10 Watt LED Driver

Discussion in 'The Projects Forum' started by RodneyB, Feb 12, 2015.

  1. RodneyB

    Thread Starter Active Member

    Apr 28, 2012
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    I sometimes just despair, for months I have been trying to build an LED driver and having such limited success.

    Attached is my latest attempt. Whilst the LED comes on its not at all bright. The resistor R1 is 18R for current limiting

    The voltage across the LED is 7.74 Volts the Supply voltage is 26.3 Volts. The IRF520 without heat sink is just warm to the touch.

    PLEASE can someone assist me and point me in the right direction.
     
  2. #12

    Expert

    Nov 30, 2010
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    R1 should be about 0.56 ohms.
     
  3. RodneyB

    Thread Starter Active Member

    Apr 28, 2012
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    Thank You !
     
  4. #12

    Expert

    Nov 30, 2010
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    At least 1 watt.
     
  5. RodneyB

    Thread Starter Active Member

    Apr 28, 2012
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    I used a 1R 5 Watt resistor and turned up the voltage Slowly at 14 Volts the LED Switched on VERY bright ,felt like all my Christmases came at once. Short lived. The IRF520 was VERY hot and the LED went off.

    Not sure what's blown
     
  6. RodneyB

    Thread Starter Active Member

    Apr 28, 2012
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    Please can you show me how you worked this out. I used the LED calculator at http://ledcalculator.net/

    I also have a 30 Watt LED which I would like to build the same driver for But need to Know how to calculate that resistor correctly.
     
  7. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    It is no wonder that the fet is getting hot.
    The powersupply is 24 volts, the led takes 7.2 volts, the voltage accross the resistor will be about 0.7 volts.
    When we take the 24 volts minus the voltage accross the led and resistor, you will have about 16 volts accross the fet.
    As the current is about 1 Ampere, the power in the fet is roughly 16 Volts times 1 Ampere is 16 Watts. (so more that the led itself).
    Both the led and fet must have a decent heatsink for cooling.

    Bertus
     
  8. John P

    AAC Fanatic!

    Oct 14, 2008
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    "R1 should be about 0.56 ohms."

    I don't see how that "LED calculator" helps at all. Your 2-transistor circuit limits current when the voltage across R1 becomes enough to cause current flow through the base of Q2, and that will occur at about 0.6V. For a 1A current, you'd want approximately the resistor that #12 suggested. Your resistor is oversized in both resistance and wattage rating.

    Your circuit is very inefficient and will get extremely hot. You could help things a lot by putting 3 of those LEDs in series and running them at 1/3 the current, which should give you about the same amount of light. Don't even think about a 30W LED unless you use a different design!

    Also--that 100K resistor is theoretically OK, but it means the current through Q2 will be tiny, and I'm not sure how that would work in a linear mode. I'd use a 10K resistor, at most.
     
  9. ScottWang

    Moderator

    Aug 23, 2012
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    The best way is to using another circuit to test how high of the Vce could get into the saturation state first, and then calculate the R1, V_R1=1.05A*R1.

    If you want to have a light current then you could try to using I_R2 around 1mA~5mA, and then the R2 will be about 24K↘4.7K.

    If you want to have a normal current as the datasheet mentioned then you could try to using I_R2 as 10mA, and then the R2 will be about 2.4K.
     
  10. BobTPH

    Active Member

    Jun 5, 2013
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    For LEDs of that power, you want a switching type driver. Or if you insist on linear, use and input voltage of more like 9V.

    Edit: By the way, after seeing your circuit and #12's suggestion I was going to post that your MOSFET would give up it's smoke if you did that, but I was too late.

    Bob
     
  11. #12

    Expert

    Nov 30, 2010
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    The way I figure this, you have 100k on the collector of Q2. That transistor has to get rid of about .00025 amps. That puts the Vbe a little below .6 volts and you want more than 1 amp in the mosfet.
    R= E/I
    R = .59V/1.05 amps
    R = .562 ohms

    As the circuit gets hot, the Vbe will diminish and not cause thermal run-away.
    This is a constant current circuit. It has nothing to do with an LED calculator.
    Believing an IRF520 is a TO-220 package, I did not assume you had no heat sink on the mosfet. That was at 4 AM for me. Now that the time here is 3 PM I can see that 20+ watts on a TO-220 is not going to work.
    (Didn't read the rest of the thread, just answered the question that was put to me.)
     
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  12. RodneyB

    Thread Starter Active Member

    Apr 28, 2012
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    Thank you for all the advice. I have used a 0R68 resistor. Current draw is 0.86Amps and the power of the LED is 7.2 Watts.

    I use the housing of the lamp as the heat sink for the Mosfet. This is working perfectly, or would it be better to say its working!

    I have set the LED onto a big heat sink and secured this to the housing. I ran it for 36 hours from29 Volts no problems.

    I don't insist on using this type of driver. I am trying to learn about LED drivers and how they work and how the various components are chosen and calculated.

    This particular driver I found on the internet and I thought it would be a good place to start. When I don't understand something firstly I look for the answer or an explanation. If I fail I come back to the forum and ask my question. To date I always get the answers.

    So in moving on, if this LED driver is not suitable for 20 Watts and 30 Watts LED's can someone point me in the right direction to suitable diagrams for drivers. In saying that if I can improve on the driver for the 10 watt LED I will be only to grateful to accept the advice.
     
  13. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    You could power two of the 10 Watt leds in series with the current driver.
    That way the voltage accross the mosfet will be less and the mosfet will stay cooler.

    Bertus
     
    RodneyB likes this.
  14. #12

    Expert

    Nov 30, 2010
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    Nice to notice that your .68 ohm resistor and .86 amps of current show that the Vbe of Q2 is 0.5848 volts. This demonstrates that the current regulator circuit works, but applying 25 volts to an 8 volt LED is a big waste of power, and that wasted power shows up in the mosfet.
     
  15. RodneyB

    Thread Starter Active Member

    Apr 28, 2012
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    Would it be better to build a voltage regulator or DC to DC converter to Provide 12 Volts. The choice of 24 Volt supply is that what my inverter batteries are for my security system. In Zimbabwe if you don't have inverters and solar you have no power
     
  16. #12

    Expert

    Nov 30, 2010
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    My preference would be to put (3) LEDs in series to spend that extra power for something useful.
    Then the LEDs would need cooling fins.
    If you don't HAVE (3) LEDs you could run this as low as 8V or 9V.
     
  17. RodneyB

    Thread Starter Active Member

    Apr 28, 2012
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    Thanks for the advice! I imported 10 x 10Watt, 10 x 20 Watt and 5 x 30 Watt. Took six weeks to get here and cost a fortune about US$2.00 a watt so I really want to get full use out of them
     
  18. ScottWang

    Moderator

    Aug 23, 2012
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    Wow, spend six weeks to waiting ... :eek: :(
    The led cost US$2.00/Watt ?
     
  19. RodneyB

    Thread Starter Active Member

    Apr 28, 2012
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    Yip order from China. By the time I had paid VAT,duty, bank charges, agents fees and anything else they could pile on that's what it worked out at.
     
  20. ScottWang

    Moderator

    Aug 23, 2012
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    I also bought some meters and other stuffs through other people from China, it's quite cheap for us.
     
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