50 Watt LED driver

Discussion in 'The Projects Forum' started by RodneyB, Sep 12, 2013.

  1. RodneyB

    Thread Starter Active Member

    Apr 28, 2012
    633
    13
    I am wanting to build a security light using the LED shown in the attached datasheet.

    I realise that the LED is going to need a serious heat sink and possibly a fan.

    I want it to work from 220 Volts ac. I know how to build a Linear voltage controller. My thinking automatically goes to a 220 Volt to 40 Volt Transformer and rectified. Smoothed and regulated to 26 Volts. In order to get the recommended forward current of 2.1A I will need a current limiting Resistor of 12R 55 Watts. This is not practical.

    How do I drive this LED? I am a bit lost here
     
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,729
    759
    Try to buy one LED driver from ebay
    OR
    Build PWM driver for the LED. This will eliminate the need of the 55watts useless heat waster when using DC.
     
    RodneyB likes this.
  3. RodneyB

    Thread Starter Active Member

    Apr 28, 2012
    633
    13

    I am going to have to try and build one because in Zimbabwe I will never be able to buy one and we don't have access to e bay.

    Could you point me in the direction of a suitable circuit diagram
     
  4. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,729
    759
    Can you get primary 220V to 20VAC 2.5A transformer ?
     
  5. RodneyB

    Thread Starter Active Member

    Apr 28, 2012
    633
    13
    Yes I can. We have a company here that winds transformers.
     
  6. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,729
    759
    Then u are done.
    Next u would need a 5A bridge. U need a standard bridge tht can handle more than 3A. Voltage should be 50V minimum PIV. Or 4 diodes with rating of 50V PIV at 3Amps or so.
    Next will be one capacitor of around 3300uf or more rated @ 35V or more.
    Next is 1 ohm 5W resistor. More power the better. The resistor will last longer. 5W will suffice if you can clamp it to the LED heat sink.

    If the heat sink u have gets too hot, try to get a computer power supply fan. If u dunno how to power the fan I will explain later.

    U will need forced cooling if the sink it too small or if u decide to run the LED more than few hours.

    If you are going to wind a transformer than u can get an extra secondary winding of 9VAC at 500mA. With this u can use smaller heatsink for the LED with a 12V PC fan. For the fan u can get a smaller bridge or diodes like 1N4001 and 1000uf 25V for smoothing cap.
     
    Last edited: Sep 12, 2013
    RodneyB likes this.
  7. RodneyB

    Thread Starter Active Member

    Apr 28, 2012
    633
    13

    I have all these components in my draws the resistor is 10 Watt ceramic. I will get the Transformer wound tomorrow. Do you just connect the LED across the smoothed Supply? The data sheet says that the typical forward voltage is 26 Volts and the minimum is 24 Volts. Am I jumping the gun a bit here
     
  8. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,729
    759
    lemme draw a schematic for u
     
  9. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,729
    759
    Check the PDF.
     
  10. ronv

    AAC Fanatic!

    Nov 12, 2008
    3,281
    1,226
    This is not a real good way to drive a 50 watt LED because of the resistor size. But having said that a 20 volt transformer may not be high enough voltage as the voltage drop on the LED could be as high as 30 volts. The peak voltage from the transformer will be only 28.3 volts minus the diode drops - say 27 volts. So maybe you use a 24 volt one but now the resistor is big again.
     
  11. John P

    AAC Fanatic!

    Oct 14, 2008
    1,632
    224
    I wonder how well this circuit would work:
    https://dl.dropboxusercontent.com/u/28291527/leddriver.jpg

    The transformer should produce some voltage that's comfortably more than the LED voltage, maybe 36V. The idea is that the output of the transformer is not filtered and so the cathode end of the diodes sees a rectified sine wave. When the waveform is high, Q1 is turned on by Q2, which is turned on via R3. Current flows through the inductor, and L1 makes it continue to flow between cycles, with D2 as a freewheeling diode. If current exceeds a threshold (defined by one Vbe drop being generated by the current through R4) Q3 is turned on, and this steals current from the base of Q2, and so limits the operation of Q1, which reduces the current through L1 and the diode string. All the values need to be written in, but it seems like a plausible design. Will someone please try it?
     
  12. ronv

    AAC Fanatic!

    Nov 12, 2008
    3,281
    1,226
    Now we're going down the right path. I can simulate something like it later this afternoon.
     
