voltage regulator to drive LEDs

Discussion in 'The Projects Forum' started by dude33333, Jul 24, 2012.

  1. dude33333

    Thread Starter New Member

    Jul 24, 2012
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    Hello, I'm new to this website, seems like a pretty great resource.
    I'm having some problems with getting my project to work. I am familiar with electronics.

    I'm trying to drive two high power LEDs (3W each, 3.6v (4vmax) forward @1000mA) with an LM338 chip. the input voltage varies from about 6-9VDC.

    using the data sheet for the LM338 I used a 120ohm for R1 (between Vout and Adj) and a 330ohm for R2 (adj to ground)

    this should give me about 5volts right?
    I dont have much for equipment, but the voltage seems inconsistent. and when I put it in use, it only takes a couple minutes for the LEDs with 1ohm 10W resistors in series with them to either blow or get so hot they desolder themselves.

    I bought these on ebay for cheap. could they be fakes? does my simple design make sense? thanks!
     
  2. #12

    Expert

    Nov 30, 2010
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    You might have missed the point that capacitors must be placed very close to the regulator or it will oscillate.
     
  3. dude33333

    Thread Starter New Member

    Jul 24, 2012
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    how large will the oscillations be? i noticed the light pulsing a bit but i dont mind the pulsing effect, I almost prefer it. but I dont prefer the LED's to blow! so looking at that pdf in section 7.1 it reccomends a 0.1uF disk capacitor, I may give that a try. any other thoughts about this circuit?
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    If the LEDs are rated for 3W, you must have them mounted to heat sinks before you power them up, or they will quickly overheat and burn up.
     
  5. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    You NEVER drive LED's from a voltage source, you drive them from a current source.
     
  6. Wendy

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  7. #12

    Expert

    Nov 30, 2010
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    I thought that you intentionally built a 5 volt source for a 4 volt LED because you knew you would need a resistor to use up the extra voltage.
     
  8. crutschow

    Expert

    Mar 14, 2008
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    You should use the LM338 as a current regulator, not a voltage regulator to drive the LEDs. To do this you connect a 1.2 ohm resistor (for 1A current) to the regulator output and connect the sense terminal to the other end of the resistor (at the load). The regulator will maintain 1.2V across the resistor, generating the desired 1A of current. For a 6-9V source you will need one LM338 and resistor for each LED.
     
  9. RamaD

    Active Member

    Dec 4, 2009
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    O/p voltage is ~4.7V (tolerances!), and with the LED at 3.6V - 4V, the current can be anywhere between 0.7A to 1.1A. Power to LED can become more than 3W and that is the reason for blowing.
    You need to sufficiently heatsink the LED to dissipate 3W to keep the temperature low. Same applies for the regulator too!
    Fig 22 of the datasheet gives schematic for using as current source. There is only 1 resistor - 1.7Ohm Res gives 735mA, and if the LED is 4V then makes it 3W. That is about the max current you can use. It is better to start with a higher resistance, say 2.5Ohm (500mA), 1 to 2W is ok. If you are using 2 LEDs in series, then you need to provide atleast 8V+Vdropout as minimum input. 6V can be bareley sufficient for one LED.
    You cant use the LEDs in parallel!
     
  10. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    Yep.. problem is 100% overheating of the LED's..nothing else.. 1A into an LED like that will QUICKLY destroy them if not heat sinked properly . "properly" is the key term here.
    I have run some Cree Xr-E/Xp-g LED's at 750mA and needed a fairly beefy heat sink to keep them cool.

    Something like this (1 per LED) with the fan running should do alright (and might not even be enough) http://www.mpja.com/Duron_Athlon-Socket-A-CPU-Heatsink-Fan/productinfo/17045+HK/

    In fact I would NEVER run a 1A LED at 1A.. Typically the datasheets 1A rating is assuming an infinite heat sink (0degC/W) which you won't be able to reproduce in the real world.. So drop the supplied current down to no more than 750mA and use a suitably large forced air heatsink like the one I posted.
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    One of the problems with using the LM338 and relatives (like the LM317) as a current regulator is the relatively large voltage dropout required in this mode of operation.

