EE 338_HW6_Solutions

EE 338_HW6_Solutions - 10/29/2007 EE 338 Fall 2007 HW#6...

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10/29/2007 EE 338 Fall 2007 HW#6 Solutions 1 (a) Using Eq. 4.42 on pg. 151, we can get the electric field for the given non-uniform doping distribution 0 18 3 00 0 () 1 since the doping distribution is given by ( ) where 10 cm and 0.1 μ m If we substitute ( ) into the given equation for electric field, we get d x d x x dd d d x dN x kT eNx d x Nx Ne N x Nx kT e ε ⎛⎞ =− ⎜⎟ ⎝⎠ == = 0 0 0 0 3 4 1 0.0259V V 2.6 10 cm 0.1 10 cm x x x x x x dN x N kT e kT e N x dx e N x x e = × × (b) We know that diffusion current density is 0 |0 0 | 0 since holes does not contribute to the diffusion current diff n n d i x x diff n n n dn Je D N n dx dN x N J x eD eD e dx x = ± (c) The only other current will rise from the fact that there is a built in electric field due to the non-uniform doping concentration. The drift current is; 0 | 0 0 drift n n x x x nx d n d n kT eN x e N e ex μ με = 1
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10/29/2007 Then the net current density becomes, 00 || 0 0 0 0 Since we didn't apply any external voltage there shouldn't be any net current density. 0
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This note was uploaded on 02/27/2008 for the course EE 338 taught by Professor Dapkus during the Fall '07 term at USC.

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EE 338_HW6_Solutions - 10/29/2007 EE 338 Fall 2007 HW#6...

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