  13. ronv

    AAC Fanatic!

    Nov 12, 2008
    3,281
    1,226
    I was having so much fun I did 2. One with transistors and one with a comparator. I used a PFET so you might get by without a heatsink if you can find a good one. I still filtered the supply so everything could turn on fully. It's not as pretty as my first girlfriend, but it looks like it works.:D

    Good idea, John!
     
  14. John P

    AAC Fanatic!

    Oct 14, 2008
    1,632
    224
    Replacing my PNP Darlington with a FET is probably a good move, though it increases the component count slightly. But must the resistor R2 on the gate be so small? That loses over 1W in the on state. I was hoping that the circuit could use the 120Hz rectified sine wave (actually the original message came from a place with British electricity, so it's more likely 100Hz) to generate pulses to drive the LEDs via the inductor. However, with a large current flowing and a low frequency, the inductor might get unreasonably large. The comparator option might be best to add some hysteresis to the circuit--the transistors shouldn't run in linear mode.

    Edited to say that I've done some rough calculations to figure out how big the inductor would have to be, and I came up with something like 0.1 Henry. And it has to carry over 2 Amps! That is probably not the best design.
     
    Last edited: Sep 12, 2013
  15. RodneyB

    Thread Starter Active Member

    Apr 28, 2012
    633
    13
    Thank you so much for all your help. I will build the circuit and get back to you. The LED costs US$60.00 so I hope I don't blow it up.

    Best wishes

    Rodney
     
  16. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,729
    759
    Who said the LED takes 36V?

    Data specifically says 26V nominal @ If of 2.1A. Max is 30V.

    Correct me if I am wrong.

    The resistor is reasonable to drop the voltage and limit current.

    Calculations I did.
    20VAC X 1.414 = 28.28VDC. .........Voltage at smoothing cap
    LED current is 2.1A.
    Resistor drop is V=IR so V=2.1A X 1Ω = 2.1V, Power dissipated by resistor =2.1V X 2.1A = 4.2W
    LED Voltage = VDC-R drop = 28.28V - 2.1V = 26.18V @ 2.1Amps.

    Specs are well with in the LED tolerance.

    The circuit will work. I suggested the PWM drive for efficiency. Building PWM is for experienced hobbyist. But I do not think OP is. Since he cannot calculate values to drive a single LED.

    Note: The value of the resistor are calculated and are theoretical. In practical world the component behave rather differently. So to keep the LED current at 2.1A you might need to experiment with the resistor value. for this u would need to measure the LED current with an ammeter while it is lit. If this is higher than 2.1A, you might have to increase the resistance a little bit to reduce current or viceversa. The best way to operate the LED is to keep the forward current stable. And for your LED it is 2.1Amps.
    If in doubt use a higher value resistor like 4.7Ω or so. This way u can be sure the LED will not burn out due to high current. But remember the voltage should not be more than 30VDC. So a 20VAC transformer will be safe. Measure the current and reduce the resistor value by 1Ω at a time until u can get the proper current. Current can be any value from 1.9A to 2.1A. You will notice the LED brightness change. But remember..
    DO NOT DECREASE THE RESISTOR VALUE TOO MUCH OR INCREASE THE APPLIED VOLTAGE IN ORDER TO GET HIGHER BRIGHTNESS. U WILL DECREASE THE LIFE OF THE LED. PROPER HEATSINK IS ALSO ESSENTIAL FOR LONGER LIFE OF THE LED.
     
    Last edited: Sep 13, 2013
  17. ronv

    AAC Fanatic!

    Nov 12, 2008
    3,281
    1,226
    Rodney,
    John is right. Lets make R2 2.7K and R1 1.5K. Should also mention R1 should probably be 5 watts or so.

    Put 5 ohms or so in series with the LED just in case. At least 20 watts.
     
  18. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,729
    759
    Where are u leading OP to ?

    Is the OP qualified for this type of circuit?
     
  19. ronv

    AAC Fanatic!

    Nov 12, 2008
    3,281
    1,226
    Just trying to protect his leds if the circuit doesn't regulate for some reason.
     
  20. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,729
    759
    The only regulation with my circuit is how stable the line voltage is.

    I went for 20VAC so tht the DC cannot go above 30VDC, which is the maximum the LED can handle. There is no way the LED will blow with the given values unless OP makes a mistake. Which by the way is out of our hands.
     
Loading...