Electrical efficiency (watts->photons) of an LED

I'm busy with feasibility on an SSL project and I'm trying to do it as scientifically valid as possible. I know that basically the most knowledgeable people when it comes to LED tech live on this forum (and AVSforums), so as a change from my normal lurking behaviour I decided to register and ask my question here.

I want to calculate the electrical efficiency, not efficacy, of a Cree XM-L LED. Actually, I did it already. It's 37.16%. I think it's not right, but I don't know where it might have gone wrong. Or maybe I do. I'll walk you through the process. It's about the cree neutral white (5000K CCT) XM-L LED. Here's the datasheet.

Cree provide in their datasheets a graph of relative spectral power density (page 3, top). It's not unambiguously telling me, but I assume this is in (normalized to 100%) W/m.

This means that I should be able to scale this to the CIE photopic curve and get photopic lumens. Again I assume photopic, but I'm pretty damn sure they're not using scotopic or hybrid lumens. So I extracted the spectral power density for neutral white as well as the CIE photopic curve and read them into Matlab. Here they are:


Red=XM-L datasheet curve, blue=CIE photopic curve

Then I multiplied them together to get the (unscaled/relative) photopic luminous spectral density:


Now I reasoned that I should scale this curve by a factor such that the integral of the curve becomes the amount of lumens quoted in the flux bin, e.g. I should scale it such that it sums up to 260 lumen for a T5 bin.

After that, I can go back: if I divide again by the CIE photopic curve, taking into account that the peak should be 683.002 lm/W and not 1 as I showed in the first picture (yes, I took that into account), I will get a scaled curve for the spectral power density in absolute W/m. Right?

So when I integrate that curve, I should get photon watts. The actual photons leaving the device, and their associated power. Well, if I do that for the XM-L neutral white T5 at 700mA, I get a photon power of 0.7543W. The current is 700mA and if we look at forward voltage vs. current, you can see the forward voltage at 700mA is about 2.9V. That means the input power is 2.03W. I can now draw the conclusion that 0.7543/2.03=37.16% of electricity is radiated as photons and 2.03-0.7543=1.2757W is dissipated as heat (62.84%, but that seems to obvious to write down (damn, I did it anyway)). Mind you, I'm really just talking about efficiency, not efficacy, obviously a UV LED may be very efficient but not effective because we can't see any of it's radiation.

Is this the correct method and the correct outcome? If so, this is absolutely remarkable. Also, thanks to Cree's datasheets we can actually derive this.



I'd like to extend this derivation to maximum theoretical efficiency and other general silicon-oriented performance metrics (and thus a way to calculate the internal droop/extraction efficiency/etc.) but first I want to know if this is correct.

I wish Cree would just include their emitters' efficiency data in the datasheets.
I can't affirm or deny your calculations, but I have a suspicion that temperature has an impact on the efficiency also.
Did you factor that into your calculations? Or did you calculate it all as being at 25 degrees C?

I have a separate worksheet that takes into account degradation due to current droop and temperature to get 'true' luminous efficacy at higher drive currents. This one is just at 25C though.

What I want to do, is to plug this efficiency-calculator into a separate matlab function, and use it to do thermal calculations in the aforementioned worksheet. Right now, I just assume all electrical power goes into heat dissipation, because I was under the impression that radiative power was not significant (<10%). However, if it is actually 40+% at low drive currents, I should take this into account and it will improve my lighting fixture calculations quite a bit.

I have a separate worksheet that takes into account degradation due to current droop and temperature to get 'true' luminous efficacy at higher drive currents. This one is just at 25C though.<br>
<br>
What I want to do, is to plug this efficiency-calculator into a separate matlab function, and use it to do thermal calculations in the aforementioned worksheet. Right now, I just assume <em>all</em> electrical power goes into heat dissipation, because I was under the impression that radiative power was not significant (<10%). However, if it is actually 40+% at low drive currents, I should take this into account and it will improve my lighting fixture calculations quite a bit.

This info may help:
T5 and T6 bins:


and T5 bin, at 25, 50, and 100 degrees C respectively:


so the efficiency would vary dependent on junction temperature, and current, and bin.
If you take the theoretical limit for perfect visible white light efficacy as being an average of ~330Lm/W, and divide the chosen point of temp/bin/current LED efficacy by it, you should get a ballpark efficiency. Eg: T5 XM-L @700mA @25 C tj =128.1 Lm/W / 330 = ~38.8% efficiency.

