0506
SOLUTIONS MANUAL
to
INTRODUCTION TO
MICROELECTRONIC
FABRICATION
SECOND EDITION
by
RICHARD C. JAEGER
-1-
© 2002 Prentice Hall
CHAPTER 1
1.1
Answering machine
Alarm clock
Automatic door
Automatic lights
ATM
Automobile:
Engine controller
Temp. control
ABS
Electronic dash
Automotive tune-up equip.
Bar code scanner
Battery charger
Calculator
Camcorder
Carbon monoxide detector
Cash register
Cellular phone
Copier
Cordless phone
Depth finder
Digital watch
Digital scale
Digital thermometer
Digital Thermostat
Electric guitar
Electronic door bell
Electronic gas pump
Exercise machine
Fax machine
Fish finder
Garage door opener
GPS
Hearing aid
Inkjet & Laser Printers
Light dimmer
Musical greeting cards
Keyboard synthesizer
Keyless entry system
Laboratory instruments
Model airplanes
Microwave oven
Musical tuner
Pagers
Personal computer
Personal planner/organizer
Radar detector
Radio
Satellite receiver/decoder
Security systems
Smoke detector
Stereo system
Amplifier
CD player
Receiver
Tape player
Stud sensor
Telephone
Traffic light controller
TV & remote control
Variable speed appliances
Blender
Drill
Mixer
Food processor
Fan
Vending machines
Video games
Workstations
Electromechanical Appliances*
Air conditioning
Clothes washer
Clothes dryer
Dish washer
Electrical timer
Thermostat
Iron
Oven
Refrigerator
Stove
Toaster
Vacuum cleaner
*These appliances are historically based only
upon on-off (bang-bang) control. However,
many of the high-end versions of these
appliances have now added sophisticated
electronic control.
-2-
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
1.2
(a) A = π d2/4
d (mm)
A
(mm2)
25
49
1
50
196
0
75
442
0
100
785
0
125
1230
0
150
1770
0
(b) n = π (450)2/(4)(12) = 159043
1.3
200
3140
0
300
7070
0
450
159000
(b) n = π (450)2/(4)(252) = 254
(a) n = π (300)2/(4)(202) = 177
(b) n = 148
0.1977( 2020−1960)
13
1.4
B = 19.97 x 10
1.5
N = 1 0 2 7 x 1 00.1505( 2020−1970) = 3 4.4 x 1 09 tr a n s is tor
s
1.6
= 1.45 x 10 bits


lo g B2 
 B1 
0.1977( Y −1960)
B = 1 9.9 7 x 1 0
Y2 − Y1 =
0.1 9 7 7
lo g2
( ) = 1.5 2 y e a r s b Y − Y = lo g1
( 0) = 5.0 6 y e a r s
(a ) Y2 − Y1 =
( ) 2 1
0.1 9 7 7
0.1 9 7 7
1.7
-3-
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition


lo g N2 
 N1 
0.1505( Y−1970)
N = 1027 x 10
Y2 − Y1 =
0.1505
lo g2
( ) = 2.00 years b Y − Y = lo g10
( ) = 6.65 years
(a ) Y2 − Y1 =
( ) 2 1
0.1505
0.1505
-4-
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
−0.06079 ( 2020 −1970 )
1.8
Using 5 Å for the
diameter of an atom, this feature size is only 15 atoms wide. However, this narrow
width can probably can be achieved.
1.9
(3 x 108 tubes)(0.5 W/tube) = 150 MW!
1.10
(a) L = (25mm)(18mm/0.5mm) = 0.90 m !
F = 8.214 x 10
µm = 7.50 x 10
−3
µm = 75 Å.
IRMS = (150 MW)/(220 VRMS) = 685 kA
(b) L = (25mm)(18mm/0.2mm) = 2.25 m !!
1.11
Two Possibilities
276 Dice
1.12
277 Dice
(a) From Fig. 1.1b , a 75 mm wafer has 130 total dice. The cost per good die is $400/
(0.35 x 130) = $8.79 for each good die. (b) The 150 mm wafer has a total of 600 dice
yielding a cost of $400/(0.35 x 600) = $1.90 per good die.
1.13
-5-
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
(a) N = 5000
2
25 (12 ) = 1 million transistors
(b) N = 5000
2
25 (0.25 2 )= 16 million transistors
(c) N = 5000
2
2
25 (0.1
) = 100
million transistors
-6-
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
1.14
Thermal oxidation
n+ diffusion mask
Oxide etch
n+ diffusion and oxidation
Contact opening mask
Oxide etch
Metal deposition
Metal etch mask
Metallization etch
Mask 1
Mask 2
Mask 3
1.15
p
n+
E
C
n+
B
-7-
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
CHAPTER 2
2.1
(a) If Y is the yield at each step, then Y25 = 0.3 or Y = 95.3 %.
(b) Y25 = 0.7 or Y = 98.6 %.
2.2
(a) Three of many possibilities
(b) Three of many possibilities
2.3
SiO2
SiO2
(a)
3 µm
(b)
-8-
3 µm
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
2.4
1) Negative resist – n+ mask
2) Negative resist – Contact mask
3) Positive resist – Metallization mask
n+ Mask
2.5
Contact
Mask
Metal
Mask
1 λ 1  193n m
= 
 = 0.536
2 F 2  180n m
(a )
NA=
(b )
DF = 0.6
(a )
NA =
2
λ
4F 2
4(180n m)
=
0.6
= 0.6
= 0.403 µm
2
NA
λ
193n m
2.6
1λ
2F
1=

1 
λ


2  0.25µm 
λ = 0.5 µm = 500 nm
λ
0.5µm
= 0.6
= 0.3 µm
2
NA
12

1λ
1
λ

NA =
0.5 = 
λ = 250 nm
2F
2  0.25µm 
DF = 0.6
(b )
DF = 0.6
2.7
Fmin ≅
λ
0.25µm
= 0.6
= 0.6 µm
2
NA
0.5 2
λ 193nm
=
= 96.5 nm or Fm in ≅ 0.1 µm
2
2
-9-
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
2.8
Fmin ≅
λ 13nm
=
= 6.5 nm or Fm in ≅ 0.0065 µm
2
2
- 10 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
CHAPTER 3
3.1
Using Fig. 3.6 with 100 nm = 0.1 µ m: (a) Wet O2 yields 0.15 hours or approximately
9 minutes. (b) Dry O2 yields 2.3 hours. Nine minutes is too short for good control, so
the dry oxidation cycle would be preferred.
3.2
Using Figure 3.6: The first 0.4 µ m takes 0.45 hours or 27 minutes. The second 0.4
µ m takes (1.5-0.45) hours or 63 minutes. The third 0.4 µ m takes (3.2-1.5) hours or
102 minutes.
3.3


d Xo  D No 
1
 X o + D  d Xo = D No t
=

or
dt  M  X + D
ks
M

o
ks
Integrating and rearranging where α is an integration constant yields:
 M 


 + X o  M  + Mα
t = X 2o 
 2DN o 
 No k s  DN o
B=
Assuming τ = 0 at Xo = Xi:
2D No
M
A=
2D
ks
τ=
Mα
D No
X 2i
Xi
+
=τ
B ( B/ A )
Problems 3.4 through 3.10 evaluate the following equations with spreadsheets.


