Solutions Manual to Accompany FUNDAMETALS OF SEMICONDUCTOR FABRICATION G. S. May Motorola Foundation Professor School of Electrical & Computer Engineering Georgia Institute of Technology Atlanta, GA, USA S. M. SZE UMC Chair Professor National Chiao Tung University National Nano Device Laboratories Hsinchu, Taiwan 1
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Solutions Manual to Accompany
FUNDAMETALS OF
SEMICONDUCTOR
FABRICATION
G. S. May Motorola Foundation Professor
School of Electrical & Computer Engineering Georgia Institute of Technology
Atlanta, GA, USA
S. M. SZE UMC Chair Professor
National Chiao Tung University National Nano Device Laboratories
Hsinchu, Taiwan
1
John Wiley and Sons, Inc
New York. Chicester / Weinheim / Brisband / Singapore / Toronto
If the initial oxide thickness is 20 nm = 0.02 µm for dry oxidation, the value ofτcan be obtained
as followed:
(0.02)2 + 0.166(0.02) = 0.01 (0 +τ)
or
τ= 0.372 hr.
For an oxidation time of 20 min (=1/3 hr), the oxide thickness in the window area is
x2+ 0.166x = 0.01(0.333+0.372) = 0.007
or
x = 0.0350 µm = 35 nm (gate oxide).
For the field oxide with an original thickness 0.45 µm, the effectiveτis given by
τ= .hr 72.27)45.0166.045.0(01.01)(1 22 =×+=+ Axx
B
x2+ 0.166x = 0.01(0.333+27.72) = 0.28053
or x = 0.4530 µm (an increase of 0.003µm only for the field oxide).
3. x2 + Ax = B )( τ+t
)(4
)2
22 τ+=− tBAAx( +
++= )(
4 )
2
22 τt
BABAx +(
8
when t >> τ , t >> B
A4
2
,
then, x2 = Bt
similarly,
when t >> τ , t >> B
A4
2
,
then, x = )( τ+tAB
4. At 980(=1253K) and 1 atm, B = 8.5×10-3 µm 2/hr, B/A = 4×10-2 µm /hr (from Figs. 3.6 and 3.7).
Since A ≡2D/k , B/A = kC0/C1, C0 = 5.2×1016 molecules/cm3 and C1 = 2.2×1022 cm-3 , the diffusion
coefficient is given by
.s/cm104.79
hr/m101.79
hr/m 102.5102.2
2105.8
222
29-
23
216
223
0
1
0
1
×=
×=×××
=
=
⋅==
−
µ
µ
CCB
CC
ABAAkD
5. For impurity in the oxidation process of silicon,
segregatio =t coefficein n2SiO inimpurity of ionconcentrat mequilibriu
silicon inimpurity of ionconcentrat mequilibriu .
6. . 006.0500
3105103
13
11
==××
=κ
7. The SUPREM input file for the first oxidation step is:
TITLE Problem 3-7a
COMMENT Initialize silicon substrate
INITIALIZE <100> Silicon
COMMENT Oxidize the wafer for 60 minutes at 1100 C in dry O2
DIFFUSION Time=60 Temperature=1100 DryO2
PRINT Layers
STOP End Problem 3-7a
The result of the PRINT Layers command is:
9
layer material type thickness dx dxmin top bottom no. (microns) (microns) node node 2 OXIDE 0.1088 0.0100 0.0010 798 804 1 SILICON 1.9521 0.0100 0.0010 805 1000 This indicates an oxide thickness of 0.1088 µm, which means 0.44 * 0.1088 µm = 0.0479 µm of
silicon has been consumed during the process (i.e., the Si/SiO2 interface is 0.0479 µm below the
original Si surface). The figure below is a graphical representation.
