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It is evident from the above V-I graph that longer the wavelength of the light emitted lower is the turn on voltage. I know that turn on voltage is voltage required to have significant current that can give a detectable brightness.And light is emitted in a diode when electron hole recombination occurs and energy of the photon emitted equals the bandgap which is a characteristic of the material used. But why does the turn on voltage is greatest for shorter wavelength.Does that mean there is relation between the bandgap and applied voltage ?

Equations of potential barrier :

$$V=\frac{kT}{e} \ln\left(\frac{N_a N_d}{n_i^2}\right)$$

From this equation, it's evident that as bandgap increases $n_i$ decreases. Resulting in a increase in barrier potential thus necessiting higher external voltage for significant current . Thus there is some relation between turn on voltage and colour of the LED.Is this correct reasoning?

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enter image description here The shorter wavelength the more voltage required because the photon that is needed to interact with the semiconductor has to have more energy. So colors like blue will require more voltage than colors such as red. Different colors use different materials and the materials that produce the shorter wavelengths require more energy to move the electron to a higher valence shell making it require more voltage.

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  • $\begingroup$ I add that a relationship as above holds also for Oled, being the band gap, in a first approximation, the energy gap between the HOMO and LUMO of the molecules. Though the I/V or luminescence/V might be affected by other parameters. $\endgroup$
    – Alchimista
    Commented Mar 7, 2020 at 9:58
  • $\begingroup$ @PM2Ring thanks ill fix that $\endgroup$
    – Noah Cristino
    Commented Mar 9, 2020 at 20:23
  • $\begingroup$ I don't know where you obtained that chart, but there is no such thing as a white LED, and the association with the wavelength 450nm with "white" makes no sense at all. The devices that are sold as "white LEDs" actually are deep blue LEDs in a package that also contains a phosphor that emits broadband, yellowish light when excited by the blue emission from the LED. The combination of the light emitted by the phosphor with blue light from the chip that is not absorbed gives an overall white appearance. $\endgroup$
    – Solomon Slow
    Commented Mar 11, 2020 at 0:12
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According to diode theory, there is no such thing as a turn-on voltage. The current-voltage characteristic of the junction is given by $$I = I_0 (e^{\frac{qV}{kT}}-1),$$ where $kT/q \approx 25$ mV at room temperature. At voltages larger than 100 mV, the unity term can be ignored.

In this equation, $I_0$ contains the dependence on band gap. It also depends on for example the area of the junction.

In practice, LEDs are designed to be used with currents of a few tens of milliampères. In the real $I-V$ characteristics, also the ohmic resistance of the semiconductor material starts to contribute.

The "turn-on voltage" is around the transition from the exponential to the linear voltage dependence, at around 10 mA.

So there is no real relation. The trend is not a coincidence, but it cannot be used as a cheap replacement for the photoemission experiment to determine Planck's constant.

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The voltages you see there are based on what is required to make the material emit light. Its based on the particular material properties that are required to make each color.

And if you look, you can see that yellow is nestled between green and orange, so they aren't in spectrum order.

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