4B
X o = 0.5 A  1 + 2 (t + τ) − 1


A
2
τ = X i B + X i (B A )
2
t = Xo B + X o (B A ) − τ
3.4
T
1150
1150
1150
3.5
B/A
5.322
5.322
5.322
<100> Silicon - Wet Oxygen
B
Xi
tau
0.667
0
0.000
0.667
1
1.688
0.667
2
6.375
Xo
1
2
3
t (hrs)
1.688
4.687
6.687
(a)
<100> Silicon - Wet Oxygen
- 11 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
T
850
B/A
6.116E-02
B
1.219E-01
Xi
0
tau
0
Xo
0.01
t (hrs)
1.643E-01
0.164 hours represents only 9.86 minutes and is too short a time for good control.
- 12 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
3.5
(b)
T
1000
3.6
B/A
4.478E-02
<100> Silicon - Dry Oxygen
B
Xi
tau
1.042E-02
0.025 6.182E-01
Can't grow 0.01 um (< 0.025 um)
Xo
0.01
t (hrs)
---
(a) Slightly over six hours
(b)
T
1150
3.7
B/A
5.322
<100> Silicon - Wet Oxygen
B
Xi
tau
0.667
0.000
0.000
(a) Approximately 3 hours in wet oxygen
Xo
2.000
t (hrs)
6.375
(b) Over 70 hours in dry oxygen
(c)
T
1050
T
1050
3.8
<100> Silicon - Wet Oxygen
B
Xi
tau
4.123E-01
0
0
<100> Silicon - Dry Oxygen
B/A
B
Xi
tau
8.920E-02
1.592E-02
0.025
3.195E-01
B/A
1.504E+00
Xo
1
t (hrs)
3.090
Xo
1
t (hrs)
73.71
(a)
<100> Silicon - Dry Oxygen
T
B/A
B
A
Xi
tau
t
Xo (µ m)
1100
0.169
0.024
0.140
0.025
- 13 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
0.174
0.500
0.074
<100> Silicon - Wet Oxygen
T
B/A
B
A
Xi
tau
t
Xo (µ m)
1100
2.895
0.529
0.183
0.074
0.036
2.000
0.950
<100> Silicon - Dry Oxygen
T
B/A
B
A
Xi
tau
t
Xo (µ m)
1100
0.169
0.024
0.140
0.950
43.931
- 14 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
0.500
0.956
(b)
<111> Silicon - Dry Oxygen
T
B/A
B
A
Xi
tau
t
Xo (µ m)
1100
0.284
0.024
0.083
0.025
0.115
0.500
0.086
<111> Silicon - Wet Oxygen
T
B/A
B
A
Xi
tau
- 15 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
t
Xo (µ m)
1100
4.865
0.529
0.109
- 16 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
3.9
(a)
T
1000
B/A
4.478E-02
B
1.042E-02
T
1100
B/A
2.895
B
0.529
<100> Silicon - Dry Oxygen
A
0.233
<100> Silicon - Wet Oxygen
A
0.183
Xi
0.025
tau
0.618
t
1.000
Xo (µ m)
0.058
Xi
0.058
tau
0.026
t
5.000
Xo (µ m)
1.542
(b) From Fig. 3.6, 1 hr at 1000 oC in dry oxygen produces approximately 0.053 µ m
oxide, and 5 hours at 1100 oC in wet oxygen produces a 1.5 µ m thick oxide. The
0.053-µ m oxide would grow in less than 0.1 hour in wet oxygen at 1100 oC and has
a negligible effect on the wet oxide growth.
3.10
(a)
T
1100
B/A
0.284
B
0.024
T
1100
B/A
4.865
B
0.529
<111> Silicon - Dry Oxygen
A
0.083
<111> Silicon - Wet Oxygen
A
0.109
Xi
0.025
tau
0.115
t
1.000
Xo (µ m)
0.126
Xi
0.126
tau
0.056
t
5.000
Xo (µ m)
1.582
(b) From Fig. 3.7, 1 hr at 1100 oC in dry oxygen produces approximately 0.12-µ m
oxide, and 5 hours at 1100 oC in wet oxygen produces a 1.5 µ m thick oxide. The
0.12-µ m oxide would grow in less than 0.1 hour in wet oxygen at 1100 oC and has a
negligible effect on the wet oxide growth.
3.11
To make a numeric calculation, we must choose a temperature – say 1100 oC. Using
the values from Table 3.1 for wet oxygen at 1100 oC on <100> silicon yields (B/A) =
2.895 µ m/hr and B = 0.529 µ m2/hr. In the oxidized region, the initial oxide Xi =
2
0.2 µ m which gives τ = X i B + X i /(B/ A ) = 0.144 hrs. The time required to reach
a thickness of 0.5 µ m = 0.52/0.53 + 0.5/2.9 - 0.144 = 0.50 hrs. In the unoxidized
region, 0.5 hours oxidation yields
[
2
]
X o = 0.5 (0.183 ) 1 + 4 (0.53 )(0.5 )/(0.183 ) −1 = 0.43 µm
This result can also be obtained using Fig. 3.6 in a manner similar to the solution of
Problem 3.13.
20 nm
50 nm
Original
43 nm
Final
- 17 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
Note that this result is almost independent of the temperature chosen. The growth in
the unoxidized area ranges from 41 nm at 1000 oC to 44 nm at 1200 oC.
3.12
To make a numeric calculation, we must choose a temperature – say 1100 oC. Using
the values from Table 3.1 for wet oxygen at 1100 oC on <100> silicon yields (B/A) =
2.895 µ m/hr, B = 0.529 µ m2/hr and A = 0.183 µ m. In the unoxidized region, we
2
desire Xo = 1 µ m which gives t = Xo B + Xo /(B/ A ) = 2.24 hrs. In the oxidized
2
region, the initial oxide Xi = 1 µ m which gives τ = X i B + X i /(B/ A ) = 2.24 hrs.
The final thickness in the oxidized region is


2.895
X o = 0.5 (0.183µm ) 1 + 4
(4.472 ) − 1 = 1.45 µm


0.183
1 µm
1.45 µm
1 µm
Original
Final
Note that this result is almost independent of the temperature chosen. The total
growth in the oxidized area ranges from 1.49 µ m at 1000 oC to 1.43 µ m at 1200 oC.
The 1-µ m region will appear carnation pink in color, and the 1.45-µ m region will
appear violet.
3.13
Using Fig. 3.6: At 1100 oC, 1.4 µ m of oxide could be grown in 4 hours. However,
the wafer has 0.4 µ m oxide already present and appears to have already been in the
furnace for 0.45 hours. Thus, 3.55 hours will be required to grow the additional 1
µ m of oxide. The oxide will appear to be orange in color.
T
1100
B/A
2.895
(100) Silicon - Wet Oxygen
B
Xi
0.529
0.400
tau
0.441
Xo
1.400
t (hrs)
3.749
3.14
Using Figure 3.10: A four-hour boron diffusion at 1150 oC requires 0.07 µ m of
oxide. A one-hour phosphorus diffusion at 1050 oC requires 0.4 µ m SiO2.
3.15
Using Figure 3.10: A 15-hour boron diffusion at 1150 oC requires a minimum of
approximately 0.15 µ m of oxide as a barrier layer.
- 18 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
3.16
Using Figure 3.10: A 20-hour phosphorus diffusion at 1200 oC requires a minimum of
3 µ m of oxide as a barrier layer.
3.17
Using Table 3.2: The 1-µ m thick oxide region will appear carnation pink in color.
The 2-µ m thick oxide region will also appear carnation pink in color.
3.18
2Xox = kλ /n = 0.57k/1.46 = 0.39k µ m yielding 0.39, 0.78, 1.17 and 1.56 µ m.
3.19
Computer program – Implement oxidation equations.
3.20
Computer program – Implement oxidation equations.
- 19 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
3.21(a) TITLE
INITIALIZE
DIFFUSION
DIFFUSION
DIFFUSION
PRINT
PLOT
STOP
PROBLEM 3.21
<100> SILICON, BORON CONCENTRATION=1E15
THICKNESS=5.0 XDX=0 DX=0.02 SPACE=200
TEMP=1100 TIME=30 DRY02
TEMP=1100 TIME=120 WET02
TEMP=1100 TIME=30 DRY02
LAYERS
CHEMICAL NET LP.PLOT
(b) Change the second statement:
INITIALIZE
<100> SILICON, ARSENIC CONCENTRATION=1E15
For (a) and (b), XO = 0.92 µ m. Problem 3.8 yielded 0.96 µ m. Boron is slightly
depleted at the silicon surface in (a) and arsenic pile-up is exhibited at the surface in
(b).
3.22
TITLE
INITIALIZE
DIFFUSION
DIFFUSION
DIFFUSION
PRINT
PLOT
STOP
PROBLEM 3.22
<111> SILICON, BORON CONCENTRATION=3E15
THICKNESS=5.0 XDX=0 DX=0.02 SPACE=200
TEMP=1100 TIME=30 DRY02
TEMP=1100 TIME=120 WET02
TEMP=1100 TIME=30 DRY02
LAYERS
CHEMICAL NET LP.PLOT
XO = 0.96 µ m. Boron is slightly depleted at the silicon surface. Problem 3.8 yielded
0.99 µ m.
3.23
TITLE
INITIALIZE
DIFFUSION
PRINT
PLOT
STOP
PROBLEM 3.23
<100> SILICON, BORON CONCENTRATION=2.7E15
THICKNESS=5.0 XDX=0 DX=0.02 SPACE=200
TEMP=1150 TIME=408.7 WET02
LAYERS
CHEMICAL BORON LP.PLOT
The result is XO = 2.0 µ m. Boron is slightly depleted at the silicon surface and
approximately uniform in the oxide. Problem 3.6 yielded 2.0 µ m in 6.375 hours
(382.4 min). The simulation requires more time to reach 2 µ m. SUPREM yields a
- 20 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
1.93-µ m oxide in 382 min. The oxidation coefficients are slightly different in
SUPREM.
For phosphorus, change the second statement to:
INITIALIZE
<100> SILICON, PHOSPHORUS CONCENTRATION=2.7E15
The result is unchanged: XO = 2.0 µ m. The phosphous concentration in the oxide is
much lower than for the boron doped substrate.
3.24
TITLE
INITIALIZE
DIFFUSION
PRINT
PLOT
STOP
PROBLEM 3.24
<100> SILICON, BORON CONCENTRATION=5E15
THICKNESS=5.0 XDX=0 DX=0.02 SPACE=200
TEMP=1050 TIME=197.2 WET02
LAYERS
CHEMICAL BORON LP.PLOT
XO = 1.0 µ m.
For dry oxidation:
DIFFUSION
TEMP=1050 TIME=4419 DRY02
For phosphorus, change the second statement to:
3.25
INITIALIZE
<100> SILICON, PHOSPHORUS CONCENTRATION=5E15
TITLE
INITIALIZE
PROBLEM 3.25 Region 1
<100> SILICON, THICKNESS=5.0 XDX=0 DX=0.02
SPACE=200
TEMP=1100 TIME=141.5 WET02
LAYERS
CHEMICAL NET LP.PLOT
DIFFUSION
PRINT
PLOT
STOP
XOX = 1.0 µ m.
TITLE
INITIALIZE
DIFFUSION
DIFFUSION
PRINT
PLOT
STOP
PROBLEM 3.25 Region 2
<100> SILICON, THICKNESS=5.0 XDX=0 DX=0.02
SPACE=200
TEMP=1100 TIME=141.5 WET02
TEMP=1100 TIME=141.5 WET02
LAYERS
CHEMICAL NET LP.PLOT
XO = 1.44 µ m.
- 21 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
If the oxidation times are changed to 134.2 min., the oxide thicknesses are 0.97 µ m
and 1.40 µ m.
- 22 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
CHAPTER 4

