HW 4 - EE 212 FALL 09-10 HOMEWORK ASSIGNMENT#4 ASSIGNED...

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EE 212 FALL 09-10 HOMEWORK ASSIGNMENT #4 ASSIGNED: TUESDAY OCT. 27 DUE: FRIDAY NOV. 6 SOLUTION SHEET This assignment must be turned in by 5PM on Friday Nov. 6. A solution sheet will be posted on the class website at that time. Any HW assignments turned in late will receive no credit. The class midterm will be on Tuesday November 10 at the usual class time. #1. Two implants are done into a 1x10 15 cm " 3 doped P type Si substrate. The implants are boron ( 15 cm " 2 @ 10KeV ) and arsenic ( 2x10 13 cm " 2 @ 200KeV ). After the implants, the wafer is annealed at 1100 ˚C for 30 minutes. Calculate the location of any PN junctions that are in the substrate after this anneal. You may neglect TED and you may assume only intrinsic diffusion for the As and B. State any other assumptions. (10 points) Answer: After the implants the doping profiles will be roughly as shown below. If we assume only intrinsic diffusion at 1100 ˚C, then the B and As diffusivities are given by D B = 1.0exp " 3.5eV 8.62x10 " 5 eVK " 1 ( ) 1373K ( ) # $ % % & ( ( = 1.43x10 " 13 cm 2 sec " 1 D As = 9.17exp " 3.99eV 8.62x10 " 5 eVK " 1 ( ) 1373K ( ) # $ % % & ( ( = 2.09x10 " 14 cm 2 sec " 1
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Thus the Dt values for B and As for a 30 minute 1100˚C diffusion are 2 Dt ( ) B = 5.15x10 " 10 cm 2 and 2 Dt ( ) As = 7.52x10 " 11 cm 2 . These correspond to " R P 2 = 1.99x10 # 15 cm 2 for As and " R P 2 = 1.21x10 # 14 cm 2 for B. Thus " R P 2 << 2Dt in both cases and so the implants can likely be treated as delta functions and the final profiles as one sided Gaussians. Presuming this is true, the final “junction depths” are: x J 2 (B) = 4Dtln Q Dt 1x10 15 cm # 3 ( ) $ % & & ( ) ) = 4 2.575x10 # 10 ( ) ln 15 2.575x10 # 10 ( ) 15 cm # 3 ( ) $ % & & & ( ) ) ) " x J (B) = 1.04 μ m x J 2 (As) = Q " Dt 15 cm # 3 ( ) $ % & & ( ) ) = 4 3.76x10 # 11 ( ) ln 2x10 13 " 3.76x10 # 11 ( ) 15 cm # 3 ( ) $ % & & & ( ) ) ) " x J (As) = 0.335 μ m The peak concentrations in the two profiles are given by: C S (Boron) = Q Dt = 15 2.575x10 # 10 ( ) = 3.5x10 19 cm # 3 C S (Arsenic) = Q Dt = 2x10 13 3.76x10 # 11 ( ) = 1.8x10 18 cm # 3 Thus the final profiles look approximately as shown below: Thus there are no PN junction in the final profile.
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#2. An OED experiment is done similar to the test structure shown on page 18 of the class notes on diffusion. The buried layer is arsenic doped. The OED increases the interstitial concentration by 3 times. Consider two cases, one in which interstitial-vacancy recombination is very fast, in which case, C I C V = C I * C V * . The other extreme is when there is no interstitial-vacancy recombination. Which case results in more OED enhancement of the buried arsenic layer? Calculate a quantitative answer. (5 points). Answer: The enhancement in the buried layer diffusivity is given by D A = D A * f I C I C I * + f V C V C V * " # $ $ % & In the first case, C V will be reduced by 3 times. In the second case, C V is unchanged. Thus First case: D A = D A * f I C I C I * + f V C V C V * " # $ $ % & = D A * 0.4 3C I * C I * + 0.6 C V * V * " # $ $ % & = D A * 1.2 + 0.2 ( ) = 1.4D A * Second case: D A = D A * f I C I C I * + f V C V C V * " # $ $ % & = D A * 0.4 I * C I * + 0.6 C V * C V * " # $ $ % & = D A * 1.2 + 0.6 ( ) = 1.8D A * Thus the enhancement is larger in the second case. This occurs because the interstitial component is enhanced in both cases, but the vacancy component is reduced only in the first case.
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This note was uploaded on 06/12/2011 for the course EE 212 at Stanford.

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HW 4 - EE 212 FALL 09-10 HOMEWORK ASSIGNMENT#4 ASSIGNED...

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