Sol HW2 ECE 255 Prof Ye - ECE255Prof.P.Ye SolutionstoHW# 2...

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ECE 255 Prof. P. Ye Solutions to HW #     2    1.-  Prob. 3.1. a) Using the given values of N A , N D  and the following known quantities:  V = 26mV,  ni=10 10 cm -3  , q=1.6x10 -19  C and  ε s =1.03x10-12 F/cm. When these values are  plugged in into Eqs. (3.6) and (3.7) it is obtained that   w do = 37.3nm. b) Using the expressions in page 79 it is found that:  xp= 3.39 nm and xn =33.9 nm. c) From Eq. (3.6) it is obtained that the built-in potential is 0.979V. d) Using any of the expressions for Emax found in page 79 it is calculated that  Emax=5.24x10 5  V/cm. 2.- Prob. 3.24 a) Using Eq. (3.9) with Vt = 0.026V, n=1 and V D = 0.675V it is found that Id = 1.88  μ A. b) Assuming the exponential factor in Eq. (3.9) is much greater than one (neglect the - 1 added to the exponential), then V D  can be worked out and found to be V D =nVt ln  (Id/Is). Therefore V D  =1x 0.026x ln(3x1.88x10
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This note was uploaded on 02/18/2010 for the course ECE 255 taught by Professor Staff during the Fall '08 term at Purdue University-West Lafayette.

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Sol HW2 ECE 255 Prof Ye - ECE255Prof.P.Ye SolutionstoHW# 2...

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