xj  2
 where D t= 10-8 cm2
(a) 10 = 5x10 exp −
 2 D t
15
4.1
18
(
)
x j = 2.92 2 Dt = 5.8 µm
 1015 
−1


x
=
2
D
t
erfc
(b) j
18 
 5 x10 
(
)
and
x= 5.3 µm
j
(c) Using Fig. 4.16 (b) with a surface concentration of 5 x 1018/cm3 and a background
concentration of 1015/cm3 yields Rs xj = 270 ohm-µ m. Dividing by the junction depth
of 5.8 µ m yields Rs = 47 ohms/. For the erfc profile, use Fig. 4.16(a) yielding 320
ohm-µ m and 60 ohms/ with xj = 5.3 µ m.
10 20
10 19
10 18
10 17
10 16
10 15
10 14
0
1
2
3
4
DistanceFromSurface(um
)
5
6
7
8
(d)
4.2
Using Fig. 3.10: (a) approximately 0.05 µ m (b) 1 µ m
4.3
(a) Using Fig. 4.8, a 1 ohm-cm n-type wafer has a background concentration of 4.0 x
[
2
]
1015 /cm3. So: 5 x1018 exp −(x j / 2 Dt ) = 4.0 x 1015. Solving for Dt with xj = 4 x
- 23 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
10-4 cm yields Dt = 5.61 x 10-9 cm2. 1100oC = 1373K, and D = 10.5 exp (-3.69/kT)
= 2.99 x 10-13 cm2/sec yielding t = 1.88 x 104 sec or 5.21 hours (313 min.).
(b) Using Fig. 4.16(c) with a surface concentration of 5 x 1018/cm3 and a background
concentration of 4.0 x 1015/cm3 yields Rs xj = 330 ohm-µ m or Rs = 83 ohms/❏ for xj
= 4 µ m.
18
−9
14
2
(c) Q = N o πDt = 5x10 π(5.61x10 ) = 6.64 x 10 / cm
(d) Assume a solid-solubility limited constant source predeposition with Q =
2No Dt / π . Try T = 1000 oC. No = 1 x 1021/cm3 and D = 10.5 exp (-3.69/8.617 x
10-5 x 1273) = 2.58 x 10-14. Solving for Dt yields Dt = 3.46 x 10-13 and t = 13.4 sec
which is a too short to control. Try T = 900 oC. No = 5.5 x 1020/cm3 and D = 1.47 x
10-15. Solving for t yields 13.0 minutes which is short but probably usable.
4.4
(a) An 1 ohm-cm n-type wafer has a doping NB = 4 x 1015/cm3 from Fig. 4.8. For the
boron profile, N(x) = 5 x 1018 exp -(x2/4Dt). Setting N(4µ m) = NB yields 2 Dt =
1.5 x 10-4 cm. For phosphorus at 950 oC, Ns = 7 x 1020/cm3, and D = 6.53 x 10-15
cm2/sec. Using t = 1800 sec yields 2 Dt = 6.86 x 10-6 cm. The junction occurs for:
4x1015+7x1020erfc (xj/6.86x10-6) = 5x1018exp [-(xj/1.5x10-4)2]
This equation can be solved approximately by realizing that the boron profile is
almost constant near the surface. Thus, 7 x 1020 erfc (xj/6.86 x 10-6) ≈ 5 x 1018.
Solving for xj yields a junction depth of 0.154 µ m. Checking the boron profile at
this depth yields N = 4.95 x 1018 /cm3 so that the approximation is justified.
(b) Working iteratively with Fig. 4.21, one finds that the phosphorus and boron
profiles each have a value of approximately 4 x 1018/cm3 at a depth of 0.75 µ m
which is the junction depth.
(c) Using Fig. 4.12, we find that the 30 min curve reaches 5 x 1018/cm3 at a depth of
slightly over 0.7 µ m.
(d) From Prob. 4.3, Dt = 1.14 x 10-12 cm2 for the predeposition step, and Dt = 5.61 x
10-9 cm2 for the drive-in step. The total is Dt = 5.61 x 10-9 cm2. The Dt product for
the phosphorus step is 1.18 x 10-11 cm2, which is much smaller than the total Dt
product for the boron step. Thus, the assumption is justified.
4.5
(a) From Fig. 4.8, a 5 ohm-cm n-type wafer corresponds to NB = 9 x 1014/cm3, and Rs
xj = 7500 ohm-µ m. A p-type Gaussian diffusion gives NS = 5 x 1016/cm3. So 9 x
1014 = 5 x 1016 exp -(7.5 x 10-4/ 2 Dt ). Solving for Dt yields = 3.5 x 10-8 cm2.
Using Fig. 4.5 to find an appropriate temperature: at 1100 oC, D is of the order of
10-13cm2/sec which gives a time over 25 hours - so we will try 1150 oC. For Do = 10.5
- 24 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
and EA = 3.69 eV, D = 9.05 x 10-13cm2/sec at T = 1423 K, yielding t = 3.87 x 104 sec =
10.7 hours. The diffusion schedule would be 1150 oC for 10.7 hours. A similar
calculation using 1100 oC yields t = 32.1 hours which is a little long.
(b) As found above, NS = 5 x 1016/cm3
(c) Q = NS πDt = 1.66 x 1013/cm2
(d) Using 900 oC for 15 minutes (about as short as can be controlled), yields D = 1.49
x 10-15 cm2/sec for boron. Q = 2NO Dt π = 6.93 x 1014/cm2. This dose is almost
two orders of magnitude too high. It is very difficult to get a low enough dose by
direct diffusion.
4.6
(a) At 1000 oC, Fig. 4.6 indicates the arsenic surface concentration will be 1021/cm3.
The Dt product can be found from Eq. 4.10:
Dt =
xj
2µm
=
N

016 
2 ln B 2 ln 3x1 21

NO
 10 
→
-11
2
D t= 1 .1 8 9 x 1 0c m
(b) At 1000 oC, D = 0.32 exp(-3.56/(8.62 x 10-5)(1273)) = 2.603 x 10-15 cm2/sec, and t
= 4567 sec or 1.27 hrs, a satisfactory time.
(c) From Fig. 4.11 and Table 4.2, x j = 2.29 NODt n i .
 EG 


