Nowadays we see LEDs with 300 lumen per watt luminous efficacy. They still waste a lot of power as heat. Now what is the limit of electricity to light conversion? Say we build an LED with 100% efficiency, which doesn't waste any power as heat or sound or UV/infrared. All its spectrum is in visible range. What will be its lumen/watt? What is the theoretical roof we would hit in distant future?


3 Answers 3


Nowadays we see LEDs with 300 lumen per watt luminous efficacy.

Maybe in the laboratory. Right now commercially available white LEDs are about 200 lm/w. 400 lm/w is the practical upper theoretical limit. US DOE estimated over 250 lm/w by 2025.

us doe led efficacy roadmap

Commercial Blue LEDs have up to about 80% efficacy (Cree XP-3G and OSRAM Oslon SSL).
White LEDs use blue LEDs covered with a yellow phosphor.
Theoretically phosphors could produce better than 100% efficacy.

If the cost of manufacturing a UV LED dropped dramatically, efficacy could skyrocket. LEDs emitting near 400 nm with remarkably high efficiencies have been reported (Morita et al., 2004)

What is the theoretical roof we would hit in distant future?

1 photon per electron is the ideal quantum efficiency of unity.

Internal, extraction, external, and power are the four basic LED efficiencies.

Internal: number of photons emitted from the active region per electron crossing the band gap. Dependent on materials and temperature

Extraction: Photons emitted by the active region that escape from the LED die. If all photons emitted by the active region are also emitted into free space, the extraction efficiency is unity.

External: the ratio of the number of useful light particles emitted to the number of electrons crossing the band gap.

Power efficiency is the "wall plug" efficiency. The number of photons emitted per watt used.

If the technology reached the theoretical maximum of one photon per electron The theoretical lm/w efficacy is based solely on the spectral output.

There are three common units of measure for LED flux. Luminous (lumens), radiometric (watts), and quantum (number of photons, SI unit = Moles).

Visible spectrum is 380 nm (770 THz) to 770 nm (390 THz)
A 380 nm violet photon carries the most radiometric power.
You will get most lumens from a watt of 555 nm (540 THz) green photons.

602,214,085,700,000,000 electrons in 1 watt.
624,200,000,000,000,000 photons in 1 µMole.
which gives 1.036508469 µmol/w

Converting Photon Energy to Photon Flux

A photon has an energy quanta Ep which is defined by:

Ep= h•f 

where h = Plancks constant 6.63 x 10^-34, and

f = Frequency = c/λ
where c = speed of light = 299,800,000 m/s and <br>
λ = wavelength in meters<br>


Ep = h•(c/λ)

The number of photons, Np, can be calculated by

Np = E/Ep
   = E•((λ•10^-9)/h•c)     
   = E [W/m2]•λ•10^-9[m]•/ (1.988•10^-25) 
   = E•λ•5.03•10^15 [1/(m²•s)]   (with Irradiance[W/m²])  

NOTE: above λ is in nm

The photon flux can be determined by:

Ef = Np/NA 

where NA = Avogadro number 6,022•10^23 / mol)

Together gives the equation for converting irradiance [W/m2] to quantum flux [µE]:

Ef = Np/NA = (E•λ•5.03•1015[1/(m2•s)])/(6.02•1017[1/µmol])
   = E•λ•0.836•10-2 [µmol/(m2•s)]

1 µMole of 555 nm photons = 0.2234 watts
Photopic Luminous Efficacy @ 555 nm = 147.2

147.2 x 1.036508469 µmol/w = 152.574 lm
152.574 lm / 0.2234 w = 683 lm/w


Light Emitting Diodes, 2nd Edition, E. Fred Schubert

candela (cd): The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540 THz hertz (555 nm) and that has a radiant intensity in that direction of 1/683 watt per steradian.
source The International System of Measuring Units

candela to lumen calculator
Photopic Luminous Efficacy, Relative Sensitivity Curve for the C.I.E. Standard Observer Irradiance Conversion Factors


It depends on what spectrum of light you want. Our eyes are most sensitive to green light at 555 nm, so a perfectly efficient light source that only emitted that color would reach 683 lm/W. If you want to include other colors (say, to get white light), you'll need more power for the same apparent brightness. Wikipedia has a great summary at https://en.wikipedia.org/wiki/Luminous_efficacy. Depending on how closely you want to match daylight in appearance, the theoretical maximum efficacy is somewhere between 251 and 348 lm/W.

  • \$\begingroup\$ 1 candela = 1/683 watt per steradian, not lumen. Conversion from candela to lumen requires the apex angle of the steradians. \$\endgroup\$ Commented Mar 7, 2018 at 1:43
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    \$\begingroup\$ @Misunderstood I don't think that's correct. 1 candela is indeed (for 540 THz light) equivalent to 1/683 W/sr, and 1 lm = 1 cd * 1 sr, so a 540 THz source emitting over a solid angle of 1 sr with an intensity of 1 cd would be emitting 1 lm, and radiating (1/683 W/sr) * 1 sr = 1/683 W, for an efficacy of 683 lm/W. \$\endgroup\$ Commented Mar 7, 2018 at 3:43
  • \$\begingroup\$ 1 watt = 683 cd, 1 cd @ 360° = 12.566 lm, 683 cd = 8583 lm. See the candela to lumen calculator at rapidtables.com/calc/light/candela-to-lumen-calculator.html. Use 360° (4 x pi x r) for full spherical surface area. Use 683 cd. \$\endgroup\$ Commented Mar 7, 2018 at 3:52
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    \$\begingroup\$ @Misunderstood But to get 683 cd over a full sphere takes 12.566 watts, so it cancels out. 1 watt only gets you 683 cd over 1 steradian. \$\endgroup\$ Commented Mar 7, 2018 at 6:57
  • \$\begingroup\$ You are correct. I need to find the errors in my math. Okay found it. I was off 1 decimal point for the number of electrons in a coulomb. I multiplied by 12.566 which was converting from lux to lumens when I already had lumens, where I should have divided by by .22 which is the number of watts in a µMole @ 555 nm. That got me 693 lm/w but .22 was rounded. My big mistake was converting lux to flux when I already had flux. Thanks for setting me straight. \$\endgroup\$ Commented Mar 7, 2018 at 22:57

1 cd = 1/683 W/sr = 4π/683 W/sphere = 4π lm ⇒ 1 W = 683 lm then the maximal, theoretic efficiency is 683 lm/W.

D. I.


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