LED Grow Light PLEASE HELP!!!!

Discussion in 'The Projects Forum' started by CrAzYHaMm, Jun 26, 2011.

  1. CrAzYHaMm

    Thread Starter New Member

    Jun 26, 2011
    Alrighty soo this is my 1st forum post ever, so please bare with me. I was researching LED grow lights and there really expensive for the size i'd like to buy, so I was hoping all you WONDERFUL people would help me.... please.

    So here's what i got so far, i would like to arange 120 3w LED's in a 8x15 layout on a sheet of aluminum, consisting of 100 3w RED LED's http://cgi.ebay.com/100pcs-3W-Red-H...987?pt=LH_DefaultDomain_0&hash=item1c1bb28e6b and 20 3w BLUE LED's http://cgi.ebay.com/20pcs-3W-Blue-H...675?pt=LH_DefaultDomain_0&hash=item1c1bb27d93. I've researched how to connect them in a series, but I'm having trouble figuring out how to power it and I dont want to blow the whole thing up.

    I have a two questions.... 1. How should I power this thing? & 2. Can you install a dimmer on something like this?

    Thanks for your time.
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    Powering depends on what you can get.

    They are 3 Watts but are low voltage. Which means you have better options.
    Like using a really high current 3V supply or use a standard 12V and wire Led's in series ( 3 per string ) and parallel them, or use a higher supply and series them accordingly.

    And of course u will have to use current limiting resistors too.

    Wire them without a resistor and u will blow them. Period ( Unless it is a small battery like Li-On)

    Questions is, how many leds do you want to use and how do yuo want them to light.
    After that find what voltage of DC supply you can get.

    An option for a PSU is a normal PC supply. U can use the 12V rail very nicely and is around every where if it fails.

    And no ...You cannot use standard dimmers to dim led's. You have to PWM drive it to adjust brightness

    Oh! U said 120 leds. Do you want them all to light together or in bunches.
  3. CrAzYHaMm

    Thread Starter New Member

    Jun 26, 2011
    Thanks for your help, Ok so I've been using this site http://led.linear1.org/led.wiz to do the math for me.

    Just please let me know if this is gonna blow up on me or not, just type these numbers into the site listed above.

    12 Source voltage
    2.4 diode forward voltage
    700 diode forward current (mA)
    100 number of LEDs in your array


    12 Source voltage
    3.8 diode forward voltage
    700 diode forward current (mA)
    20 number of LEDs in your array

    RUNNING off of a 12V 30A regulated power supply

    So is this a failure or success?
  4. SgtWookie


    Jul 17, 2007
    I don't like that particular "wizard". It's stupid about the resistors; there is no "overhead" left for them.

    You didn't mention the resistors; they were shown in the schematic.

    Try it again using 11.99v instead of 12v. You'll see a big difference with the red LEDs, winding up with something like this:
    Code ( (Unknown Language)):
    2. Solution 0: 4 x 25 array uses 100 LEDs exactly
    3.     +----|>|----|>|----|>|----|>|---/\/\/----+  R = 3.9 ohms
    4.     +----|>|----|>|----|>|----|>|---/\/\/----+  R = 3.9 ohms
    5.     +----|>|----|>|----|>|----|>|---/\/\/----+  R = 3.9 ohms
    6. ... etc...
    7.     +----|>|----|>|----|>|----|>|---/\/\/----+  R = 3.9 ohms
    9. The wizard says: In solution 0:
    10.   each 3.9 ohm resistor dissipates 1911 mW
    11.   !! the wizard thinks the power dissipated in your resistors is a concern
    12.   together, all resistors dissipate 47775 mW
    13.   together, the diodes dissipate 168000 mW              
    14.   total power dissipated by the array is 215775 mW      
    15.   the array draws current of 17500 mA from the source.
    Last edited: Jun 26, 2011
  5. Kerim


    Mar 3, 2011
    Once the LED current of each colour will be decided on, the design will become easy.
    It seems you are content with 700mA.

    I usually drive small LEDs up to 1W. So I tried to search for some data concerning the 3W type. I started with the red LED, I didn't get similar values for its nominal current even all references refer to 3W red.
    One of these reference "3W RedLED Lambertian Emitter" said the absolute maximum current is 700mA for 3W red LED. At this current the forward voltage would be about 2.7V. It is amazing that the product I*V is less than 2W!

    Returing to your project;

    First solution:

    For your RED:
    25 branches x (4 red LEDs in series + 3.3 Ohm 3W )
    Red current:
    25 * 0.7 = 17.5 A

    For your Green:
    10 branches x (2 grn LEDs in series + 3.3 + 3.3 Ohm each 3W)
    Green current:
    10 * 0.7 = 7A

    Total current:
    17.5 + 7 = 24.5 A

    Note: I let 3.3 + 3.3 for green instead of 6.8 Ohm 5W, to avoid buying 2 resistor types :)

    Second solution, higher power efficiency (lower dissipation in resistors):

    20 branches x (3 red LEDs + 1 green LED + 1.5 Ohm 2W)
    10 branches x (4 red LEDs in series + 3.3 Ohm 3W )
    Total current:
    30 * 0.7 = 21 A


    Note: I usually use resistors rated at 2 to 3 time their dissipation :)


    Formulas (approximated):
    Rs = (Vcc - n*V_led) / I_led [Ohm]
    P_rs = I_led*I_led*Rs [W]


    Example: if we like to use 3 green LEDs in series:
    Rs = (12 - 3*3.8) / 0.7 = 0.6 / 0.7
    Rs = 0.86 Ohm => 1 Ohm
    P_rs = 0.7*0.7*1 = 0.49 W
    So Rs is in this case 1 Ohm 1W


    Mistyping : Green => Blue
    Last edited: Jun 27, 2011
  6. #12


    Nov 30, 2010
    You can also get the red and blue spectrum by mixing fluorescent tubes.
    A long time ago, it was a Daylight tube with a Warm White Deluxe.
    It's probably easier now with stuff like 6500 Kelvin tubes available.

