# How can I estimate the optical power that a single-color LED generates?

This excellent answer addresses the Luminous efficiency and Luminous efficacy of LED lighting, but these are based on human vision and it's variability with wavelength. It's a good idea to read it.

I'd like to be able to convert these to absolute efficiency, which would be the ratio of optical power out to electrical power in. If I put 5W of electrical power into an LED and it produces 3W of light, then the absolute efficiency is 60%, and incidentally I will also know that I have about 5 - 3 = 2W of heat to manage.

Question: How can I at least estimate the optical power that a single-color LED radiates given the typical parameters - say luminous intensity in Candela or Luminous efficiency - from a data sheet? Assume this is a narrow-band LED and can be characterized by a single wavelength, $\lambda$, to keep the solution from getting too complicated.

• In general, the better LED arrangements (those actually designed to produce light for human "seeing") will be roughly 10% efficient. An 8-10 W LED light bulb will provide about 1 W of radiant power. You might also look here: hyperphysics.phy-astr.gsu.edu/hbase/vision/efficacy.html – jonk Aug 26 '17 at 16:14
• @jonk I'm asking about single-color LEDs that don't use a phosphor to help convert blue or UV to white light. They will probably have a higher efficiency than that. See this comment and the one below it. – uhoh Aug 26 '17 at 17:01
• I use a couple of spectrophotometers I have (one is from Ocean Optics), plus software I wrote, and an optical bench. But that takes time. I've also used an integrating sphere, before. The irradiance efficiency wasn't particularly good. But it has been 10 years and more. The one comment from Russel, below yours in the link you provide, should be about 30%. But you say narrow band. In any case, there are very nice procedures for figuring these things out. Integrating spheres, or as I prefer, an optical setup with specific distances and a calibrated spectrophotometer. – jonk Aug 26 '17 at 17:12
• @jonk thanks! It's hard work but interesting for sure. In this case I'm hoping to start with whatever specs are provided with a product. Since Candela is power per unit solid angle, I suppose multiplying by some estimate of the solid angle would do it, especially if an angular distribution or cone angle is given. – uhoh Aug 26 '17 at 17:33

So in some cases, you have it easy. As soon as the emitted light is out of range of the visible spectrum, the only measure for the manufacturer is the radiant flux which is the optical power, and exactly what we are after.

For example the NCSU033B is a UV-LED by Nichia and they give the following data:

• Forward current: 500 mA
• Forward voltage: 3.8 V

The electrical power is 1.9 W, the light output is 450 mW. So it has an efficiency of 23.7%.

That's pretty simple right?

Now with visible light LED, you usually get a figure like lumen or candela. Which are based on the weighting of our eye. And things will get complicated.

So the next easy step is when the luminous flux in lumen is given. Under the assumption that we have a single wavelengh light source, which we of course never have.

We take again the weighting curve:

In this diagram the weighting is given in a range from 0 to 1. What's missing is that at the peak you have $683~\frac{\text{lm}}{\text{W}}$. And that factor is actually radiant flux to lumens. So we can write:

$$\Phi_e = \frac{\Phi_v}{683~\frac{\text{lm}}{\text{W}} V(\lambda)}$$

Where: $\Phi_e$ is the radiant flux in W, $\Phi_v$ is the luminous flux in lm and $V(\lambda)$ is the luminosity function.

The index e is usually there to denote energy, while v denotes visual, visual quantities are always weighted with the luminosity function.

Let's plug in an example: I've chosen the Osram Top LED and we can have a look at the different colors they are offering. The nice thing about this datasheet is, they give the luminous flux and luminous intensity, which we will visit in the next step. I wasn't able to find a datasheet which gives luminous flux and radiant flux - I guess that wouldn't be worth the trouble for the manufacturer.

Actually, and that's quite remarkable, Osram provides two wavelength figures. The one of the emitted wavelength and the dominant wavelength. As you can see the dominant wavelength is smaller, because of the form of the weighting function, our eye will "prefer" the wavelength nearer to 555 nm. We have to take the emitted wavelength for calculation, but I'm not sure how other manufacturers handle this. Well actually as long as we assume a single wavelength we always end up with some approximation.

And to do the calculation, we need the table for the luminosity function (sheet 1931 col observer column C).