For the half of the wafer, the oxide is removed, and the wafer is re-oxidized in wet O2. For this half,
we use the following SUPREM input file:
TITLE Problem 3-7b
COMMENT Initialize silicon substrate
INITIALIZE <100> Silicon
COMMENT Oxidize the wafer for 60 minutes at 1100 C in dry O2
DIFFUSION Time=60 Temperature=1100 DryO2
COMMENT Remove the oxide
ETCH Oxide All
COMMENT Oxidize the wafer for 30 minutes at 1000 C in wet O2 DIFFUSION Time=30 Temperature=1000 WetO2 PRINT Layers
STOP End Problem 3-7b
The result of this PRINT Layers command is: layer material type thickness dx dxmin top bottom no. (microns) (microns) node node 2 OXIDE 0.2291 0.0100 0.0010 803 814 1 SILICON 1.8513 0.0100 0.0010 815 1000
This indicates a final oxide thickness of 0.2291 µm on the etched side, which means an additional 0.44 10
* 0.2291 µm = 0.1008 µm of silicon has been consumed during the process. The total distance from
the Si/SiO2 interface to the original Si surface on this side is therefore 0.0479 µm + 0.1008 µm =
0.1487 µm.
The unetched side is simulated using the SUPREM input file:
TITLE Problem 3-7c
COMMENT Initialize silicon substrate
INITIALIZE <100> Silicon
COMMENT Oxidize the wafer for 60 minutes at 1100 C in dry O2
DIFFUSION Time=60 Temperature=1100 DryO2
COMMENT Oxidize the wafer for 30 minutes at 1000 C in wet O2 DIFFUSION Time=30 Temperature=1000 WetO2 PRINT Layers
STOP End Problem 3-7c
The result of this PRINT Layers command is: layer material type thickness dx dxmin top bottom
4. (a) The resolution of a projection system is given by
178.065.0
mμ193.06.01 =×==NA
klmλ µm
== 222 )65.0(
m193.05.0)(
µλNA
kDOF = 0.228 µm
(b) We can increase NA to improve the resolution. We can adopt resolution enhancement
techniques (RET) such as optical proximity correction (OPC) and phase-shifting Masks
(PSM). We can also develop new resists that provide lower k1 and higher k2 for better
13
resolution and depth of focus.
(c) PSM technique changes k1 to improve resolution.
5. (a) Using resists with high γ value can result in a more vertical profile but throughput decreases.
(b) Conventional resists can not be used in deep UV lithography process because these resists
have high absorption and require high dose to be exposed in deep UV. This raises the
concern of damage to stepper lens, lower exposure speed and reduced throughput.
6. (a) A shaped beam system enables the size and shape of the beam to be varied, thereby
minimizing the number of flashes required for exposing a given area to be patterned.
Therefore, a shaped beam can save time and increase throughput compared to a Gaussian
beam.
(b) We can make alignment marks on wafers using e-beam and etch the exposed marks. We can
then use them to do alignment with e-beam radiation and obtain the signal from these marks
for wafer alignment.
X-ray lithography is a proximity printing lithography. Its accuracy requirement is very high,
therefore alignment is difficult.
(c) X-ray lithography using synchrotron radiation has a high exposure flux so X-ray has better
throughput than e-beam.
7. (a) To avoid the mask damage problem associated with shadow printing, projection printing
exposure tools have been developed to project an image from the mask. With a 1:1
projection printing system is much more difficult to produce defect-free masks than it is with
a 5:1 reduction step-and-repeat system.
(b) It is not possible. The main reason is that X-rays cannot be focused by an optical lens. When
it is through the reticle. So we can not build a step-and-scan X-ray lithography system.
8. All of the above values can be entered from the Parameters menu or by clicking on the appropriate
14
icon on the toolbar. The resulting resist profile is shown in the figure below.
In comparison to Example 3, we see that no resist feature is printed under the modified process
conditions. The combination of the long pre-bake time and low exposure dose prevents the feature
from being defined.
15
CHAPTER 5
1. As shown in the figure, the profile for each case is a segment of a circle with origin at the
initial mask-film edge. As overetching proceeds the radius of curvature increases so that the
profile tends to a vertical line.
