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EE143 Fall 2004 Midterm 1 solutions
Problem 1 (i ) Three lithography steps are used in this process flow. Mask 1: Pattern Poly-1 hinge plate Mask 2: Pattern the staple anchor openings through bottom PSG Mask 3: Pattern the Poly-2 staple (ii) Four CVD steps are used: CVD1- bottom PSG deposition CVD2- poly-1 deposition CVD3- top PSG deposition CVD4- poly-2 deposition (iii ) Four thin-film etching steps are used: Etch 1: Etch Poly-1( hinge plate) Etch 2: Etch staple anchor openings through bottom PSG Etch 3: Etch Poly-2 (staple) Etch 4: Etch all PSG sacrificial layers to release hinge
(iv) NO. Part of the hinge structure is underneath the staple structure. If staple is formed first, there is no way to pattern the hinge outline (litho and etch) AND no way to deposit the part of poly-1 which is underneath the staple structure.
(v) AGREE. There is no high-temperature processing steps used after Al deposition, Al melting is a non-issue. [Optional answer: DISAGREE. One has to find a proper wet/plasma etching recipe with the right selectivity to etch PSG without attacking Al.]
(vi) Major differences (schematic shown below: red is structure using poly-1 oxidation) : (1) poly-1 plate thinner ( poly-Si is consumed during thermal oxidation) (2) cross-section of poly-1 under staple area is smaller (affects poly-1 & poly-2 gap spacing under staple) (3) staple cross-section has more vertical sidewalls because there is no PSG (layer 1).
Thinner and smaller
Thinner and smaller
Lower and more vertical
Thinner and smaller
Thinner and smaller
Lower and more vertical
Problem 2 (a) (i) A= B/(B/A) = 0.027 / 0.178 = 0.152 μm x (^) ox^2 + Axox = B( t + 0)
For xox = 0.2 μm , 0.2 2 + 0.152 ×0.2 = 0.027 × t
⇒ t = 2.6 hrs
(ii) From 1000 °C, wet oxidation curves : With no initial oxide, it takes ~ 0.3 hrs to form 0.2 μm of oxide. With no initial oxide, it takes ~ 1.4 hrs to form 0.5 μm of oxide. Starting with 0.2 μm of oxide, it takes (1.4-0.3) = 1.1 hrs to form an additional 0.3 μm of oxide.
(b) If final thin oxide thickness is < 100nm, one empirical way to correct the Deal-Grove model prediction is to used an initial oxide thickness xi ~20nm instead of zero even when the starting wafer has no initial oxide.
For very thin oxides (tens of nm) , one can also modified the Deal-Grove growth rate to : dxox
dt =^
B
A+2x (^) ox+ C^ •^ exp [–xox^ / L] where is L ~7 nm.
Note that the empirical exponential term is only significant for small xox. This additional term goes to zero when x (^) ox is large.
(c ) HCl reacts with oxygen to generate water molecules: 4HCl + O2 → 2H 2 0 + 2Cl 2.
The created H 2 0 adds a faster oxidation mechanism (i.e. steam oxidation) to the dry-
oxygen oxidation mechanism. Cl also increases the interface reaction with an increase of ks.
Problem 3 (a) (0.6 - ∆) • 2.17 = 0.6 ⇒ ∆ = 0.324 μ m
(b) (i) If Rp = 0.6 μm , half of the implanted dose will be inside Si and half of dose will
be inside SiO 2. For B +ions, we need ~ 220 keV ion energy. The corresponding ∆Rp is ~ 0.1 μm
(ii) Cp =
2 × 1013
2 π∆Rp
= 8×1017/cm
(xj - Rp)2 = 2 (0.1) 2^ • ln(
8 × 1017
∴ xj 2 = Rp + 0.365 μm = 0.965 μ m
(iii ) Using μp ~ 160 cm^2 /V-sec at Cp = 8×1017/cm and RS ~ 1/ (q μ φSi ) = 1 / (1.6 × 10 –19^ × 160 × 1 × 1013 ) = 3900 ohm /square.
