143f2010-07-soln

# 143f2010-07-soln - EE143 Fall 2010 HW#7 Solutions Problem 1...

This preview shows pages 1–2. Sign up to view the full content.

EE143, Fall 2010 HW#7 Solutions Problem 1 All etchimg rates have only vertical components ( completely anisotropic) Poly etech rate = 0.1 m /min Oxide etch rate = 1/5 poly etch rate = 0.02 m /min Resist etch rate = 1/ 2 poly etch rate = 0.05 m /min Slope of resist before etching = tan -1 (0.5/0.15) = 73.3 o After 5 minutes of etching (a) Poly-Si will be etched to a depth of 0.1 5 = 0.5 m Photo resist will be etched to a depth of 0.05 5 = 0.25 m Bottom of resist opening will increase by 2 x ( 0.05 5 cot 73.3 o ) = 0.15 m (b) Oxide slope angle = tan -1 ( 0.1/ 0.075) = 53.1 o Problem 2 i.)The poly etching is completely isotropic and the mask etching is amistropic Note that the lateral etching rate of poly > lateral removal rate of mask! (Ii) x max = 5 m iii) x min = (5 - 2) m = 3 m iv) mask etched = 2x = 2 (0.01 m/min 1 m 0.1 m/min cot 60ｰ = 2 0.058 m = 0.116 m mask max = 5 - 0.116 = 4.88 m v.) t mask (max) = 0.5 m - ( 1 m 0.1 m/min ) 0.01 m/min = 0.4 m poly mask 1 m x max X Y x min before after m 5 mask max x x mask

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 03/03/2012 for the course EECS 142 taught by Professor Ee142 during the Spring '04 term at Berkeley.

### Page1 / 3

143f2010-07-soln - EE143 Fall 2010 HW#7 Solutions Problem 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online