{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ee606_s04_soln9-11

ee606_s04_soln9-11 - Solution to Selection Problems...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solution to Selection Problems Homework 9 to 11 17.11 (a) J N = J N r = qμ n n E = − qμ n n dφ dr If the z-direction points from the surface into the bulk, I D = − integraldisplay integraldisplay J N r dzdθ = − integraldisplay integraldisplay z c 2 πrJ N r dz = ( − 2 πr dφ dr )( − q integraldisplay z c 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 dφ 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 Q N dφ and I D = − 2 π ln( r 2 /r 1 ) μ n integraldisplay V D Q N dφ Since Q N = − C ( V G − V T − φ ), I D = 2 π ln( r 2 /r 1 ) μ n C [( 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 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...
View Full Document

{[ snackBarMessage ]}

Page1 / 5

ee606_s04_soln9-11 - Solution to Selection Problems...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online