

Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Main points of this past exam are: Simple Process Sequence, Cross-Sections, Fabricating, Jaeger, Substrate, Various Voltages, Additional Lithography
Typology: Exams
1 / 3
This page cannot be seen from the preview
Don't miss anything!


EE143 Midterm #1 Solutions Sp 2003 Problem 1 (a) Either one of the following solutions will be acceptable. Solution 1 (best) : One additional lithography mask is needed to protect the MOSFET region when doing the poly-Si removal over ohmic contact area and p+ implant Solution 2 (acceptable) : Two additional lithography masks are needed : 1) protect the MOSFET region when doing the p+ implant for the ohmic contact regions (without poly-Si on top) , and 2) to protect the ohmic contact region when doing the n+ implant for source/drain.
(b) All process steps same as MOSFET process. The key steps to add the p+ ohmic contact regions are marked in italic Starting wafer, p- Si LOCOS process steps (Mask #1)
Gate oxide growth (thermal oxidation) Undoped poly-Si deposition (CVD) Pattern poly-Si (Mask#2)
Blanket As implant to form n+ poly-Si and n+ source/drain regions
Pattern photoresist to cover all MOSFET regions (Mask #3)-Extra mask Etch away poly-Si over ohmic region Boron implant to form p+ ohmic contact regions
Strip resist Deposit CVD SiO (^2)
Metal contact opening (Mask #4) Al deposition Al interconnect patterning (Mask #5)
p+
SiO p+
SiO
p- substrate
p+
SiO p+
SiO p+
SiO
p- substrate
p+
SiO
p+
SiO p+
SiO
p- substrate
n+ n+ p+
p+ SiO Gate oxide
photoresist
High Dose Boron Ion Implantation
p+
SiO p+
SiO
p- substrate
n+ n+ p+
p+ SiO Gate oxide
photoresist
High Dose Boron Ion Implantation
p+
SiO
CVD SiO
p+
SiO
Al
p- substrate
n+ n+
n+ poly Si CVD SiO
p+
p+ SiO
CVD SiO2 CVD SiO
MOSFET
Ohmic Contact to substrate
Gate oxide
p+
SiO
CVD SiO
p+
SiO
Al
p- substrate
n+ n+
n+ poly Si CVD SiO
p+
p+ SiO
CVD SiO2 CVD SiO
MOSFET
Ohmic Contact to substrate
Gate oxide
p+
SiO p+
SiO
p- substrate
poly Si
p+
SiO
Gate oxide
p+
SiO p+
SiO
p- substrate
poly Si
p+
SiO
Gate oxide
p+
SiO p+
SiO
p- substrate
poly Si
p+
SiO
Gate oxide
p+
SiO p+
SiO
p- substrate
n+ n+
n+ poly Si
p+
SiO
Gate oxide
Arsenic Ion Implantation
p+
SiO p+
SiO
p- substrate
n+ n+
n+ poly Si
p+
SiO
Gate oxide
Arsenic Ion Implantation
Problem 2
(a) x 2 2 + A x 2 = B (t + τ) ;x 1 2 + A x 1 = B (t )
( x 22 - x 12 ) + A (x 2 – x (^) 1) = B τ
(x (^) 2–x (^) 1)(x (^) 2+ x (^) 1) + A (x 2 – x 1 ) = B τ
(x (^) 2–x (^) 1) =
B τ A + x (^) 2+x 1 =
xo^2 + A xo A + x (^) 2+x 1 = = x^ o^ •^
A + xo A + x 1 + x (^2)
(b ) No. Since ( x (^) 2+ x 1 ) is always larger than xo ,
A + xo A + x (^) 1+ x 2 is always < 1. (c ) For large t, both x2and x 1 will approach ∞. Therefore (x 2 –x (^) 1) → 0. (d) ∆ = x (^) o + 0.54 (x 2 –x (^) o ) - 0.54 x 1 = 0.46 x (^) o + 0.54 (x (^) 2–x (^) 1)
(e) Effect on ∆ Brief Explanation Initial oxide thickness xo ↑ + Original step height is larger
Oxidation time t ↑ - x 1 gets closer to x (^2) Oxidation temperature ↑ -^ Faster oxidation rate, x 1 gets closer to x^2 Oxidant gas pressure ↑ -^ Faster oxidation rate, x 1 gets closer to x^2 (f) At lower dopant concentration, we have mainly neutral vacancies. At higher doping concentrations, we have more electrons (or holes) to create extra charged vacancies V+ , V-, V= etc with concentrations proportional to (p/n (^) i ), (n/n (^) i ) , (n/n (^) i ) 2 etc from Law of Mass Action. When charge carrier concentration
is larger than n (^) i ( ~10 19 /cm^3 at 1000C), the increase of charged vacancies can be substantial. The B/A
parameter is related to the interfacial reaction term ks. More vacancies (broken Si bonds) available at the
SiO2/Si interface will increase k (^) s.
Problem 3 (a) For 200 keV P → Si, Rp =0.254 μm , ∆Rp =0.0775μm
Peak concentration Cp = (0.4 × 10 13 )/(0.0775 × 10 -4^ ) = 5.2 × 10 17 /cm^3
From the mobility curve for electrons (using peak conc as impurity conc), μn = 350 cm^2 /V-sec
Rs =
qμnφ =^
≈ 1780 Ω /square.
(b) Cp exp[ -( x (^) j -0.254) 2 / 2 ∆Rp^2 ]= N (^) B with x (^) j in μm
∴ ( x (^) j - 0.254) 2 = 2 ×(0.0775) 2 ln [ 5.2 × 10 17 /10 16 ] or x (^) j = 0.254 ± 0.22 μm ;
x (^) j 1 = 0.032 μ m and x (^) j 2 = 0.474 μ m
(c ) For implant profile with annealing , C(x,t) =
Np
[ 1 +
4Dt 2(∆Rp) 2
exp [
Since the Dt product (10 -12^ cm^2 ) is << ( ∆ R (^) p) 2 (6.4 × 10 -11^ cm^2 ) , the implant profile is basically unchanged after this thermal annealing step. (d)The Phosphorus peak is (0.5+0.254) = 0.754 μm below the poly-Si surface. We therefore need ~320 keV Boron to make the Boron peak position coincides with the Phosphorus peak. (e) ∆Rp for Boron is 0.105 μm. Required Boron dose φ = 5.2× 10 17 × 0.105× 10 -4^ /0.4 = 1.4 × 10 13 /cm^2. (f) Since ∆Rp for Boron is 0.105 μm which is larger than the Phosphorus implant ∆Rp (=0.0775 μm) and
the Boron implant peak concentration equal to to that of Phosphorus, the Boron will overcompensate the Phosphorus at all depths.