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Trying to identify the LED in the Coleman Slim-Lantern key-fob

Discussion in 'Electronic Basics' started by wylbur37, Apr 14, 2004.

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  1. wylbur37

    wylbur37 Guest

    I recently bought one of those cute key-fob LED flashlights at K-Mart.
    The brand name is Coleman, and the flashlight is flat and shaped like
    one of their typical Coleman lanterns. It cost $3.49 and runs on two
    CR2016 (3V lithium) disk batteries (6V total). There's a resistor
    inside in series that has a value of 51 or 52 ohms. (The bands on the
    resistor were green, brown, black and gold, although I wasn't sure if
    the second band was brown or red).

    Anyway, I'm trying to identify the specs (voltage, amperage, mcd,
    etc.) of the LED used in this flashlight (in case I want to transplant
    it for use in another application).

    Since no information appears on the LED itself, does anyone know how I
    could go about finding out what the specs are? Better yet, if someone
    has one of these flashlights and already knows what the specs are,
    please mention them.

    By the way, there's an evaluation (with photos) done on this item at
    the following URL (but no information about the LED specs) ...
    http://ledmuseum.jpoproductions.com/clant.htm

    Thanks.
     
  2. Jerry G.

    Jerry G. Guest

    I would use any of the standard type high output LED's that are in the
    colour white area. I am sure if you venture in to some of the catalogue
    from the major electronics parts suppliers, you will find something similar.
    Considering that most of these LED's have a 1.6 to 2.2 Volt drop, I would
    suspect that the current pull is about 60 to 80 ma. I have a feeling that
    if you buy these LED's in small quantity they will be on the expensive side.

    A number of years ago, I had one of these myself, but it had a red LED in
    it. It looks interesting and more realistic, that they are now using a white
    LED in these. I will have to go get one for curiosity. We have some
    sporting goods stores around here that sell these.


    --

    Greetings,

    Jerry Greenberg GLG Technologies GLG
    =========================================
    WebPage http://www.zoom-one.com
    Electronics http://www.zoom-one.com/electron.htm
    =========================================


    I recently bought one of those cute key-fob LED flashlights at K-Mart.
    The brand name is Coleman, and the flashlight is flat and shaped like
    one of their typical Coleman lanterns. It cost $3.49 and runs on two
    CR2016 (3V lithium) disk batteries (6V total). There's a resistor
    inside in series that has a value of 51 or 52 ohms. (The bands on the
    resistor were green, brown, black and gold, although I wasn't sure if
    the second band was brown or red).

    Anyway, I'm trying to identify the specs (voltage, amperage, mcd,
    etc.) of the LED used in this flashlight (in case I want to transplant
    it for use in another application).

    Since no information appears on the LED itself, does anyone know how I
    could go about finding out what the specs are? Better yet, if someone
    has one of these flashlights and already knows what the specs are,
    please mention them.

    By the way, there's an evaluation (with photos) done on this item at
    the following URL (but no information about the LED specs) ...
    http://ledmuseum.jpoproductions.com/clant.htm

    Thanks.
     
  3. Since you didn't say what color the LED was, I had to go to the website
    above to find out. Now that that issue is resolved and the LED is given
    as a white LED, you can get a replacement from
    http://www.whitelightled.com/ for a dollar plus postage. The white LED
    is commonly known as a Nichia NSPW500BS. Go to the above website or
    else to www.nichia.com to get specifications.
     
  4. I can say that the resistor is probably 51 and not 52 ohms since 51 is a
    standard value and 52 is not.

    As for the LED: Based on the LEDMuseum page, I am guessing that this is
    a plain old ordinary common (by today's standards) "super red" or "super
    bright red" or "ultrabright red" GaAlAsP LED with a typical forward
    voltage drop at 20 mA around 1.85-1.9 volts, maximum rated current of 30
    mA, brightness probably 3,000 mcd (possibly less), and peak wavelength
    probably around 660 nm.

    - Don Klipstein ()
     
  5. would

    I think you need to read up on current LED technology. White LEDs use a
    blue LED with phosphor, and their forward V drop is _not_ 1.5 to 2.2V,
    but 3.1 to 3.6V.
    side.

    I've been running white LEDs at 50 to 60 mA, and I'm finding them dying
    in a matter of days to weeks. The 20 mA that they are rated for is
    okay, but 30 mA is iffy and and above that will start to overheat them,
    and they lose their long lifetime.
    I got one of those Coleman key fob lanterns, but it was not flattened
    out, it was just like a regular lantern, with two button cells in the
    base and an amber LED in the globe. It sucked, because the LED shined
    up into plastic, not out to the location where the light was needed.
    It's now stuck away somewhere where it won't be a pain in the butt to
    someone.
     
  6. wylbur37

    wylbur37 Guest

    No, it's not a red LED, it's a *WHITE* LED.
    Sorry I neglected to mention that.
     