2. (a) 20 sec
0.6 × 20/60 = 0.2 µm…..(100) plane
0.6/16 × 20/60 = 0.0125 µm……..(110) plane
0.6/100 × 20/60 = 0.002 µm…….(111) plane
22.12.025.120 =×−=−= lWWb µm
(b) 40 sec
0.6 × 40/60 = 0.4 µm….(100)plane
0.6/16 × 40/60 = 0.025 µm…. (110) plane
0.6/100 × 40/60 = 0.004 µm…..(111) plane
4.025.120 ×−=−= lWWb = 0.93 µm
(c) 60 sec
0.6 ×1 = 0.6 µm….(100)plane
0.6/16 ×1 = 0.0375 µm…. (110) plane
0.6/100 ×1= 0.006 µm…..(111) plane
=×−=−= 6.025.120 lWWb 0.65 µm.
3. Using the data in Prob. 2, the etched pattern profiles on <100>-Si are shown in below.
(a) 20 sec l = 0.012 µm, W µm 5.10 == bW
16
(b) 40 sec l = 0.025 µm, W µm 5.10 == bW
(c) 60 sec l = 0.0375 µm W µm. 5.10 == bW
4. If we protect the IC chip areas (e.g. with Si3N4 layer) and etch the wafer from the top, the width of the bottom surface is
18846252100021 =×+=+= lWW µm
The fraction of surface area that is lost is
( × 100%=(1884221
2 /) WWW − 2-10002) /18842× 100% = 71.8 %
In terms of the wafer area, we have lost
71.8 % × =127 cm2)2/15(π 2
Another method is to define masking areas on the backside and etch from the back. The width of each square mask centered with respect of IC chip is given by
6252100021 ×−=−= lWW = 116 µm
Using this method, the fraction of the top surface area that is lost can be negligibly small.
6. The higher the temperature, the faster defects anneal out. Also, the solubility of
electrically active dopant atoms increases with temperature.
7. ox
t CQ
V 1V 1 ==∆
where Q1 is the additional charge added just below the oxide-semiconductor surface by ion implantation. COX is a parallel-plate capacitance per unit area
given by d
C sox
ε=
(d is the oxide thickness, sε is the permittivity of the semiconductor)
8. The discussion should mention much of Section 7.3. Diffusion from a surface film
avoids problems of channeling. Tilted beams cannot be used because of
shadowing problems. If low energy implantation is used, perhaps with
preamorphization by silicon, then to keep the junctions shallow, RTA is also
necessary.
9. From Eq.11
84.022.0
6.04.0erfc21
=
−=
SSd
The effectiveness of the photoresist mask is only 16%.
023.022.06.01erfc
21
=
−=
SSd
The effectiveness of the photoresist mask is 97.7%.
10. 510 2
1 −==u
e2-u
πT
02.3=∴u
d = Rp + 4.27 pσ = 0.53 + 4.27 × 0.093 = 0.927 µm
29
11. The SUPREM input file for this problem is:
TITLE Problem 7-11 COMMENT Initialize silicon substrate INITIALIZE <100> Silicon Phosphor Concentration=1e14 COMMENT Implant Boron IMPLANT Boron Energy=30 Dose=1e13 PRINT Layers Active Concentration Phosphorus Boron Net PLOT Active Net Cmin=1e11 STOP End Problem 7-11
The resulting doping profile is shown in the figure below.
Examining this figure and the SUPREM output file gives: (a) The peak of the implanted boron occurs at a depth of 0.11 µm. (b) The boron concentration at the peak is 8.59e17 cm-3. (c) The junction depth is 0.4492 µm.
30
12. The SUPREM input parameters that must be determined are the dose and
implant energy. The dose can be determined from Eq. 11 in Chapter 6 as
DtCtQ s13.1)( ≅
where Cs can be read directly from the SUPREM output file for Example 3 in
Chapter 6 as 4.6e19 cm-3, 163.2 −≈ eD cm2/s for boron at 850 oC (see Figure
6.4), and t = 900 s. Substituting these numbers into the above expression gives
a dose of 2.36e13 cm-2.