Actually, the theoretical max for white LEDs might have been incorrect based on this chart:


As the datasheet indicates that a T5 cool white XM-L can be anywhere from 5000-8300K, this chart seems to indicate that the theoretical max possible luminous efficacy for that colour temp range is from about 400-450 Lm/W.
So that changes the equation somewhat.
128.1/400= 32%.
128.1/450= 28%.
So this indicates that the efficiency of the T5 cool white XM-L, @5-8.3K, @700mA, @25 degrees C tj, @2.9V FV =28-32%?

Terribly sorry about the double post there, but as these are my first posts and something went wrong during submission, I don't have any way to correct it...

Anyway, that type of estimate doesn't work. There's a big difference between CCT and blackbody color temperature that will throw that kind of ballpark estimate off. As the spectrum between manufacturers and individual LEDs varies appreciably, you cannot say that one 5000K CCT 120lm/W LED has the same efficiency as the other. Early white LEDs produced a significant amount of UV because of the high bulk losses causing locally high voltages on the die. That's light completely ignored by lumen estimates, but still lots of photon energy (and thus less heat output than you'd expect - higher efficiency).

However, that figure of 28-32% is interesting. It is low compared to my calculation (possibly meaning I did something wrong), but that may very well be explained by the huge blue peak (high energy photons, and lots of them) in the XM-L LED spectrum.

(this makes me think that it would be *very* interesting from a heat output point of view to find a way to convert the energy that is normally dissipated as heat into low-energy photons, so that the LED inherently stays cooler)

The vast majority of efficiency is lost in the phosphors used to make white LEDs, the bare blue die are much more efficient.

Yes. Cree XT-E puts out 550mW @ 350mA (1.05W) at 75 degrees, or even 625mW at 25 degrees die temperature. That means that most of the electric power is converted into light. Of course, the basis of a white LED is that exact same blue die but instead of the extracted photons radiating outward immediately, the phosphors absorb most of those photons and emit them at a lower energy, converting the rest of that energy into phonons (heat). We can actually calculate how much loss that causes and that is certainly something I would like my program to automate, but first I want to know if what I did was correct.

By the way, these datasheets are astonishingly complete. I can quantify almost all losses directly from the data. Bulk losses, current droop, extraction efficiency, it's marvelous. KUTGW, Cree.

mux said:

Right now, I just assume all electrical power goes into heat dissipation, because I was under the impression that radiative power was not significant (10%). However, if it is actually 40+% at low drive currents, I should take this into account and it will improve my lighting fixture calculations quite a bit.

Click to expand...

Radiative power for LEDs is quite a bit less than 10% under ALL operating conditions. Remember that radiative power is proportional to absolute temperature to the 4th power. Let's say a lamp filament with the same area of an LED die emits 100 watts of radiative power at a temperature of 3200K. An LED die with the same surface area operating at 60*°C (343K) will only emit around 0.01 watts. For all intents and purposes radiative power of LEDs is totally negligible.

Your procedure is essentially correct, mux. You just made a false assumption, that the SPD graph on the datasheet is scaled to W/nm. It is just a relative SPD, as it is just a "typical" one, give or take in the center of the color range: each color bin have a different SPD. Using that graph is a good way of having an idea of the typical efficiency of those LEDs (and color rendering, etc), that's all. The exact output in W/nm varies depending of the radiant efficiency of the LED, so higher bins scores higher, but the relative SPD continues being the same. You need to normalize SPD figures in a way its integral is equal to 1 (Watt).

Actually, what you obtained when multiplying photopic curve and SPD is the LER (Luminous Efficacy of Radiation) of such SPD (more exactly, LER is the integral of such curve given an SPD of area=1W). Then you just need to divide luminous Efficacy in lm/W (which will vary depending of the flux bin you analyze) by the LER figure, and you get the radiant efficiency, the figure you are looking for (Wout/Win).

