LED Efficiency (complete with rabbits)

[fredfish01]

Idle minds like rabbit holes. Tonight I started out wondering about how much heat, in watts, I would need to dissipate for the clusters I want to try.

 

Now, to do this, the internets tell me I need to know the efficiency of the LEDs in question along with various thermal resistances (hence the multiple rabbitsess).

 

LED efficiency took a while to understand how to calculate, but I think I've got it.

 

Efficiency = Typical Luminous Flux (in mW) / volts * amps (mA)

 

So, for the Rebel ES royal blue (Steve's bin) I get:

 

1120 mW / 2.96 v * 700 mA = 54%

 

Hey, thats pretty good!

 

So, do I have that right??

 

If so: Lime = 17% not so good, Warm White = 7% ouch , Blue = 3% .

 

That blue, also Steve's bin, is not much more efficient than an incandescent lighbulb. Boy do we pay a penalty for adding extra more colours to our LED arrays!

 

As for those thermal resistance rabbitsess, I'll leave those for another day.

 

 

[tako]

You'd have to use radiant flux to calculate it that way, not luminous flux. The other thing you're probably doing is using units of lumens mistakenly. Notwithstanding that, the equation you've written still needs parentheses, .i.e. / ( volts * amps).

 

I'd probably just go look up the detailed product specifications from the manufacturer.

[Horerczy]

Lime, warm white, and blue are done in lumens so they're luminous efficacy (lumens/watt) not efficiency (mW radiance/watt).

[fredfish01]

You'd have to use radiant flux to calculate it that way, not luminous flux. The other thing you're probably doing is using units of lumens mistakenly. Notwithstanding that, the equation you've written still needs parentheses, .i.e. / ( volts * amps).

 

I'd probably just go look up the detailed product specifications from the manufacturer.

Hmm, yes. Looking at the Luxeon data sheet (table 1), it is quite confusing. They have a column labeled as Luminous Flux (lm) or Radiometric Power (mW). I missed the 'or' before.

 

The royal blue are clearly identified as mW, the rest are not identified as either mW or lm. Perhaps they are pixie dust units pD??

 

I did miss the brackets didn't I. That's the story of highschool math for me. Part marks 'cuz I got the answer right, but missed half the stuff in the middle.

[fredfish01]

Lime, warm white, and blue are done in lumens so they're luminous efficacy (lumens/watt) not efficiency (mW radiance/watt).

Having worked a number of years in marketing, that can mean only one thing. Its even worse than I calculated...

 

Now I'm doubly curious as to what the efficiency of the other LEDs is. It would not surprise me that the efficiencies of the whites are similar to what we see for fluorescent T5. We are, after all, dealing with phosphors.

[tako]

Now I'm doubly curious as to what the efficiency of the other LEDs is. It would not surprise me that the efficiencies of the whites are similar to what we see for fluorescent T5. We are, after all, dealing with phosphors.

Without getting too pedantic: Sure, the emission from the phosphors will be the same if it's energized by the same radiation from inside the bulb/LED. The efficiency gain comes from how you generate that radiation.

[fredfish01]

Without getting too pedantic: ...

Well, the pedant in me must point out that I wrote "would not be surprised" suggesting that it was a distinct possibility, but not a certainty.

 

Once you're down the rabbit hole, might as well check out the warren. It don't look purdy.

 

The only hard numbers I could find (someone who actually measured stuff instead of vaguely referencing some other source) shows that as of 2009 T5 still had greater luminous efficacy. I've found a few 2013 and 2014 sources that claim, without providing a source, that the luminous efficacy of white leds is up to 110-120 lm/w. That would put it in the range of or slightly better than T5.

 

While luminous efficacy and radiant efficiency are two different things, I suspect there is some correlation between the two for white LEDs.

 

Given all that, I think that it is reasonable to say that white LEDs are in the same range of efficiency (radiometric) as a white T5 tube, but probably less efficient given manufacturers unwillingness to provide hard numbers.. I'd love for someone to provide hard numbers showing otherwise, but I don't expect to see anything.

 

What I posted for other colours is probably dead wrong.

[jedimasterben]

The Rebel ES lime is around 190lm/w at 85C. It uses the same diode in the Luxeon T as a base, which is around 62% efficient at 85C (versus 25C for the Rebel ES royal blue). There will be some efficiency loss with the application of phosphor, but realistically lime should still be at least 40% efficient.

