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LED's for dummies

Discussion in 'Electronic Basics' started by [email protected], Aug 16, 2006.

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  1. Guest

    I'm apparently an incredible dummy, because I can't wrap my mind around
    this concept yet. So I'm sorry to ask such a rudimentary set of
    questions.

    I have been given a bowling pin. I am supposed to take this bowling
    pina dn turn it into a work of art which will be auctioned off for a
    local charity. My idea as a budding electronics hobbyist is to
    basically decoupage the pin and then install a bunch of LED's driven by
    a wall wart.

    So I will cut the pin in half at its widest point and hollow out the
    inside as much as I can so that the electronics will fit in there, then
    I will drill holes or wiring and try to place the LEDs in little
    recesses that I will create.

    ANYWAY, I will be using 18-24 LEDs of various colors. Here is a list
    (i'd be using 3-4 of each color):

    Blue - 3.6V, 20mA
    Orange - 1.7V, 20mA
    Amber - 1.7V, 20mA
    Green - 3.3V, 20mA
    Red - 2.3V, 20mA
    Sea Green - 3.3V, 20mA

    My first problem is that I don't know for sure if those voltage specs
    are just the voltage required to power the diode or if it is a
    measurement of the voltage drop, or even if there is a difference
    between the two.

    But since the current for all of them is the same, I considered wiring
    them in series. The problem there is that I was hoping to use a 9V,
    500mA wall wart that I already have, and if those voltages are voltage
    drops, I can't wire too many at a time.

    So clearly I have to use a combination of parallel and series circuits
    in order to control the current while keeping the voltage under, say
    7.2 volts. And I know that I will have to incorporate resistors
    somewhere along the line. But that's it. I can't make it to the next
    step. I'm not sure how I should distribute the LEDs so as to create
    the most efficient circuit. I'm not sure where to put the resistors.
    I want to draw a wiring diagram, of course, but I can't visualize this
    for some reason.

    If anyone can point me in the right direction here, I'd really
    appreciate it. I'm not looking for someone to bail me out as much as I
    need someone to show me what my mental obstacle is, maybe helpe me get
    past it. Thanks in advance for any help you can share!
     
  2. PeteS

    PeteS Guest

    The voltages look like the typical forward voltages at the specified
    currents. Those currents, incidentally, will make the LEDs quite
    bright.

    Assuming you are using a 9V wall wart, you could:

    Power + ------ resistor1 -------- two blue leds ----- power -
    ------ resistor2 -------- two blue leds ------ power -

    and so on so that the typical forward drops add up to about 7.2V in
    each chain

    I don't know wall wart you would use. Let's take 9V for argument's sake

    so Resistor 1 = (9V - 7.2V) / 20mA = 1.8 / 2E-2 = 90 ohm. Power
    dissipation in the resistor = 36mW. A typical (and easily available)
    value would be 100 ohm.

    For another string, take 4 orange LEDs, 1.7V @20ma. Vf total ~ 6.8V. So
    R = (9 - 6.8) / 0.02 = 110 ohm. Once more I would use 100 ohm.

    The actual currents won't be exactly 20mA - it's going to depend on the
    *actual* forward voltage drops of the LEDs. The current then will be
    given by [V(power) - V(LEDs)] / R(string). Add up all the currents to
    see what wil get drawn from the wall wart - shouldn;t be a problem but
    make sure it can handle it.

    Hope that helps with the idea!

    Cheers

    PeteS
     
  3. Chris

    Chris Guest

    Hi. You're stuck because you haven't chosen a wall wart voltage yet.
    Just for grins&giggles, try a 12VDC wall wart -- they're fairly common.

    Now figure you want to have about 1/2 of the voltage in LEDs, and about
    1/2 in series resistors. That will give you some headroom for
    variations in DC voltage due to line and load.

    Now it's fairly easy. Combine two or three LEDs of the same type to
    get something between 4.5 and 7.5V (just add the nominal voltages), and
    then choose a series resistor so that (12V - LED voltages) / R = about
    20mA.

    Example: 3 red = 2.3V * 3 = 6.9V
    12V = 6.9V = 5.1V
    5.1V / R = 0.02A
    R = 5.1 / 0.02 = 255 ohms.

    Choose 220 ohms or 270 ohms (standard values).

    The reason you should choose LEDs of the same type is that you'll
    probably have to do some tweaking to get the LEDs to look nearly the
    same brightness.

    Now put these strings of two or 3 LEDs with a series resistor in
    parallel, and you're good to go.

