EE212 HW 2

# EE212 HW 2 - EE 212 FALL 09-10 HOMEWORK ASSIGNMENT#2 ASSIGNED THURSDAY OCT 1 DUE THURSDAY OCT 8 SOLUTION SHEET Reading Assignment Chapters 3 and 4

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EE 212 FALL 09-10 HOMEWORK ASSIGNMENT #2 ASSIGNED: THURSDAY OCT. 1 DUE: THURSDAY OCT. 8 SOLUTION SHEET Reading Assignment: Chapters 3 and 4 in the text. #1. Spend 30 min or so scanning the information in the 2007 ITRS Front End Processes (on the class website or at the website given in the class notes). You don’t need to turn anything in for this HW problem. #2. An experimental structure like that shown on page 9 of the class notes for Chapter 3 is prepared. The Ge in the microcrucible has a cross-sectional of 0.2 μm X 0.2 μm and is 50 μm long. Presuming that the maximum “pull rate” equation derived in class applies to the LPE (Liquid Phase Epi) freezing process, estimate the time required for the 50 μm Ge thin film to recrystalize. You should be able to find physical parameter values for Ge by searching on the web, for example, http://en.wikipedia.org/wiki/Germanium . The emissivity for Ge may be hard to find, so you can use the same value as is used for Si in the text (0.55). If you need to make assumptions about other Ge parameters, explain your reasoning. (10 pts). Answer: The maximum pull rate equation derived in class is: v PMAX = 1 LN 2 "# k M T M 5 3r The physical parameters needed (from http://en.wikipedia.org/wiki/Germanium ) are as follows: Physical Parameter Si Value Ge Value L – Latent heat of fusion 430 cal gm -1 36.94 kJ mole -1 N - density 2.328 gm cm -3 5.323 " - Stefan Boltzmann constant 5.67x10 " 5 erg cm " 2 sec " 1 K " 4 same " - emissivity 0.55 0.55 k M - thermal conductivity at T M 0.048 cal sec " 1 cm " 1 K " 1 60.2 W m -1 K -1 @ 300K T M - melting temperature 1690 K 1211

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The thermal conductivity given in Wikipedia at 300K and we need this value at the melting temperature. This can be estimated using the formula in the class notes (or text) as follows: k S = k M T M T " k M = k S T T M = k S 300K 1211K = 14.94 W m # 1 K # 1 Thus we have: v PMAX = 1 LN 2 "# k M T M 5 3r = 1 36.94x10 3 J mol \$ 1 ( ) 72.64gmmol \$ 1 ( ) \$ 1 5.323 gm cm \$ 3 ( ) 2 5.67x10 \$ 5 erg cm 2 sec K 4 % ( ) * 0.55 ( ) 14.94 J sec m K % ( ) * 1211 K ( ) 5 10 \$ 7 J erg % ( ) * 10 \$ 2 m cm % ( ) * 3 10 \$ 5 cm
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## This note was uploaded on 06/12/2011 for the course EE 212 at Stanford.

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EE212 HW 2 - EE 212 FALL 09-10 HOMEWORK ASSIGNMENT#2 ASSIGNED THURSDAY OCT 1 DUE THURSDAY OCT 8 SOLUTION SHEET Reading Assignment Chapters 3 and 4

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