    There is a (very roughly) 1.7v dropout from the IN terminal to the OUT terminal; and since Vref (V(out) referenced by the ADJ terminal) is nominally 1.25v, and you're taking the current from the ADJ terminal, there's ~1.7v+1.25v = ~2.95v voltage drop minimum across the regulator - that just for starters. You'll also need some more voltage for "headroom", or else the regulator won't regulate over supply and temperature variations; say at least an additional 1 volt.

    So, if an LED has a Vf (forward voltage at the rated current) of ~3.8, and your regulator has a total of around ~2.95v + 1v = ~3.95v across it, the supply voltage would need to be at least 3.8v + 3.95v = 7.75v, and the efficiency will be about 3.8v/7.75v*100 = ~49%. That might be OK if you're just testing things, but you really would not want to operate it long term like that, as it's not very efficient - and your regulator would need a bigger heat sink than the LED would.
     
  12. dude33333

    Thread Starter New Member

    Jul 24, 2012
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    Wow, you guys rock!

    ok so #12, you are right, my original schematic was to produce 5v and have the two LEDs in parallel with a 1ohm resistor in series with each.

    I didnt know the lm338 could work as a current regulator, it seems like maybe I should redesign the circuit to utilize current limiting instead of voltage limiting. But now looking at sgt wookies math, I will need 7+ volts, which is not an option for me. I am stuck with an unreliable 6+ volts, I have no control of the voltage, it varies from 6-9 volts on its on and the LEDs need to work throughout that range. the heat sinks need to be able to fit through a hole about 1.5 inches in diameter.

    this is looking trickier and trickier, but I will keep on trying. I wonder how much current I could put through them just using a piece of angle aluminum as a heat sink?

    also, do you all have a different/better method of how to drive two high powered LED's from a crappy voltage source ranging from 6-9 volts (maybe even 5.5-9.5)
     
    Last edited: Jul 26, 2012
  13. Wendy

    Moderator

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    If you want to get into something more advanced you could build a Switching Mode Power Supply constant current source, it will power the LEDs with a very high efficiency. It does this by converting power. So if you have 12VDC powering a 1W LED at 3.6V, the 12V will be providing around 0.1A while the LED gets its 0.35A. It (not the LED) runs cooler as a result.

    The LM317 and LM337 have the advantages of being simple, but Wookie has explained the problems.
     
  14. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    Not very much at all. (I wouldn't go over 350mA) with anything that isn't intended as a heat sink.. You need high surface area not just a mass of metal. (hence fins/pins,etc... on heat sinks)
     
  15. #12

    Expert

    Nov 30, 2010
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    How about an LDO chip for a current regulator? It would still waste more power (as heat) than the LEDs but it would fit the voltage requirements. (The Roman Black current regulators are built for 12 volts, so they aren't the right thing for this.)
     
  16. dude33333

    Thread Starter New Member

    Jul 24, 2012
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    So my original design should work with proper heat sinking? Can someone confirm this? does R1 or R2 need to be a certain wattage rated?

    also, someone said I cant run LEDs in parallel. I found some warnings against this online, but dont really understand why?
     
  17. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    You could use an unbalanced current mirror where each LED was driven by a single PNP power device current source "base mirrored" over to a reference and feed the whole thing from a voltage reg. That way you could adjust the voltage to minimize power dissipation in the pass PNP devices. Would be a very cheap solution.

    (See figure) The left resistor sets the bias current, the right resistor is the LED. The trick is to unbalance it by putting a resistor in each emitter lead which will ratio the currents. You could run 20 mA in the bias side and 250 mA in the LED by selecting the emitter resistor values.
     
  18. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    No, you can't drive LEDs from a voltage source. You can set up linear regs as a current source but as listed above, large voltage drop across it wastes power.
     
  19. bountyhunter

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    current mirror with emitter resistors:
     
  20. Wendy

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    Mar 24, 2008
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