1.1 2
3
2
31 3
31
n i = 1.0 8x1 0 T e x p−

 = 1.0 8x1 0 (1 2 7 3) e x p −

 k T
 8.6 2x1 0−5 x1 2 7 3
and ni = 9.07 x 1017/cm3.
 0.2x10 −4 cm  2  9.07x1017 
 
 = 6.92 x 10−14 cm2
D t = 
21

2.29
  10

The calculation in (c) is a much smaller value.
4.7
Using Fig. 4.10 for a constant-source diffusion with N/NO = 10-4, the normalized
vertical xj = 2.75 units, and the normalized horizontal xj at the surface = 2.25 units.
Thus horizontal xj = (2.25/2.75) x vertical xj. The lateral diffusion = 0.5 µ m x
(2.25/2.75) or 0.41 µ m. L = Lox - 2∆ L = 3 – 0.82 = 2.18 µ m.
4.8
(a) As drawn in the figure, the body of the resistor is L/W = 100 µ m/10µ m or 10
squares, and each resistor terminal will contribute 0.35 squares for a total or 10.7
squares.
- 25 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
100 µm
90 µm
30 µm
(b) Lateral diffusion is 5 µ m, so the length and width of the resistor body become 90
µ m and 20 µ m respectively, and L/W = 4.5 squares. Each terminal now contributes
0.65 squares for a total of 5.8 squares.
- 26 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
(c)
A base diffusion is usually a Gaussian diffusion. Using Fig. 4.16(d) with a surface
concentration of 5 x 1018/cm3 and a background concentration of 1015/cm3 yields RS xj
= 400 ohm-µ m. For xj = 6 µ m, RS = 67 ohms/❏. At the mask level, the resistor
appears to have a resistance of 710 ohms. The resistor will actually have a resistance
of 330 ohms when fabricated.
4.9
(a) N = (110 µ m/20 µ m) + 2(0.14) = 5.78 ❏. A surface concentration of 5 x 1018
can be achieved by a two step-diffusion or an implant; either yields a Gaussian
profile. Using Fig. 4.10(b) with N/NO = 1016/5x1018 = 2 x 10-3, we find the ratio of
lateral to vertical diffusion to be 2.1/2.6, and the lateral diffusion = 3 µ m(2.1/2.6) =
2.4 µ m.
(b) Now N = (110 µ m/24.8 µ m) + 2(0.14) = 4.72 ❏ where the ends still contribute
approximately 0.14 ❏ each.
(c) We find RS xj = 250 Ω -µ m using Fig. 4.16(d) with NO = 3 x 1018 and NB = 1016.
For xj = 3 µ m, RS = 83 Ω /❏. R = (7.72 ❏)(83 Ω /❏) = 390 Ω .
4.10
(a)
2λ
3λ
2λ
2λ
3λ
7λ
3λ
(b) There are 3 long legs, 2 shorter legs, 4 vertical links, 8 corners and 2 contacts.
N = 3(22/2) + 2(20/2) + 4(3/2) + 8(0.56) + 2(0.35) = 64.2 ❏.
- 27 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
3λ
2λ
2λ
3λ
19λ
3λ
(c) N= 3(21/3) + 2(19/3) + 4(2/4) + 8(0.56) +2(0.5) = 41.2 ❏ where the contacts have
been estimated to contribute 0.5 squares each.
- 28 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
(d) Assume a Gaussian profile. Using Fig. 4.10(b) with N/NO = 1016/1019 = 10-3, we
find the ratio of lateral to vertical diffusion to be 2.2/2.7, and the lateral diffusion = 2
µ m (2.2/2.7) = 1.63 µ m.
N=3
22(2 ) − 2(1.63 )
20 (2) − 2 (1.63)
6 − 2 (1.63 )
+2
+4
+ 8(0.56) + 2(0.65 ) N =
2 (2) + 2(1.63 )
2(2 ) + 2(1.63 )
2(2 ) + 2(1.63 )
34.2 ❏.
(e) In this case, the lateral diffusion = 3 µ m(2.2/2.7) = 2.45 µ m, and
N=3
22(2 ) − 2(2.45)
20(2 ) − 2(2.45)
6 − 2 (2.45)
+2
+4
+ 8(0.56) + 2 (0.65)N =
2 (2) + 2(2.45)
2(2) + 2(2.45)
2(2 ) + 2(2.45)
27.3 ❏.
4.11
D(t) = DO exp -EA/k(To-Rt) ≈ DO exp -EA/kTo)(1+Rt/To) for Rt/To << 1. The integral
becomes
to
(Dt)eff = Do exp(−EA kTo )∫ exp(− EA Rt kTo2 )dt
0
and
(Dt )eff
2
= D(To )(kT o E AR ) for large enough to.
4.12
x j = 2 D o exp(− E A kT ) t ln(NO N B ) = 2 exp(− EA kT ) t ln(NO N B )
dx j dT =2 exp(−E A kT )(E A 2kT 2 ) t ln (N O NB )
x
S Tj =
dx j  T  EA
  =
x j  dT  kT
For boron or phosphorus, EA = 3.69 eV:
xj
ST =
3.69
= 31.2
8.62x10−5 (1373)
and
dx j
xj
 10 
= 31.2
 = 0.227
 1373
A 10-K error in temperature results in a 23% error in junction depth!
4.13
From Fig. 4.8, a 10 Ω -cm p-type wafer corresponds to a doping of NB = 1.2 x
1015/cm3. The two-step diffusion results in a Gaussian profile. For NO = 5 x 1016/cm3
with xj = 5 µ m, 1.2 x 1015/cm3 = 5 x 10 16/cm3 exp -( 5 µ m/2 Dt )2, and Dt = 1.676
x 10-8 cm2. Choose phosphorus as the impurity (As and Sb diffuse too slowly). After
- 29 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
several attempts, T = 1125 oC is found to give a satisfactory time. The results are D =
10.5 exp (-3.69/(8.62x10-5)(1398)) = 5.17 x 10-13 cm2/sec which then yields t = 3.24 x
104 sec or 9.00 hrs. The drive-in step is 1125 oC for 9 hours.
16
−8
13
3
The dose Q = N O πDt = 5x10 π(1.676 x10 ) = 1.147 x10 /cm . For the pre-
13
2
deposition step, N O Dt = Q π 2 = 1.017x10 / cm . This is a low dose, so let us
try 900 oC, the lowest temperature in Fig. 4.6. From Fig. 4.6, N O = 5.3 x 1020/cm3
which yields Dt = 3.682 x 10-16 cm2. At 900 oC, D = 10.5 exp (-3.69/(8.62x10-5)
(1173)) = 1.449 x 10-15 cm2/sec, and t = 0.254 sec! Even at this low temperature, we
cannot achieve a controllable time. We will have to drop to 800 oC and try again. At
800 oC, NO ≈ 4 x 1020/cm3 which yields Dt = 8.223 x 10-16 cm2. At 800 oC, D = 4.954 x
10-17 cm2/sec, and t = 16.6 sec – still not workable. We will have to use an ionimplantation step (discussed in the next chapter). Another possibility is a liquid
“spin-on” doping source with a fixed concentration.
4.14
NB = 1016/cm3 and xj = 2 um,
−4
18
16
2 Dt = 2x10
ln (2x10 10 )= 8.689 x 10-5 cm. Using Eq. 4.13,
x j
 −1 
2x10 −4
  x  2   −1
R S =  ∫ qµ p N A (x)dx  = 1.602x10−19 (300)(2x1018 ) ∫ exp − 
  dx 

  
2
Dt


0

0
For
NO
=
2
x
1018/cm3,
Integrating with the QUAD function in MATLAB® yields RS = 135 Ω /❏.
4.15
(a) From Fig. 4.8, a 0.3 Ω -cm p-type wafer corresponds to NB = 5 x 1016/cm3. For NO
−4
20
16
ln (
10
5x10 )= 7.254 x 10-5 cm.
= 1020/cm3, and xj = 2 um, 2 Dt = 2x10
−1
0.5x10−4

  x 2 
−19
20
R S = 1.602x10 (100) 10
exp − 
  dx  = 14.5 Ω / s q.
∫

 2 Dt   

0
1.0x10 − 4

  x  2   −1
−19
20
R S = 1.602x10 (100) 10
exp − 
  dx  = 34.9 Ω /sq.
∫


 2 D t   
0.5x10 − 4
1.5x10−4

  x  2   −1
R S = 1.602x10−19 (100) 1020 ∫ exp − 
  dx  = 203 Ω /sq.


 2 D t  

1.0x10−4
2.0x10−4

  x  2   −1
R S = 1.602x10−19 (100) 1 020 ∫ e xp − 
  dx = 2890 Ω /sq.


 2 Dt   

1.5x10 −4
(b) Putting these four sheet resistance values in parallel yields RS = 9.72 Ω /❏.
( )
( )
( )
( )
(c) Irvin’s curves for an n-type Gaussian layer with NB = 5 x 1016/cm3 and NO =
1020/cm3 gives RS xj = 25 Ω -µ m, and Rs = 12.5 Ω /❏.
- 30 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
4.16
Find the surface concentration and junction depth after predeposition and drive-in for
boron with a 900 oC 15-minute pre-deposition and a 5-hr 1100 oC drive-in. From
Table 4.2: xj = N ODt ni ; D = 3.17exp(-3.59/kT), NO = 2.78 x1017(RS xj)-1 and Q =
0.67 NO xj. At 900 oC, NO = 1.1x1020/cm3 from Fig. 4.6, and D = 1.21 x 10-15cm2/sec.
x j = 2.45 (1.1x10 20 )(1.21x1015 )(900)/ 4x1018 = 0.134 µm
and Q = 9.9x1014/cm2 (7 times larger than in Ex. 4.3).
From Ex. 4.2, NB = 3 x 1016/cm3, D = 2.96 x 10-13 cm2/sec and t = 18000 sec. After
the drive-in, x j = 2 Dt ln(NO N B ) = 3.44 µm (24% greater).
4.17
Using Laplace Transforms, we get an ordinary differential equation:
N(x,t = 0 − )
d 2 N(x,s) sN(x,s)
(1)
−
=−
dx2
D
D
where N(x, t = 0-) = 0 for no impurities in the wafer until t > 0. For this case the
solution is N(x,s) = A (s)exp (−x s D )since N must be finite at d = ∞.
Constant Source Diffusion: For this case, the boundary condition is N(0,t) = N Ou(t),
and

 x 
N
s 
N(x,s ) = O exp  −x
.
. Using transform tables, N(x,t ) = N Oerfc

 2 Dt 
s
D

Limited Source Diffusion: For this case the boundary condition is N(x, t = 0 -) =
Qδ (x) where Q is the impurity dose in atoms/cm2. Integrating equation (1) for x = 0to x = 0+ yields dN(x,s)/dx = -Q/D, and therefore A (s) = Q/ Ds . From the
transform tables,
 x2 
Q