  7. wylbur37

    wylbur37 Guest

    How would you reduce the current of the power source to make it compatible
    with the rated current of the LED?
    After all, the standard formula for the desired resistance is ...

    R = (V8 - Vf)/If

    R = the resistor
    V8 = dc supply voltage
    Vf = rated forward voltage of the LED
    If = rated forward current of the LED at specified forward voltage

    Notice that nowhere in the formula is the current of the power source.
    So if the voltage of the source is already equal to the rated voltage
    of the LED, how would you reduce the current?
     
  8. That last paragraph, above, has got me bumfuzzled. I'm not sure what
    you mean.
     
  9. Well, you wouldn't, but in the real world it's never that simple.
    First, the "voltage" of the LED isn't a specific value, it's a range
    of values depending on manufacturing tolerances, temperature, and
    phase of the moon. Likewise, the voltage of your power source isn't
    exactly <so many> volts, it's a battery with a nominal voltage, which
    in the real world will start out high and end up low.

    Many tiny flashlights don't use a dropping resistor at all, they just
    depend on the internal resistance of the battery to limit the current
    to one that will not burn out the LED immediately, though it may show
    some degredation in light output over time.

    There are lots of solutions, from higher voltages and current limiting
    resistors to complex switching buck/boost converters. You pretty much
    get what you pay for.
     
  10. "Watson A.Name - \"Watt Sun, the Dark Remover\""
    Wylbur may not understand that the supply current and the LED current
    are identical...
     
  11. wylbur37

    wylbur37 Guest

    I'd like to take the LED from the key-fob flashlight and make it into
    a booklight (so I can read a book or use a computer in the dark so as
    not to disturb others in the room).

    However, since I plan to have this LED on continuously for several
    hours at a time, it's not practical or economical to run it on the
    original lithium disk batteries (because the batteries would quickly
    be exhausted).

    So I figured on using an AC adapter that puts out the same 6V that the
    two batteries in series had put out. But if the AC adapter puts out
    100mA of current (as it says on the unit) and if the LED is rated at
    20mA (as someone had suggested), then wouldn't I be overloading the
    LED? (After all, you said in your original posting that running your
    LEDs at a higher current than their rating would prematurely burn
    them out).

    So, assuming that's true, the idea is to find some way to reduce the
    current that the LED is exposed to from 100mA to 20mA when running it
    on the AC adapter.
     
  12. To reduce the current in an LED, use a larger resistor.

    Ohm's Law is:

    R = E / I

    where R is resistance, E is voltage, and I is current.

    In the case of an LED circuit, E is the voltage drop across the resistor.


    Example:

    A good old-fashioned red LED whose internal voltage drop is 1.8 volts.

    A 6-volt power source.

    You want 20 mA through the LED.

    R = (6 - 1.8) / 0.020 = 210 ohms.



    We should also check the power rating the resistor, since in some cases a
    1/8-watt resistor will not be big enough. (The power rating of the resistor
    is a maximum, so anything larger than the actual power dissipation will be
    safe.)

    P = E * I = (6 - 1.8) * 0.020 = 0.084 watt

    This is well under 1/8 watt, so a common 1/8-watt resistor will do fine.



    Do similar calculations for any LED and any voltage source.

    Or, if you like to experiment, simply hook up a milliammeter in series with
    the power source. Start with a resistor that you know is too large (like
    1000 ohms) and measure the current flow. Change resistors until you get
    what you want.


    Crucially, unlike a light bulb, and LED does not limit its own current. The
    voltage drop across an LED is nearly constant regardless of the current.
    That's why I treated it as constant in the calculation above. That's also
    why an LED requires a resistor. If you connect it directly to a voltage
    source, it will generally burn out, and if it doesn't, its current will vary
    tremendously with small fluctuations in the voltage source.


    This must be one of the most frequently asked questions in all of
    electronics. I haven't seen anyone post a good answer recently.



    --
     
  13. [snip]
    I did exactly the same thing, I put several white LEDs in a wall wart AC
    adapter to let them run for the last 5 or 6 months. I used a 6VDC, 100
    mA adapter, which had a voltage of about 7.7 to 8VDC with no load on it.
    I used 150 ohm resistors in series with the LEDs, but if you have only
    one LED, then use a 180 or 220 ohm resistor. For a 220 ohm resistor,
    you should measure 4.4VDC across it at 20 mA. For a 180 ohm resistor,
    you should measure 3.6VDC across it at 20 mA. A bit higher or lower
    should be okay. To find the current, just divide the voltage you
    meassure by the resistance.

    View the following with courier font.

    220 ohm
    + o----/\/\/\--------+
    |
    Wall Wart --- --> white LED
    6VDC 100 mA \ / -->
    --- Cathode is flattened point
    |
    = o------------------+
     
  14. Or to save having to change resistors, just put a 1000 ohm pot in there,
    and adjust to get the correct current. Then measure the pot and use the
    next higher standard value.