[fredfish01]

The Rebel ES lime is around 190lm/w at 85C. It uses the same diode in the Luxeon T as a base, which is around 62% efficient at 85C (versus 25C for the Rebel ES royal blue). There will be some efficiency loss with the application of phosphor, but realistically lime should still be at least 40% efficient.

How do you figure 40% efficiency? Do you have radiometric output numbers for the lime? I don't think there is a way to get from lm/w to radiometric efficacy, but I could be wrong.

 

The Luxeon data sheet lists the lime as 142 to 184 lm/w typical @85C, so I would say 190 lm/w is a little optomistic .

 

My understanding is that 40% efficiency is huge. MH is in the 15-20% range, and T5 may be higher than that. Again, I can find no numbers for radiometric efficacy for T5.

 

If I were an LED manufacturer and had an LED that was that much more efficient than competing technologies, I'd be shouting it from the rooftops the way Phillips is with royal blue.

 

I'm geeky pedantic curious about this and would really love to know.

[Horerczy]

When measuring lumens from a light it's important to remember that is a weighted scale towards green and red radiation.

 

All florescent lamps have large spikes in both areas.

 

LEDs do not.

 

So from a lumens efficacy point of view florescent lamps well often appear brighter.

[fredfish01]

When measuring lumens from a light it's important to remember that is a weighted scale towards green and red radiation.

 

All florescent lamps have large spikes in both areas.

 

LEDs do not.

 

 

 

Agreed, however...

 

So from a lumens efficacy point of view florescent lamps well often appear brighter

 

Maybe. If we are comparing to a narrow band direct emitter like blue or red, I would agree.

 

The Luxeon lime emits broadly, but [almost] all of its output is in the 500 - 600nm region where our eyes are most sensitive, so which one is more skewed and to what degree? [Almost] the same could be said about any white LED because they all emit a lot of their light in that same region.

 

That's why I want actual radiometric measurements. I expect that all white LEDs have such measurements. How else would a designer figure out heat sinking for a given design or application?

 

Its getting dark here. Did we just go deeper into the rabbit hole?

[jedimasterben]

That's why I want actual radiometric measurements. I expect that all white LEDs have such measurements. How else would a designer figure out heat sinking for a given design or application?

Voltage times current

[fredfish01]

Voltage times current

You'll need to spell it out for the dummy. How do you get from measured voltage and current to voltage used for light vs voltage converted to heat?

 

Edit: Did I forget to mention I know just enough to be dangerous?

[fredfish01]

Well, todays dive down the rabbit hole was more interesting, but no more iluminating (on energy efficiency) than yesterday's.

 

Coming soon to an LED near you (or not): quantum dot phosphors, super hard phosphors that work much better at LED operating temperatures, narrow band phosphors and more...

 

http://www.journalofsolidstatelighting.com/content/pdf/s40539-014-0011-8.pdf

 

Cheap LED luminairs (I'm looking at you Orbit Marine) have a useful life of maybe 10,000 hours. Where's the finger wag smiley when you need it?

 

I'm pretty sure that the Luxeon lime LED is just the phosphor used to radiate the green/yellow portion of a white LED and the PC amber is the red portion. Put a royal blue, lime and PC amber on a 3 up and you can tune your white to whatever you want it to be. I wonder if they used the same Luxeon T diode in the PC amber as the lime? None the less, now I understand why a warm white LED is less efficient (15-30%) than a cool white. It's the green to amber phosphor ratio.

 

All the industry rags peg white LED efficiency to be on par with that of linear fluorescent tubes. The lime may be a smidge more efficient because of the diode used, but its not likely to be out-of-this-universe more efficient.

 

Super T8? I did not know...

[fredfish01]

OK, time to revisit the rabbits (because the alternative is cleaning my apartment).

 

Ben. I know v * a = w

 

For the lime at 700ma that's 1.89 watts, but that is total power consumed. I suppose that's enough to figure out heat sink requirements if you want to overbuild a fixture, but would not an engineer want to know specifically how many watts in heat they need to dissipate. No?

 

I still want to know the actual efficiency of the Luxeon lime. How many watts go to light and how many to heat. Lets call them x and y.

 

We get 1.89 = x + y

 

Unless I reeeeally wasn't paying attention in high school, that's not a solvable equation. I don't see anything on the Luxeon data sheet that helps me resolve either x or y.

 

There will be some efficiency loss with the application of phosphor, but realistically lime should still be at least 40% efficient.

 

 

Is this a guess on your part or do you have some additional info? Everything I've seen shows that there is a huge efficiency loss with the use of phosphors.