N(x, t ) =
exp  −
πDt
 4Dt 
4.18
RS = 1/σ t where t is the layer thickness. From Fig. 4.6, the maximum electrically
active concentration for boron is 4.3 x 1020/cm3 and 5 x 1020/cm3 for arsenic. From
the expressions in Prob. 4.24, the limiting mobilities at high concentration are 48 and
92 cm2/V-sec for holes and electrons, respectively.
Boron: RS = 1/qµ Nt = 1/(1.6x10-19 x 48 x 4.3x1020 x 10-4) = 3.0 ohms/❏ for t = 1
µ m, and 12.1 ohms/❏ for t = 0.25 µ m.
- 31 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
Arsenic: RS = 1/(1.6x10-19 x 92 x 5x1020 x 10-4) = 1.4 ohms/❏ for t = 1 µ m, and 5.4
ohms/❏ for t = 0.25 µ m.
4.19
We must find the time t such that 1016 = 1018erfc(xj/2 Dt ) for xj = 4 x 10-2 cm. This
yields Dt
= 1.82. From Fig. 4.5, gold has a diffusion coefficient of
approximately 4 x 10-7 at 1000 oC, and t = 300 seconds. Only 5 minutes is required
for gold to completely diffuse through the wafer!
- 32 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
4.20
$SUPREM IV -- Use Default Grid
INITIALIZE
<100> SILICON PHOSPHORUS=3E16 WIDTH=6.5
$Diffusion Barrier
DEPOSITION
OXIDE THICKNESS=0.5
ETCH
OXIDE RIGHT P1.X=4
$Predeposition
DIFFUSION
TEMP=900 TIME=15 BORON GAS.CONC=1.2E21
$Reflect about right edge to complete structure
STRUCTURE REFLECT RIGHT
$Plot Boron Contours
SELECT
Z=LOG10(BORON) TITLE=”Contours of Boron Concentration”
PLOT.2D
SCALE Y.MAX=13 Y.MIN=0
FOREACH
X (16 17 18 19)
CONTOUR VAL=X COLOR=2
END
CONTOUR
VALE=3E16 LINE.TYP=2 COLOR=2
4.21
TITLE
INITIALIZE
DIFFUSION
PRINT
PLOT
DIFFUSION
PRINT
PLOT
DIFFUSION
PRINT
PLOT
PLOT
ETCH
DIFFUSION
PRINT
PLOT
PLOT
STOP
PROBLEM 4.21 TWO STEP DIFFUSION FROM EXAMPLE 4.3
<100> SILICON, PHOSPHOUS CONCENTRATION=4E15
THICKNESS=6.0 XDX=0 DX=0.015 SPACE=400
TEMP=900 TIME=15 BORON GAS.CONC=1.2E21
LAYERS
CHEMICAL BORON LP.PLOT
TEMP=1100 TIME=304 DRY02
LAYERS
CHEMICAL BORON LP.PLOT
TEMP=1100 TIME=76 WET02
LAYERS
CHEMICAL BORON LP.PLOT
CHEMICAL NET LP.PLOT
OXIDE
TEMP=950 TIME=30 PHOSPHORUS SOLIDSOL
LAYERS
CHEMICAL PHOSPHORUS LP.PLOT
CHEMICAL NET LP.PLOT CMIN=1E13
Boron concentration is high in the oxide and becomes somewhat depleted below the
1018 level at the surface of the final profile. The boron junction depth is predicted to
be 4.3 µ m, slightly greater than the 4 µ m calculated by hand. The second pn
junction occurs at a depth of 0.48 µ m. Using Fig. 4.12, the 30 min. curve intersects
a level of 1018 at a greater depth of 0.9 µ m.
- 33 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
4.22
V=
IRS ln(2)
ln(2 )
= (1 0−5 )(3 0 0)
= 0.6 6 2 m V
π
π
- 34 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
4.23
(a) The cylinder contains 100 ft3 x 0.001 = 0.1 ft3 of diborane. The volume of the
room is 10 x 12 x 8 or 960 ft3. 0.1 ft3/960 ft3 = 1.04 x 10-4 or approximately 100 ppm.
(b) Life threatening exposure is 160 ppm for 15 min. Evacuate rapidly!
(c) Life threatening exposure is 6-15 ppm for 30 min. Evacuate immediately!
4.24
From Prob. 4.15, 2 Dt = 7.254 x 10-5 cm.
 2x10 −4
 −1
  x  2
20


RS =
qµ N(x))N(x)dx
where N(x) = 10 exp − 
 
 ∫ (


 
2
Dt

0
and µ n(N) is given in this problem. Using QUAD integration in MATLAB® yields
RS = 9.81 Ω /❏ for the n-type diffusion. Repeating with µ p(N) yields RS = 19.3
Ω /❏ for a p-type layer.
- 35 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
CHAPTER 5
5.1
From Figs. 5.3 (a) and (b), 60 keV through 0.25 µ m SiO2 yields Rp = 0.19 µ m and
∆ Rp = 0.09 µ m. For the Gaussian implant, N(x) = Np exp[-(x-Rp)2/2∆ Rp2] with NP =
Q/∆ Rp 2π = 4.43 x 1018/cm3.
(a) N(0.25 µ m) = 4.43 x 1018exp [-(0.25-0.19)2/2(0.09)2] = 3.5 x 1018/cm3.
∞
 (x − 0.19µm)2 
18
 dx = (0.751) 2πN ∆R
(b) QSi = ∫ 4.43x10 exp −
p
p
 2(0.09µm )2 
0.25µm
QSi = 7.5 x 1013/cm2
(c) 3 x 1015 = 4.43 x 1018exp [-(xj –0.19)2/2(0.09)2], and xj = 0.34 + 0.19 = 0.53 µ m
from the implant peak; 0.47 µ m from the Si-SiO2 interface.
5.2
From Fig. 5.3 for boron at an energy of 10 keV, Rp = 0.031 µ m and ∆ Rp = 0.015
µ m. In this case, Rp is only two times ∆ Rp, and the full Gaussian profile will not be
completely below the surface. The dose Q is given by
∞
  x − 0.031 2 
Q = ∫ NP exp− 
  dx with x inµm


 
0.015
2

0
Numerical integration with MATLAB® shows that only 98.06% of the profile is in the
silicon, so that
Q=
[ 2πN
P
(0.015µm )](0.9806 ) and NP = 5.42 x 1020/cm3 for Q = 2 x 1015/cm2.
  x − 0.031 2 
j
  = 1016 yields xj = 0.10 µm which agrees
5.42x10 exp − 
Then:
  0.015 2  
well with the graph of the profile given below.
20
- 36 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
10 21
10
20
10
19
10 18
10 17
10
16
10 15
0
0.05
0.1
0.15
0.2
0.25
0.3
Distance From Surface (um)
0.35
0.4
0.45
0.5
5.3
1 µ m SiO2 is equivalent to 1 µ m of Si. From Fig. 5.3, an implant of phosphorus to
a depth of 1µ m requires an energy of 900 keV, and yields ∆ RP = 0.13 µ m.
5.4
First, calculate the total oxide thickness needed to ensure that the implanted impurity
concentration is less than 1015/10 = 1014/cm3 at the Si-SiO2 interface. We know that
RP = 0.05 µ m, the thickness of the oxide. For arsenic, this requires E = 80 keV from
Fig. 5.3, and ∆ Rp = 0.017 µ m. NP = Q/∆ RP 2π = 2.35 x 1017/cm3. From Eqn.
(5.9) with NP/NB = 235, XO = RP + 3.94∆ RP. XO = 0.05 + 3.94(0.017) = 0.117 µ m of
oxide. The additional oxide required is 3.94(0.017) = 0.067 µ m. However, Xnitride =
0.85 XO so only 0.057 µ m of silicon nitride is required.
5.5
Using Irvin's curves for a p-type Gaussian layer, Fig. 4.16(d), with R S xj = 625 ohmµ m yields NP = 2.7 x 1018/cm3. At x = xj, 2.7 x 1018exp[-(5x10-4/2 Dt )2] = 1016,
and Dt = 1.06 x 10-4 cm. For the final layer, the dose in silicon is Q Si = NP πDt
= 5.1 x 1014/cm2. With the pre-deposition implant peak at the surface, the implanted
dose will be 2 x QSi or 1.0 x 1015/cm2. D = 10.5 exp (-3.69/kT) = 2.96 x 10-13 at T =
1373 K, and the drive-in time is t = 3.80 x 104 sec or 10.5 hours.
5.6
Using Irvin's curves for a p-type Gaussian layer, Fig. 4.16(d), with R S xj = 400 ohmµ m yields NP = 4 x 1018/cm3. At x = xj, 4 x 1018exp[-(2x10-4/2 Dt )2] = 1016, and
Dt = 4.09 x 10-5 cm. For the final layer, the dose in silicon is Q Si = NP πDt =
2.9 x 1014/cm2. With the pre-deposition implant peak at the surface, the implanted
dose will be 2 x QSi or 5.8 x 1015/cm2. D = 10.5 exp (-3.69/kT) = 2.96 x 10-13 at T =
1373 K, and the drive-in time is t = 5.64 x 103 sec or 1.57 hours.
- 37 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
5.7
Using Irvin's curves for a p-type Gaussian layer, Fig. 4.16(d), with R S xj = 62.5 ohmµ m yields NP = 6 x 1019/cm3. At x = xj, 6 x 1019exp[-(2.5x10-5/2 Dt )2] = 1016, and
Dt = 4.24 x 10-6 cm. For the final layer, the dose in silicon is Q Si = NP πDt =
4.5 x 1014/cm2. With the pre-deposition implant peak at the surface, the implanted
dose will be 2 x QSi or 9.0 x 1014/cm2. D = 10.5 exp (-3.69/kT) = 2.96 x 10-13 at T =
1373 K, and the drive-in time is t = 60.7 sec! This very short time will require rapid
thermal annealing, or the drive-in temperature could be reduced.
5.8
From Fig. 5.3 for phosphorus at an energy of 20 keV, R p = 0.025 µ m and ∆ Rp =
0.012 µ m. In this case, Rp is only two times ∆ Rp, and the full Gaussian profile will
not be completely below the surface. The Dose Q is given by
∞
  x − 0.025  2 
Q = ∫ NP exp− 
  dx with x inµm
  0.012 2  
0
Numerical integration with MATLAB® shows that only 98.14% of the profile is in the
silicon, so that
Q=
[ 2πN
P
(0.012µm )] (0.9814 ) and NP = 3.39 x 1020/cm3.for Q = 1015/cm2.
  x − 0.025 2 
j
  = 1016 yields jx= 0.80 µm which agrees
3.39x10 exp − 
Then:

0.012
2
 

well with the graph of the profile given below.
20
10
21
10
20
10
19
10 18
10
17
10
16
10
15
0
0.05
0.1
0.15
Distance From Surface (um)
- 38 -
0.2
0.25
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
5.9
(a) Irvin's curves can be used for each half of the distribution. We have 10 15 =
1019exp[-(xj-1)2/2(.11)2] with xj in µ m. Solving this equation yields (xj - RP) = 0.47
µ m which corresponds to xj normally used in Irvin's curves. From Irvin's curves
with NP = 1019/cm3, RS xj = 230 ohm-µ m, and RS = 490 ohms/❏ for xj = 0.47 µ m.
We effectively have two of these regions in parallel, so the total RS = 245 ohms/❏.
(b) Q = NP∆ RP 2π = 2.76 x 1014/cm2.
(c) RP = 1 µ m requires 470 keV for boron from Fig. 5.3.
(d) From (a) xj = 1 ± 3.03 ∆ RP 2π = 1.47 µ m and 0.53 µ m.
5.10
(a) Using Fig. 5.3 for 50-keV boron, Rp = 0.15 µ m and ∆ Rp = 0.050 µ m. Since the
concentrations are not known, let us assume a worst-case situation with NP = 1021/cm3
and NB = 1014/cm3. Also, remember that photoresist also requires 80% more
thickness than Si or SiO2. Thus the minimum photoresist thickness will be
X PR = 1.8 (R P + 6.1∆RP ) = 0.82 µm
(b) For 50-keV phosphorus, Rp = 0.060 µ m and ∆ Rp = 0.025 µ m. XPR = 0.38 µ m.
(c) For 50-keV arsenic, Rp = 0.033 µ m and ∆ Rp = 0.012 µ m. XPR = 0.19 µ m.
- 39 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
5.11
mv2
2E
or v=
; Also, remember 1 Joule = 1 kg-m2/sec2.
2
m
(a) For B+, m = 10.811(1.673 x 10-27 kg) = 1.809 x 10-26 kg.
E=
v=
2(5keV)(1.602x10 −19 J/ eV)
1.809x10
−26
kg
= 2.98 x 105 m /sec
(b) For (BF2)++, m = [10.811+2(18.998)](1.673 x 10-27 kg) = 8.17 x 10-26 kg.
v=
2(10keV)(1.602x10 −19 J /eV )
8.165x10
−26
kg
5
= 1.98 x 10 m /sec
(a) For (B10H14)+, m = [10(10.811)+14(1.079)](1.673 x 10-27 kg) = 2.05 x 10-25 kg.
v=
5.12
2(5keV)(1.602x10 −19 J/ eV)
2.045x10 −25 kg
= 8.85 x 10 4 m/ sec
From Fig. 5.4 we see that the concentration is largest at x = R P. Using Eqn. 5.7, N(y)
= 0.5 x 1020{erfc [(y-a)/ 2 (.022)]} = 1016 for the junction edge. (y-a) = 2.63 2
(0.022) = 0.082 µ m. ∆ L = 2(.082) = 0.16 µ m. A graph of the concentration at x =
RP is given below. A blowup of the junction region agrees with the above
calculation.
- 40 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
10
21
10 20
10 19
10 18
10 17
10
16
10 15
10 14
0
0.2
0.4
0.6
0.8
DistancefromCenter of Opening(um)
- 41 -
1
1.2
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
5.13
100 keV: RP = 0.30 µ m and ∆ RP = 0.07 µ m.
200 keV: RP = 0.55 µ m and ∆ RP = 0.09 µ m.
The doses are given by Q = 2πNP ∆R P and are 8.77 x 1013/cm2 and 1.13 x
1014/cm2 for the two implants with NO = 5 x 1018/cm3.

  x − 0.3   2
  x − 0.55  2 
  + exp − 
   as plotted below.
So: N( x) = 5x10  exp− 

  0.07 2  
  0.09 2   
18
From the graph above, it is clear that the shallow profile controls the position of the
first junction and the deep profile controls the second junction.
  x − 0.3  2 
j1
18
  = 1 016 yie ld s j1x = 0.0 53 µm
5x1 0 e xp− 
  0.0 7 2  
  x − 0.5 5 2 
j2
  = 1 016 yie ld s j2x = 0.87 µm
5x10 e xp− 
  0.0 9 2  
These two junction values agree well with the graph of the profile.
18
5.14
Approximately (5 x 1022) x (0.2 x 10-4) = 1 x 1018 silicon atoms/cm2 are in the layer to
be formed. We need to implant two oxygen atoms per silicon atom for a total of 2 x
1018 oxygen atoms/cm2. The 125 mm wafer has an area of 123 cm2, so the total
- 42 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
number of atoms required is 2.45 x 1020 oxygen atoms. If this number is implanted in
15 minutes, 2.45 x 1020/(15 x 60) = 2.73 x 1017 atoms/sec are required. If each atom
carries a single charge, the beam current will be 43.6 mA, and the power in the 5MeV beam is 218 kW. The wafer will melt away!
5.15
∆ VT = 0.75 volts; ε ox = 3.9 x 8.854 x 10-14 F/cm; Cox = ε ox/4 x 10-6 cm = 8.63 x 10-8
F/cm2. ∆ VT = qQ/Cox and Q = Cox ∆ VT/q = 4.0 x 1011/cm2.
5.16
(1015/cm2)(π )(20/2)2 cm2 = 3.14 x 1017 total boron atoms. 10-5 A corresponds to (10-5
Coul/sec)/(1.6 x 1019 Coul/charge) or 6.25 x 1013 charges/sec. 3.14 x 1017 atoms/6.25
x 1013 atoms/sec = 5030 sec or approximately 84 minutes.
5.17
The sheet resistance is found by evaluating Eq. (4.13). N(x) is a Gaussian profile and
the mobility can be modeled by the mathematical approximations given in Prob. 4.24.
For given values of RP and ∆ RP, xj1 and xj2 can be found and the integral can be
evaluated with the QUAD function in MATLAB®, for example. Xj1 will be zero if
the Gaussian profile interests the semiconductor surface.
 x j2
 −1
R S = q ∫ µ (N) N(x ) dx 
 x j1

