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EE143 Midterm Exam #2 Solutions Sp
Problem 1
(a) The center only has translational errors: 0.5 μm along x, -0.5 μm along y.
After substracting the translational error, we have:
Top Right Center Left Bottom x -0.5 +0.2 0 -0.2 +0. y +0.2 +0.5 0 -0.5 -0.
The run out error is 0.2 μm
The rotational error is 0.5 μm (counter-clockwise)
(b) DOF/R =
0.8 NA
= 2 → NA= 0.
( c)
Example 1
Resist openings at the oxide/resist interface will have different sizes on thicker and thinner oxide regions. If the resist is used as the etching mask for oxide, we end up with different oxide feature sizes.
Example 2
(0.2, 0.5)
(-0.5, 0.2)
(-0.2,-0.5)
(0.5,-0.2)
Oxide
Reflecting substrate
Photoresist
Photon Intensity due to standing wave
min
max
min
min
Oxide
Reflecting substrate
Photoresist
Photon Intensity due to standing wave
min
max
min
min
After development
Positive After development Photoresist.
Positive
Photoresist.
The resist slope angles have large variations if the underlying film has topography variations. For etching recipes which have mask erosion problems, the protective mask with end up with large variation of dimensions during the etching.
(d) (i) contrast γ = 1/ log 10 (100/20) = 1.
(ii) The exposure energy density versus position are shown below for the 1sec and 3sec exposures. The dashed lines are the normalized remaining resist thickness from the resist contrast curve. Intercepting points show position of that particular remaining thickness.
The resist cross-sections are shown below:
0
1
0 0.5 1 1.5 2 2.5 3
RESIST
1 sec
0
1
0 0.5 1 1.5 2 2.5 3
3 sec
RESIST
Position x in um
Exposure Energy Density
(mJ/cm2)
1 sec
3 sec
(iv) Area of sputtering target can be made much larger than that of an evaporating source. A larger area can be considered as a superposition of many small-area sources. By adding the flux from all the sources, a large area source will provide better film deposition uniformity.
(b) (i) With small flow rates, deposition is limited by gas transport , Grove model ⇒.
F 1 ≡ DG ⋅
CG - CS
δ
⇒ hG =
DG
δ
where DG = diffusivity
dy dt ∝^ hG^ ∝^
δ
∝ U
(ii) With high flow rates, deposition is limited by surface reaction , dy/dt ∝ ks which is independent of flow rate. (iii) For small flow rates, deposition rate is limited by gas transport. From Grove model,
dy / dt ≈ hG
CG
ρ At a growth rate of 0.1 μm /minute = 1.67 × 10 -7^ cm/sec.
hG =1.67 × 10 -7^ ×
5 × 1022
4 × 1016
= 0.208 cm/sec
(iv) (v) Using a higher flow rate with same pressure will supply more reactant gas /unit time into the CVD reactor. Mass depletion due to consumption of reactants will be less severe.
Problem 4
(i) The pockets of SOG are used for partial planarization. (ii)Because Al is already present when nitride is deposited, the nitride is put down by PECVD which does not have perfect conformal coverage when deposition temperature is relatively low. They form the “keyholes” especially when the gap has higher aspect ratios (iii) A typical Al alloy composition used for interconnects is Al-2%Si- 2% Cu. (iv) MoSi 2 acts like a diffusion barrier to minimize Al reaction with substrate Si. (v) Silicides do not electromigrate. The silcide/Al/silicdie sandwich provides a local shunting path in case the Al forms a local void due to electromigration. (vi) Approach 1 : Use trench oxide isolation instead of LOCOS. This gives a better planarized surface early on in the process flow. Approach 2 : Use W-plugs to partially or fully planarize contact holes before interconnect deposition Approach 3: Fill up contact holes with thick layer deposition. Then use etchback or CMP to planarize back to dielectric surfaces.
Sputtering Target
Profile due to one small-area source
Superposition of all small-area sources
Sputtering Target
Profile due to one small-area source
Superposition of all small-area sources