The implant energy required can be approximated by matching the diffused
and implanted concentration profiles at the surface (x = 0) and at the junction
and using Eq. 1 to solve for Rp and σp simultaneously. Note from the
SUPREM output file corresponding to Example 3 in Chapter 6 that the
junction occurs at xj = 0.0555 µm, at which point the doping concentration is
1016 cm-3. As stated before, the surface concentration is 4.6e19 cm-3.
These equations cannot be solved analytically, but after several iterations, the
approximate values of Rp and σp required are found to be 0.011 µm and 0.008
µm, respectively. These values correspond to an implant energy of 5 keV
(extrapolating from Figure 7.6a). The require SUPREM input file is therefore:
TITLE Problem 7-12 COMMENT Initialize silicon substrate INITIALIZE <100> Silicon Phosphor Concentration=1e16 COMMENT Implant Boron IMPLANT Boron Energy=5 Dose=2.36e13 PRINT Layers Active Concentration Phosphorus Boron Net PLOT Active Net Cmin=1e11 STOP End Problem 7-12
The resulting doping profile appears in the figure below.
31
32
CHAPTER 8
1. MkT
dvf
dvvf
v
v
av πν 8
0
0 ==∫∫
∞
∞
Where
−
=
kTM
kTMf
2exp
24 2
22/3 νν
πν
M: Molecular mass
k: Boltzmann constant = 1.38×10-23 J/k
T: The absolute temperature
ν: Speed of molecular
So that
cm/sec1068.4m/sec 4681067.129
3001038.122 427
23
×==××
×××= −
−
πν av .
2. cm)Pain (
66.0P
=λ
Pa 104.4150
66.066.0 3−×===∴λ
P .
3. For close-packing arrange, there are 3 pie shaped sections in the equilateral
triangle. Each section corresponds to 1/6 of an atom. Therefore
dd
N s
23
21
613
triangle theof area trianglein the contained atoms ofnumber
×
×==
=282 )1068.4(3
232
−×=
d
= . 214 atoms/cm 1027.5 ×
33
dd
4. (a) The pressure at 970°C (=1243K) is 2.9×10-1 Pa for Ga and 13 Pa for As2. The
arrival rate is given by the product of the impringement rate and A/πL2 :
Arrival rate = 2.64×1020
2L
AMTP
π
= 2.64×1020
×
×
× −
2
1
125
124372.69109.2
π
= 2.9×1015 Ga molecules/cm2 –s
The growth rate is determined by the Ga arrival rate and is given by
(2.9×1015)×2.8/(6×1014) = 13.5 Å/s = 810 Å/min .
(b) The pressure at 700ºC for tin is 2.66×10-6 Pa. The molecular weight is 118.69.
Therefore the arrival rate is
s⋅×=
×
××
×−
2102
620 cmmolecular/ 1028.2
125
97369.1181066.21064.2
π
If Sn atoms are fully incorporated and active in the Ga sublattice of GaAs, we
have an electron concentration of
. cm 1074.12
1042.4109.21028.2 3-17
22
15
10
×=
×
××
5. The x value is about 0.25, which is obtained from Fig. 8.7.
34
6. The lattice constants for InAs, GaAs, Si and Ge are 6.05, 5.65,5.43, and 5.65 Å,
respectively. Therefore, the f value for InAs-GaAs system is
066.005.6)05.665.5( −=−=f
And for Ge-Si system is
. 39.065.5)65.543.5( −=−=f
7. (a) For SiNxHy
2.11NSi
==x
∴ x = 0.83
atomic % 2083.01
100=
++ yyH =
∴ y = 0.46
The empirical formula is SiN0.83H0.46.
(b) ρ= 5× 1028e-33.3×1.2 = 2× 1011 Ω-cm
As the Si/N ratio increases, the resistivity decreases exponentially.
