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Gate Dielectric - Chemical Engineering - Previous Solved Exam, Exams of Engineering Chemistry

Main points of this exam paper are: Gate Dielectric, Materials, Avoiding Tunneling, Maintaining, Necessary, Electrical, Including Leakage Currents

Typology: Exams

2012/2013

Uploaded on 04/01/2013

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Name:___Key_____
CHEMICAL ENGINEERING 179
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 1023 (mole)-1; e = 1.602
x 10-19 C; mp = 1.673 x 10-27 kg ; 1 liter = 1000 cm3 ; 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.
1. What is ‘high k gate dielectric?’
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).
2. What is considered the most likely limitation to future semiconductor device scaling?
Heat must be dissipated from currents/resistances in devices, including leakage currents.
Not mention heat dissipation : -1
3. What is the mathematical form for solid state diffusivity? Is this an 'activated process?'
0
exp( )
d
E
DD kT
=
+3
This is an activated process. +2
4. Write the expression for radiative heat flux, defining each term.
4
rad
jT
ε
σ
=
, where
ε
is emissivity,
σ
is Stefan-Boltzmann constant, T is absolute temperature.
pf3
pf4

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Name:___ _Key______

CHEMICAL ENGINEERING 179

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.

  1. What is ‘high k gate dielectric?’

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).

  1. What is considered the most likely limitation to future semiconductor device scaling?

Heat must be dissipated from currents/resistances in devices, including leakage currents.

Not mention heat dissipation : -

  1. What is the mathematical form for solid state diffusivity? Is this an 'activated process?'

0 exp(^ ) D D Ed kT

This is an activated process. +

  1. Write the expression for radiative heat flux, defining each term.

4 j rad = εσ T ,^ where

ε is emissivity, σ is Stefan-Boltzmann constant, T is absolute temperature.

  1. What is the key dimensionless quantity in the Deal-Grove model?

k D

δ

, where k is oxidation rate coefficient, δ is film thickness and D is diffusivity.

  1. What happens when the quantity is (a) small compared to 1 and (b) large compared to 1? Sketch the O-species concentration profiles in each case.

(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.

  1. Write the equation for mean speed in terms of the speed distribution function (hint: the integral form).

2 3/ 2 2 0

( ) 4 exp( ) 2 2

m mv v v dv kT kT

= ∫ • v

  1. What is the difference between mean thermal speed of a gas and the average gas velocity?

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).

  1. What is a ‘chemically amplified resist?’

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.

  1. Estimate the kinematic viscosity (

= ) of Ar at 1 Torr and 300K if its diffusivity at 1 atm

(760 Torr) and 300K is 0.1 cm 2 /s.

D ൌ

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 ଵ଴כଷכସ 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: -