An explanation of the logic behind it:

ηv=Φv/P

where

ηv=Luminous Efficacy (lm/W)
Φv=photometric flux (lm)
P=power (Input Watts=Vf*If)

Multiplying by Φe/Φe, where Φe is the radiometric flux, you get

ηv=(Φv/Φe)*(Φe/P)

as Φv/Φe=k=LER (lm/Wout), constant for a given SPD

and Φe/P=ηe= radiant efficiency (radiometric output/input power, Wout/Win), the figure you want to know. So ηe=ηv/k

When you know the SPD, you can calculate LER of it, by the procedure you used: multiply SPD of integral=1 by photopic curve and by 683,002. It mathlab does it directly, great, if not you need an Excel sheet with the coefficient of the photopic curve in 1nm increments and multiply it by the SPD in 1nm wavebands (you can use whatever figures you have for the relative SPD, and normalizing them to 1, just dividing each waveband by the total).

mux said:

However, that figure of 28-32% is interesting. It is low compared to my calculation (possibly meaning I did something wrong), but that may very well be explained by the huge blue peak (high energy photons, and lots of them) in the XM-L LED spectrum.

Click to expand...

Kinnza explained in detail how to calculate LER and then WPE(Wall Plug Efficiency).
Here are some basic ballpark numbers:

XM-L Cool White - T6 or higher
Junction temp 25 C
350 mA
CCT 5000K or higher

Underlying Royal Blue LED pump - 60-65% efficient
Phosphor conversion losses 20-24%
XM-L Cool White LED - 50% efficient
LER range(bin dependant) 280-330 lumens per watt

Stephen Lebans

jtr1962 said:

Radiative power for LEDs is quite a bit less than 10% under ALL operating conditions. Remember that radiative power is proportional to absolute temperature to the 4th power. Let's say a lamp filament with the same area of an LED die emits 100 watts of radiative power at a temperature of 3200K. An LED die with the same surface area operating at 60*°C (343K) will only emit around 0.01 watts. For all intents and purposes radiative power of LEDs is totally negligible.

Click to expand...

I think we're talking about different definitions of radiative power. I mean radiation in the exact meaning of the word: radiated photons. All light emitted. You mean purely blackbody radiation, which indeed is negligible and totally invisible.

Kinnza said:

Your procedure is essentially correct, mux. You just made a false assumption, that the SPD graph on the datasheet is scaled to W/nm.

Click to expand...

Well, no, I did recognize that the SPD graph in the datasheet was relative. However, because only the luminous flux is given I need to first convert to luminous density, then scale that appropriately, then convert back to the actual SPD, which I believe is exactly what you're telling me to do. Yay.

Is there any way to analytically/numerically predict or estimate the change in spectral density over different tints and CCTs? Also, you say that essentially the spectral power density 'shape' does not vary with flux bins (given a chromaticity and CCT) but it does change with drive current, right? As the die voltage inevitably scales with forward current, there will be some shift towards blue going from low to high drive currents. Plus there will be some parasitics like fringe effects (green halos on some dies). Can you say anything meaningful about that?

@slebans: thank you very much. So cool white LEDs are 50% efficient eh... that's a very compelling thought. Is this including extraction losses and optical losses (just the on-package optics, not external optics)?

jtr1962 said:

Radiative power for LEDs is quite a bit less than 10% under ALL operating conditions. Remember that radiative power is proportional to absolute temperature to the 4th power. Let's say a lamp filament with the same area of an LED die emits 100 watts of radiative power at a temperature of 3200K. An LED die with the same surface area operating at 60*°C (343K) will only emit around 0.01 watts. For all intents and purposes radiative power of LEDs is totally negligible.

Click to expand...

Not all radiated power is blackbody - and he is right that he should only consider the thermal fraction of the power consumed.

IMSabbel said:

Not all radiated power is blackbody - and he is right that he should only consider the thermal fraction of the power consumed.

Click to expand...

I was only referring to blackbody radiation in my post. Now that LEDs have gotten much more efficient, visible radiation obviously needs to be accounted for when calculating the true amount of waste heat which needs to be dissipated, or you can end up overdesigning the heat sink.

mux said:

Is there any way to analytically/numerically predict or estimate the change in spectral density over different tints and CCTs? Also, you say that essentially the spectral power density 'shape' does not vary with flux bins (given a chromaticity and CCT) but it does change with drive current, right? As the die voltage inevitably scales with forward current, there will be some shift towards blue going from low to high drive currents. Plus there will be some parasitics like fringe effects (green halos on some dies). Can you say anything meaningful about that?

@slebans: thank you very much. So cool white LEDs are 50% efficient eh... that's a very compelling thought. Is this including extraction losses and optical losses (just the on-package optics, not external optics)?

Click to expand...

