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Unformatted text preview: Question 1 Consider an n-type 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 build-in 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 built-in potential of 0.65V (the barrier height) one finds x d = 148.7 nm and the corresponding density using the same estimate for the built-in potential (0.65V) is: 2 2 ) ( d s a i d qx V N = 2.98 x 10 17 cm-3 This doping density is then used to repeat the process after calculating the built-in potential:...
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This note was uploaded on 11/07/2011 for the course ECEN 5355 at Colorado.