Help with my LED circuit

[Fingersdeep]

      I've got 3 3W RGB LEDs that I need to power through a 15V 65W power supply. I'm having trouble with voltage dividers since all my resistors are rated for 1/4 watt and I don't know how to power all these LEDs without sending to much voltage through them.
      The red has a voltage drop of 2.2 - 2.6, green 3.2 - 3.8, blue 3.3 - 3.8 and each led is rated for 350 mA. So 1050 mA for each 3W led.

[mariush]

Resistors and voltage dividers are not the way to go. 
The forward voltage of each led will change considerably as each led heats up, so if you pick the wrong current limiting resistors your circuit may eventually send too much current through the leds.

You need proper led drivers, here's for example a very simple one : http://www.digikey.com/product-detail/en/AL8808WT-7/AL8808WT-7DICT-ND/3882102

See the datasheet, you can configure the maximum current using a single resistor and optionally, you can also adjust the current (therefore how bright the led will be) by sending a voltage on the CTRL pin (a voltage between 0.5v and 2.5v will adjust brightness between 20% and 100%).

It's  a surface mount chip, but it's easy enough to solder on prototyping boards or you could buy from eBay adapter boards that convert that shape to 6 or 8 pin DIP that would be easy to solder by hand.

The simplest way would be to build three small led driver boards with this chip and each of these boards will drive the two leds of same color (you connect them in series). This would allow you to control the brightness of each color independently at a later time, if you would want to.

[Brumby]

If you are constrained (or determined) to stick to a resistor solution, you are going to have to choose your resistors for safety (as in not exceeding voltage ratings).  This will result in you sacrificing brightness.  Also, as actual voltage varies from chip to chip, you might end up with differing brightness between the R, G & B LEDs across the 3 devices, resulting in colour variance (if that's a concern).

Constant current power supply (as found in LED drivers) address the issues and will give you access to maximum output.

[hamster_nz]

You could always build a reasonable constant current source using a few "junk box" parts.

Something like

from https://en.wikipedia.org/wiki/Current_source - put the LED where "LOAD" is.

A 300 Ohm for R1, replacing DZ with a string of two diodes (the other way around), which should hold the transistor's base at about 1.2V or so,  and a 1 Ohm resistor for R2 should get you something between 500mA and 1000mA.

(edit) R2 should be 3 or 4 1/4W 3.3Ohm resistors in parallel as it will also dissipate 0.5W to 1W, depending on how it all works out.

Most of the excess power is converted to heat by the transistor. This could be around 12V@1A = 12W for the red LED, so don't forget to add a heat sink!

[TerminalJack505]

Here's an alternative to a dedicated LED driver IC--the good ol' 555 timer.

[ocw]

Some simple relatively low cost constant current regulators are made by ON Semiconductor.  One appropriate for your use might be a NSI50350AST3G.  See:
http://www.mouser.com/ds/2/308/NSI50350AS-D-114166.pdf   Mouser has them for $0.69 each or ten for $0.538 each.  Unless you are sure that you need the maximum amount of light you might try a  NSI50150ADT4G instead.  It will permit you to adjust the current level from 150 - 350 ma so that you can control the light output.  See:
http://www.mouser.com/ds/2/308/NSI50150AD-D-272592.pdf  You'll be around that one's dissipation limit when feeding 350 ma however.  Selecting a lower current won't decrease the LED's light intensity that much while the LED and regulator will be cooler.  I have eight different model ON Semiconductor regulators in stock which I've used without any problems.

[pmbrunelle]

Yikes, we are suggesting LED driver ICs (either linear or PWM) for a beginner who can't figure out how to drive a LED with series resistance?  One must start by understanding the fundamentals...

The key to using a resistor current source is to drop most of the voltage across the resistor, not the load. With a 15 V supply, a 1 V change in LED voltage should not significantly change the voltage across the resistor, therefore, the current should not significantly change.

I would suggest one series resistor (made of a series/parallel combination of identical resistors) per LED. Possibly compromise for now on the current to limit dissipation, depending on the number of 1/4 W resistors you have. So 9 resistors (series/parallel combinations) for 9 LEDs.

I would try to submerge the resistors in water; this would increase (how much, I don't know, try it out!) the power handling until you get the proper parts. These would need to be wirewound resistors of the proper rating.

I find it more convenient to dissipate "high" power in resistors than transistors, because you get full rated power dissipation from an axial-leaded resistor "out-of-the-box". With TO-220-style parts, you need to add heatsinking to have any sort of power handling, which is really a pain...

If you consume a total of 3.15 A to drive the LEDs, this would be compatible with your 15 V, 65 W supply.

[mariush]

Yes, I am suggesting a led driver IC because the schematic for the whole circuit is already there in the datasheet, and a few pages further the datasheet explains how the chip works and how to choose the parts around the chip.

Resistors just won't work for this application, because the current is too high and the voltage drop is also too high.

Here's some basic lesson about limiting currents in led.

Leds has forward voltage (let's say 2.7v) and maximum current (350mA).  In order to have only 2.7v reach the led, the difference between the input voltage and this 2.7v has to disappear and that's why you put a resistor in series with the led.

The formula to is simple  :  Vinput - Forward Voltage LED  = Resistance / Current  ... so Resistance = (Vin - Vled) / Current  .. in your case, it's (15v - 2.7v) / 0.35A  = 35.14 ohm

The power dissipated in the resistor will be equal to P = IxIxR  = 0.35x0.35x35.14 = 4.3 watts.  So it would be wise to use a resistor rated for 7w or 10w, just so that it wouldn't overheat. Even so, with 4w dissipated, the resistor will probably be quite hot. 
Alternatively, you can use several resistors in series instead, for example 4x 8-10 ohm resistors each with a 3w rating.

But you'd still have two problems:

1. as the resistor heats, the resistance will change slightly, usually the resistance will decrease, which means the current going through the led will go up.  You can counter this by picking a much higher resistor in the first place (ex use above formulas using 330mA instead of 350mA, to leave 20mA as a safety margin)

2. as the LED itself heats up, the forward voltage will change... for example let's say instead of 2.7v it will be 2.5v  ... so now if you go again through the formula, you have 35.17 ohm instead of 35.14 ohm if you do the math for 350mA ... even more if you play it safe for less current.

A LED driver only needs a couple of carefully chosen parts (how to pick them is written in the datasheet) and gives you about 85-95% efficiency, and the chip will constantly monitor the current going through the led and will keep it 350mA no matter how the forward voltage fluctuates, and there's no resistor to waste a few watts on, and make extra heat.

[Fingersdeep]

Thank you guys for all the help, is it possible to use PWM with these led drivers though?

[Landrew2390]

http://www.diodes.com/_files/datasheets/AL8808.pdf

Check on Page 2 under "Pin Descriptions" for an overview of some of the options.  The driver that mariush suggested has a couple of different modes depending on what you're wanting to do.

As a tip, when in doubt, read the datasheets.  I tend to read 5-10 datasheets before I find a part that I like for a project.