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I have a quantum formula describing what kind of photon should be emitted by an LED depending on its voltage. Of course the colour is depending on the material, but every type of LED also needs its specific voltage.

My formula uses 2.5V as an example and tells me that an LED working with 2.5V should emit photons with a wavelength of approximately 470nm, which is blue.

\begin{align} \lambda &= \frac cf = \frac{ch}{E} = \frac{ch}{eU} \\ &= \frac{\rm 3\times10^8 \,\frac ms \cdot 6.626\times10^{-34}\,J\,s} {\rm1.602\times10^{-19}\,C \cdot 2.5\,V} \approx \rm 469\,nm \end{align}

But in reality, blue LEDs need about 3.0V - 3.5V while 2.5V is enough for green LEDs!

Why does the equation not fit the reality and where goes my additional energy of about 0.5eV per photon? Is it converted to thermal energy (why and how?) or what happens with it?

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    $\begingroup$ I suspect you are seeing the usual forward voltage drop of a p-n junction: maybe at zero current your calculation would be valid but with sufficient current to see the light, the voltage needs to be higher. See for example figure 4 in this paper - there is an additional forward voltage that is a function of current. $\endgroup$
    – Floris
    Commented Apr 20, 2015 at 13:52
  • $\begingroup$ Let me record my thoughts from chat here: I don't think this is on topic because it's a question about the internal workings of an electronic device (an LED). We handle some questions about simple circuits - resistors, capacitors, and inductors - to the extent that the questions are about the physics of electric current flow, but not diodes. (Though considering we have a tag for them, maybe it's not so clear....) I think the first part of this, about the equation, might be on topic if it can be divorced from LEDs. $\endgroup$
    – David Z
    Commented Apr 20, 2015 at 14:05
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    $\begingroup$ @DavidZ - that is a very strange cutoff. The physics of diodes is a big topic - and so is the energy of electrons in a solid state device (heck, "solid state physics" is a whole field in itself). I cannot understand that you consider this off topic. "How much energy does an electron need in order to generate photon emission in a solid state junction, and why is it more than just the energy needed to create the photon" is very much physics. $\endgroup$
    – Floris
    Commented Apr 20, 2015 at 14:09
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    $\begingroup$ I've just received a lovely freshman physics project using exactly this physics (LED cutoff voltage vs. emitted wavelength) to measure the Planck constant. This is solidly a physics question even though it has engineering applications. It's on-topic and shouldn't have been closed. $\endgroup$
    – rob ♦
    Commented Apr 20, 2015 at 14:30
  • $\begingroup$ @Floris if this were a solid state physics question, I'd expect to see some mention of the materials involved, as solid state physics is all about the properties of materials, isn't it? My view is that this falls under the category of how some device works, and those kinds of questions I consider off topic in general (except for advanced physics experiments). Also, I was (perhaps mis)remembering this meta post. $\endgroup$
    – David Z
    Commented Apr 20, 2015 at 17:55

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What your formula is actually meant to give is the emission wavelength as a function of the material bandgap:

$$\lambda \approx \frac{\rm1240\,eV\,nm}{E_{gap}}.$$

In your example, when carriers recombine in a semiconductor with a bandgap of 2.5 eV, photons having a wavelength of 496 nm (blue-green) will be emitted.

The forward drop of an LED whose active-region material has a bandgap of 2.5 eV is usually significantly higher than 2.5 V. This means that the energy given to any electron injected in the active region is higher than the energy of the photon that you get in exchange. As a result you are heating up the device.

This happens when there is a significant series resistance. Only with ideal contacts, and p- and n-regions with zero resistivity, the forward drop of an LED can be equal to $E_{gap}/e$, so that no heating is caused.

By driving the LED with very small currents, when the voltage drop is smaller than $E_{gap}/e$, it is theoretically even possible to cool down the device. It is very difficult to put a real device in this condition: while the photon emission is cooling it down the small current is still heating it up, but it has been experimentally proven.

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