    Good luck
    Chris
     
  4. Alan B

    Alan B Guest

    On 16 Aug 2006 13:34:16 -0700, in message
    <>,
    scribed:

    These are the voltages and currents that you can expect the LEDs to use
    when they are operating.
    Your second problem is that you are limited to a reasonable maximum of 20
    LEDs, 0.020 X 20 = 0.400, or 80% of your supply's rated maximum. I
    wouldn't go over that. Which, since your plan is 18-24, I suppose isn't a
    problem after all.
    The resistors that you choose will be based on a rather simple mathematical
    formula. You want the optimum current to be 0.020A. Your supply is rated
    at 9.0V. For this design, it appears that two LEDs in series is optimal.
    Put two diodes in series, calculate the expected voltage drop and subtract
    from 9.0. Take the remainder and use Ohm's Law to calculate what value
    resistor will give you that voltage drop at 20mA. Voila! You have two
    diodes and a resistor in series that give 9.0V at 20mA. Rinse, repeat.

    9.0 = Vd1 + Vd2 + Vr1, where Vr1 = r1 * 20mA.

    The resistor can go anywhere in the series combination.
     
  5. jasen

    jasen Guest

    there won't be any bulky electronics
    if you can drill it from the base up the centre you may be able to do it all
    from the bottom, start with a 1" (or larger) spade bit and hollow out the
    bottom half then switch to a 0.5" and go up further
    more would be better...
    it's basically the same thing.
    yup. if you can, get a 12v or 24v wall-wart,
    efficiency isn't real important. there's not enough power
    there to start a fire or anything.

    if the wall-wart is 9V regulated:

    two blues 7.2V use resistor 91 ohms (should be 90)
    a green and two oranges, 6.9V use resistor 110 ohms (105)
    a sea green and two ambers, 6.9V 110 ohms (105)
    three reds, 6.9V 110 ohms (105)
    two greens 6.6V 120 ohms (120)

    all those will be close enough.
    hook the leds together in series (head to tail) and at one end put

    (+) ---/\/\/\--->|--->|--->|---- (-)

    use heatshrink tubing to insulate the connections.

    the resistor. user oneresistor for each series group of leds

    (+) ---/\/\/\--->|--->|--->|---- (-)
    (+) ---/\/\/\--->|--->|--->|---- (-)
    (+) ---/\/\/\--->|--->|--->|---- (-)
    (+) ---/\/\/\--->|--->|--->|---- (-)
    (+) ---/\/\/\--->|--->|--->|---- (-)

    if you thread the wires into the holes first and then attach the LEDs that
    may give the tidiest result. "wirewrap" wire is made for this sort of thing.

    you can grind the flange off the LEDs to make them fit drilled recesses better
    a sanding disk is good for this.

    you'll need to be sure to connect them the right way round.

    Bye.
    Jasen
     
  6. Guest

    Thanks, everyone. This is pretty much what I thought I would need to
    do. Time to break our the breadboard and see if I can't make it work
    before I actually start making things light up!
     
  7. redbelly

    redbelly Guest

    Just to add to the others' good suggestions: it would be good to
    measure the Wall Wart with a voltmeter, to verify it runs at the stated
    voltage. It could be a couple of volts higher than the rating, which
    would figure into your resistor calculation.

    Mark
     
  8. ehsjr

    ehsjr Guest


    Just to give you another option:
    Create 3 or 4 strings of LEDs in series, using
    1 of each color led you have. Use a 24 volt
    wall wart such as DCTX-2460 from Allelectronics
    http://www.allelectronics.com/
    Add an LM317 and two resistors (3.6K & 240 ohms)and a
    capacitor to provide a regulated 20 volts output, and
    use a 56 ohm resistor in series with each string.

    Option 2 is probably best:


    ..------. -----
    | +|----+--Vin|LM317|Vout---+----+------+------+
    | | | ----- | | | |
    | 24V | | Adj | [LEDs] [LEDs] [LEDs]
    | DC | | | | | | |
    | | [.1uF] +--[240R]--+ | | |
    | Wall | | | [56R] [56R] [56R]
    | Wart | | [3.6K] | | |
    | | | | | | |
    | -|----+--------+---------------+------+------+
    ..------.

    Voltage at the Vout pin will be 20.0
    Each string of LEDs has a nominal Vf of ~18.9 volts.
    The 56 ohm resistor will limit the current through
    the string to ~19.6 mA

    Ed
     
  9. I'd go down to the dollar store and see what they have. They often have all
    sorts of weird and wonderful LED flashers. Easy, quick and cheap AND you can
    buy some spares.
     
  10. Guest

    That's a lot of really great ideas. Thanks for the creative options!
     
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