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# Problem 2:

Consider a pn junction GaAs LED. Assume that photons are generated uniformly in all directions in a plane perpendicular to the junction.

Part A: Taking into account total internal reflection calculate the fraction of photons that have the potential of being emitted from the semiconductor.
Part B: Using the results of part (a) and including Fresnel loss, determine the fraction of generated photons that will be emitted from the semiconductor into air (neglect absorption losses).

• Part A

For the following problem we can base on Snell's law for the critical angle, before we have total internal reflection, which is:

 (4)

Thus as we have been given the refractive indexes or GaAs - 3 and for air - 1 we have:

 (5)

Having the critical angle we can thus relate it to the total circle of 360 , which will give us the fraction of photons having the potential of being emitted:

 (6)

However we should take the other side of the plane possibility, thus we should simply multiply the result x2. Therefore the end potential of photons being emitted is 8,809 %.

• Part B

The fraction of generated and emitted photons (escaping from the LED) will depend on the reflections inside the device, thus a very commonly used formula (not only for optics but for electronic transmission lines as well) for the reflection coefficient is defined as:

 (7)

which will be for our case:

 (8)

Now having the fraction of photons that have the potential of being emitted and the reflection coefficient, then the fraction of photons emitted will simply be:

 (9)

A short comment on the results, it seems that probably a proper geometry and configuration of the pn junction is crucial for the delivered efficiency from the device. Getting a reflection loss of roughly 3% while the photons with potential of being emitted is 8 % speaks by itself that due to the difference in refraction indexes of the materials used, the device loses a lot of its efficiency.

Next: Problem 3: Up: TFYA39_HW_SET4 Previous: Problem 1:
Deyan Levski 2013-06-21