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Unformatted text preview: 3.2 GaAs GaAs has an effective density of states at the conduction CB N c of 4.7 10 17 cm3 and an effective density of states at the VB edge N v of 7 10 18 cm3 . Given its bandgap E g of 1.42 eV calculate the intrinsic concentration and the intrinsic resistivity at room temperature (take as 300 K). Where is the Fermi level? Assuming that N c and N v scale as T 3/2 , what would be the intrinsic concentration at 100 C? If this GaAs crystal is doped with 10 18 donors cm3 (such as Te), where is the new Fermi level and what is the resistivity of the sample? The drift mobilities in GaAs are shown in Table 3Q2 Table 3Q2 Dopant impurities scatter carriers and reduce the drift mobility ( e for electrons and h for holes). Dopant concentration (cm3 ) 0 10 15 10 16 10 17 10 18 e (cm 2 V1 s1 ) 8500 8000 7000 4000 2400 h (cm 2 V1 s1 ) 400 380 310 220 160 Solution The intrinsic concentration is n i = N c N v ( ) 1/ 2 exp E g 2 k B T so that n i = 4.7 10 17 cm 3 ( ) 7 10 18 cm 3 ( ) [ ] 1/ 2 exp 1.42 eV 2 8.6174 10 5 eV K 1 ( ) 300 K ( ) = = 2 . 223 10 12 m 3 = 2 . 223 10 6 cm 3 The conductivity is = en i e + h ( ) = 1.608 10 19 C ( ) 2.223 10 6 cm 3 ( ) 8500 cm 2 V 1 s 1 + 400 cm 2 V 1 s 1 ( ) = 3 . 17 10 9  1 cm 1 The resistivity is = 1 = 1 3.17 10 9  1 cm 1 ( ) = 3 . 16 10 8 cm The energy distance between the bottom of the conduction band E c and Fermi level of the intrinsic GaAs E Fi is E c E Fi = k B T ln N c n i so that E c E Fi = 8.617 10 5 eVK 1 ( ) 300 K ( ) ln 4.7 10 17 cm 3 ( ) 2.223 10 6 cm 3 ( ) = 0.675 eV . If N c and N v scale with temperature as T 3/2 then the intrinsic concentration at a given temperature T is given by the expression n i ( T ) = N c (300 K) N v (300 K) [ ] 1/ 2 T 300 K 3 /2 exp E g 2 k B T , so at 100 o C (373 K) we will have n i (373 K) = 4.7 10 17 cm 3 ( ) 7 10 18 cm 3 ( ) [ ] 1 /2 373 K 300 K 3/ 2 exp 1.42 eV 2 8.6174 10 5 eVK 1 ( ) 373 K ( ) = 6.61 10 8 cm3 . For the case when the same GaAs crystal is doped with 10 18 cm3 donors, one might be tempted to think that n = N d and to apply the usual expression for the calculation of the Fermi level position: E c E Fn = k B T ln N c n = k B T ln N c N d The result from such calculation will be negative (since N d &gt; N c , degenerate semiconductor), indicating that the Fermi level will be somewhere in the conduction band. Actually, the upper expression is derived on the basis of the assumption that the Fermi level is several kT below the bottom of the conduction band, which allows us to replace the FermiDirac statistics with the more simple Boltzmann statistics, so it is not applicable to degenerate semiconductors....
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 Spring '09
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