Driving a 1.4A LED with LM317 - feasible?

Discussion in 'The Projects Forum' started by letocharm, Jun 2, 2012.

  1. letocharm

    Thread Starter New Member

    Jun 1, 2012
    7
    0
    Hi All

    I've got a high-power 1.4A, 3.4V LED I need to light up. There are designs all over the net for LED driving with this circuit, but I have some concerns since my LED is so high power.

    FIrst off, I'm going pretty close to the current limit. I bought some 2 W 0.91 ohm (close as I could get to 0.89) resistors, I'm not (too) worried about them. But for the IC, I'm guessing I'll need a heatsink? Should I go with the 3 A version of this IC instead?

    Also, what should the input voltage be? I wasn't understanding this too well... it says the ref voltage is 1.25 V. Since my LED is 3.4 V... will a ~ 5 v supply be ok? How much current will it need? I'm guessing >1.4 A.

    Thanks; EE isn't my strong point. ;)
     
  2. wayneh

    Expert

    Sep 9, 2010
    12,156
    3,063
    Yes and yes. If your continuous current is really 1.4A.
    Depends which side of the squiggle you're on. My guess is no, because of the dropout in the regulator.

    Current in the supply and through the regulator will be the same as in your load.

    Why are you driving an LED with a voltage regulator? Are you planning to use one of the current regulator configurations, I hope?
     
    letocharm likes this.
  3. #12

    Expert

    Nov 30, 2010
    16,354
    6,852
    The graph on page 4 (not including the cover page) shows you need about 2 volts more than the LED to get 1.4 amps through this regulator.

    You can use slightly more resistance in order to get slightly less current because 1.4 amps is the MAXIMUM the LED can use.

    I expect you'll need a heatsink on the LED, too. Did it come with a heat sink?
     
    letocharm likes this.
  4. letocharm

    Thread Starter New Member

    Jun 1, 2012
    7
    0


    You think I should be using the 138? Right, I'll order that.

    About the graph; you mean the dropout voltage? Had to google this. So I'll need about 6 V to be safe. Good, this is doable.

    We're going to machine an aluminum heatsink for the LED, and use some thermal grease.

    Waynech: I'm going to use it in this configuration:

    [​IMG]
     
  5. #12

    Expert

    Nov 30, 2010
    16,354
    6,852
    138 is just the title of the datasheet for 138, 238, and 338 chips.
     
  6. letocharm

    Thread Starter New Member

    Jun 1, 2012
    7
    0
    THanks for the help guys. I'll update this thread in a week or so when the parts come in, for posterity.
     
  7. wayneh

    Expert

    Sep 9, 2010
    12,156
    3,063
    Good, just wanted to be sure you were using constant current, not voltage. ;)
     
  8. panic mode

    Senior Member

    Oct 10, 2011
    1,321
    304
    what exactly you want to do? is there anything else (you mention 5V supply)?

    linear regulators have min. voltage drop so to meet that you need higher supply which means more dissipated power and more heat.

    constant current is great but may not be needed if your load is fixed (LED) and if your plan was to use fixed voltage (like 5V), in which case you can just use resistor.

    is the LED always at full brightness when on, or you expect to use constant current source as dimmer too?
     
  9. letocharm

    Thread Starter New Member

    Jun 1, 2012
    7
    0
    We just want it to light up at full brightness. We don't want to use a resistor because this particular LED is going in a product and it can NOT fail due to fluctuations in source voltage - especially since these LEDs have forward voltages specified in a "range" from 2.9V to 4.3V, with 3.5V being "typical" (http://www.thorlabs.com/Thorcat/20700/20780-M01.pdf). The forward voltage drop of LEDs furthermore decreases with temperature; as it gets hotter, more current would run through it - positive feedback in the worst way. Granted these ICs aren't going to work forever, but it's my feeling this is the best solution.
     
    Last edited: Jun 3, 2012
  10. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Sorry that I didn't see this sooner, but I have been busy.

    You will need at least 3v more than the LED requires.

    LM317 Datasheet: https://www.national.com/ds/LM/LM117.pdf
    Look on page 6 at the graph titled "Dropout Voltage", at the Iout=1.5 line. You'll see that the dropout is at least 2.3v, and can be as high as 2.5v over temperature.
    Power dissipation in the regulator itself will be approximately the dropout voltage times the desired 1.4A current.

    The dropout voltage is nominally 1.25v, but can be anywhere from 1.2v to 1.3v and still be within manufacturer's specifications.

    The total drop across the regulator & resistor therefore can be as high as 2.5v + 1.3v = 3.8v.

    You say that your LED has a rated Vf of 3.4v. You may find that it's actually somewhat higher when running at 1.4a; around 3.8v.

    So, you may need as much as 7.6v from your power supply.

    Remember, any voltage that is not dropped across the LED will be dropped across the regulator. You have an interest in keeping the voltage drop across the regulator reasonably low, while still allowing some "headroom" so that the regulator is actually regulating. You can use low-value fixed resistors to drop some of the voltage if necessary - or even some 1N540x series 3A rectifier diodes.
     
  11. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    Continuous current of 1.4A is high, you would be better off with the 3A regulator instead.

    Alternatively if your PSU voltage is reasonably stable you can put a power resistor between Vin and Vout pins of the LM317, with the resistor carrying 0.7A or so. The regulator will conduct the rest, whatever amps necessary so the total is still regulated to 1.4A.
     
  12. letocharm

    Thread Starter New Member

    Jun 1, 2012
    7
    0
    Hi Guys - some results, and questions. I ended up going with the LM138k.

    So, using a 2W 0.91 ohm 1% resistor, I'm getting a max of 1.16 A at ~6.3 Volts (lab power supply). This doesn't really make sense - considering that at worst the reference voltage is 1.19V, yielding 1.19/.91=1.3A.

    I thought at first this was temperature related; when I was driving it at 10V the current started dropping rapidly, but when I brought it back down to 6V it's stable. I measured the temperature with a infrared gun - ~130F, well within operating limits (150C or ~300F).

    In search of why this might be, I saw this in the manual:

    What exactly do they mean by "near the load"? Define "near"...?

    I have it connected exactly as in the picture on page 1 of this thread.
     
    Last edited: Jun 11, 2012
  13. letocharm

    Thread Starter New Member

    Jun 1, 2012
    7
    0
    So it has decreased to 1 A now I've let it run for a while. I think this is a thermal problem; I as yet don't have a heatsink on it, but it's a little weird because the case IS the output.
     
  14. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    Your resistor is a modern 2W type? They are pretty small.

    You said 1.16A and the Vref is 1.25v so the resistor is dissipating 1.45W and a modern 2W resistor will get real hot! That causes it's resistance to rise with the result of the current dropping.

    For something disspating 1.45W for extended periods I would use a 5W resistor, or at least two of those 2W types.
     
  15. letocharm

    Thread Starter New Member

    Jun 1, 2012
    7
    0
    Turns out I was soldering poorly. Current is roughly stable now at 1.3A

    Thanks for the tip RB. I'm going to go with a trio of 300 ohm 5W in series (5W isn't very convenient - couldn't find any at 0.9 ohms)
     
Loading...