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Unformatted text preview: ) 3. a) μ n =1000 (cm 2 /V.s), μ p =500(cm 2 /V.s) D n =(kT/q) μ n =26 (cm 2 /s) D p =(kT/q) μ 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)=nqμ n E=10 11 x1.6x10 19 x1000x10 5 =1.6 (A/cm 2 ) J p (drift)=pqμ 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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This note was uploaded on 03/28/2011 for the course EE 121b taught by Professor Bjt-gamma during the Winter '08 term at UCLA.
- Winter '08