It's easy to get 'efficacy' (how well something works) and 'efficiency' (ratio of useful work) confused. As we'll see, knowing this difference helps us make LED efficiency calculations and understand why red LEDs are used in grow lights.
Luminous efficacy of pure blackbody radiation as a function of temperature
As lighting efficiency improves, it is useful to understand the theoretical limits to luminous efficacy for light that we perceive as white. Independent of the efficiency with which photons are generated, there exists a spectrally imposed limit to the luminous efficacy of any source of photons. We find that, depending on the acceptable bandpass and—to a lesser extent—the color temperature of the light, the ideal white light source achieves a spectral luminous efficacy of 250–370 lm/W. Thomas W. Murphy, Maximum spectral luminous efficacy of white light
Key Efficacy Concepts
The following points mean that we can theoretically get about 47% more light for the energy input with 660nm LEDs versus 450nm LEDs. It explains why red LEDs are breaking the 4 µmol/joule barrier, and white LEDs based on blue LEDs with a phosphor never will. Green LEDs are electrically inefficient and is a physics/semiconductor issue (our eyes are most sensitive to green light so we don't notice).
Comparison of various light source technology's luminous efficacy (source)
- 1240/wavelength of light in nanometers = energy of a photon in eV (electron volts).
- 10.37/eV of photon = µmol/joule or the maximum possible PPE (photosynthetic photon efficacy).
- Max possible PPE * LED efficiency = the PPE for the specific LED
- Example: 660nm photon. (1240/660=1.88eV) (10.37/1.88=5.52 µmol/joule). At 100% efficiency, a red 660nm LED would have a PPE of 5.52 µmol/joule.
- Example: 450nm photon. (1240/450=2.76eV) (10.37/2.76=3.76 µmol/joule). At 100% efficiency, a blue 450nm LED would have a PPE of 3.76 µmol/joule.
- Question: what is the electrical efficiency of a 660nm LED with a PPE of 2.8 µmol/joule? (1240/660=1.88eV) (10.37/1.88=5.52 µmol/joule) (2.8 µmol/joule/5.52 µmol/joule=50.7% efficient)
- Question: what is the electrical efficiency of a 450nm LED with a PPE of 2.8 µmol/joule? (1240/450=2.76eV) (10.37/2.76=3.76 µmol/joule) (2.8 µmol/joule/3.76 µmol/joule=74% efficient)