Arduino High Power LED

Discussion in 'The Projects Forum' started by Plecc, Jun 4, 2012.

  1. Plecc

    Thread Starter New Member

    Aug 26, 2010
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    HI all,

    I would like dim a blue high power LED using an arduino uno.
    The LED is 3W blue 4.8v 700mA.
    Does anyone know of a simple constant current circuit that can be dimmed using the PWM from the arduino?
     
  2. MrChips

    Moderator

    Oct 2, 2009
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  3. Plecc

    Thread Starter New Member

    Aug 26, 2010
    22
    0
    Thanks for the reply,
    I was hoping to knock up something simple with bits around the house.
    I found this site -
    http://bildr.org/2011/03/high-power-control-with-arduino-and-tip120/
    Would the third circuit on this page be suitable?
    I have a TIP122 instead of the TIP120.
    The only bit I'm unsure about is how to work out what resistor is required to run this LED on a 12V 1200mA power supply?
     
  4. MrChips

    Moderator

    Oct 2, 2009
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    You can use the circuit diagram shown in the DLD101 data sheet to roll your own using discrete components.
     
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  5. Plecc

    Thread Starter New Member

    Aug 26, 2010
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    Sorry - Forward Voltage is 3.8v not 4.8.
     
  6. Plecc

    Thread Starter New Member

    Aug 26, 2010
    22
    0
    I'm a bit too much of a novice to be able to choose the correct components to recreate the IC circuit the resistors and the NPN transistor I have but Q1 I have no idea what it is.
    It looks a bit like a variable resistor with a diode strapped across the back of it.
     
  7. MrChips

    Moderator

    Oct 2, 2009
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    Q1 is a HEXFET
    R1 = 4.7kΩ
    R2 = 47kΩ
    Rc = try 2.2kΩ
    Rs = Vbe/ILED = about 1Ω for 700mA LED
     
  8. wayneh

    Expert

    Sep 9, 2010
    12,118
    3,042
    Huh? What looks like that?

    That 3rd (lightbulb) circuit will work for you. Your TIP122 is rated to a higher voltage than the TIP120 but is otherwise identical, so it'll be fine too.

    For the LED, I'd aim to start with about half the rated current and only raise it once I was convinced everything is hunky dory. So you need 300mA from an 8V drop in voltage (supply - Vf). Ohms law ∆V = I•R 8v = 0.3A•x x = 27Ω

    You need to consider heat. Your LED needs a heat sink and your resistor will dissipate I^2•R watts, so 0.3^2•27 = 2.43 watts. I'd use a resistor rated to at least 5W. If you eventually get up to, say, 600mA, you'll need 0.6^2•15=5.4W (rough numbers). So a 10W rated resistor, to be safe.

    This all assumes continuous operation. That's the safe way to design it. If you KNOW the duty cycle will never exceed, say, 50%, you could think about skimping on heat management. Not the way I'd do it, I'm just sayin'.
     
    Last edited: Jun 4, 2012
  9. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,387
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    Here's two basic methods. One of these should work nice for you:

    [​IMG]

    Both are for 3.3V or 5V logic inputs. Most any NPN transistor will work, such as a 2N2222A.

    The first is pretty simple and best when you're running your LED off a voltage larger then 5 volts. The transistor will be in the active mode to help dissipate off power. I used this method on a job where my power varied between 10 and 40 volts. You select R1 by the following:

    R1 = ( Vin - Vbe) / Iled = ( 5 - .7) / .7 = 6.14 ohms

    Power(R1) = I^2/R = .7 x .7 / 6.14 = .0798

    The closest standard value I found was 6.19 ohms in 1/8 W or 1/4 W (better) versions.

    The next version is a bit more typical and uses the transistor as a saturated switch. R1 is selected by:

    R1 = (Vdd - Vled) / Iled = (5 - 3.8 ) / .7 = 1.74 ohms

    Power(R1) = I^2/R = .7 x .7 / 1.74 = .28

    Unbelievably 1.74 ohms is a standard value available in 1/2W (not recommended) to 1W (Mikey likes) versions.

    (Parts found off Digikey)
     
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