

    Mitch Guest

    Excellent info! I can dig some heatsinks out of my old
    computers. The power supply is a Kelmar 8604-B 5-10vdc
    5a. It is made for sound "exciter" lamps, so I'm fairly
    certain it is regulated.

    I also got 4 LXHL-LD3C Luxeon Star III LEDs to
    experiment with. They are 140 max lumens, 3.51v 1400ma.
    If I can figure out the proper resistors tomorrow
    morning, I may make the lamp with those instead. I was
    just planning on using the older LXHL-MD1Ds because
    others had done it and it seemed safer to follow in
    their footsteps.
     
  11. Four 15 ohm ones in a 2-by-2 series-parallel arrangement is a 15 ohm 2
    watt resistor.

    Put two 10 ohm 1 watt ones in parallel with each other and put that
    combo in series with a third 10 ohm one, and you get a 10 ohm 2 watt
    resistor.

    Put four 10 ohm 1 watt ones in series-parallel 2-by-2 arrangement and
    you get a 10 ohm 4 watt resistor.

    With a 2-by-2 series-parallel arrangement of four identical resistors,
    it does not matter whether or not you put two parallel pairs in series or
    put two series pairs in parallel.

    - Don Klipstein ()
     
  12. John Fields

    John Fields Guest

     
  13. John Fields

    John Fields Guest

    ---

    Two 10 ohm half watters in series, in parallel with two 10 ohm half
    watters in series, like this:


    +---[10R]-----[10R]---+
    | |
    +---[10R]-----[10R]---+

    |<--10ohms, 2 watts-->|


    Or, two 10 ohm half watters in parallel, in series with two 10 ohm
    half watters in parallel, like this:



    +---[10R]--+--[10R]---+
    | | |
    +---[10R]--+--[10R]---+

    |<--10ohms, 2 watts-->|
     
  14. John Fields

    John Fields Guest


    If the're out of 10 ohm half watters when you get there you can do
    this:

    +---[15R]--+--[15R]---+
    | |
    +---[15R]--+--[15R]---+
    | |
    +---[15R]--+--[15R]---+

    |<--10ohms, 3 watts-->|

    Or this:

    +---[15R]--+--[15R]---+
    | | |
    +---[15R]--+--[15R]---+
    | | |
    +---[15R]--+--[15R]---+

    |<--10ohms, 3 watts-->|
     
  15. John Fields

    John Fields Guest

    ---
    Oops...

    You get a 15 ohm 1.5W resistor.


    +--[10R]--+--[10R]---
    | |
    +--[10R]--+

    equals

    ---[5R]---[10R]---

    so, since:


    P = I²R

    rearranging and solving for the current in the 10 ohm resistor when
    it's dissipating 1 watt:

    P 1W
    I = sqrt ---(PR) = sqrt ----- = 0.316 amperes
    R 10R


    Now, with 316mA through 5 ohms,


    P = I²R = 0.316A² * 5R = 0.5W

    So, since the 10 ohm resistor is dissipating one watt and the set of
    paralleled resistors is dissipating half a watt, the array is
    dissipating 1.5 watts.
     
  16. John Fields

    John Fields Guest

    ^
    --- |
    Oops again... (PR)----+ shouldn't be there.
    ---
     
  17. Oops, better keep my eye on the screen better. It's obviously 15 ohms,
    and that's what I meant to type.

    - Don ()
     
  18. Thanks... I did screw that one up.

    - Don ()
     
  19. Rich Grise

    Rich Grise Guest

    This is the kind of question we really like around here - you gave
    us what information you have, and said, "This is the goal".

    Yes, a power rheostat (variable resistor) would work here, but
    a new one would be astronomical:
    http://dkc3.digikey.com/PDF/T052/1102-1103.pdf for an example.

    But you should be able to find a 12 ohm, 2 watt resistor! I'm
    sure that will be fine.
    http://dkc3.digikey.com/PDF/T052/1078.pdf

    Break a Leg!
    Rich
     
  20. Rich Grise

    Rich Grise Guest

     
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