# HW 5 - EE 212 FALL 09-10 HOMEWORK ASSIGNMENT#5 ASSIGNED...

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EE 212 FALL 09-10 HOMEWORK ASSIGNMENT #5 ASSIGNED: TUESDAY NOV. 17 DUE: Thursday DEC. 3 SOLUTION SHEET This assignment must be turned in not later than 5PM on Thursday Dec. 3. A solution sheet will be posted on the class website at that time. Any HW assignments turned in late will receive no credit. Final Exam is Thursday Dec. 10, 7 -9 PM. #1. The figure below shows a SPEEDIE simulation of an LPCVD deposition over a narrow trench. The various contours are for equal time steps during the deposition. Explain why the simulation shows a time independent deposition rate on the top surfaces, but the deposition rate decreases with time on the sidewalls and the bottom of the trench. (5 points). Answer: As the film deposits, the opening at the top of the trench narrows, closing off the entrance for the depositing gases to enter the trench. Thus the deposition rate on the sidewalls and bottom of the trench decreases with time. This could also be explained in terms of a narrowing viewing angle of points on the sidewall and trench bottom, to the depositing flux. -1.00 1.00 0.0 -0. 5 0.0 0.5 1. 0 1. 5 2. 0 a) m i c r o n s microns

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#2. A silicon epitaxial layer is deposited nominally at 700 ˚C. The system being used operates in the surface reaction controlled regime in which temperature control is very important (E A = 1.6 eV). If the deposition rate needs to be controlled to ± 5%, what temperature control (± how many ˚C) is needed in the epitaxial reactor? State any assumptions. (5 points). Answer: If the growth rate is in the surface reaction controlled regime, then the deposition rate is given by v C T N Yk S = C T N Yk 0 exp E A kT Assuming that the temperature dependence is primarily in the exponential term, we have = 2 2 0 1 exp 1 T k E v kT E T k E Yk N C dT dv A A A T But Δ v = ± 0.05v so that ( ) ( )( ) C ˚ 2.5 or 5 . 2 6 . 1 973 10 62 . 8 05 . 0 05 . 0 2 1 5 2 ± ± = ± = ± = Δ K eV K eVK x E kT T A Alternative Method: exp A E v kT Solving for T, ln( ) A E T k ν At T = 700C (973K), and Ea = 1.6 eV, 1.6 exp 5.156 9 (8.617 5)(973) v e e = At v = 0.95v, using second eqn to solve for T gives T =970.5K At v = 1.05v, using second eqn to solve for T gives T=975.5K Giving a delta T of 2.5 degrees C (or K)
3). For CVD deposition of a film, it is found that the mass transfer coefficient h G = 100 cm/sec and the surface reaction rate coefficient k S = 1x10 7 exp 1.9eV /kT ( ) cm/sec. For a deposition at 800˚C, which CVD system would you recommend using: (a) a cold-walled, graphite susceptor type: or (b) a hot- walled, stacked wafer type? Explain your answer. (5 points). Answer:

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## This note was uploaded on 06/12/2011 for the course EE 212 at Stanford.

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HW 5 - EE 212 FALL 09-10 HOMEWORK ASSIGNMENT#5 ASSIGNED...

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