Building LED lights from facts, no theories

[redline]

I got a bunch of LM317s kicking around, but wasn't planning on using them.
I figured resistors would do the job as long as I will willing to do a 200 hour array break-in and readjust resistance if needed. I am also going to have a couple of panel current meters monitoring 6 arrays each. I am also going to have a couple more meters where I can spot check current to individual arrays.

Does an electronic regulator have about the same power loss as a resistor?

Also, I figured using 24 volt power supply is better then using a lower voltage since any power supply voltage variation is distributed across a larger number of LEDs.

I was trying to find info on how large arrays are set up in the sign industry, but could find very little on the web.

I don't know how to calculate thermal efficiency (yet), so I have another question. I figured mounting bare emitters directly to the heat sink is preferable to stars, since you eliminate an extra thermal barrier. But you need to use an adhesive instead of a grease which cuts down a bit on thermal efficiency. Any thoughts on best way to go...stars or emitters?



I am getting ready to start constructing the individual modules. I will run the design specs by you for criticism before I start inhaling solder fumes.

I am going with 10 to 12 watt arrays. so will be running around .5 amp. I am going to try and keep the resistor around the .5v to 1.5v drop range. Do you think this will provide adequate secondary regulation? I hate to see it eating up more then 10% of the power.

[knna]

I got a bunch of LM317s kicking around, but wasn't planning on using them.
I figured resistors would do the job as long as I will willing to do a 200 hour array break-in and readjust resistance if needed. I am also going to have a couple of panel current meters monitoring 6 arrays each. I am also going to have a couple more meters where I can spot check current to individual arrays.

Does an electronic regulator have about the same power loss as a resistor?

No, it has lower losses. But it depends of how much voltage it need to adjust. If its little, the power loss is kept small. The closer the voltage of the array to 24V (supplied by th PS), the less volts drop.

You will notice it for how much hot get the LM317 (if i remember well LM317 have a max current rating of 0.5A, so you could use them).

Also, I figured using 24 volt power supply is better then using a lower voltage since any power supply voltage variation is distributed across a larger number of LEDs.

Yep, but the issue here is not variation on the PS, but different current flowing for same voltage with the changing chips temperature. So the larger the array power, the larger the variation:

say Tj is about 80ºC, for example. The K2 drops between 2 a 4 mV for each ºC over 25ºC. So 80-25=55ºC of increase on Tj lead to 165mV less (at 350mA, average of -3mV/K). The more LEDs in the array, the higher voltage drop along the array, so you need an higher resistance resistor to compensate for it, but lower than the sum of two resistors if you split the string in two series, each one with its own resistor (so better a 24V PS than a 12V one)

In this example, with 165mV less the associated increase in current is about 250mA. If you dont compensate that with the resistor, the array will work at 750mA (in the practice, still higher, because at 750mA and only 0.165V less, the power being burn is over 20% (near 25%) higher than initially, and that lead to the chips getting still hotter, and so on.

So you need a resistor that when all leds on the array drops 165mV, as current increases it increases the voltage dissipation as closer to that figure as possible. For example, if the string is 8 leds long, 8*165mV=1.32V. At 500mA, it implies 2.64 ohms. Power dissipated by the resistor would be 0.5A*1.32V=0.66W.


But obviously, you will determine the value of the resistor according to how much difference of voltage are between that supplied and the string requirements at the target current. In this case, its going to be similar, as you say its going to be about 1 and 1.5V, so its going to be very similar of that calculated before (about 2.5 Ohm, which drops 1.25V at 0.5A).

So if you use a 2.5Ohms resistor to regulate the current at 0.5A, then as chips heats, current across the circuit raises. But each 0.05A of increased current does the resistor drops 0.05A*2.5 Ohm=0.125V more. At a increase of 250mA, it would drops 625mV more. In the first example, we calculated a drop in each led of 165mV, so the resistor would compensate fully a string 5 leds long.

