Minimum Number of LEDs on 42 VDC driver

Discussion in 'General Electronics Chat' started by swap20, May 10, 2012.

  1. swap20

    Thread Starter New Member

    Apr 24, 2012
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    Hi

    I have 42 vDC LED driver, input is 12 vDC with Constant current 350ma. I want to run 8 LEDs in series. Those are of 1 watt each with 350 MA.

    Will this overheat the LED and decrease the life span of LED?

    Your inputs please
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    Welcome to AAC!

    How much voltage each LED drops is completely dependent on color. We need the Vf (Forward dropping voltage) of your parts to be able to answer the question. We will also need the specs of the 42VDC driver.

    You will need heatsinking with these parts though, no way around it. It can be a convienent chunk of metal but it has to be there.
     
  3. swap20

    Thread Starter New Member

    Apr 24, 2012
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    Hi Bill,

    Vf of LEDs are 3v to 3.8 V.

    LED driver is DC - DC constructed using irf540n mosfet, input is 12v DC. no name chinese brand, that's what we get in India ;)

    Thanks
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    The range you give is way too wide to work well with, but I can give you the basics.

    I have a tutorial for low power LEDs, much of the basics is the same, such as how you calculate a resistor.

    LEDs, 555s, Flashers, and Light Chasers

    Here is the problem, LEDs must have current limiting, no options. That transistor only turns the current on/off. Going through the math, if the LEDs are 3.8V Vf then you will have a total drop of 30.8VDC. The resistor would be:

    (42VDC - 30.8VDC) / 0.35A = 33.7Ω ≈33Ω , at 11.8VDC * 0.35 ≈ 4W (should be 8W)

    Not good, but it will work.

    The amps for 33Ω will be ≈ 0.358A

    If the LEDs Vf is 3.0 then everything changes. Total Vf = 24VDC. Current = (42VDC - 24VDC) / 33Ω = 0.545 which could dramatically shorten the life of the LEDs.

    To figure out the correct resistor you must know the true Vf of the LEDs.

    One way to correct this it to use a constant current source, which will compensate nicely. You want to go through the logic train on this?
     
  5. jaclement

    Active Member

    Apr 15, 2009
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    If what you have is device that takes 12 volts input and outputs 350 ma as a current source up to 42 volts, then you want to have a string of leds that has a voltage drop of close to but less than 42 volts. White leds have a voltage drop of a little less than 4 volts, so 10 leds in series should be about right. The current source may have a minimum voltage output because of power dissipation. You should test the current source with a rheostat or resistors to simulate the led load. So 120 ohms across the output (42/.35 ~ 15 watts) would be the maximum resistance. Measure the voltage across the resistor and the current though it. Drop the resistance gradually the current should stay at 350 ma. and the voltage should drop. Run the circuit at that voltage that the led string would be dropping and see if the current source doesn't get to hot. With star type leds at 350 ma. minimal extra heat sinking is necessary . I use bathub sealer to sheet metal or other heatsink.
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    Minimal is not the same as none, which goes back to what I said, some heatsinking required. If the heatsinks and LEDs are too hot to touch after a while, you will need more heatsinking, or a fan.

    A common trick I use is not a rheostat (which tends to be expensive), but parallel resistors, using inexpensive and easy to find low wattage resistors. As long as the resistors have some air space between them they will dump their excess heat.

    Take the 18V, which is the worst case voltage differential. You could use a bunch of smaller resistors, but it could take as many as 50 resistors.

    There are other ways however.

    A LM317 makes a really dandy constant current source. The case is reasonably easy to heat sink (it is gong to get really, really hot, so you will need to heat sink it well).

    Here is how to do it.

    [​IMG]

    Look familiar? It was in the tutorial link I showed. The resistors are ¼W types.

    The LM317 is a very cheap part, but it will be disapating around 6.3W worst case (the more voltage the LEDs drop, the less wattage the LM317 has to handle). You can buy it from Radio Shack, it is that common.

    There are other high tech ways to decrease this heat radically, but I suspect it is bit too advanced (some circuitry required). Basically you have to build a SMPS (switching mode power supply) current regulator.
     
  7. kuza

    Member

    May 11, 2012
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  8. Audioguru

    New Member

    Dec 20, 2007
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    Your power supply is 12VDC at 350mA which is 4.2W.
    The voltage stepup circuit is probably 80% efficient so its output power is 4.2W x 0.8= 3.36W.
    If the output is 42V then its current is only 3.36W/42V= 80mA, not 350mA.
     
  9. Wendy

    Moderator

    Mar 24, 2008
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    Good point, I had missed some of that.
     
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