This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Solution to Selection Problems Homework 9 to 11 17.11 (a) J N = J N r = q n n E = q n n d dr If the z-direction points from the surface into the bulk, I D = integraldisplay integraldisplay J N r dzd = integraldisplay integraldisplay z c 2 rJ N r dz = ( 2 r d dr )( q integraldisplay z c n ndz ) The value of the second parenthesis is just n Q N , we have I D = 2 r n Q N d dr Integrating over the r-width of the channel, integraldisplay r 2 r 1 I D 2 r dr = I D 2 integraldisplay r 2 r 1 dr r = I D 2 ln( r 2 /r 1 ) = n integraldisplay V D Q N d and I D = 2 ln( r 2 /r 1 ) n integraldisplay V D Q N d Since Q N = C ( V G V T ), I D = 2 ln( r 2 /r 1 ) 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 ln( r 2 /r 1 ) 2 r...
View Full Document
This note was uploaded on 12/12/2010 for the course ECE 606 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.
- Spring '08