This preview shows pages 1–2. 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: Question 1 Consider an ntype silicon Schottky diode with a barrier height of 0.65V. d) If tunneling occurs when the barrier width is 10nm for an applied voltage of  4.35V, what is the maximum doping density for which tunneling just does not occur? e) What is the corresponding resistivity f) Draw the corresponding energy band diagram The total potential is obtained from: s d d a i x N q V 2 2 Where the buildin potential is given by: d c t B i N N V ln The potential at 10 nm from the surface equals: s d d B a i x N q V 2 ) 10nm ( 2 One can eliminate the doping density by taking the ratio of both equations: 2 2 B ) 10nm ( 1 d d a i x x V So that a i d V x B 1 1 10nm Starting from an estimate for the builtin potential of 0.65V (the barrier height) one finds x d = 148.7 nm and the corresponding density using the same estimate for the builtin potential (0.65V) is: 2 2 ) ( d s a i d qx V N = 2.98 x 10 17 cm3 This doping density is then used to repeat the process after calculating the builtin potential:...
View
Full
Document
This note was uploaded on 11/07/2011 for the course ECEN 5355 at Colorado.
 '08
 staff
 Volt

Click to edit the document details