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:

  $$\theta_{c} = arcsin\biggl(\frac{n_{1}}{n_{2}}\biggr)$$ (4)

  
  

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

  $$\theta_{c} = arcsin\biggl(\frac{1}{3,66}\biggr) = 15,856 \degree$$ (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:

  $$\frac{\theta_{c}}{360} = \frac{15,856}{360} = 4,4045 \%$$ (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:

  $$\Gamma = \biggl(\frac{n_{2}-n_{1}}{n_{2}+n_{1}}\biggl)^{2}$$ (7)

  
  

  which will be for our case:

  $$\Gamma = \biggl(\frac{3,66-1}{3,66+1}\biggr)^{2} = 0,3258$$ (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:

  $$(1-0,3258).0,08809 = 0,05937 = 5,937 \%$$ (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.