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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 14 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
14.28 
 For AsGaAl xx 1 system, a direct bandgap for 
 45.00  x , we have 
 xEg 247.1424.1  
 At 45.0x , 985.1gE eV, so for the direct 
 bandgap 
 985.1424.1  gE eV 
 which yields 
  871.0625.0  m 
_______________________________________ 
 
14.29 
(a) From Figure 14.24, 64.1gE eV 
  756.0
64.1
24.124.1

gE
m 
(b) From Figure 14.24, 78.1gE eV 
  697.0
78.1
24.124.1

gE
m 
_______________________________________ 
 
14.30 
 85.1
670.0
24.124.1


gE eV 
 From Figure 14.23, 35.0x 
_______________________________________ 
 
14.31 
 85.1
670.0
24.124.1


gE eV 
 From Figure 14.24, 38.0x 
_______________________________________ 
 
14.32 
 (a) For GaAs, 66.32 n and for air, 0.11 n . 
 The critical angle is 
 













  86.15
66.3
1
sinsin 1
2
11
n
n
C 
 The fraction of photons that will not 
 experience total internal reflection is 
 
 
%81.8
360
86.152
360
2
C
 
 (b)Fresnel loss: 
 3258.0
166.3
166.3
22
12
12 


















nn
nn
R 
 The fraction of photons emitted is then 
    %94.50594.03258.010881.0  
_______________________________________ 
 
14.33 
 We can write the external quantum efficiency 
 as 
 21TText  
 where 11 1 RT  and where 1R is the 
 reflection coefficient (Fresnel loss), and the 
 factor 2T is the fraction of photons that do not 
 experience total internal reflection. We have 
 
2
12
12
1 










nn
nn
R 
 so that 
 
2
12
12
11 11 










nn
nn
RT 
 which reduces to 
 
 2
21
21
1
4
nn
nn
T

 
 Now consider the solid angle from the source 
 point. The surface area described by the solid 
 angle is 2p . The factor 1T is given by 
 
2
2
1
4 R
p
T


 
 From the geometry, we have 
 











2
sin2
2
2
sin CC Rp
R
p 
 
 Then the area is 
 






2
sin4 222 CRpA

 
 Now 
 






2
sin
4
2
2
2
1
C
R
p
T



 
 From a trig identity, we have 
  C
C 

cos1
2
1
2
sin 2 





 
 Then 
  CT cos1
2
1
1  
 The external quantum efficiency is now 
 
 
 Cext
nn
nn
TT  cos1
2
14
2
21
21
21 

 
 or 
 
 
 Cext
nn
nn
 cos1
2
2
21
21 

 
_______________________________________