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Spring 2003 UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences
EE143 Midterm Exam #
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Make sure the exam paper has 7 pages + 3 pages of data for reference
Instructions: DO ALL WORK ON EXAM PAGES This is a 90-minute exam (4 sheets of notes allowed)
Grading: To obtain full credit, show correct units and algebraic sign in
answers. Numerical answers which are orders of magnitude off will
receive no partial credit.
Problem 1 (20 points)________________
Problem 2 (25 points)_______________
Problem 3 (30 points) ________________
Problem 4 (25 points) ________________
TOTAL (100 points) __________________
Problem 1 Simple Process Sequence (20 points total)
The following process sequence and cross-sections are taken from Chapter 1 of the EE143 textbook by Jaeger on fabricating a N-channel MOS transistor.
(a) (5 points) Suppose we would like to add an ohmic contact structure to the p-substrate (see schematic below) so that the substrate can be biased at various voltages. How many additional lithography masks will be needed? Briefly describe the function of the extra mask(s).
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
Problem 2 Thermal oxidation (25 points total) We are interested in the oxide thickness difference ( x 2 -x (^) 1) after thermal oxidation of an oxide window on a Si wafer.
(a ) (4 points) Use the Deal-Grove Model result x (^) ox^2 + A x (^) ox = B (t + ฯ) to show that :
( x 2 - x (^) 1) = x (^) o โข
A + xo A + x (^) 1+ x (^2)
(b) (3 points) Can ( x 2 - x (^) 1) be larger than xo? Explain.
(c ) (3 points) For very long oxidation time ( t โ โ), what will be the value of ( x 2 - x (^) 1)?
(d) (3 points) Expressed the surface oxide step height โ in terms of xo , x (^) 1, and x (^) 2.
(e) (8 points) By maintaining other variables the same , how will an increase of the following parameter affect the value of the surface height difference (โ) in the following table. Use โ+โ to represent increase, โ- โ to represent decrease, and โ0โ to represent no change. Effect on โ Brief Explanation Initial oxide thickness xo โ Oxidation time t โ Oxidation temperature โ Oxidant gas pressure โ (f) (4 points) Use charged vacancy concentration to explain why the B/A term increases rapidly with doping concentrations above 10^19 /cm^3.
Si
xo SiO2^ x
x
SiO
Si
Thermal
Oxidation
Si
xo SiO
Si
xo SiO2^ x
x
SiO
Si
Thermal
Oxidation
Dry oxidation, 900Dry oxidation, 900ooCC
Problem 3 Ion Implantation (30 points total) (A) (a) ( 4 points) Phosphorus is implanted into a p-Si with a uniform background concentration of 10^16 /cm^3.
The Phosphorus (P+) dose is 10 13 /cm^2 and the energy used is 200 keV. Find the sheet resistance of the phosphorus implanted layer using Rs โ1/(qฮผฯ).
(b) (4 points) Find the junction depths x (^) j1 and x (^) j2 formed by the Phosphorus implantation.
(c ) (3 points) After phosphorus implantation, a thermal annealing step is carried out with a Dt product of
10 -12^ cm^2 to restore the crystalline damage and to activate the dopants. You find the implanted profile does not change much. Explain.
(d )( 3points) Poly-Si (thickness = 0.5 ฮผm) is then deposited on top of the Si substrate described in part (c ), followed by a Boron (B+) implantation step. The Boron energy is chosen such that the Boron peak position coincides with the Phosphorus peak position in the Si substrate. What is the chosen boron (B+) ion energy ( in keV)?
(e) (3 points) For conditions used in part (d), the Boron (B+) dose is chosen such that the boron peak concentration is equal to the phosphorus peak concentration. What is the chosen boron (B+) dose?
(f) ( 5 points) Sketch qualitatively the hole concentration versus depth after both phosphorus and boron implantations in a semilog plot.
Phosphorus Implant
p-substrate (1E16 /cm3)
x
x
j
j
R p
x =
Phosphorus Implant
p-substrate (1E16 /cm3)
x
x
j
j
R p
Phosphorus Implant
p-substrate (1E16 /cm3)
x
x
j
j
R p
x =
Problem 4 Diffusion and Sheet Resistance (25 points total)
(A) A Boron drive-in step gives a half-gaussian depth profile. The n-type Si wafer has a background concentration = 4 ร 10 15 /cm^3. The drive-in profile has a surface concentration of 5 ร 10 18 /cm^3 and a
junction depth of 4 ฮผm.
(a) (3 points) Calculate the Dt product of the drive-in profile?
(b ) (3 points) Use the Irvins curves to find the sheet resistance of the boron diffused layer?
( c ) (3 points) What is the boron dose in the diffused layer?
(B) Suppose the boron dose in part (A) was introduced by solid-solubility-limited diffusion (i.e. the predeposition process) prior to the drive-in step. (a)( 6 points) You have to choose either a 1000 oC or a 900 oC predeposition temperature. Which temperature will you choose and explain why? Justify your answer quantitatively.
Given : Solid solubility of boron (1000C) =1 x 10^21 /cm^3 Solid solubility of boron (900C) = 5.5 x 10^20 /cm^3 Diffusion constant of boron D = 10.5 exp (-3.69 eV/ kT) cm^2 /sec
(b) (4 points) Calculate the junction depth with the predeposition conditions you choose in part(a).
(C) (6 points) Considering high-concentration diffusion effects of dopants in Si. (i) Do you think the junction depth of the boron drive-in step in Part (A) will be larger, smaller, or no change? Explain.
(ii) Do you think the junction depth of the boron predeposition step in Part(B) will be larger, smaller, or no change? Explain.
Information for reference Electron charge q= 1.6 ร 10 -19 coulombs; Boltzmann constant k = 8.62 ร 10 -5^ eV/K ni of Si = 3.69 ร 1016 ร T 3/2^ exp [- 0.605eV/kT] cm-
z
0.
0.
0.
0.
0.
0.
0.
1
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.
erfc(z)
exp(-z^2)
Irvin Curves