 

Say, this isn't one of those "I could tell ya, but then I'd have to kill ya" NDA kinda things is it?

 

In other random LEDness...

 

I just realized that Nanobox is using the Luxeon T package for their royal blue and whites. Looking at the a typical luminous flux for a 5000K led for the T vs Rebel ES @700ma (as close to apples to apples as I can get) it looks like the T gets you around a 30% efficiency boost... which puts gives you a good boost over the luminous efficacy of a typical T5 (which is probably higher than the T5 HO aquarium bulbs).

 

Wow, do those amber/red Rebel leds lose efficiency as junction temperature increases!

 

Sigh, cleaning. Guess I can't avoid it any longer...

[jedimasterben]

OK, time to revisit the rabbits (because the alternative is cleaning my apartment).

 

Ben. I know v * a = w

 

For the lime at 700ma that's 1.89 watts, but that is total power consumed. I suppose that's enough to figure out heat sink requirements if you want to overbuild a fixture, but would not an engineer want to know specifically how many watts in heat they need to dissipate. No?

 

I still want to know the actual efficiency of the Luxeon lime. How many watts go to light and how many to heat. Lets call them x and y.

 

We get 1.89 = x + y

 

Unless I reeeeally wasn't paying attention in high school, that's not a solvable equation. I don't see anything on the Luxeon data sheet that helps me resolve either x or y.

http://www.labsphere.com/uploads/technical-guides/The%20Radiometry%20of%20Light%20Emitting%20Diodes%20-%20LEDs.pdf

 

I just realized that Nanobox is using the Luxeon T package for their royal blue and whites.

You're welcome

[fredfish01]

Geez, I want answers, you give me homework.

 

Thanks.

Oh yeah, that NanoBox chip is cool beans.

[afyounie]

So, I started reading all of this and I'm kinda confused by the direction you are taking. All this talk about lumens per watt, but you want to know percentage of heat dissipated and percentage that is light. Resistive loads produce heat. Everything has a resistive load. For diodes, they are modeled with a voltage bias and resistance in series. The voltage bias is the minimum voltage needed to turn on the diode, usually about 7mV.

 

The resistance is calculated as r_d = V_T/I₀. V_T = kT/q which is approx. 26mV. I₀=I_s*exp(V₀/V_T).

 

I_s is specific to the diode. You should be able to use the current drawn to power a single diode as I₀.

 

Power = Current*Voltage or P_R=I²/R.

 

With the power dissipated due to the resistive load of the diode, now you can calculate the percentage dissipated as heat. Just resistive power consumed divided by total power.

 

Total power is P=I*V.

1-P_R will give you approximately what is used to generate light.

 

I'm a Mechanical Engineer getting a masters in EE and just getting into the thick of junction semiconductors. So I may be oversimplifying. Wish I could share my electronic devices book with you. It explains how semiconductors work. Which is very cool.

[fredfish01]

So, I started reading all of this and I'm kinda confused by the direction you are taking....

 

Well, that would be a reflection of my confusion, or lack of knowledge. It's no help that I have pretty much zero electronics knowledge.

 

As far as all the lumen stuff, LED's are, after all, a light source. The vast majority of efficacy data presented for lighting products are in terms of lumens per watt.

 

Thanks for the explanation and formulas. I figured there must be some straight forward way to figure out energy dissipated as heat, just didn't know where to look. So, this would also tell me how efficiently a diode generates light.

 

For phosphor coated LEDs there is a secondary process though to get from blue light to something more broad spectrum. There is energy conversion to heat there as well and I don't think that would be reflected in what you have given would it?

 

Ben. Point taken. Its complicated.

 

As to what direction I am taking, what I want to do is cut through all the fud relating to LEDs that we see in the hobby, from both the lovers and haters, to get a good understanding of the efficacy of LEDs for reef aquarium lighting. My personal reason for doing so is that I cannot run the reef I want using metal halide lighting, and probably not T5 either.

[afyounie]

Heat generated due to phosphor reaction should be negligible. You can assume it is 0 or so small it doesn't make an impact.

[fredfish01]

Aha, progress!

 

http://www.journalofsolidstatelighting.com/content/1/1/19

 

Energy losses can be categorized into quantum loss, stokes shift loss and package loss. I think this paper is saying that the maximum theoretical efficiency that can be obtained is 73%. That would yield a total theoretical radiant efficiency of around 42% using the 53% radiant efficiency number for Luxeon T LEDs at 85C stated in the data sheet.

 

Current real world would be some number less than that.

 

There be rabbits in this hole.