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Main points of this exam paper are: Gate Dielectric, Materials, Avoiding Tunneling, Maintaining, Necessary, Electrical, Including Leakage Currents
Typology: Exams
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Exam 1 Wednesday, March 18, 2008 Closed Book with 3x5 Card
kB = 1.381 x 10-23 J K-1; R = 8.314 J (mole K)-1 = 1.987 cal (mole K)-1 ; NA = 6.022 x 10^23 (mole)-1; e = 1.
x 10-19 C; mp = 1.673 x 10 -^27 kg ; 1 liter = 1000 cm^3 ; STP = 273 K, 760 torr (1 atm); 1 atm = 1.013 x 10 5 Pa; 1 Pa = 1 J/m^3.
Short Answer. 5 pts. each.
These are materials used to replace SiO2 with high dielectric constant. This allows the gate dielectric to be thicker, avoiding tunneling, while maintaining the necessary electrical properties (capacitance).
Heat must be dissipated from currents/resistances in devices, including leakage currents.
Not mention heat dissipation : -
0 exp(^ ) D D Ed kT
This is an activated process. +
4 j rad = εσ T ,^ where
k D
δ
(a) Reaction is rate-limiting step; diffusion is fast compared to reaction: O profile flat and nearly equal to O concentration at surface of SiO2. +2.
(b) Diffusion or mass transfer is rate-limiting; diffusion is slow compared to reaction; O profile linear with O concentration near 0 at Si-SiO2 interface. +2.
2 3/ 2 2 0
( ) 4 exp( ) 2 2
m mv v v dv kT kT
Mean thermal speed is the average random speed (a scalar), whereas the average gas velocity is the mean speed in a particular direction (a vector).
Chemically amplified resists contain a photoacid group that is released when struck by a photon. This acid diffuses and releases other acid sites, thereby 'amplifying' the effect of the original photon. The resist is developed by rinsing in an aqueous base, resulting is dissolution of the exposed region. Unexposed regions are not affected.
= ) of Ar at 1 Torr and 300K if its diffusivity at 1 atm
(760 Torr) and 300K is 0.1 cm 2 /s.
ழ௩வ ఒ ଷ + μ ൌ ୫୬ழ௩வఒ ଷ υ ൌ μ ൌ^
୫୬ழ௩வఒ ଷ כ^
RT P୫ ൌ^
ழ௩வఒ ଷ ൌ D ~^
ଵ P + D 1 (0.1 cm 2 /s)*P 1 (760 Tor) = D 2 *P 2 (1 Torr)
D 2 = 76 cm^2 /s +
Problem.
(30) 1. A new dopant has been developed and is tested using drive-in diffusion into a
silicon wafer. The diffusivity (D) of the dopant in Si is 3 x 10 -13^ cm 2 /s. How long does it take for
the dopant concentration to reach 1% of the surface concentration at a depth of 1 micron below
the surface?
Recall that for drive-in diffusion, the concentration profile at depth z and time t, is given
by, 2 ( , ) exp( ) 4
C z t Q^ T^ z π Dt Dt
Cs ൌ QT √D୲
C ൌ C ሺ0.0001 cm, tሻ ൌ QT √D୲^ exp ቀെ ଵ షఴ ଵכଷכସ షభయ^ ୲כ ቁ ൌ 0.01 Cs
Misusing QT: -
Take Eqn. 1 into Eqn. 2
t = 1809 sec Wrong unit conversion or answer with correct step: -