1270
µ n = 92 +

 0.091
N
1+ 

 1.3x1017 
 (x − R )2 
P

N( x) = N P exp  −
 2∆R 2P 
µ p = 48 +
447

 0.076
N
1+ 

 6.3x1016 
5.18
For a boron dose of 1015/cm3, 1000/T = 6.5 or T ≤ 154 K. For a phosphorus, 1000/T
= 3.25, and T ≤ 308 K.
5.19
Integrating numerically using a spreadsheet yielded 5.60 x 10-12 cm2. At 1050 oC
(1323 K), D = 9.10 x 10-14 cm2/sec, and Dt = 5.46 x 10-12 cm2.
5.20
Integrating numerically using a spreadsheet yielded 5.97 x 10-13 cm2. At 1050 oC
(1323 K), D = 9.10 x 10-14 cm2/sec, and Dt = 4.55 x 10-13 cm2.
5.21
Integrating numerically using a spreadsheet yielded 1.82 x 10-12 cm2. At 980 oC (1253
K), D = 1.49 x 10-14 cm2/sec, and Dt = 1.79 x 10-12 cm2.
- 43 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
5.22
Integrating numerically using a spreadsheet yielded 2.51 x 10-13 cm2. At 980 oC (1253
K), D = 1.49 x 10-14 cm2/sec, and Dt = 2.24 x 10-13 cm2.
- 44 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
CHAPTER 6
6.1
Φ = 2.63 x 1020 (1.013 x 105)/ 32 • 300 = 2.73 x 1023 molecules/cm2-sec. For close
packing with radius r, the area of one atom is 4r2cos(30o). For oxygen with r = 3.6 Å
(See Ex. 6.2), there are 2.23 x 1014 molecules/cm2 which yields t = 820 psec.
6.2
Φ = 2.63 x 1020 (10-3)(1.013 x 105)/ 32 • 300 = 2.73 x 1020 molecules/cm2-sec. For
close packing with radius r, the area of one atom is 4r2cos(30o). For oxygen with r =
3.6 Å (See Ex. 6.2), there are 2.23 x 1014 molecules/cm2 which yields t = 0.82 µ sec.
6.3
M = 32, T = 300 K, P = 10-4 Pa.
Φ = (2.63 x 1020)(10-4)/ 32 • 300 = 2.68 x 1020 molecules/cm2-sec
λ=
kT
(1.38x10 −23 J /K )(N − m / J )(10 2 cm/ m )(300K ) = 7190 cm
=
2
2
2πpd
2π (10 −4 N/ m2 )(10 −4 m 2 /cm 2 )(3.6x10 −8 cm )
or 71.9 m. (10-4 Pa)(0.0075 torr/Pa) = 7.5 x 10-7 torr
6.4
n=
10 −8 Pa(1 N/ m 2 /Pa)
P
=
= 2.42x1012 /m 3
kT 1.38x10−23 J /K (300K )(1 N - m/J)
n = 2.42x106 molecules/cm 3
6.5
P = nkT = (1000/ cm3 )(10 6 cm3 / m 3 )(1.38x10 −23 J/ K)(300K)(1 N - m/J)
P = 4.14x10 −12 N/ m 2 = 4.14x10−12 Pa
6.6
From Prob. 6.1, close packing of 5-Å spheres will yield = 4.62 x 1014 Al atoms/cm3.
100 nm/min = 10-5 cm/min. (10-5cm/min)/(5x10-8cm x cos(30o)/atom) = 231 atomic
layers/min = 7.70 atomic layers/sec.
Φ = (7.70 layers/sec)(4.62 x 1014 atoms/cm2-layer) = 3.56 x 1015 atoms/cm2-sec. M =
27 for Al and using T = 300 K, P = (3.56 x 10 15) 27 • 300 /2.63 x 1020 or P = 1.22 x
10-3 Pa.
6.7
Choosing φ = 0 for simplicity, cos φ = 1. At the edge of the wafer, r = 802 + 5 2 ,
and cos φ = (80/80.16). So at the wafer center, G 1 = m(12)/π ρ (802). At the wafer
- 45 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
edge G2 = m(80/80.16)2/π ρ (80.16)2. G2/G1 = (80/80.16)4 = 0.992 or 0.008 µ m
thickness variation.
6.8
Using the result from Prob. 6.7, G2/G1 = (2ro/r)4. r = 100 2 + 10 2 , = 100.5 cm. G2/G1
= (100/100.5)4 = 0.980. A 0.02 µ m thickness variation will occur.
6.9
Using the result from Prob. 6.7, G2/G1 = (2ro/r)4. G2/G1 = 0.55µ m/0.6µ m and d/r =
0.9785. r = d 2 + 15 2 and d/ d 2 + 15 2 = 0.9785. d = 71.1 cm.
6.10
4 hours = 1.44 x 104 sec. Assume oxygen molecules for example. From Ex. 6.2, NS
= 2.23 x 1014 molecules/cm2 with M = 32. t = N S/Φ and P = (2.23x1014) 32 • 300 /
(2.63 x 1020)(1.44 x 104) = 5.77 x 10-9 Pa = 4.33 x 10-11 Torr - an ultrahigh vacuum
(UHV) system.
6.11
v = (Ng/N) ks hg/(ks+hg). ks = 2 x 106 exp (-1.9)(1.6 x 10-19)/(1.38 x 10-23)(1473) = 0.64
cm/sec. hg = 1 cm/sec, Ng = 3 x 1016 atoms/cm3. v = (0.64 x 1/1.64)(3 x 1016/5 x 1022)
= 2.3 x 10-7 cm/sec or 0.14 µ m/min.
(b) At T = 1498 K, ks becomes 0.82 cm/sec, and v increases to 0.16 µ m/min. The
change is 0.02 µ m/min, a 14% increase.
(c) Setting ks = 1 yields T = 1520 K or 1247 oC.
(d) From the SiH4 curve, ks = 0.2 µ m/min at 1000/T = 0.93 and 0.01 µ m/min at
1000/T = 1.1. EA = -1000 k (∆ ln ks/∆ (1000/T)) = -1000 x 8.617 x 10-5 x (ln(.2)ln(.01))/(.93-1.1) = 1.52 eV.
6.12
The graph on the next page is generated with MATLAB ® using equations 6.31 and
6.32. For both boron and phosphorus at 1200 oC, D = 10.5 exp (-3.69/(8.617x10-5)
(1473) = 2.49 x 10-12cm2/sec. xepi = 10 µ m and vx = 0.2 µ m/min give the growth
time t = 3000 sec. 2 Dt =1.73 µ m. The two profiles are given by
 x − xepi  
1018 
 
N1 (x ) =
1 + erf 
2 
 1.73µm  
 x − xepi 


 x + xepi  
1016 
x
 + exp 
 erfc 
 
N 2 (x) =
erfc 
2 
 1.73µm 
 0.0747µm 
 1.73µm  
From a blow-up of the graph, xj = 7.2 µ m.
- 46 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
6.13
Evaporation and sputtering require good vacuum systems, whereas CVD can be done
at higher pressure or in some cases even at atmospheric pressure. Evaporation is
limited to elemental materials that can be melted. On the other hand, sputtering
replicates the target material, and metals, dielectrics and composite materials can all
be sputtered. Evaporation and sputtering tend to be low temperature processes. CVD
systems can do large wafer lots at one time, but elevated temperatures are often
involved and limit the points that CVD can be introduced into a process.
- 47 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
Graph for Problem 6.12.
10 19
10 18
10 17
10 16
10 15
0
2
4
6
8
DistanceFromSurface(um
)
10
12
14
16
6.14
The volume of aluminum required is π r2t = π (5cm)2(10-4cm) = 7.85 x 10-3 cm3.
Only 15% of the Al actually is deposited on the wafer, so the total volume of Al
required is 5.24 x 10-2 cm3. Aluminum has a density of 2.7 x 10-3 kg/cm3. One kg of
Al has a volume of 370 cm3 and can be used to deposit a film on approximately 7100
wafers.
6.15
(a) Using the result from Prob. 6.7, G2/G1 = (2ro/r)4. r = 200 2 + 50 2 = 206.2 mm.
G2/G1 = (200/206.2)4 = 0.886. A 0.11 µ m thickness variation will occur.
(b) For a 200 mm wafer 40 cm above the source: r = 400 2 + 100 2 = 412.3 mm.
G2/G1 = (400/412.3)4 = 0.886. A 0.11 µ m thickness variation will occur. For a 300
mm wafer 40 cm above the source: r = 400 2 + 150 2 = 427.2 mm. G2/G1 =
(400/427.2)4 = 0.769. A 0.23 µ m thickness variation will occur.
6.16
From Fig. 6.10, the growth rate at 1100 oC in SiCl4 is 0.1 µ m/min. The 1-µ m film
takes 600 sec to grow. D = 0.32 exp (-3.56/(8.617x10 -5)(1373)) = 2.74 x 10-14
cm2/sec, and 2 Dt = 0.081 µ m. N(x) is given by (x in µ m – see graph on next
page)

 x −1 
N(x ) = 5x1019 1 + erf 
  cm-3


0.081 
- 48 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
6.17
From Fig. 6.10, the growth rate at 950 oC in SiH4Cl2 is 0.15 µ m/min. The 2-µ m
film takes 800 sec to grow. D = 10.5 exp (-3.69/(8.617x10-5)(1223)) = 6.45 x 10-15
cm2/sec, and 2 Dt = 0.045 µ m. N(x) is given by (x in µ m)

