Electronic – Is the LED drop voltage difference between colors linked to the different wavelength energy

led

To me appears that LEDs that emitt light with less energy (e.g. IR and red) have less voltage forward drop than the ones with more energy associated to their wavelength (such as blue or UV).

That would be fascinating.

Is this a true correlation or is it dependent solely on the technology available?

Best Answer

The energy level of photons is not the reason Vf rises with the energy level of the photons.

Why? Because that does not always happen.

Here is the 100 µmol energy level for four wavelengths of InGaN LEDs and their Vf.

Notice how as the Vf rises, the energy decreases.

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Source Vf: Lumiled Rebel Color Datasheet
Source Energy: How do I convert irradiance into photon flux?
and Photometric, Radiometrtic, Quantum Conversions


A photon cannot be measured with a volt meter.
The photon and the energy it carries has been emitted from the LED.
So how could a photon's energy possibly be included in the Vf when it is off traveling at the speed of light away from the LED?


Photon energy does not directly contribute to Vf.
The instantaneous resistivity of materials used are what determines Vf


More Energy = Less Photons

This question is based on the fact that a longer wavelength photon carries less energy than a shorter wavelength photon.
A 660 nm deep red photon carries 66% as much energy as a deep blue photon.

But that is only part of the equation.

3.76 µmols of 450 nm deep blue photons will carry 1 watt of energy.
5.52 µmols of 660 nm deep red photons will carry 1 watt of energy.

That's 56% more red photons than blue per watt.

It takes one electron to create 1 photon.
1 µmol = 602,214,076,000,000,000

So it's kind of a wash.
While blue carries more energy, less blue photons are generated per watt.
While red carries less energy, more red photons are generated per watt.
Source: Photometric, Radiometrtic, Quantum Conversions

Regarding the claim

a certain voltage is required for the electrons to get them across the depletion region. The electron releases its energy as a photon.
...the bandgap of the material gives the characteristic wavelength. Higher bandgaps give shorter wavelengths.

While the energy in the bandgap approximates the released optical energy,
the bandgap energy is not represented in Vf

The bandgap energy approximates the released optical energy only if the LED's thermal characteristics are overlooked.
Source: Light Emitting Diodes by E. Fred Schubert

If you were to go to Digikey and sort (ascending) white LEDs by Vf
You will find in the adjacent column, the efficacy (lm/W), the LEDs with very high efficacy. Then if you sort by efficacy (ascending) you will find higher Vf.

With more electrons being converted to photons (higher efficacy) there are less electrons that make it through the bandgap to the conduction band. The electrons in the conduction band will add to the Vf whereas those converted to photons are not included in the Vf.

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