143f2010-04-soln

143f2010-04-soln - 1 N.CHEUNG EE143 Fall 2010 Homework...

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1 N.CHEUNG EE143, Fall 2010 Homework Assignment #4 Solutions Problem 1 a.)100 keV Phosphorus Si R p = 0.12 m and R p = 0.045 m Peak concentration C p = 2.5 R p = 1.2 10 13 2.5 0.045 10 -4 = 1.05 10 18 /cm 3 Using n (at C p ) 270 cm 2 /V-sec R s 1 q  = 1 1.6 10 -19 270 1.2 10 13 = 1929 / square b.) C p exp[ - (x j - R p ) 2 2( R p ) 2 ] = N B = 10 15 [ - (x j - R p ) 2 2( R p ) 2 ] = ln 10 15 1.05 10 18 = -6.96 x j = 0.288 m Problem 2 140 keV boron will have R p =0.4 m and R p =0.081 m With N p =0.4 / R p , R s 1/q  and the mobility curve for holes, we can solve the required dose by trial- and-error to obtain: N p =10 19 / cm 3 , p = 60 cm 2 /V-sec and =2 10 14 / cm 2 to give R s =520 /square. Problem 3 (a) Peak concentration C p = 0.4 ion dose (# atoms/ion) R p Ion Energy Effective B+ Implant energy Atoms/Ion Rp (Angstrom) Cp (atoms/cm3) (1) B(+) 50keV 50 keV 1 500 8.0E+19 (2) B (2+) 100keV 100keV 1 700 5.7E+19 (3) B2 (+) 50keV 25 keV 2 330 2.4E+20 (b) x j = R p + [ 2( R p ) 2 ln C p N B ] 1 2 The B 2+
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This note was uploaded on 03/03/2012 for the course EECS 142 taught by Professor Ee142 during the Spring '04 term at Berkeley.

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143f2010-04-soln - 1 N.CHEUNG EE143 Fall 2010 Homework...

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