In order to calculate it accurately, you need to estimate the raise of Tj. Thermal resistance of your setup is about 20 K/W, eyeballing. Maybe 25 K/W. At 0.5A, blue K2 draws 3.51V (average, you should measure it for your batch), that may drop to 3.35 after 200h. Its 3.51V*0.5 A=1.575W, for 25K/W, increase of Tj= 39K. So at -3mV/K, you may expect a max drop of voltage at each led of 117mV. In 24V, you may fit 6 blue leds (at 3.51V each), up to 7 (at 3.35V). So you get a max drop in a blue string of 117*7=819mV.

Ive done this calculations so may may know how to do it.

But the right way to do them is by first calculating the number of LEDs on each string. To do it accurately, you should measure the voltage drop of your leds at you current target. Ill use the average Vf from the K2's datasheet: 3.51V for blues and 3.22V for reds (at 0.5A).

The PS gives 24V. So in each string fits 6 blues or 7 reds, and rest 2.94V (blues) and 1.46V (red) to dissipate with the resistor. At 0.5A, then you need to use a resistor of 2.94V/0.5A=5.88 Ohms (remember, R=V/I) for blues and 1.46/0.5=2.92 Ohms for red strings.

In order to compensate for the voltage drop due to the increased temperature, always choose the closer value of resistance over the calculated figure. That way, you set the current in cold below the current target, so as leds gets hotter, the final value is close with your actual target.

For the average figures of Vf of the K2, it would be a 6 ohm resistor for blue strings and a 3 Ohm for the red strings.

Would this minimize the current swing? Lets go to calculate it, although with the calc done in the example, i know the answer is yes.

At 25 K/W (estimated thermal resistance junction to ambient temp), blues may increase Tj up to 39K, as seen before. At -3mV/K, its 117mV less. For 6 leds in the string, its 702mV. How much the resistor will let increase the current before compensate it? As I=V/R, I=0.7V/6 ohms=0.117A, about 100mA.

As the setting for cold LEDs were regulated below 0.5A, final current on operating conditions is going to be lower than 600mA. Likely, about 550mA.

The problem of this design is that a 6 ohm resistor is going to dissipate about 550mA and 3.3V, or 1.8W wasted for each string.

Notice that the excess voltage on operating conditions for that Vf figure is pretty close to the Vf of another blue led. If the Vf of your leds is exactly that, then you could use a string of 7 blue leds connected directly at the 24VDC PS without any resistor and get a current very close to 500mA. As far as the PS is solid giving the 24V, it should work perfectly. But in order to do it you need the Vf of your leds were exactly that (and likely, it wont , but who knows?).

Measure accurately the Vf of your leds is a must if you are going to control them by voltage.

I was trying to find info on how large arrays are set up in the sign industry, but could find very little on the web.

They mostly use strings of 3 leds with a resistor on a 12V PS, and prefer to forget about reliability of the system. Mainly because they rarely use LEDs over 200mW, in which thermal effect is way less important than when using high power leds.

I don't know how to calculate thermal efficiency (yet), so I have another question. I figured mounting bare emitters directly to the heat sink is preferable to stars, since you eliminate an extra thermal barrier. But you need to use an adhesive instead of a grease which cuts down a bit on thermal efficiency. Any thoughts on best way to go...stars or emitters?

If you keep the thermal adhesive layer thin enough (by applying some pressure during the curing process), the thermal resistance added is kept below 1.5 K/W, and very often, below 1K/W. Of course, a good thermal adhesive makes a difference. The Arctic Silver is common and very good.

If possible, better the emitters. Mounting is more complicated, but it worth the thermal enhancement.

The problem of many stars is they are of poor quality. Thick layers of dielectric, poor soldering jobs, bad adhesives. Very often they penalize thermal path. With good ones, there is little difference with the emitter glued on the heatsink directly.

But if you know to mount directly the emitters, its cheaper and better.

[redline]

Knna,
I have an idea that I have also been working on in order to get more growth in Z-plane. This would also help if you are wanting to grow taller plants.

I have a design for a vertical plug-in module that drops down below the top of the canopy. The problem is with cooling. Heat sinks block too much light, plus they are dissipating heat onto the plants.
My design low profile water cooled module, where I am mounting emitters right onto a piece of U bent copper tubing about 6 inch long. I have only built one and it seems to do okay. I don't know how ganging up mulitple units will work.

