>>106861963
Letβs start from fundamentals.
1. Define what 1 lumen means.
Luminous efficacy measures how well radiant energy (in watts) is converted into visible light as perceived by the human eye.
The maximum theoretical luminous efficacy for any light source is 683 lumens per watt at 555 nm (green light) β thatβs where the human eye is most sensitive.
2. Warm white light (2700 K)
A βperfectβ LED at 2700 K would not emit all its power at 555 nm. It would emit a full-spectrum (roughly blackbody-like) distribution centered near 2700 K, but with photons only in the visible range (so no wasted IR or UV).
The theoretical limit of luminous efficacy for a perfect 2700 K white light (zero infrared, perfectly shaped to match human eye sensitivity) is around 370 lumens per watt.
This comes from integrating the Planck spectrum weighted by the photopic response curve and then normalizing.
3. Calculate power required for 800 lumens.
P = rac{800 ext{lumens}}{370 ext{lumens per watt}} = 2.16 ext{watts}}
So, under physically perfect conditions β no wasted photons outside the visible range and no losses β youβd need about 2.2 W of radiant power to produce 800 lumens of 2700 K light.
For context:
A real high-quality LED at 2700 K achieves ~120β160 lm/W, so it would need about 5β7 W.
The blackbody limit (emitting full IR) is far worse β about 15 lm/W, so an incandescent bulb needs ~50 W for 800 lm.
In summary:
Perfect 2700 K LED β2.2 W minimum possible radiant energy for 800 lm.