20111ee121B_1_EE121B_HW1_soln

20111ee121B_1_EE121B_HW1_soln - ) 3. a) n =1000 (cm 2...

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EE 121B HW1 Solution (Winter 2011) 1. N D =10 15 cm -3 a) T=0, freeze-out, n=0, p=0, intrinsic b) T=300K, n= N D =10 15 (cm -3 ), p= n i 2 /N D =10 5 (cm -3 ), n-type c) T=1000K, n i (T=1000K) >> n i (T=300K) since n i is a strong function of temperature ) 2 / exp( ) 2 / exp( 2 / 3 kT E T kT E N N n g g V C i = (Note ) 2 / 3 2 / 3 , T N T N V C So, 7 2 / 3 10 15 . 2 )] 026 . 0 * 2 /( 12 . 1 exp[ ))] 3 / 10 ( * 026 . 0 * 2 /( 12 . 1 exp[ ) 3 10 ( ) 300 ( ) 1000 ( x K T n K T n i i = = = = Hence, (cm -3 ) 17 10 15 . 2 ) 1000 ( x K T n i = = Since , the semiconductor is intrinsic. n=p=2.15x10 17 (cm -3 ) D i N K T n >> = ) 1000 ( d) Electron conc. (cm -3 ) 150 0 300 freeze-out intrinsic extrinsic 10 15 T (K) 1000 600 e) Add N A =10 16 cm -3 Since N A >N D, it is p-type after compensation p= N A -N D =9x10 15 (cm -3 ) n= n i 2 /p=1.1x10 4 (cm -3 )
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2. Voltage=1 (V), Si piece length=1 (um), N A =10 15 cm -3 , μ n =1000 (cm 2 /V.s), μ p =500(cm 2 /V.s) Electric field, E=1/10 -4 =10 4 (V/cm) Drift current J= σ E=(nqμ n + pqμ p )E ~ pqμ p E= 10 15 x1.6x10 19 x500x10 4 =800 (A/cm 2
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Unformatted text preview: ) 3. a) n =1000 (cm 2 /V.s), p =500(cm 2 /V.s) D n =(kT/q) n =26 (cm 2 /s) D p =(kT/q) p =13 (cm 2 /s) b) E=10 5 (V/cm), p=10 9 (cm-3 ) n= n i 2 /p=10 11 (cm-3 ) J n (drift)=nq n E=10 11 x1.6x10 19 x1000x10 5 =1.6 (A/cm 2 ) J p (drift)=pq p E=10 9 x1.6x10 19 x500x10 5 =8x10-3 (A/cm 2 ) J(drift)= J n (drift)+ J p (drift)~ 1.6 (A/cm 2 ) Direction: toward the p-doped side c) |J(diffusion)|=|J(drift)| but the opposite direction. So, J(diffusion)= 1.6 (A/cm 2 ) Direction: toward the n-doped side d) |J n (diffusion)|=|J n (drift)| qD n (dn/dx)=1.6, so (dn/dx)=3.8x10 17 (cm-4 ) |J p (diffusion)|=|J p (drift)| qD p (dp/dx)= 8x10-3 , so (dp/dx)=3.8x10 15 (cm-4 )...
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20111ee121B_1_EE121B_HW1_soln - ) 3. a) n =1000 (cm 2...

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