LED Flashlights on 12v

Discussion in 'General Electronics Chat' started by SteveDouglas, Apr 10, 2012.

  1. SteveDouglas

    Thread Starter Active Member

    Apr 30, 2009
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    I'd like to use the LED flashlights innards from Harbor Freight on the kid's PowerWheels cars for headlights. The cars are 12v, the flashlights require 3 AAA batteries each (4.5v).

    What is the best way to power these from 12VDC ?

    I considered using an LM317 regulator but seems like there's probably an easier, cheaper way ???

    Thanks for any advice,

    Steve
     
  2. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    Wire two lights in series for 9V and use a resistor to drop the rest.
     
  3. SteveDouglas

    Thread Starter Active Member

    Apr 30, 2009
    42
    2
    right, thanks, I considered that but how do I know what wattage resistor to use ? I have a lot of 1/2 watt and 1/4 watt resistors on hand, I guess I could try the 1/2 watt.

    Thx
     
  4. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    the lights won't be using very much current. I would be surprised if it exceeded 50mA.
    1/4 watt should do very well.
     
  5. SteveDouglas

    Thread Starter Active Member

    Apr 30, 2009
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    OK, Simple enough, I'll go that route. Thanks for the fast help !!!
     
  6. KJ6EAD

    Senior Member

    Apr 30, 2011
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    The light has 32 LEDs in parallel and will draw 640mA max if you give it 4.5V. These lights are grossly inefficient so it would do no harm to remove the internal limiting resistor and use an LM317 as a current regulator. It's easier to heat sink an LM317 than a resistor.

    If I'm right about the current, you're going to need a 5W resistor (50% derated) doing it the way you plan (3V X 640mA = 1.92W).
     
    Last edited: Apr 11, 2012
  7. panic mode

    Senior Member

    Oct 10, 2011
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    my flashlight has fewer LEDs than yours and current draw is about 230mA. Best way is to measure, but i'm guessing you will have something in the 400mA.
    Connecting two flashlights in series requires 9V. remaining 3 or so volt need to be dropped using resistor. I agree about easier mounting of LM317 but I don't think that would be best option because LM317 has 1.2V output plus about 2.5V drop which is already over 3V we have to deal with. Even if it is "there", it would be border line and offer no adjustment.
    Resistor would be R=3V/current
    if current is 360mA for example (0.36A) then resistor is 3/0.36=8.33 Ohm.
    Standard value is 8.2 Ohm.

    Minimum Power is P=I^2*R=0.36*0.36*8.2=1.063W so technically 1.5 or 2W could do the job. I'd rather stick with larger one (5W or 10W) just because they have larger area and will heat less.
     
  8. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    They are pulling over 250mA from an AAA battery?

    If they are that power hungry then the internal battery resistance will cause the voltage to drop lower than 4.5V on three AAA. (new ones might hold up for a few minutes)
     
  9. KJ6EAD

    Senior Member

    Apr 30, 2011
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    The LEDs have a Vf of 3.4 to 3.6 so when you get rid of the low value limiting resistor in each you can wire the two lights in series using a regulator with a battery that is discharged to 10.5V.

    The lights as "designed" will not draw 640mA because of the limited capacity and internal resistance of the AAA cells but when you give them a nice juicy gel cell to drink from they will.

    Don't even get me started on the fact that each light has 32 LEDs in parallel!

    An LM317T, a 68Ω resistor (>1/16W) and a heat sink is all you need.
     
    Last edited: Apr 11, 2012
  10. Audioguru

    New Member

    Dec 20, 2007
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    The LEDs in the cheap flashlight are all matched pretty well. If they were not matched then some LEDs would be very bright and burn out soon and other LEDs will be dim.
     
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