Question about LEDs (less question, more like whether what I've written is correct)

Mr.Meerkat · March 19, 2017

So, my physics assignment is to investigate brightness of a LED against current supplied to it.

I have to write an underlying physics section and well is this correct or have I just written a load of garbage?

Spoiler

There are conductors, semiconductors and insulators. Conductors and insulators do exactly what their name suggests, allow and prevent electricity from flowing. Semiconductors on the other hand are a bit different as it only allows electricity flow once certain conditions has been met. In an conductor, there is no gap between the valence band and the conduction band while in a Insulator, there is a large gap between the 2 bands meaning an extremely large amount of energy has to be supplied to the electrons so they can “jump” from the Valence band to the conduction band. However, the energy required is either not possible or it disintegrates the material. A semiconductor on the other hand has a band gap that’s in between the size of the non-existent band gap of a conductor and the very large gap of a insulator. This means that once the electrons has been supplied with enough energy, they will be able to make the “jump” from the Valence band into the Conduction band. The energy can be supplied by both increasing the temperature of the material (which lowers its resistance) or by supplying a voltage higher than the semiconductor’s forward voltage (the voltage required by the semiconductor for the electrons to be able to make the jump).

A LED (light emitting diode) is a PN junction diode. This is created by joining a P-type semiconductor (has an extra hole, a positive charge carrier) and a N-type semiconductor (has an extra electron, a negative charge carrier) together. When they are joined together, a depletion region is created. This means that the charge carriers (i.e. the holes and electrons) in both the P-type and N-type regions are unable to combine. However, when the negative terminal of a cell or power supply is connected to the N-type region of the diode and the positive terminal is connected to the P-type region, the depletion layer becomes smaller and the electrons in the N-type region moves towards the P-type region and the holes in the P-type region moves towards the N-type region. This is known as a forward-baised junction. If the voltage supplied is high enough, it allows for an electron to easily move between the 2 regions where an electron and a hole can combine in the junction which completes an atom and with a LED, the combination emits a photon. The photon is the light we see emitted by LEDs.

Due to P=IV (Power=Current x Voltage), if the voltage stays constant, increasing the current will result in more energy being supplied to the LED which means more holes and electrons are able to combine, emitting more photons. However, E=hf states that more energy should result in higher frequency photons being emitted rather than a larger number of photons being emitted however due to the materials that makes up the LED, only a single colour can be emitted (including RGB LEDs as that’s just a LED with 3 different PN junctions), limiting what frequency the photons can be.

kingkickolas · March 19, 2017

Everything looks really good. The only thing I would change is some stuff in the last paragraph.

34 minutes ago, Mr.Meerkat said:

Due to P=IV (Power=Current x Voltage), if the voltage stays constant, increasing the current will result in more energy being supplied to the LED which means more holes and electrons are able to combine, emitting more photons. However, E=hf states that more energy should result in higher frequency photons being emitted rather than a larger number of photons being emitted however due to the materials that makes up the LED, only a single colour can be emitted (including RGB LEDs as that’s just a LED with 3 different PN junctions), limiting what frequency the photons can be.

Yes increasing the current increases the power to the LED, emitting more photons. But it doesn't change the energy level difference in the atoms that emit the photons, so E=hf is the same regardless of the current (since E is the atomic energy level difference). It's just that increasing the current increases the dq/dt, so more atoms every second are being excited.

So yeah this results in an approximately linear relationship between current and brightness. There is some loss in the material which makes the curve fall off from linear slightly at higher currents (and power). Also, note that not all of the power into the LED (P = IV) is converted to light. Good LEDs typically have an efficiency of about 30% (although there's variation up and down), and the rest of the P turns into heat.

Otherwise, yeah it looks good.