I chose an LED for my application but I need optical power, in mW.

In the data sheet, luminous intensity is shown in mcd but I did not find formulas for calculation although there are graphs, etc.

Where luminous intensity is 180 mcd and viewing angle 120°, etc.

Please provide any necessary formulae for my case.

I could not find the exact formula on this website.

  • 1
    \$\begingroup\$ This app note provides an overview. thorlabs.de/catalogPages/506.pdf For your monochromatic case it's not too complex. For ballpark you can assume an optical efficiency of ~20% and calculate from the electrical power dissipated. \$\endgroup\$
    – Mark
    Commented Apr 24, 2019 at 22:12
  • \$\begingroup\$ How much power do you need 5mW? 50mW? \$\endgroup\$ Commented Apr 25, 2019 at 1:07
  • \$\begingroup\$ @Mark, every "source" that claim to calculate emitted optical power (in units of power) from luminosity starts with certain "efficacy" based on different type of sources. Your assumption about 20% is totally arbitrary, I have a UV LED that specifies 54% efficiency. The question is clearly articulated in the title. I believe the luminous intensity can't be translated into "energetic intensity" (aka "radiant intensity") without knowing spectral density of the source, but I am waiting for someone more knowledgeable to provide good concise answer. \$\endgroup\$ Commented Apr 25, 2019 at 2:08
  • \$\begingroup\$ This question has the same unknown, and therefore the same answer applies: Optical power cannot be computed when the beam dispersion angle is unknown. beam dispersion angle =/= viewing angle (definition wise) \$\endgroup\$
    – Huisman
    Commented Apr 25, 2019 at 6:28
  • \$\begingroup\$ I need a SMD LED with optical power of 5-6mW . I noticed that some of you mention an efficiency of 20% (Mark) Thank you for your link . From data sheet we can find : Power dissipation - max. 70mW ; Luminous intensity 180mcd \$\endgroup\$
    – Patrick19
    Commented Apr 25, 2019 at 16:48

1 Answer 1


Okay, let me try to solve this problem.

Parameters: peak wavelength 527 nm, Luminous Intensity 0.18 cd at 5 mA and 2.9 V, emission angle 120 °.

From Wikipedia article "Candela" :

The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540×10^12 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian.

So 1cd = 1/683 W/sr = 1.464 mW per steradian.

Luminous intensity measures (lm and cd) are linked to human eye sensitivity curve, which complicates re-calculations into optical power. The maximum in light frequency in the above definition corresponds to wavelength of 555 nm, which is pretty close to the green LED emission, so we can ignore the difference along the "Photopic Spectral Luminous Efficiency Curve".

The LED emits into 120°, but if we want to calculate total emission, we need to integrate over all angles, so the "effective" (ballpark estimate) angle for "flat emission" will be about +- 50°. One steradian is +-33°, so the led illuminates an area of about (50/33)^2 = ~2.3 sr. The narrower is viewing angle, the brighter is LED appearance, but the emitted power is obviously the same.

Therefore, the emitted power is 1.464 mW/sr * 0.18 cd = 0.263 mW/sr, times 2.3 sr, or about 0.6 mW total.

The consumed power is 5 mA * 2.9 V = 14.5 mW, and therefore "luminous efficacy" of this LED is about 4%.

Corrections and comments are welcome.

POINT TO TAKE HOME: If you need a 6 mW radiated green LED, you should search for a 1000 mcd rated LED with viewing angle of around 30°.

  • \$\begingroup\$ You make the assumption here that the viewing angle (which is twice the angle where the brightness is 50%) is OP's emission angle, which not necessarily has to be the case. \$\endgroup\$
    – Huisman
    Commented Apr 25, 2019 at 6:27
  • \$\begingroup\$ Next, you didn't take the spatial distribution into account, which reduces the average brightness. Or is this how you estimated the "effective" angle of 50° ? \$\endgroup\$
    – Huisman
    Commented Apr 25, 2019 at 6:38
  • \$\begingroup\$ @Huisman, yes, my estimate of the 50 deg (instead of 60 from specs) is meant to account for the spatial non-uniformity. \$\endgroup\$ Commented Apr 25, 2019 at 14:13
  • \$\begingroup\$ Dear Ale..chenski : Shall I understand that optical power that it result from data sheet is from your calculation =0.6 mW . You mention consumed power 14,5mW- So what that 70mW - power dissipation means , from data sheet of SMD LED ? Thanks. I do not know to make optical calculations, sorry for question-is for better understanding.Please detail. \$\endgroup\$
    – Patrick19
    Commented Apr 25, 2019 at 17:01
  • \$\begingroup\$ Dear Huisman and anybody who read this : Is the same situation with LED SMD and LED related viewing angle and different brightness along it ? Is the same situation like in the link : lamphq.com/beam-angle \$\endgroup\$
    – Patrick19
    Commented Apr 25, 2019 at 17:08

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