8. Set Ta2O5 thickness = 3t, ε1 = 25
SiO2 thickness = t, ε2 = 3.9 Si3N4 thickness = t, ε3 = 7.6, area = A then
tA
3C 01
OTa 52
εε=
AAA 020302ONO
tttC
1εεεεεε
++=
( )tA
32
032ONO 2
Cεε
εεε+
=
( ) ( ) 37.56.79.33
6.729.3253
2C 32
321
ONO
OTa 52 =××
×+=
+=
εεεεε
. C
35
9. Set BST thickness = 3t, ε1 = 500, area = A1 SiO2 thickness = t, ε2 = 3.9, area = A2 Si3N4 thickness = t, ε3 = 7.6, area = A2 then
( )tA
tA
32
2032101
23 εεεεεεε
+=
.0093.0AA
2
1 =
10. Let
Ta2O5 thickness = 3t, ε1 = 25 SiO2 thickness = t, ε2 = 3.9 Si3N4 thickness = t, ε3 = 7.6 area = A then
dA
tA 0201
3εεεε
=
.468.03
1
2 tt
d ==εε
36
11. The deposition rate can be expressed as
r = r0 exp (-Ea/kT)
where Ea = 0.6 eV for silane-oxygen reaction. Therefore for T1 = 698 K
−== 116.0 exp2
)()(
211
2
kTkTTrTr
ln 2 =
− 300
698300
0259.06.
2T0
4 T2 =1030 K= 757 .
12. We can use energy-enhanced CVD methods such as using a focused energy
source or UV lamp. Another method is to use boron doped P-glass which will
reflow at temperatures less than 900 .
13. Moderately low temperatures are usually used for polysilicon deposition, and
silane decomposition occurs at lower temperatures than that for chloride reactions.
In addition, silane is used for better coverage over amorphous materials such
SiO2.
14. There are two reasons. One is to minimize the thermal budget of the wafer,
reducing dopant diffusion and material degradation. In addition, fewer gas phase
reactions occur at lower temperatures, resulting in smoother and better adhering
films. Another reason is that the polysilicon will have small grains. The finer
grains are easier to mask and etch to give smooth and uniform edges. However,
for temperatures less than 575 ºC the deposition rate is too low.
15. The flat-band voltage shift is
FBV∆ = 0.5 V ~ 0C
Qot
288
14
0 F/cm109.610500
1085.89.3 −−−
−
×=×
××==
dC oxε
.
∴ Number of fixed oxide charge is
37
2-1119
80 cm101.2
106.1109.65.05.0
×=×
××=
−
−
qC
To remove these charges, a 450 heat treatment in hydrogen for about 30
minutes is required.
16. 20/0.25 = 80 sqs.
Therefore, the resistance of the metal line is
5×50 = 400 Ω .
17. For TiSi2
30 × 2.37 = 71.1nm For CoSi2
30 × 3.56 = 106.8nm.
18. For TiSi2:
Advantage: low resistivity
It can reduce native-oxide layers
TiSi2 on the gate electrode is more resistant to
high-field-induced hot-electron degradation.
Disadvantage: bridging effect occurs.
Larger Si consumption during formation of TiSi2
Less thermal stability
For CoSi2:
Advantage: low resistivity
High temperature stability
No bridging effect
A selective chemical etch exits
Low shear forces
Disadvantage: not a good candidate for polycides
38
Ω×=×××
××==−−
− 344
6 102.3103.01028.0
11067.2ALρR19. (a)
F..
...STL
dAC 13
4
64414
109210360
1010110301085893 −−
−−−
×=×
×××××××===
εε
ns 93.0109.2103.2 155 =×××= − RC
(b) Ω×=×××
××==−−
− 344
6 102103.01028.0
1107.1LA
R ρ
F101.21036.0
1103.01085.88.2dA 13
4
414−
−
−−
×=×
×××××===
SC εε
TL
ns 42.0101.2102 133 =×××= −RC
(c) We can decrease the RC delay by 55%. Ratio = 093