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?

  • 3
    \$\begingroup\$ Yes. It is a true correlation. Note: some LED's may employ phosphors. If so, they may be, for example, UV LED's with phosphors in the lens. The color seen by the observer will be determined by the phosphors. But otherwise, yes, the photon energy and forward voltage are closely related. \$\endgroup\$
    – user57037
    Sep 16, 2018 at 20:39
  • 5
    \$\begingroup\$ Not only is it correct, you can use it to calculate Planck's constant! Get a bunch of LEDs of known wavelengths. Calculate their frequencies using c=fλ. Measure their forward voltages. Calculate the energy by multiplying by elementary charge: E=Vq. Now plot E versus f and the slope will be Planck's constant, h. \$\endgroup\$
    – DrSheldon
    Sep 16, 2018 at 22:30
  • \$\begingroup\$ @mkeith if what you say is true, why is it when an InGaN LED's wavelength shortens from deep blue to blue to cyan to green the energy carried by the photons decreases and Vf increases? \$\endgroup\$ Sep 16, 2018 at 22:53
  • \$\begingroup\$ @DrSheldon You would have to measure the bandgap energy, compensate for thermal characteristics, then you'd have the total amount of energy. You also need the spectral distribution, and the number of photons art each wavelength. See this link and try to calculate backwards from knowing only the total energy. berthold-bio.com/service-support/support-portal/knowledge-base/… -- I used the formulas from that link to create this page and tested with a spectrometer: growlightresearch.com/ppfd/convert.html \$\endgroup\$ Sep 16, 2018 at 23:09
  • 1
    \$\begingroup\$ @Misunderstood, it will take me a long time to digest all you wrote in your answer. But if I look at your graph excerpted from the textbook, it sure looks like the bandgap energy and forward voltage are pretty well correlated, even if some of the AlGaInN points are above the trend line. \$\endgroup\$
    – user57037
    Sep 17, 2018 at 6:01

3 Answers 3


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.

enter image description here

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.


The wavelength range of commercially available LEDs with single-element output power of at least 5 mW is 360 to 950 nm. Each wavelength range is made from a specific semiconductor material family, regardless of the manufacturer. Source: Photonics - Light-Emitting Diodes: A Primer.

The article is worth a read.

enter image description here

Figure 1. The LED color guide from Lumex gives a good overview of the various LED types, chemistry and wavelengths. For some explanation, if required, see LEDs and colour (mine).

Like all diodes (the D of LED), a certain voltage is required for the electrons to get them across the depletion region. The electron releases its energy as a photon. Your hunch is correct and the bandgap of the material gives the characteristic wavelength. Higher bandgaps give shorter wavelengths.

enter image description here

Figure 2. The forward voltage drops vary with current. What is an LED?.

This data for this graph was taken from various datasheets and carefully plotted. The LEDs, however, were from different manufacturers and there is some variation in the forward voltages.

White LEDs, for example, are 450nm deep blue LEDs covered with wavelength converting phosphors. When a deep blue photon is absorbed by the phosphor it is reemitted at a longer wavelength (e.g. blue-cyan-green-red). So the white IV curve will be the same as the deep blue curve within the same product line. I'm still working on this.

  • 1
    \$\begingroup\$ While the following text is true, the energy in the bandgap is not represented in the forward voltage. Vf is a result of the resistivity of the n,p and dopants. THIS IS TRUE, BUT...: Your hunch is correct and the bandgap of the material gives the characteristic wavelength. Higher bandgaps give shorter wavelengths. \$\endgroup\$ Sep 16, 2018 at 22:36

It's linked, with some details that mean you can't draw a straight line through all the points.

The energy needed to create a photon of any particular wavelength sets the absolute minimum Vf that a diode requires when running. In addition to that, there are further small voltage drops dependent on the particular technology, the particular materials that go to make a particular bandgap semiconductor.

IIRC, yellow and green require a very similar voltage, which is probably technology dependent. But generally, red and IR do require less, and blue and UV more, due to photon energy requirement.

  • \$\begingroup\$ Can you add in some detail on what might be included in the technology dependency? As mentioned in my answer, I'm having trouble getting good data for my LED IV curves. There are differences in LEDs from different manufacturers and as a result my yellow curve appears to have higher voltage than the green whereas one might expect it to lie between orange and green. \$\endgroup\$
    – Transistor
    Sep 16, 2018 at 12:37
  • \$\begingroup\$ @Transistor the fact that a diode is 3 junctions, two with metal to semi, and only one with semi to semi, means that the metal to semi junctions are going to have an influence on total forward voltage. I was shooting from the hip there, trying to recall results from way back, but from your results it looks like I was spot on with the yellow/green thing. I did wonder whether to mention argon/potassium, as where the periodic table generally follows atomic weights, except for a few places where it doesn't, but it's not too helpful. \$\endgroup\$
    – Neil_UK
    Sep 16, 2018 at 12:56
  • \$\begingroup\$ @Transistor The photon energy has little to do with Vf. Wire bonding has nothing to do with Vf. The forward voltage is more electron related than photon. With greater efficacy (photons per watt) there are fewer electrons in the bandgap as more electrons have been converted to electrons. Once an electron becomes a photon it's electrical energy can no longer be measured. Less electrons means lower voltage, lower voltage means lower thermal power generated (electrical converted) and therefore higher efficacy. The rest is the width of the bandgap and the energy required to cross it. \$\endgroup\$ Sep 16, 2018 at 14:03
  • \$\begingroup\$ @Neil: I hadn't considered the metal to semi junctions. I don't think it was mentioned in my studies, many decades ago, nor in my reading of the hobby electronics magazines. I'll follow up on that. Thank you. \$\endgroup\$
    – Transistor
    Sep 16, 2018 at 14:17
  • \$\begingroup\$ @Transistor sure the bond and wire will have resistance but it is minimal (mOhms) and it is the same amount of resistance across all colors within the same product line. Just as the resistance of the heterostructures, and bulk resistance of the materials adds to internal serial and parallel resistances but not related to wavelength energy. But heterostructures, and bulk resistance are very much related to Vf. \$\endgroup\$ Sep 16, 2018 at 15:30

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