Oxidation - Microfabrication Technology - Solved Exam, Exams of Materials science

Main points of this past exam are: Oxidation, Lightly Doped, Processed, Processing Steps, Fixed Temperature, Fixed Oxidizing, Oxidation Time

Typology: Exams

2012/2013

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EECS143,Sp 99 N. Cheung
Midterm Exam #1 Solutions
Problem 1
(a) The observed indicates the growth rate is slower than the Deal-Grove model after 2000 Å of oxide is
grown. Even if we take the limit that growth rate is proportional (time)1/2 , the oxide will be thicker
than 4000 Å after 4 hours of oxidation. The Si surface layer must have a faster oxide rate than the bulk
of the wafer.
Conjecture 1 : The processed Si wafer was oxidized first to an oxide thickness of 100 Å and then have the
oxide dissolved in HF. FALSE
Dopant segregation can enhance oxidation rate but the small oxide growth cannot change the surface layer
dopant concentration due to dopant segregation.
Conjecture 2 : The processed Si wafer has a highly doped surface layer ( doping > 1019/cm3) which is less
than 1000 Å thick. TRUE
The Si wafer has a highly doped surface layer ( N > 1019/cm3) which is less than 1000 Å thick ( i.e., 0.46 ×
2000 Å 1000 Å). The underneath substrate is lightly doped. The Si wafer has a highly doped region
underneath a lightly doped surface region. Oxidation rate is higher when the doping concentration is higher
than 1019/cm3, mainly through the linear term B/A. After this layer of highly doped Si is consumed, the
growth rate slows down.
Conjecture 3 : The processed Si wafer has a thin layer of poly-Si layer on top surface. TRUE
Initial oxidation of poly-Si is very fast. After poly-Si is all consumed, the oxidation rates slows down.
(b) SiO2volume
Si volume = 5×1022
2.3×1022=2.17
Volume of sphere radius3 or radius of SiO2
radius of Si = (2.17)1/3 = 1.29
Therefore , radius of SiO2 sphere =1.29 µm.
(c) LOCOS Advantages :
(i) Self-aligned channel stop ; (ii) oxide topography more planar than opening an oxide window.
LOCOS Disadvantages: The "bird's beak" region wastes device area.
Problem 2 (a)
Parameter Electrical Channel Length L
Implant Dose
Substrate conc. NB
Sidewall Angle θ
Gate material changed from poly-Si to Tungsten
Implant ions changed from Phosphorus to Arsenic
(same energy)
(b) Restore crystallinity of damage Si caused by ion bombradment
Position implanted dopants into substituitional sites of Si lattice so that the dopants gives "shallow"
dnor or acceptor energy levels [ i.e. dopant activation]
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EECS143,Sp 99 N. Cheung Midterm Exam #1 Solutions Problem 1 (a) The observed indicates the growth rate is slower than the Deal-Grove model after 2000 Å of oxide is grown. Even if we take the limit that growth rate is proportional (time)1/2^ , the oxide will be thicker than 4000 Å after 4 hours of oxidation. The Si surface layer must have a faster oxide rate than the bulk of the wafer.

Conjecture 1 : The processed Si wafer was oxidized first to an oxide thickness of 100 Å and then have the oxide dissolved in HF. FALSE Dopant segregation can enhance oxidation rate but the small oxide growth cannot change the surface layer dopant concentration due to dopant segregation.

Conjecture 2 : The processed Si wafer has a highly doped surface layer ( doping > 10^19 /cm^3 ) which is less than 1000 Å thick. TRUE

The Si wafer has a highly doped surface layer ( N > 10^19 /cm^3 ) which is less than 1000 Å thick ( i.e., 0.46 × 2000 Å ≈1000 Å). The underneath substrate is lightly doped. The Si wafer has a highly doped region underneath a lightly doped surface region. Oxidation rate is higher when the doping concentration is higher

than 10^19 /cm^3 , mainly through the linear term B/A. After this layer of highly doped Si is consumed, the growth rate slows down.

Conjecture 3 : The processed Si wafer has a thin layer of poly-Si layer on top surface. TRUE Initial oxidation of poly-Si is very fast. After poly-Si is all consumed, the oxidation rates slows down.

(b)

SiO 2 volume Si volume =^

5 × 1022

2.3× 1022

Volume of sphere ∝ radius^3 or

radius of SiO 2 radius of Si = (2.17)

Therefore , radius of SiO 2 sphere =1.29 μm.

(c) LOCOS Advantages : (i) Self-aligned channel stop ; (ii) oxide topography more planar than opening an oxide window. LOCOS Disadvantages: The "bird's beak" region wastes device area.

Problem 2 (a)

Parameter Electrical Channel Length L

Implant Dose ↑ ↓

Substrate conc. NB ↑ (^) ↓

Sidewall Angle θ ↑ ↑

Gate material changed from poly-Si to Tungsten (^) ↑

Implant ions changed from Phosphorus to Arsenic (same energy)

(b) Restore crystallinity of damage Si caused by ion bombradment Position implanted dopants into substituitional sites of Si lattice so that the dopants gives "shallow" dnor or acceptor energy levels [ i.e. dopant activation]

(c) Tilt the crystal by about 7 degrees with respect to beam incidence direction plus wafer rotation to avoid axial and planar channeling Pre-amorphise the Si substrate first by Si implanattion, followed by dopant implantation.

Problem 3 (a)

Profile 1 : CS erfc [ xj /2 Dt] = CS erfc [ 0.7/(2 × 0.1] = CS erfc [3.5] = 10^16 /cm^3

Therefore CS = 1.3 × 10 22 /cm^3 [ 26% boron !!!]

Profile 2 : RS = 1/ { q μp CS xj ) or CS = 1/ ( q μp xj RS ) = 3 × 10 20 / cm^3

The solid solubility of boron in Si is less than 10^21 /cm^3. Profile 1 is unrealistic. Profile 2 is a better approximation.

[Note: If you use the Irvin Curves for Profile 1 with RSxj = 3.5 ohm-um, the extrapolated surface

concentration will also be in the 10^22 /cm^3 range]

(b)

During the initial stage of diffusion, the doped region is intrinsic ( n = p) so we won’t expect to see any high concentration diffusion effect.

However, Boron diffuses faster than As and the near surface region becomes highly n-doped and the deeper region becomes p-doped. Since the net carrier concentration can still be higher than ni. We will start to

observed high concentration diffusion effects in both regions for longer diffusion times.

Problen 4 (a) Let translational error be (xt,yt).

The run in/out errors will contribute +δx to A1 and -δx to A The rotational errors will contribute +δy to A1 and -δy to A Therefore,

x 1 = xt+δx

x 2 = xt - δx

y 1 = yt+δy

y 2 = yt - δy

Rearranging terms,

δx =

x 1 -x 2 2 (run in/out error) = (0.2-0.4)/2 = - 0.1 um [ run-in error]

δy =

y 1 -y 2 2 (rotational error) = (0.2-0.4)/2 = -0.1 um or 2E-6 radians [ clockwise]

xt =

x 1 +x 2 2 = (0.2+0.4)/2 = 0.3 um

yt =

y 1 +y 2 2 = (0.2+0.4)/2 = 0.3 um

(d) lm ∝ λ/NA. NA reduces by 2X, lm = 0.5um x 2 = 1um DOF ∝ λ/NA^2. NA reduces by 2X, DOF = 1um x 4 = 4 um