# I Relationship between the depletion region and the band gap of a diode?

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1. Oct 12, 2017

### Griffin Kowash

Hi,

Lately I've been reading about a lab exercise in which Planck's constant is estimated using LEDs. Every procedure I've encountered states that the energy of an emitted photon (and by extension the potential across the semiconductor's band gap) is equal to the threshold voltage multiplied by the elementary charge.

My understanding is that the voltage applied across a diode serves to counteract the electric field created by the depletion region, allowing current to begin flowing. Considering this, it seems like the threshold voltage is related more to the depletion region than it is to the band gap of the semiconductor. I don't understand the connection there, and so far I haven't had any success trying to research it.

When I asked my physics professor, he told me that electric field of the depletion region depends directly on the semiconductor's band gap, which explains the Ephoton = eVthreshold equation. Unfortunately he didn't have time to say more, and I won't get a chance to talk with him for another week or so.

Does anyone have any insight to offer about why this equivalence is true?

Thanks,
Griffin

Here are a couple examples of the sort of lab exercise I'm referring to:
http://www.phys.uconn.edu/~hamilton/phys258/N/led.pdf
http://laser.physics.sunysb.edu/~kegan/Poster/poster.pdf

2. Oct 13, 2017

### Lord Jestocost

There are some misconceptions regarding this lab experiment. I have checked this paper, [PDF]99. Measuring Planck's constant by means of LED's, and the references given therein. I think it provides a good description of the relations.

Ephoton = eVthreshold isn't true!

3. Nov 15, 2017

Whenever light is emitted from an electron, there has to be a change in the electron's energy. In general, the electron must go from a state with higher energy to lower energy. An electric field allows an electron to gain energy as it traverses the field, or loose it if the electron is moving against that field. In general, acceleration of an electron when strong enough will result in the emission of radiation. Richard feynman, on the lectures on physics, discusses the relationship between acceleration and radiation.

Inside a solid state semiconductor, electrons have a range of energies ... and small increments or decrements in speed represent insufficient energy to emit light of visible range. Red light, at 700nm, has about 1.65eV of energy in it. So, in order for an electron to emit red light it must "suddenly" change it's momentum so that a difference in energy of about 1.65 (or more) electron volts is available to give to the photon. In practice, electrons on an e-k (approximately parabolic curve) will gain and loose energy *very* often due to thermal vibration of the atoms around them resulting in electrons rapidly changing their energy states by a few milli-electron volts very rapidly.

Because the electrons "bump" into each other so often because there is a high density, it's difficult for there to be sudden accelerations that significantly change a single electron's total energy. Even if an electron scatters off an atom or other object, it's energy is usually conserved but it's direction changes. The critical issue is that in order for electrons to emit radiation, they must not only change direction but also loose energy.

That's why an electron typically only emits visible light when it is abruptly "stopped" (decelerated) by being captured by an atom.
The band gap, in a semiconductor, is one possible way for an electron to decelerate; it's actually a measure of the amount of decelerating energy required for an atom (or impurity) to capture a moving electron and bind it into an orbital around the atom. Because electrons can not have any energy level found inside the band gap (or forbidden gap) any electron which crosses the bandgap must have it's energy increase or decrease by at least the band-gap energy. eg: it must abruptly change from an energy above the gap, to one below the gap. That's why it radiates, for it is not just changing direction but ALSO slowing down.

For example:
In silicon that band gap is about 1.1(7)eV ... so that the minimum color of light that silicon electrons ought to emit is around 1050nm.
The silicon atoms (which vibrate) can emit much longer wavelength radiation; but will be a wide frequency range where nearly no light is emitted below 1050nm, before IR from the silicon atoms can be detected.

4. Nov 15, 2017

When a diode is in thermal equilibrium, and no bias voltage is applied to it; the statistics of recombination makes a voltage appear across the diode. This voltage, in theory, should be directly proportional to the bandgap voltage of the semiconductor involved (except when heterojunctions, schottkey, etc. are used.) In the links you posted, it's the equation for P(E) in the second link which is very important. Note: The standard diode equation above the P(E) equation is based on an approximation, called the boltzman approximation to the fermi-dirac equation and not applicable to your question; BUT: If you look at the equation for P(E), you'll see that it's based on a voltage known as the "Fermi" Energy (or potential); This is an actual measure of energy in compatible units as the voltage from a battery (volts). 1 electron Volt, is the energy of an electron acclerated through a 1 Volt potential (from a battery). That's why voltage is known as "potential" energy, because it only becomes energy when it acts on an electron. The important thing to grasp is that the actual BATTERY voltage applied to the diode in the equation P(E) is SUBTRACTED from the fermi-energy, before being plugged into the exponential. Looking at P(E), you can see that when the battery voltage applied is just equal to the fermi energy, the exponent equals exactly one.
exp(0) = 1.

If a person knows the doping of a semiconductor, and other factors, you can compute how much current would flow if the voltage applied was exactly equal to the fermi energy (voltage). It just so happens, that the fermi level is almost exactly halfway through the bandgap in a semiconductor. (It exact at absolute zero temperature but is close at all temperatures.) So, that fermi voltage (or built in voltage of a diode) must increase directly in proportion to the bandgap of the semiconductor. It's an estimate, but it's usually pretty close.

With that in mind, realize that the first paper is basically relying on an an unexplained estimate of what amount of current in a silicon diode would flow if the fermi-voltage (which is dependent on the bandgap voltage) were to be neutralized. The estimates of planks constant, then, would only be as accurate as the estimate of current that the teacher gives for neutralizing the bandgap voltage in silicon. I think since the diode function is exponential, the error a student will get will be proportional to the logarithm of the error for the current that the teacher tells the students to use.

There is probably a bit better way of making the estimate of planks constant.
Qualitatively; For example, if you were to find two totally different diodes that had exactly the same brightness for a given amount of current, one diode being red and the other being blue, the difference in light energy would be directly proportional to the difference in energy being supplied by the batteries; eg: If you put 0.01A through a blue CREE silicon carbide you will find that the voltage across the blue diode is much higher than if you put the same amount of current through a RED silicon LED.

The reason is simple, the energy required to produce blue light is higher than that required to produce red. The voltage drop across the diode must increase to supply more Energy to generate the blue light. In electronics, Power (energy/second) = volts * current; so, since I specified that the current was the same and the brightness of diodes is the same; then then the voltage tells you the relative energy difference of the two colors of light.

Last edited: Nov 15, 2017