    BTW, I dunno about you, Mike, but most resistors (at least the thru hole
    type, anyway) that I see are 1/4W, not 1/8W.
    Hey, yours is a good one ;-)
     
  15. Soeren

    Soeren Guest

    (wylbur37) wrote in
    Just remember there is a resistor in the circuit (in between the lithium
    cells).

    (It is *not* the voltage of the batteries you need for calculating the
    resistor for use with an adapter, it is the LEDs voltage drop - measure it
    or calculate with ~3V6)

    The adapter doesn't "put out", the circuit it feeds draws from it. The
    100mA is a rating of the maximum current you can draw (without overloading
    the adapter).

    You control the current of the LED by choosing the right size resistor as
    others have allready explained.

    You need to replicate the resistor (51 Ohm) in the lantern, but you just
    need a higher value.
    In the lantern, when the battery cells is full, the current draw is around
    47mA (given a LED voltage drop of 3V6). While this is acceptable in a
    lantern used only for a few seconds (or minutes) occasionally, any long
    term use at this current will degrade and eventually kill the LED too
    fast. go for a max. current of 20..25mA.

    The resistor will then be:
    (6-3.6)/0.025 = 96 Ohm
    (6-3.6) / 0.020 = 120 Ohm
    Standard values (E12) are 100 Ohm and 120 Ohm.

    The LED will not give the same amount of light as in the lantern, of
    course, but it will last longer.
     
  16. Start with V=IR, R=V/I

    The wall wart will put out the voltage stated, at a current up to the
    value given (150ma I think you said). The voltage drop across the
    series resistor and
    the LED will have to add up to the output voltage of the wallwart. The
    LED drop is more or less constant voltage, so as you increase the
    resistance of the series resistor, the current trhough the total
    resistor and LED goes down,. So you need to measure the output
    voltage of the wallwart feeding a pure resistance at let's say 30 ma,
    that may be a little higher than the rated voltage when the output
    current is a full 150 ma (say you get 7V). Then put the LED in series
    with the resistor and see how bright it is, it should be fairly dim
    because there is less current flowing. Measure the voltage across the
    resistor (say 4V) Now, divide the voltage across the resistor (4V) by
    the current you want thru the LED, say 0.030 amperes (30 ma), 4/0.03 =
    133 ohms.
    That is the nominal valuue you want for the resistor.

    Remember, voltage is like water pressure, current is like the size of
    the pipe used to deliver the water.

    H. R. (Bob) Hofmann
     
  17. Oops, I was going from memory of an LEDMuseum page on those Coleman LED
    mini-lanterns...

    Typical voltage drop of an InGaN white LED (blue with broadband
    yellow-glowing phosphor) at 20 mA is 3.5 volts. Give or take about 1/4
    volt. The maximum voltage drop at 20 mA before the LED is considered by
    the manufacturer rejectable as defective due to excessive voltage drop at
    20 mA is 4 volts (for most white LEDs).

    - Don Klipstein ()
     
  18. wylbur37

    wylbur37 Guest

    Two questions ...

    1. I'm aware that when running an LED on a DC power source, the
    polarity must be observed or the LED won't light. But when the rated
    voltage of the LED is less than the source voltage and a resistor has
    to be used, does it make any difference whether the resistor is
    attached to the anode side of the LED or the cathode side?

    2. The AC adapter for my "itty-bitty booklight" is rated "4.8V 500mA",
    but it's actually 4.8 volts *AC*, not DC. I hooked up an LED (with a
    resistor on the "+" side of the LED) to the socket and found that it
    "works". But just because it "lights up" doesn't mean that it's
    "right" (in other words, if you overload an LED by running it on a
    higher voltage, it'll "light up" too, but you'll be damaging it). So
    the question is, is it OK to run an LED on an AC power source? and if
    "yes", is it sufficient to have one resistor or do you have to have
    one on either side of the LED?
     
  19. In series is all that matters. Either side is fine.
    LEDS don't have much reverse voltage capability. Your 4.8 volt AC
    supply will produce more than 4.8 volts when it is unloaded. 10 or 20
    percent more. It also produces a sine wave that peaks at 1.414 times
    as high a voltage in each direction than the effective voltage you
    read with your meter. so that LED may have to withstand something
    like 7 to 9 volts peak during the half cycle that it is blocking the
    current. Some LEDs will handle that and some will not.

    You may want to add components to lower this high reverse voltage.
    Some possibilities are, a series diode that will prevent any
    significant reverse current, an anti parallel diode that will conduct
    enough reverse current that the series resistor will waste all the
    reverse voltage, both of the above, another LED connected anti
    parallel to both do the job of the second diode mentioned, and also
    produce extra light, a bridge diode between the AC supply and the LED
    resistor combination to convert the AC to rectified DC.

    By the way, the transformer in the wall wart doesn't like DC loads
    much, either. It will run warmer than it normally would for the same
    output power drawn from both half cycles.
     
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