Yes, LER only depends of spectra (SPD) (as it result for multiplying it by photopic curve*683,002 on finite wavebands or integrating the resulting curve, when starting SPD is 1W power). A given SPD has fixed chromaticity coordinates and CCT, of course. All that are just spectral characteristics, which only depends of SPD. But beware, although a given SPD has an unique CCT, LER and chromaticity, you can find same figures for SPDs very different. You can achieve same chromaticity coordinates with different SPDs, or same CCT, or same LER. The three at the time is difficult, but not impossible.

Specifically, you may think that once you has measured an spectra of a given color bin, all LEDs of that bin is going to be the same. But its not true. Due the way phosphor converted white LEDs works, it strongly depends of the peak wavelength of the blue LED used as pump. Manufacturer may use blues with different wl peak but with different phosphor (usually, thickness or density), resulting on a SPD falling on the same area but different from the other. Usually they are not big differences, but they could be so.

Any thing that changes SPD changes spectral derived figures as those. Drive current and operating temperature affects SPD, so affects spectral derived figures. It is almost impossible to derive the effect theoretically, as drive current affects SPD on some different ways, some working on one direction and some in the opposite (at least, for PC white LED). Al lower current,for example, the blue peak is at slightly shorter wl, but shape of the blue emission is narrower, and so on. Check this thread with lots of detailed info about it, old but highly informative: http://www.candlepowerforums.com/vb/showthread.php?70073-Constant-Current-vs.-PWM-dimming-Revealed

Ring and halos are other question. All depend of the spectra you are studying. If you get it using an integrating sphere, SPD recorded correspond to the average of the whole emission. If you record it with a probe, it will vary, depending on how much homogeneous is the light beam. Although this has improved largely along the last years, still the best white LEDs show some heterogeneity along the beam. But for studying the radiant efficiency, you need to consider the full emission and the average SPD of it.

But although SPD changes with current and temperature, fortunately LER dont change very much due it, so you can still derive radiant efficiencies at different drive currents with a decent margin error. Of course, the best way of doing it is with an spectrometer and IS, measuring both emission and SPD.

About the 50% efficiency of best white LEDs, always take in mind that a given LED efficiency depends of the drive current (and temperature). That figure is running them soft, and you can get higher efficiencies at lower currents. Once they are running hotter or at higher currents, efficiencies drops. But yes, top while LEDs can get it. Thinking in light output/input watts at the whole package.

slebans said:

... Underlying Royal Blue LED pump - 60-65% efficient ...
Stephen Lebans

Click to expand...

Which LEDs are these? Looking at the datasheet for XT-E Royal Blue, I see about 48% to 55% efficiency.

http://www.cree.com/products/pdf/XLampXT-E_ROY.pdf

Would this practical experiment allow for a calculation of efficiency/power wasted as heat:

Take 2 jugs of exactly 1 litre of water. Temperatures of water in both measured and logged.
Connect up 1 emitter and seal it with epoxy resin to make the electrical connections waterproof.
Put the emitter in 1 jug of water and turn it on.
Measure and log the current drawn and the forward voltage.
Log the temperature increase in the jug with the emitter in it over time, until it reaches equilibrium.
Calculate the power needed to increase 1 litre of water by that amount, with the temperature logged from the other jug as a control ambient temp.
Calculate the power drawn by the emitter over the time period using the logged current and forward Voltage.
From those calculations, measure the power wasted as heat, and therefore the efficiency of the emitter.

Would that work?
I use water and 1 litre, because they are easy to calculate, known constants.

The same test could be repeated at various constant currents, and using different emitters to get a comparison.

The jugs have no insulation so you would lose a lot of heat to the environment which would be difficult to account for accurately. If you were going to try something like this then the second jug would have to have a resistive heater in it and then you would adjust power to the resistive heater until you found the power level that kept this reference jar at the same temperature as the test jug, this power rate would approximately match the power being put into the test jug.

But I think that the insulated black cylinder apparatus that I described the other day in another thread would be much more accurate.

http://www.candlepowerforums.com/vb...ED-intensity&p=3810764&viewfull=1#post3810764

David_Campen said:

Which LEDs are these? Looking at the datasheet for XT-E Royal Blue, I see about 48% to 55% efficiency.

http://www.cree.com/products/pdf/XLampXT-E_ROY.pdf

Click to expand...

Cree does not sell the Royal Blue XM-Ls used in the production of White XM-L LEDs.

Stephen Lebans