Another issue is that you would have to reduce top lighting in order to keep within your watts/sq. ft. parameter. So the question that needs to be answered is "where will a watt of LED power need to be positioned to get the best results?" On the top?, On the side? or a combo?
I hate to introduce more complexity into the array if it is not beneficial

[Weezard]

Aloha Knna,

Beeg mahalo fo' all dem fish.

Just a quick FYI.
The LM317 maxes at 1.5 Amps.
And they need about a 2.5V drop.
Not all that bad for 24 -36V strings.
So, 2.5W wasted as heat. and only half of it is regulator heat.
(1.25V X 1 amp in sense resistor)
Got some running @ 50 C. @ 1 amp with a 2" sq 1/2" high finned 'sink, no fan.
That help, R. L.?

ciao
Weeze

[AfricanAlien]

A basic high-power led circuit???

Say i have a 24V DC @ 1 amp power supply
6.84V @ 700ma 5W high-power LED (3 led's in total)
high-power constant-current source input 6v-24v @ 700ma

If i connected the three LED's in series with the constant-current source and connect it to the power source, would this work?

Ignoring the thermal issues, just need to know if my understand of LED array is correct using high-power LEDs

[Weezard]

A basic high-power led circuit???

Say i have a 24V DC @ 1 amp power supply
6.84V @ 700ma 5W high-power LED (3 led's in total)
high-power constant-current source input 6v-24v @ 700ma

If i connected the three LED's in series with the constant-current source and connect it to the power source, would this work?

Ignoring the thermal issues, just need to know if my understand of LED array is correct using high-power LEDs

Yes.

Weezard

[redline]

Weez,
So you are saying that I will most likely loose more energy in a LM317 then I would with a sense resistor?
I haven't had the time to really study linear regs, so I am very weak on this subject.
But it just seems to me if you are dropping 2.5 to 4 volts (LM317 instead of 1.5 volts (resistor) at equal currents you are going to dissipate more power.
Of course the trade off is you are giving up better current regulation.

Knna/Weez- If I go with resistors and 24v VR driver, do I need to be concerned about start up transients damaging LEDs?

[redline]

I mount directly and use Artic Alumina. Is the silver a better way to go?

I didn't use pressure since I was worried about shorting out the pad. I did try to minimize layer thickness.
Now it seems I would be better off using pressure and if it shorts, just tear it up and mount another one. I will custom make some kind of fixture that applies weight to the LED

[Weezard]

Weez,
So you are saying that I will most likely lose more energy in a LM317 than I would with a sense resistor?

Little bit.

I haven't had the time to really study linear regs, so I am very weak on this subject.

Not that much to know really. Take you about 15 minutes to learn all you need for practical application.

But it just seems to me if you are dropping 2.5 to 4 volts (LM317 instead of 1.5 volts (resistor) at equal currents you are going to dissipate more power.

That is correct.
Don't matter how you do it. V x A = W

Of course the trade off is you are giving up better current regulation.

Exactly!
I don't stress over a Watt or 3.
And a dropping resistor will happily pass-on a voltage spike!
The LEDs may not be quite so happy to receive it.

Knna/Weez- If I go with resistors and 24v VR driver, do I need to be concerned about start up transients damaging LEDs?

In a word? Yes.
Not to mention shut-down spikes, lightening strikes and plain old everyday "gotcha" spikes.
Of course that does depend on the VR supply used, how heavily you load it, and what you feed it.
I'd just rather Pay the piper daily when considering the cost, time and effort of LED replacement.

With voltage controlled devices, I use VR.
With current driven devices ,CC is the best way to go for full control, consistency, and longevity. IMO.

I'd really like to see your multi-source lighting scheme in operation.
It sounds like a good idea.

Aloha,
Weeze

[knna]

Much time since i worked with LM317. But for what i remember, although specs usually advice to let 3V for the possible drop in the regulator, when you measure it often is around 1.5V. You know, there is wide tolerances and manufacturers obviously give safe margins of operation.

Anyway, i think the better current regulation worth using the LM317. Resistors arnt adequate to drive high power leds. Doable, but seeing the maths on my last post, is easy to understand why current control is advisable. 10%, and still 20% of variation on the current target is not a problem. But with resistors the variation may go way far from that.

