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Light-Emitting Diodes

                                                    Light-Emitting Diodes  115

          To estimate the efficiency of an LED, we can start out with the as-
          sumption that the internal quantum efficiency is unity. That is, that
          for every electron injected there is a photon created. However, the ef-
          ficiency measured in the laboratory is the external efficiency, and this
          depends on the number of photons that manage to exit from the LED
          into free space. This external efficiency is much less than unity.
            In an LED structure having the same composition throughout,
          called a homostructure, a good assumption is that about half the pho-
          tons are emitted with energy above the band gap and are absorbed be-
          fore exiting the LED. The unabsorbed light will be reflected at the in-
          terface between the LED and the air according to Fresnel’s equations.

          Example 6.2
          Calculate the reflection coefficient for light exiting a light-emitting
          diode perpendicular to the surface. Fresnel’s equations give the reflec-
          tion of light from a surface for a given angle of incidence and polariza-
          tion. For the case of light incident perpendicular to the surface, these
          equations take on the same simple form, independent of the polariza-
          tion:
                          (n 1 – n 2 ) 2
                      R =           = 0.32   (n 1 = 1, n 2 = 3.6)    (6.11)
                          (n 1 + n 2 ) 2

            From the example above, about 70% of the light can exit perpendi-
          cular to the surface. However, not all the light that reaches the sur-
          face can exit, because its angle of incidence is different from 90°. For
          light hitting the surface at an angle, Snell’s law comes into play (refer
          to Figure 6.8):
                        n 1 sin(
 1 ) = n 2 sin(
 2 )  (Snell’s Law)
                                               1
                                   n 1
                          sin(
 2 ) =    sin(
 1 ) =    sin(
 1 )
                                              3.6
                                   n 2
            The maximum value for 
 1 is 90°. At this condition,
                                          1
                                sin(
 2 ) =    = 0.28
                                         3.6
                                      
 2 = 16°                      (6.12)
          Only light that intercepts the planar interface between a semiconduc-
          tor and air with an angle less than 16° can be transmitted from the
          semiconductor LED into free space. The percentage transmitted is
          given by Fresnel’s equation. We refer to this light as lying within the
          escape cone of the structure. Any light intercepting the surface at a



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