ee606_s04_soln9-11 - Solution to Selection Problems...

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Solution to Selection Problems Homework 9 to 11 17.11 (a) J N = J N r = n n E = n n dr If the z -direction points from the surface into the bulk, I D = integraldisplay integraldisplay J N r dzdθ = integraldisplay integraldisplay z c 0 2 πrJ N r dz = ( 2 πr dr )( q integraldisplay z c 0 μ n ndz ) The value of the second parenthesis is just μ n Q N , we have I D = 2 πr μ n Q N dr Integrating over the r -width of the channel, integraldisplay r 2 r 1 I D 2 πr dr = I D 2 π integraldisplay r 2 r 1 dr r = I D 2 π ln( r 2 /r 1 ) = μ n integraldisplay V D 0 Q N and I D = 2 π ln( r 2 /r 1 ) μ n integraldisplay V D 0 Q N Since Q N = C 0 ( V G V T φ ), I D = 2 π ln( r 2 /r 1 ) μ n C 0 [( V G V T ) V D V 2 D / 2] (b) Setting r 2 = r 1 + L , we can write ln( r 2 /r 1 ) = ln( r 1 + L r 2 = ln(1 + L/r 1 ) 1
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If L/r 1 1, ln(1 + L/r 1 ) = ( L/r 1 ) 1 2 ( L/r 1 ) 2 + · · · = L/r 1 Thus 2 π ln( r 2 /r 1 ) 2 πr 1 L = Z L and one obtains the usual I D V D result.
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