9x 1W LED's parallel + series resistor calculation help needed

I am struggling with the calculations of the current limiting resistor for my circuit so I wanna ask you guys for help. I've tried doing this calculations on online calculators and by myself with math but the results i get are all different from different sources. I have 9x 1W green LED's which I want to drive using 12V 2A switching power supply but by my lame beginner electronics calculations that's not enough it seems and this is why I need your help too. This is the scheme I am trying to figure out:

Most of my calculations are suggesting that for R1 I need to use 3,9ohm 5W resistor. As noted on the image each LED should be supplied with 3V 300mA. I read many different resistor calculations but I cant figure out the right way for my circuit since I don't really know many things in electronics. I would appreciate if any of you could provide explanation how to calculate such circuit and help me out, or if you have any better idea of how I should build this thing. I will make a video about my build and post it on YouTube and I will make sure to mention all of you.

• As mentioned in other answers, take a look at electronics.stackexchange.com/q/22291/49251 Commented Apr 12, 2015 at 13:54
• thanks @Gregd'Eon for your informational link, but have in mind my green leds are meant to be driven by 3.4V at 350mA thats why i am lowering the performance and supply just to be sure that the leds will be safe. i had several circuits build with multiple leds in parallel and series in the past, the parallel ones showed to be much more efficient. same circuit build with 30 0.5mm leds 20mA white leds, ones in parallel and ones in series on 3V alkaline batteries, the parallel ones keept going after one whole month without lowering intensity and the series ones died after one week. Commented Apr 12, 2015 at 20:00

I would not use one resistor for all LED branches. Use one resistor for each LED branch.

simulate this circuit – Schematic created using CircuitLab

Uled=3V
Iled=300mA
Uledbranch=9V
Iledbranch=300mA
Vcc=12V

R= 12-9 / 3e-1 = 10 Ohms
Pr= 12-9 * 3e-1 = 0.9W

Your 12V 2A switching power supply has required current output for driving 3x300mA branches. It might flicker or get hot.

You should also try experiment with connecting 4 LEDs in series and limit the current with resistor (it will be much cheaper and smaller).

• When you put the LEDs in series, the current does not add up. He'll need a 10 Ohm resistor, which will dissipate about 1W. The total consumption of the 3 branches is 0.9A, so the power supply should be fine. Commented Apr 12, 2015 at 12:14
• My bad, will correct... Commented Apr 12, 2015 at 17:21
• thanks for suggestion and explanation. so i would need to use a resistor with 10Ohms at least 2W? Commented Apr 12, 2015 at 19:52

Each string of series LEDs needs its own current limiting resistor. This of course simplifies the problem conceptually. Trying to feed parallel strings of LEDs with a common resistor is a potential disaster because one string will hog all the current and may destroy itself with over-current.

Look at the following picture. It shows the forward voltage-current relationship of a typical white LED: -

Manufacturing tolerances might make one LED draw a wide range of currents for a fixed voltage applied to its terminals. If you applied another LED in parallel one LED might take virtually all the current whilst the 2nd LED took virtually zero - this is the problem of not using a seperate resistor for each string of LEDs.

• thanks for your suggestion but to keep it simple (for a beginner mind like mine) your saying that if one string fails the others will burn out? Commented Apr 12, 2015 at 19:54
• I'm saying that one string will draw more current than another because all LEDs are not exact devices and this will potenially burn out then, when it does the next string might burn out etc.. Commented Apr 12, 2015 at 20:33
• true, but this leds are meant to be driven at 3.4V at 350mA so i am way lower than the normal operating values... Commented Apr 13, 2015 at 2:14
• Lets say three LEDs in series on one string develpoed a voltage of 3x 3.3 volts - that's 9.9 volts. This leaves 2.1 volts across a 3R9 resistor implying a current of 538 mA. By having another string in parallel doesn't mean that all that current is equally shared. The next string (on its own) could be at 3x 3.4 volts = 10.2 volts implying a current of 462mA BUT with only 9.9 volts across the string it might take maybe 50mA leaving 488mA thru the string of LEDs that develop a smaller terminal voltage. Commented Apr 13, 2015 at 8:57
• aha now i get you. Commented Apr 13, 2015 at 9:26