    It just seems simpler and cheaper to me to use fluorescent tubes.
  7. CrAzYHaMm

    Thread Starter New Member

    Jun 26, 2011
    Thanks SgtWookie, I didnt really understand what you meant by this "It's stupid about the resistors; there is no "overhead" left for them." Could you please explain this, sorry I'm still a nOOb to all this but hopfully with your help I'll learn. Also would my pervious layout work as long as I use the proper resistors?
  8. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    Over head means that resistors must be of a high power rating so that they will last longer and heat given of is less compared to the exact power.

    For eg: a 2 W will heat up a bit hot to the touch when it dissipates 2W, but if you use say 5W instead of 2W, the 5W will heat up less when it dissipates 2W & will be cooler and lasts longer

    Are you fixed on using LED's. You did not say how you want to use them.

    Give all the info's so that anyone replying will give u full details on what to get and how to wire them

    One more thing.
    Using Red's in the same string as blue will result in brightness difference as the Vf are different.
    Always use same color Led's in the same string

    You can use 3 Blue's /string + resistor for a 12V supply

    You can use 4 Red's /string + resistor for a 12V supply
    Last edited: Jun 26, 2011
  9. Wendy


    Mar 24, 2008
  10. SgtWookie


    Jul 17, 2007
    If you are building an LED string the "cheap way", you use a number of LEDs and a resistor in series. The voltage drop across the resistor actually results in wasted power, but it controls how much current flows through the LED string.

    Resistors do not change much at all in value over a wide range of temperature. However, as LEDs heat up, their Vf decreases, resulting in more current flow through the LED, which causes more heat, etc. If there is not sufficient current limiting, you will wind up with a "Chernobyl effect" and the LED string will burn up due to excessive current.

    I start off calculating how many LEDs I can use in a string by subtracting 1v or ~10% from the supply voltage, dividing that remainder by the LED typical Vf, and using the truncated integer result, as:
    Num_LEDs = INT( (Vsupply - 1) / LED_Vf)
    Num_LEDs = INT( (12v - 1) / 2.4v)
    Num_LEDs = INT( 11 / 2.4)
    Num_LEDs = INT( 4.58333... )
    Num_LEDs = 4

    Then, get the total LED voltage drop, and subtract that from the supply voltage to get the remaining voltage that needs to be dropped across the resistor:
    LED_Vdrop = 4*2.4 = 9.6v
    Vremaining = 12v-9.6v = 2.4v

    Next step is to calculate what resistance is required to limit the current:
    Rlimit = 2.4v/ 700mA = 2.4/.7 = ~3.43 Ohms or greater
    Here is a link to a standard decade table of resistor values:
    Bookmark that page - and this one, too:
    Looking in the green E24 column on the 1st linked page, you'll see 330 (3.3) and 360 (3.6) are the closest value to 3.43. 3.3 Ohms would result in too much current; 2.4v/3.3 Ohms= 727mA, and 2.4v/3.6 Ohms = 667mA. You are much better off to err on the side of caution.

    Then you calculate the resistor wattage required; the minimum for reliability is current * voltage * 1.6, so 0.667a * 2.4 * 1.6 = 2.56 Watts or greater.
    Checking stock at a couple of vendors, Mouser has 3W 3.6 Ohm resistors available for $0.25/ea when you by 100, or $0.39/ea singly:
    Digikey's price is $0.71/ea for an equivalent:

    Now for the blue LEDs - you can actually operate three red LEDs and one blue LED in a series string, and use a different value resistor.
    3*2/4+3.8=11 volts, leaving 1v remaining. 1/.7 = 1.4 Ohms. 1.5 Ohms is the closest standard value.
    1v/1.5 Ohms = 0.667A / 667mA; about 1.07 Watts required (.667 * 1v * 1.6 ~=1.07 Watts)
    Mouser has these 2W 1.5 Ohm resistors for $0.22/ea:
    Digikey has these 1.5 Ohm 2W resistors for around $24/ea at the quantity you'll need:

    You would need 20 strings with 1 blue and three red, and 10 more strings with 4 red, for a total of 30 strings.
    30 strings * 667mA ~= 20 Amperes; ~240 Watts total dissipation including the resistors; 20 blue, 100 red. As was mentioned before, the wattage rating of these LEDs just doesn't make any sense.
    Last edited: Jun 26, 2011
  11. SgtWookie


    Jul 17, 2007
    Now that the basic array maximum current limiting is resolved, you might want to consider PWM for dimming the array.

    The PWM circuit will need to be able to switch 20+ Amperes of current, which is quite a bit. Building just the PWM circuit isn't hard using a 555 timer, a few caps, resistors, and a pot - but switching that much current will require some pretty capable MOSFETs.
  12. Wendy


    Mar 24, 2008
    Or you could use transistor current regulators as several threads showed in the post I made. It will generate a lot more heat, as Wookie said, but constant current will eliminate the other problem he mentioned for less cash.