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Sol HW1 ECE 255 Prof Ye

# Sol HW1 ECE 255 Prof Ye - ECE255Prof.P.Ye SolutionstoHW#1 1...

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ECE 255 Prof. P. Ye Solutions to HW #1 1.-  Prob. 2.7. Using Eq. (2.4) and plugging in the given values for mobility and electric field: Vn = (700 cm 2 /Vs)x(2500V/cm)= 1.75 x 10 6  cm/s   (electrons) Vp=(250 cm 2 /Vs)x(2500 V/cm)=6.25 x 10 5  cm/s   (holes) Using Eq. (2.5) and plugging in the given values for concentration of electrons and holes: Jn=(1.6x10 -19 C)x(10 17 cm -3 )x(700 cm 2 /Vs) x(2500V/cm)= 2.8 x 10 4  A/cm 2    (electrons) Jp=(1.6x10 -19 C)x(10 3 cm -3 )x(250 cm 2 /Vs) x(2500V/cm)= 1 x 10 -10  A/cm 2    (holes) 2.- Prob. 2.14 According to Table 2.1  ρ   <  10 -3  (to be a conductor). Therefore   σ  > 10 3  (since  σ  = 1/ ρ ). From Eq. (2.7): σ  = q(n μ n  + p μ p )   but since it is intrinsic silicon, then p=n=ni.  Therefore: σ  = qni( μ n  +  μ p )   but since  σ  > 10 3  then qni( μ n  +  μ p )   > 10 3

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Sol HW1 ECE 255 Prof Ye - ECE255Prof.P.Ye SolutionstoHW#1 1...

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