driving power leds

Discussion in 'The Projects Forum' started by O.K., Dec 10, 2014.

  1. O.K.

    Thread Starter New Member

    Dec 10, 2014
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    Hi!

    I need some help in driving power leds.

    I want to make a growbox, and I want to use power leds for light. I have 2 reds and 2 blues, the exact types:
    - uv blue 700 mA 3,4-3,6 V
    http://www.ebay.com/itm/251468113696?_trksid=p2059210.m2749.l2649&ssPageName=STRK:MEBIDX:IT
    - blue 700 mA 3,0-3,4 V
    http://www.ebay.com/itm/251738619924?_trksid=p2059210.m2749.l2649&ssPageName=STRK:MEBIDX:IT
    - deep red 700 mA 2,2-2,4 V
    http://www.ebay.com/itm/251738619924?_trksid=p2059210.m2749.l2649&ssPageName=STRK:MEBIDX:IT
    - red 700 mA 2,0-2,4 V
    http://www.ebay.com/itm/261527758059?_trksid=p2059210.m2749.l2649&ssPageName=STRK:MEBIDX:IT

    I have 10 of all type. I want to use a pc psu as power source. Is it possible to connect 1 uv blue + 1 blue + 1 red + 1 deep red in series and one resistor and put it to 12 V? So I will have 10 paralell circuit with 10 limiting resistors. Is it safe? I don't want to drive it with 700 mA current, just about 600 mA.

    Thanks for reply.
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,442
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    Yes, that's ok.
     
  3. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    Thats 10.6V drop across the leds, and approx 2V across the resistor makes it about 3 ohms,
     
  4. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    You need to calculate the resistor values for each LED as their Vf is different.

    And don't forget to calculate the required wattage of the resistor and multiply it by 2 or 3 for a "safety factor" to keep it from running really hot or burning up.

    Personally I would never run "high powered" LEDs with a resistor for the current limiting. A constant current source is always a "better" solution.. Any change in Vf will equal a change in current delivered to the LED..
     
  5. O.K.

    Thread Starter New Member

    Dec 10, 2014
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    Thank you for reply.
    What do you mean "You need to calculate the resistor values for each LED as their Vf is different."
    I want to add just 1 resistor to a circuit (4 kind of leds in series).
    I also scare of demaging leds. For example one led dies -> other 3 get over voltaged and also dies. Could it happen? Or when i turn it on instacionarius state cause voltage rise in a led and voltage decrease in an another -> demage?
    I prefer the resistor limiting methood because it is cheap.
    Thanks for reply.
     
  6. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    That would be the minimum total Vf. If you design for that and you happen to have higher Vf LEDs then either they won't light or will light only dimly. I make the worst-case total Vf = 11.8V for the 4-LED string. That leaves insufficient headroom for proper current control with a nominally 12V supply and a single series resistor :(.
    Then use a constant-current source to drive the LEDs properly. Probably cheaper than replacing fried LEDs ;).
    How stable is your 12V supply?
     
  7. O.K.

    Thread Starter New Member

    Dec 10, 2014
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    THe power supply is a pc psu which has a standard voltage: 12 +- 0,6 V.
     
  8. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    your problem is all the leds have different volt drops, vary from 10.6 to 11.8v, you could put the four in series without a resistor, or better still five, its better to use a separate resistor for each led all in parallel,otherwise you can use 3 in series and a constant current regulator.
     
  9. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    Sorry.
    I missed the part where you stated 4 in series
     
  10. O.K.

    Thread Starter New Member

    Dec 10, 2014
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    What do you think about this circuit?
     
  11. O.K.

    Thread Starter New Member

    Dec 10, 2014
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    or this one?
     
  12. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Q5 is upside-down in post #11. Apart from that, either circuit should work.
     
  13. O.K.

    Thread Starter New Member

    Dec 10, 2014
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    Hi!

    I have a Question about the PSU. I've made the circuit, and I wired the two 12 V rails of the psu together just for safety (current is 8 A). When I use both rails the IRFU9014 gets hot, but when I use just one rail its cold, and also leds are not so bright. What could cause this?
     
  14. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    What respective current is each 12V rail rated to carry?
    Are the two 12V rails voltage-regulated independently of each other?
    How much current are you trying to pass through one IRFU9014?
    What heatsink arrangement are you using?
     
  15. O.K.

    Thread Starter New Member

    Dec 10, 2014
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    The problem solved. It was related to the resistance of the wire, which is very little, but at 8 A it drops a noticeable voltage, and doubling the wire makes this drop less. So it is not related to the psu.
    Each IRFU9014 runs about 0,6 A, and drops about 1,2 V, heatsink not used.
     
  16. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Good to hear you've solved the problem.
     
  17. O.K.

    Thread Starter New Member

    Dec 10, 2014
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    Hi!
    I still have the heat problem. The dissipated power is about 1 W, but the transistor gets very hot. Do you think that this transistor is not suitable for this task? Or applying heatsink will solve the problem? The case is TO-251
    so it's difficult to add a heatsink, but buying new transistors makes extra cost.
    In data sheet I see the following thermal resistances:

    Maximum Junction-to-Ambient RthJA 110 C/W
    Maximum Junction-to-Case (Drain) RthJC 5.0 C/W

    What does these mean? Without heatsink 1 W makes 110 C temp. rise? And with a good heatsink I can reach 5 C temp rise per W?

    Thanks for reply.
     
  18. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Yes.
    You would need an infinite heatsink to achieve that :(. A 50C/W heatsink in free air would reduce the case temperature by about 50C.
     
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