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My book says

... But the threshold voltage are much higher and slightly different for each color. The reverse breakdown voltages of LEDs are very low.

While I understand that the reverse breakdown voltage should be low because of the heavy doping, but I do not understand reason for high threshold voltage.

I know that the band gap should be higher because the wavelength of the light emitted needs to be smaller. So does band gap have any relation with the threshold voltage?

Edit: I found out that barrier potential depends on the negative logarithm of \$n_i\$ and so does the band gap

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  • \$\begingroup\$ Yes, bandgap and forward voltage are related. The exact relationship will have to wait for someone else to answer. \$\endgroup\$ – Hearth Mar 29 '19 at 13:57
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    \$\begingroup\$ Additionally, "slightly different" isn't as true as it used to be, because now blue and violet LEDs are common, which have a much higher Vf (over three volts, in some cases up to four) than common red and green LEDs (which typically have a Vf of two-and-a-bit volts). \$\endgroup\$ – Hearth Mar 29 '19 at 13:58
  • \$\begingroup\$ The bandgap of LEDs is a very interesting subject; standard silicon didoes are group IV materials. LEDs, by contrast, are group III / V materials (two different materials are present and doped) and can be tuned for different colours by adjusting the bandgap. \$\endgroup\$ – Peter Smith Mar 29 '19 at 14:49
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The forward voltage drop of an LED is indeed related to the color of light emitted.

As electrons traverse the N-P junction they lose energy. That energy is lost in the form of light. The more energy lost, the shorter the wavelength of light emitted.

Historically, the development of LEDs has been the pursuit of higher band-gap voltages for shorter wavelengths of light. That is why early LEDs were all red or infrared, with yellow, green and finally blue coming later.

BTW White LEDs are actually blue, but also contain a light emitting phosphor layer, similar in function to that in florescent light bulbs, to convert the blue light to white light.

This article shows the relationship between color and energy. Color and Energy Scroll down to the section called "Spectral Colors", there's an excellent table there.

Of course, the other parts of the LED still have resistance, so some more voltage will be needed to get a decent current to flow. That's why it is not uncommon to need more voltage than the minimum.

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    \$\begingroup\$ In RGB why does Blue Green often have the same voltage and most often White (B) is lower than green? and why is GaP Green (lower energy) yet lower voltage than Blue? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 29 '19 at 15:13
  • \$\begingroup\$ What about ideality factor? and bulk resistance? and relative dielectric effect on reducing speed of light increasing wavelength \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 29 '19 at 17:21
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Ideally, each electron that is conducted in an LED emits a photon. Again ideall, the photon has all the energy lost from the electron.

If you plot a table of LED forward voltages vs. the photon energy of their light output in electron-volts, you'll find that the LED forward voltage is pretty much the photon energy of the light output, plus a bit extra to account for the second law of thermodynamics.

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    \$\begingroup\$ In RGB why does Blue Green often have the same voltage and most often White (B) is lower than green? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 29 '19 at 15:11
  • \$\begingroup\$ What about ideality factor? and bulk resistance? and relative dielectric effect on reducing speed of light increasing wavelength. It is not simply λ= hc/E with gap energy , E in eV \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 29 '19 at 17:23
  • \$\begingroup\$ @SunnyskyguyEE75 I was giving a rough answer, that, AFAIK, works pretty well to predict and explain in round numbers what an LED forward voltage is going to be. I welcome your clear, concise explanation that takes all the stuff you mention into account! \$\endgroup\$ – TimWescott Mar 30 '19 at 3:44
  • \$\begingroup\$ I would have but the OP seems to be satisfied with answer accepted \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 30 '19 at 4:05
  • \$\begingroup\$ Using this method gives a large error on Planck's constant. h=E/v \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 30 '19 at 4:16

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