I dont think transients are a practical problem. Although any manufacturers recommend to avoid them, there are many systems working without compensating them and work fine. So its preferable, and life test are done avoiding them, so its difficult to know if not compensating them shorten leds life. But on large strings i dont think its a mayor problem. But i dont know for sure, i use smooth regulators.

The Alumina and the Silver works pretty similar. In the practice, no difference. Any of them have low heat transmission compared to a metal. The key with them is to keep layer thin if you want it dont penalize heat transfer too much. I like the Crees because the electrically isolated bottom slug, as it allows to forget concerns about derivations if the adhesive layer not cover fully the slug.

I like to apply the adhesive directly to the slug, ensuring all is covered and then place it and apply some pressure with the hand (i use a nut to avoid pressing the dome). I let the adhesive cure enough to keep LED in place and then i put a book over the module (a nut over each led) for all the night. Using anodized aluminum is a secondary way of avoiding electrical derivations (the aluminum oxide layer is not electrically conductive).

You can try too to add kapton tape to the conductive slug. It works very well isolating. But then the problem is to glue the led itself to the board. But i think it would be possible to add adhesive (and not requiring to use thermal conductive one) on the perimeter of the base.

I think the main reason of many people to use stars is ensuring that there is no problems with conductive bottom slugs (and beware with that, ive found stars which indeed have the bottom electrically connected)

AfricanAlien, it will work perfectly as far as the constant current device is able to give 3*6.84V=20.5V @700mA. From 24V input, it should, still if its a linear regulator. But be ensure about that.

I have an idea that I have also been working on in order to get more growth in Z-plane. This would also help if you are wanting to grow taller plants.

I have a design for a vertical plug-in module that drops down below the top of the canopy. The problem is with cooling. Heat sinks block too much light, plus they are dissipating heat onto the plants.
My design low profile water cooled module, where I am mounting emitters right onto a piece of U bent copper tubing about 6 inch long. I have only built one and it seems to do okay. I don't know how ganging up mulitple units will work.

Heat from a heatsink shouldnt be any problem for plants. All thermal issues of LED arrays are about themselves perfomance and duration, not about plants. If you design well the system, the heatsink shouldnt be over 40ºC. A max, you should feel it slightly warm at the touch. That temp wont be any problem for plants (its similar to T12 fluorescents walls temperature).

Heatsinks on side walls shouldnt block any large amount of light either. But intracanopy, light losses may be noticeable, i agree with that.

I think there are two ways to fighting it. By using low power densities arrays intracanopy, able to be mounted on flat heatsinks, flat white painted (i use a highly reflective DIY mix, by adding barite to latex paint).

Or by mounting the leds on tubes (preferably, at least with some flat area, or directly squared) and aircooling them.

NASA intracanopy modules used this last option: each module has its own driver on a edge with a fan sucking thought a tube where LEDs are mounted.

Another issue is that you would have to reduce top lighting in order to keep within your watts/sq. ft. parameter. So the question that needs to be answered is "where will a watt of LED power need to be positioned to get the best results?" On the top?, On the side? or a combo?

Yep, this question is very important. I believe nobody along the Mj forums has studied this topic. And not only about LED lighting, but for any type of side lighting.

On the group buy im doing on a local forum, most people is going to use side and/or intracanopy lighting (mostly, side because cabs are small, mostly microgrow areas).

The initial rule of thumb we are using to size LED modules is to use 300uE/m2 from top and 180uE/m2 for lower areas. We think that each module should cover one cubic feet.

So we put 27uE for each upper cubic feet, so the distance covered for each LED module is about 1ft. And 16 uE on each bottom cubic feet.

But as the system is modular, at any moment each grower may change the configuration and put more or less power to up or sides.

We expect that we may learn the best power distributions from this experience, as there are 20 growers participating.

The initial rule of thumb has been extracted from the light distribution of a 400W HPS in 1 sq meter (about 11 sqft) for plants 75cm tall (2 1/2 ft). But translating some more light to bottom areas than the HPS does, and less directly on top (about 25%), so keeping light density way more even, and just about 60% higher on the top of the plants than on the bottom.

I hope that the experience of 20 growers using different styles of growing allows us to learn fast the best way of distributing light along the 3 dimensions of a grow.