HW_Problem_4_solution - HW_Problems_4_soln 1 = 4.45 1017 cm...

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HW_Problems_4_soln 1. 17 3 4.45 10 C N cm = × 18 3 7.72 10 For GaAs at 300k V N cm = × 17 17 17 17 17 17 10 Boltzmann: ln( ) (0.026)ln 4.45 10 0.039 10 10 J-D: ln( ) (0.026)[ln ] 4.45 10 8(4.45 10 ) 0. Fn C C C C Fn C C C C n E E kT E eV N E eV n E E kT E eV N E = + = + × = = + = + + × × = 037 eV 17 18 10 For holes: ln( ) 0.113 7.72 10 Fp V V E E kT E eV = = + × 2. (1). The drift velocity is 2 15 (1300 / )(100 / ) 1.3 10 / d v E cm vs v cm cm s µ = = = × The excess energy is * 2 31 3 1 1 (0.26 9.1 10 )(1.3 10 / ) 2 2 d E m v kg m s = = × × 25 6 2.0 10 1.25 10 J eV = × = × (2). This energy is relatively small compared to the electron thermal energy: 3 3 at 300k: 39 2 2 th kT E kT meV = = (3). At 5KV/cm, the drift velocity is: 3 6 (1300)(5 10 ) 6.5 10 / d v cm s = × = × The excess energy is
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