Once more, the design for my custom LED panel has changed. This is mostly due to number crunching and looking at the data sheets for the Luxeon Rebel LEDs I'll be using. The biggest influence factor here is temperature. LEDs are more efficient at lower temperatures. What affects temperature in LEDs the most? Well being that LEDs are current driven devices...you guessed it! Current. So I have decided to cut the current in half, down to 350mA. In turn, this lowers voltage required to power the LEDs. Lower voltage and lower amps means lower watts. So the 60w panel I designed before, will now only be putting out 24.5w. To combat this, I will be adding 6 more LEDs to the mix.
Now for the spectrums part. I have since decided to not use both the orange spectrum and warm white spectrum. Orange because it has a tendency to create hermies, and white is simply not needed. So in total my spectrums will be 447.5, 465nm, 626nm, and 650nm. This doesn't have the important 660 spectrum, but I would have to totally redesign my panels with a different MFG and I decided that's too much work. Enough as it is
Onto the panel itself. The panel I have chosen is 8.5" by 13". I have chosen this length so that each LED has about 1 square inch of surface to dissipate heat all by itself. I don't want them too close, but too far either. I would use 2 inches between LEDs if I could afford the heatsink that big. The one I'm looking at is only 35 bucks or so, which isn't too bad. So the total surface area of this heat sink is .7 feet.
Now a bit about the optics. The LEDs have a standard viewing angle of 120 degrees. So if I built my panel and did not use optics, I have 2 options. Lower the panel down to my plants so the light isn't dispersed all over the place, losing efficiency. Or I could leave them higher up and have really stretchy plants. I have only 2 vertical feet to work with once you add in the pot size and thickness of the LED panel. So at 2 feet and 120 viewing angle I get a beam diameter of 6.9 feet. This is huge, this is twice the square footage of my grow area. So roughly half of the light will be reflected off the walls, losing more efficiency. Now if I use optics and tighten up the viewing angle to 12 degrees, I get a .42 feet(or 3 inches) spread from each LED. So now we're looking at a coverage area of 10" x 14.5". So this light will cover about a foot and a half of INTENSELY focused light. No light will be reflected off the walls(or very little light). I have told people it's the same exact aspect as a flashlight.
You can focus it to one really narrow bright beam or one really wide beam that covers the whole room. The really intense light is what you want. I hope people can understand my reasoning here. I'm working with a little more than 2 square footage in my cabinet, so to help out with the side lighting I will be adding my 4 LED panels to the sides of the cab to help with the undergrowth and to even out the light distribution.
Now onto the power supply and electrical stuff.
The red LEDs draw 2.9v @ 350mA, blues draw 3.15v @ 350mA.
I need to multiply times six for the number of LEDs I'm using per string. I use the blue LEDs because they have a higher voltage. 3.15 * 6 = 18.9
I will be using 5 string of LEDs, 4 red and 1 blue in the middle.
5 * .35A(350mA) = 1.75A
If you use Ohm's Law to calculate wattage:
Red - 2.9v * .35A = 1.015w(rounding to 1w)
Blue - 3.15 * .35 = 1.0125w(rounding to 1w)
Since I'll be using 30 total LEDs - 30 * 1w = 30w
So I need a power supply that has constant current capabilities, puts out at least 18.9v, 1.75A, and 30w.
I have found the
Meanwell PLN-100-24. It puts out 24v at 4A, 96w. I used this one because it's current and voltage regulating. It will also be mounted near a fan to aid in cooling, and increase the life of the supply. It will also be operated at less than half of it's capabilities so I think it should last quite some time.
I have attached 4 pictures, 1 of a heat sink like the one I will be using, 1 of the power supply, 1 of the design on that heat sink, and 1 of the lens I'll be using.
The next easiest is a simple resistor. The resistor does consume power, though, but is usually needed since an 'ideal' 3.6 volt source is rarely available. Use Ohms law (Resistance(R)=Voltage(E)/Current(I)) to calculate the value and wattage needed: (R=E/I)Each white LED gives a voltage drop of 3.6 volts. As an example, for a 12 volt light, you can run a maximum of 3 white LEDs in series at full power (3.6 x 3 = 10.8 volts drop). Subtract this from your supply voltage of 12 volts to get the additional voltage that must be dropped (in this case, 12 - 10.8 = 1.2 volts of additional drop needed). In this case, 1.2 volts of additional drop / .025 amps (25 ma) = 48 ohms. Use the next highest value of resistor available, 50 ohms. You must also be sure the resistor can handle enough current. Volts x Amps = Watts; resistors are rated in watts. So in this case, 1.2 volts x .025 amps = 0.03 watts. A 1/4 watt resistor will work fine, but if you run a second string of 3 LEDs in parallel, each string would need its own 50 ohm resistor. It's important that each string has its own resistor....putting them in parallel with a single resistor is bad practice.
This method is cheap and works great, but there's one problem--voltages in a remote power system (or car, for that matter) tend to vary. In our home system, voltages range from about 12 volts when the batteries are low up to 14 volts when equalizing the battery bank. An LED lamp string designed to run at 25 milliamps at 12 volts would be pushing 64 ma at 14 volts, which would be very bright and PROBABLY last at least a few hundred hours...but then then when your batteries are low, the LEDs will pull only 10ma or so, making them very dim. If you are looking for maximum lifespan (which could be over 10 years of run time) and brightness that doesn't vary with your battery condition, try a voltage regulator circuit (below).So, we highly recommend a simple voltage regulator chip for the safety of your LEDs. White LEDs are expensive, and it would be a shame to blow them out. Parts for a current-limiting circuit are very cheap--less than $2. Use the Ohm's law calculations above to select the resistor for the voltage you choose. Or, use the regulator in a current-limiting configuration to run the LEDs. You can also use an LM317 adjustable voltage regulator set to the exact current level needed by your strings of LEDs. See the circuit diagrams
It creates a loop which will eventually overdrive the LEDs and fry them. I don't know where you got the batteries from, I have never thought about using batteries until I saw a solar panel setup... but that's another story!