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Simple LED setup for motorcycle

Discussion in 'Sensors and Actuators' started by hzuiel, Jan 8, 2012.

  1. hzuiel

    hzuiel

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    Jan 8, 2012
    I saw some LED's at autozone that are powered by a little battery so you don't have to mess with wiring them into your car or motorcycle's wiring system. They cost 20 dollars though, and it was a 4pack, powered by one small battery, I only needed two, because i'm only planning to light up 2 places on my motorcycle, and the battery was an odd size, which makes them expensive, and i have a huge AA stockpile. So i went to radioshack and bought two battery holders, a switch to keep up by the handlebars, and two 8000mcd 10mm blue LEDs.

    I just burned up both LEDs so obviously there's something missing from my equation. My understanding of electricity isn't that advanced. I no doubt needed some kind of inline resistor.

    The information on the LEDs(which i will have to go buy more of) is FW Current 20mA, FW supply: 3.2v (typical), 3.8(max).

    These battery holders have all of the batteries in series, a AA is 1.5v so I would need to wire them in parrallel to give 3 volts, which should be close enough for the LEDs right?

    So what else do I need in this equation to spare the LEDs such a horrible fate? A resistor? if so, what value would the resistor need to be? I have some resistors from doing alarm work but they're only 3.3k and 2k. I can go buy different resistors if needed though.
     
    Last edited: Jan 8, 2012
  2. davelectronic

    davelectronic

    1,079
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    Dec 13, 2010
    Hi there
    Welcome.
    Use four cells in series and an 80 ohm half watt resistor in series with the led if its a single diode. :)

    80 ohms for rechargeable 140 ohms if alkaline .
     
    Last edited: Jan 8, 2012
  3. hzuiel

    hzuiel

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    Jan 8, 2012
    Oh wait so series increases voltage. That means one of these battery packs is 6 volts? Each battery holder has slots for 4 AA's, and they're not rechargeable, they're alkaline. So if I wire the battery holders in parrallel, then I'll have 6 volts, and I need to use a half watt, 140ohm resister, inline on each LED, with the LEDs in parrallel, or one inline resister in the circuit and the LEDs in series? I'm such a noob......
     
  4. davelectronic

    davelectronic

    1,079
    12
    Dec 13, 2010
    No in parallel the voltage stays the same but the current increases.

    6 volts supply - 3.2 volts working voltage = 2.8 volts divided by the led's forward working current 0.020 milliamps = 140 ohms.

    1.5 volts is to low.

    Yes series increases voltage.
     
    Last edited: Jan 8, 2012
  5. hzuiel

    hzuiel

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    Jan 8, 2012
    So just use one battery holder, which gives me 4 AA's in series, which should be 6 volts. Then wire the LEDs in series, with a single 1watt 140 ohm resistor? I wanna make sure of this before i take another trip to radioshack.
     
  6. jackorocko

    jackorocko

    1,284
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    Apr 4, 2010
    if the voltage drop of each LED is 3.2V then wiring more then one LED in series with a 6V battery pack is gonna fail. Even worse if the forward voltage drop is closer to max of 3.8V. Not to mention a alkaline battery has a nominal voltage of 1.5 volts when it is brand new, it drops from there over the life of the battery. So even if the LED conducted under 3V, pretty soon your likely to see battery voltage sag below 1.4V per cell. http://data.energizer.com/PDFs/E91.pdf
     
    Last edited: Jan 8, 2012
  7. hzuiel

    hzuiel

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    Jan 8, 2012
    So parallel then, with a resistor on each circuit?
     
  8. davelectronic

    davelectronic

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    Dec 13, 2010
    Last edited: Jan 8, 2012
  9. hzuiel

    hzuiel

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    Jan 8, 2012
    There will be two LEDs, both 3.2v 20maA, I have (2) 4 cell, 6v, alkaline battery holder packs. I have a switch that needs to be in the circuit to turn both LEDs on and off.
     
  10. Resqueline

    Resqueline

    2,848
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    Jul 31, 2009
    We don't know anything about your battery holders or how you wired them in the first place. Two cells in series should usually not burn out a blue LED.
    Chinese made LED lamps are usually just one lithium cell or three AAA/button cells wired directly to white or blue LED's. The LED's will eventually fail being run like this.

    Three alkaline AAA cells and a resistor of (4.5V-3.2V) / 0.02A = 65 Ω in series will waste the least amount of power while still being safe.
    You can connect up to 5 LEDS in parallel on each battery pack and still retain some battery life, but remember to use one 62 or 68 Ω resistor for each LED.

    If you were to use rechargeable NiMh batteries then you'd have to use four cells in series with a 75 ohm resistor.

    If you're to use your 4-cell holders then use Dave's recommendation of 140Ω. It only needs to be a 0.06W resistor.
    Wire the switch in series with the battery and the LED's (each with its own series resistor) in parallel.

    Blue is a really bad choice for anything but a "cool lighting effect" as your eyes are unable to focus using that color, and it's also bad for your dark vision ability btw..
     
    Last edited: Jan 8, 2012
  11. davelectronic

    davelectronic

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    Dec 13, 2010
    Last edited: Jan 8, 2012
  12. hzuiel

    hzuiel

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    Jan 8, 2012
    To be even more specific, the battery holders contain 4 alkaline store bought AA batteries which are 1.5v. The holder arranges the batteries in series, so each battery holder is 6v.


    That's all it's for. I have a premade headlight assembly that had a mix of blue and white LEDs which light up the road pretty well. These LEDs are just "FX" lighting over the rear wheel and around the engine area.


    [EDIT}Nobody has responded yet so I'm gonna go ahead and do it according to what I'm understanding from that website. That is two battery packs of 4 alkaline cells each, in parallel with each diode on a parallel circuit from those batteries, with a 140ohm half watt resistor on each circuit, and the switch wired inbetween the positive side of both circuits and the positive side of both battery leads. If this is wrong, someone correct me, i'll be checking as I work.
     
    Last edited: Jan 8, 2012
  13. jackorocko

    jackorocko

    1,284
    1
    Apr 4, 2010
    Why do you want to run 12V with only one LED per parallel run? That means you are dropping 12V-3.2V = 8.8V across the series resistor. This is just waste!!!!! You want to run as many LED's in series (per parallel run) as you can as close to the max voltage, this drops the least voltage across the series resistance and maximizes battery usage.

    This is why those calculator pages suck. I would do like res. says and run 5 led's in parallel with a 4.5V (3)AAA's wired in series.
     
    Last edited: Jan 8, 2012
  14. hzuiel

    hzuiel

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    Jan 8, 2012
    No no, each holder is 6v, in parallel it'd still be 6v, and i"d have twice as much battery power to use before they go out. I already bought both battery holders, I might as well use what I have.
     
  15. davelectronic

    davelectronic

    1,079
    12
    Dec 13, 2010
    Two holder in parallel will indeed give you 6 volts and twice the capacity. :)
     
  16. jackorocko

    jackorocko

    1,284
    1
    Apr 4, 2010
    you don't need to fill up the fourth cell just to waste power. Just short it out. 4.5V double the capacity with 5 LED's or the same with 10 @ max efficiency.
     
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