Which gives us (linear interpolation, and datasheet values):

• 645 nm: 0.1382, Vf = 2.05 V, If = 20 mA, P = 41 mW, $\Phi_v$ = 2.2 lm
• 634 nm: 0.2266, Vf = 2.05 V, If = 20 mA, P = 41 mW, $\Phi_v$ = 2.7 lm
• 624 nm: 0.3330, Vf = 2.15 V, If = 20 mA, P = 43 mW, $\Phi_v$ = 3.5 lm
• 610 nm: 0.5030, Vf = 2.15 V, If = 20 mA, P = 43 mW, $\Phi_v$ = 3.7 lm
• 597 nm: 0.6693, Vf = 2.20 V, If = 20 mA, P = 44 mW, $\Phi_v$ = 2.7 lm

So we can calculate:

• 645 nm: $\Phi_e = \frac{2.2~\text{lm}}{683~\frac{\text{lm}}{\text{W}} 0.1382}$ = 23.3 mW
• 634 nm: $\Phi_e$ = 17.4 mW
• 624 nm: $\Phi_e$ = 15.4 mW
• 610 nm: $\Phi_e$ = 10.8 mW
• 597 nm: $\Phi_e$ = 5.9 mW

Which turns into these efficiencies:

• 645 nm: $\eta = \frac{\Phi_e}{P_{el}} = \frac{23.3~\text{mW}}{41~\text{mW}}$ = 56.8 %
• 634 nm: $\eta$ = 42.4 %
• 624 nm: $\eta$ = 35.8 %
• 610 nm: $\eta$ = 25.1 %
• 597 nm: $\eta$ = 13.4 %

So as the wavelength decreases the efficiency decreases, at least for this family of LED.

But what if my datasheet only supplies a value for luminous intesity which is measured in candela?

So we need a way to come from luminous intensity to luminous flux. Well for that we need something called the solid angle, which is measured in steradians. And a lumen is a candela multiplied by a steradian. A sphere has $4\pi$ steradians, so a light source emitting a candela uniformly in all directions would have 12.6 lumens. The solid angle is dimensionless like radian, but sr is often added to make clear why a candela suddenly turned into a lumen: $1~\text{lm} = 1~\text{cd} * 1~\text{sr}$

Now a LED has never a uniformly light distribution, at most it could cover a half sphere. Usually though it will have a smaller viewing angle. We take this as the apex angle of a cone. There is a neat little formula to calculate the solid angle of a cone with a given apex angle:

$$\Omega = 2 \pi (1 - \cos \theta)$$

Where: $\Omega$ is the solid angle and $\theta$ is half the apex angle of the cone.

With the Osram LED as before we get a range of candela in the datasheet and the apex angle of 120°:

• 645 nm: 0.355 ... 1.120 cd, mean: 0.7375 cd
• 634 nm: 0.450 ... 1.400 cd, mean: 0.9250 cd
• 624 nm: 0.560 ... 1.800 cd, mean: 1.1800 cd
• 610 nm: 0.710 ... 1.800 cd, mean: 1.2550 cd
• 597 nm: 0.450 ... 1.400 cd, mean: 0.9250 cd

So we first calculate our solid angle: $\Omega = 2 \pi (1 - \cos 60°) = \pi ~\text{sr}$ - that was easy for once.

And with that we get:

• 645 nm: $\Phi_v = 0.7375~\text{cd} * \pi~\text{sr} = 2.32~\text{lm}$
• 634 nm: $\Phi_v$ = 2.9 lm
• 624 nm: $\Phi_v$ = 3.7 lm
• 610 nm: $\Phi_v$ = 3.9 lm
• 597 nm: $\Phi_v$ = 2.9 lm

Which is more than what is given as typical value by Osram, but it's not very far off. I mean, the range given for the luminous intensity is huge and just taking the mean without knowing how their production varies is just a guess. Maybe keep this result in mind and take a bit less than the mean in future.

Actually, after thinking a bit more about it, I think the overestimation stems from not including the actual brightness per angle, as that will decrease from 0° to 60°.

With the lumen values, you can go and use the approach above again.

Now the above ways are an approximation based on simplifying:

• the wavelength distribution of the LED to a single wavelength
• the radiation characteristic to a simple cone

To do things right™ you will have to integrate over the actual distribution and radiation characteristic to get better values. Once you have mastered that, you can apply the same principle to any kind of light source. So you can go and do the same calculation for a white LED and see what happens there.

I'm actually inclined to do that. But it would probably explode this answer.

• Wow! This is a very thorough and valuable answer! Thank you for taking the time to work things through in each of the different scenarios. I have a hunch I will be referring back to this answer regularly. – uhoh Sep 2 '17 at 3:36
• fyi this could still use a more quantitative answer: How to make a green LED as visually bright as a 0 magnitude star? – uhoh Sep 2 '17 at 3:37