 x − 2 
20
-3
N(x) = 10 1 + erf
  cm



0.045 
Graph for Problem 6.16
10
21
10
20
10
19
10 18
10 17
10 16
10
15
10 14
0
0.5
1
1.5
Distance From Surface (um)
2
2.5
3
Graph for Problem 6.17
- 49 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
10
21
10
20
10
19
10 18
10 17
10
16
10 15
10
14
0
0.5
1
1.5
Distance From Surface (um)
2
2.5
3
CHAPTER 7
7.1
(a) RS = ρ /t = 3.2 x 10-6 Ω -cm/10-4cm = 0.032 Ω /❏.
(b) R = RS (L/W) = 0.032Ω /❏ x (500µ m/10µ m) = 1.6 Ω .
(c) Cox = (3.9)(8.854 x 10-14F/cm)/10-4cm = 3.5 x 10-9 F/cm2.
C = Cox (WL) = (3.5 x 10-9 F/cm2)(0.05 cm)(0.001 cm) = 0.175 pF.
(d) RC = 1.6Ω x (0.175 x 10-12F) = 0.28 ps.
7.2
(a) RS = 500 x 10-6 Ω -cm/10-4cm = 5 Ω /❏. R = RS (L/W) = 5Ω /❏ x
(500µ m/10µ m) = 250 Ω . Cox = (3.9)(8.854 x 10-14 F/cm)/10-4cm = 3.5 x 10-9 F/cm2.
C = Cox (WL) = (3.5 x 10-9 F/cm2)(0.05 cm)(0.001 cm) = 0.175 pF. RC = 250Ω x
(0.175 x 10-12 F) = 43.8 ps.
(b) RS = 25 x 10-6 Ω -cm/10-4cm = 0.25 Ω /❏. R = RS (L/W) = 0.25Ω /❏ x
(500µ m/10µ m) = 12.5 Ω . Cox = (3.9)(8.854 x 10-14 F/cm)/10-4cm = 3.5 x 10-9
F/cm2. C = Cox (WL) = (3.5 x 10-9 F/cm2)(0.05 cm)(0.001 cm) = 0.175 pF. RC =
12.5Ω x (0.175 x 10-12 F) = 2.19 ps.
(c) RS = 1.7 x 10-6 Ω -cm/10-4cm = 0.017 Ω /❏. R = RS (L/W) = 0.017Ω /❏ x
(500µ m/10µ m) = 0.85 Ω . Cox = (3.9)(8.854 x 10-14 F/cm)/10-4cm = 3.5 x 10-9
F/cm2. C = Cox (WL) = (3.5 x 10-9 F/cm2)(0.05 cm)(0.001 cm) = 0.175 pF. RC =
0.85Ω x (0.175 x 10-12 F) = 0.149 ps.
- 50 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
7.3
(a) RS = 1/σ t = 1/qµ Nt
For boron: RS = 1/(1.6x10-19)(75)(4.3x1020)(0.25x10-4) = 7.8 ohms/❏.
For arsenic: RS = 1/(1.6x10-19)(100)(5x1020)(0.25x10-4) = 5.0 ohms/❏.
(b) At high surface concentrations, the curves of Fig. 4.16 stop at RS xj = 10 Ω -µ m.
A junction depth of 0.25 µ m yields a sheet resistance of 40 Ω /❏.
7.4
D = 0.04 exp (-0.92/723 x 8.62x10-5) = 1.55 x 10-8cm2/sec. t = 30 min = 1800 sec and
Dt = 52.7 µ m. The volume of aluminum that will absorb silicon is V = (2 x
53µ m)(15µ m)(1µ m) = 1590 µ m3. At T = 450 oC, the equilibrium concentration
of silicon in aluminum is 0.5%, so the volume of silicon required is VSi = 5 x 10-3VAl.
Dividing by the contact area of 100 µ m2 yields a depth of 0.08 µ m.
7.5
(a) ρ c = RA = (0.5 Ω )(10-8cm2) = 5 x 10-9 Ω -cm2.
(b) R = ρ c/A = ρ c /10-10cm2 = 50 Ω . This value is large, but it is difficult to do
much better.
7.6
(MTF2/MTF1) = exp(-0.5/k)(1/T2 - 1/T1).
For T2 = 300K and T1 = 400K, the ratio is 126.
For T2 = 77K and T1 = 400K, the ratio is 2.67 x 1026!
7.7
(a) 50% of the failures have occurred at 40 time units.
(b) 50% of the failures have occurred at 430 time units.
The copper line is 10 times more resistant to electromigration than the AlCu line.
7.8
(a) For minimum resistance we would use a constant source diffusion resulting in an
erfc profile. Using Irvin's curves, RS xj = 90 Ω -µ m and RS = 23 Ω /❏ for a 4-µ m
deep junction.
(b) The surface area per unit length including the sidewall contribution is (15+4+4) =
23 µ m2 per µ m of length. For NB = 1015,
 1015 
φ bi = 0.56V + (25.8m V) ln 10  = 0.857 V
 10 
- 51 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
For VR = 0, with KS = 11.7,
C=
7.9
1.602x10−19 (1015 )(11.7)(8.854 x10 −14 )
(25x10 ) = 2.46 fF /µm
2(0.857)
−8
I = JA = (5 x 105A/cm2)(10-4cm)(4 x 10-4cm) = 20 mA.
7.10 I = JA = (106A/cm2)(0.25 x 10-4cm)(0.5 x 10-4cm) = 1.25 mA.
L
7.11 R = ρ A = (5µΩ − cm )
10 −4 cm
(0.25 x10
−4
2
cm )
= 0.80 Ω
7.12 (a) RS = ρ /t = 1.7 x 10-6 Ω -cm/0.5x10-4cm = 0.034 Ω /❏.
(b) R = RS (L/W) = 0.034Ω /❏ x (50µ m/0.5µ m) = 3.4 Ω .
(c) Cox = (2)(8.854 x 10-14 F/cm)/10-4cm = 1.77 x 10-9 F/cm2.
C = Cox (WL) = (1.77 x 10-9 F/cm2)(50x10-4cm)(0.5x10-4cm) = 0.443 fF.
(d) RC = 3.4Ω x (0.443 x 10-15 F) = 1.5 fsec.
- 52 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
CHAPTER 8
8.1
Sheet resistance, contact resistance, threshold voltage, junction depth, alignment
errors, current gain, transconductance, oxide thickness, breakdown voltage, channel
length, etc.
8.2
2115 = 4.15 x 1034 states. At 10-7 sec/state, each die will take 4.15 x 1027 sec. There
are 3.15 x 107 sec/year, so each die will require 1.3 x 1020 years. Wafer test will
require 1.3 x 1022 years.
8.3
(a) 10 mm x 1000 µ m/mm = 104 µ m. 15 mm x 1000 µ m/mm = 1.5 x 104 µ m.
Along the 10 mm edge, there will be room for (104/125) - 2 = 78 pads. Along the 15
mm edge, there will be room for (1.5x104/125) - 2 = 118 pads. The total number of
pads is 2(78+118) = 392 pads.
(b) Along the 10 mm edge, there will be room for (10 4/100) - 2 = 98 pads. Along the
15 mm edge, there will be room for (1.5x104/100) - 2 = 148 pads. The total number
of pads is 2(98+148) = 492 pads.
(c) Along the 10 mm edge, there will be room for (104/200) = 50 solder balls. Along
the 15 mm edge, there will be room for (1.5x104/200) = 75 solder balls. The total
number of solder balls is (50 x 75) - 4 = 3746.
8.4
DoA = 10 and α = 1.0
Yield Formula
8.5
Y
Number of Good Dice
100 mm: N = 254
150 mm: N = 616
exp(-DoA)
4.54 x 10-5
0
0
(1+DoA/5)-5
4.12 x 10-3
1
2-3
{[1- exp(-DoA)]/DoA}2
0.01
2-3
6
[1- exp(-2DoA)]/2DoA
0.05
12 - 13
30 - 31
1/(1+DoA)
0.09
23
56
The yield depends on the exact positioning of the die sites. The best-case die map
yields 2 good die when DoA = 4 giving a yield of 2/26 or 7.7%. The worst-case
- 53 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
partitioning would yield no good die for 0% yield. Poisson statistics predicts Y =
exp(-4) or 1.8%. For our wafer, Y = {0.43, 0.22, 0.077} for DoA = {1, 2, 4}. For Y =
α
[1 + DoA/α ]- , numerical fitting yields an excellent fit for α = 3.1.
- 54 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
8.6
(a) Do = 10 and A = 0.4 cm2. Y = [1-exp(-8)]/8 = 0.125. N = π (R-S)2/S2 =
π (62.5mm - 6.3mm)2/(6.3mm)2 = 248 die, where S has been approximated by 40
mm. Thus, there will be Y x N = 31 good dice. The total cost will be $1.60 +
$250/31 = $9.67/die.
(b) The area of each die becomes 1.15(20) = 23 mm2. S is approximated by A =
4.8 mm, and N = 455 dice. Yield Y = [1-exp(-4.6)]/4.6 = 0.215, and Y x N = 98 good
dice. The new cost per packaged die is $1.60 + $250/98 = $4.15/die. The total cost
of the two-die set is $8.30, which is more economical!
8.7
(a) Do = 5, and wafer cost = $150. A = 40 mm2, and S = 40 mm. N = 248, and Y
= 0.245. Y x N = 61 dice, and C = [1.60 + 150/61] = $4.06. For two dice, A = 23
mm2, N = 455, Y = 0.391 and Y x N = 178. The total cost is 2[1.60 + 150/178] =
$4.89 which is more expensive.
(b) For a single chip, C = [1.60 + 300/61] = $6.52. For two dice, C = 2[1.60 +
300/178] = $6.57.
8.8
(a) For the first process: A = 25 mm2, and N = π (50-5)2/52 = 254 dice. Y = [1 – exp
(-2 x 2 x 0.25)] / (2 x 2 x 0.25) = 0.632. Y x N = 161 good dice. The cost per die is
C1 = M/161 for a wafer cost of M $/wafer. For the second process: A = 12.5 mm 2. N
= π (50-3.54)2/12.5 = 543 die, and Y = [1 - exp(-2 x 10 x 0.125)] / (2 x 10 x 0.125) =
0.367. Y x N = 199 good die per wafer. C 2 = 1.3M/199 = M/153. The new die is
only slightly more expensive even at a defect level of 10 defects/cm 2. The change is
not economical at the present time, but as the defect density improves with time, the
second process will become more economical.
(b) For equal die cost in the new process, we need Y x N = 1.3 x 161 = 209 good die.
This requires Y = 209/543 = [1 - exp (-2 x 0.125Do)]/(2 x 0.125Do). Letting X =
0.25Do, results in the equation 0 = 1 - 0.385X - exp(-X) which may be solved
numerically using Newton's method or the solver on a calculator yielding X = 2.35
and Do = 9.4
(c) We would switch since we expect the new process to improve, and the die cost
will be less as we move down the learning curve.
(d) The cost in the first process is M/161. In the new process, N = π (50- A )2/A
and Y = [1 – exp(-0.2A)]/0.2A (A in mm2). For equal cost we must have Y x N =
1.3(161), and we must solve the equation below for A (A in mm2).
π (50- A )2[1-exp(-0.2A)]/0.2A2 = 1.3(161)
Using Newton's method or a calculator solver yields A = 12.17 mm2.
8.9
α
lim (1 + DoA/α )-
α
= lim 1/(1 + DoA/α ) = exp(-DoA).
- 55 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
α →∞
α →∞
- 56 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
8.10
DoA
exp(-DoA)
α =5
α = 5000
1
2
3
4
5
6
7
8
9
10
3.68E-01
1.35E-01
4.98E-02
1.83E-02
6.74E-03
2.48E-03
9.12E-04
3.35E-04
1.23E-04
4.54E-05
4.02E-01
1.86E-01
9.54E-02
5.29E-02
3.13E-02
1.94E-02
1.26E-02
8.42E-03
5.81E-03
4.12E-03
3.68E-01
1.35E-01
4.98E-02
1.83E-02
6.75E-03
2.49E-03
9.16E-04
3.38E-04
1.24E-04
4.59E-05
8.11
N = (0.1)(π d2/4). N = 17.7, 31.4, and 70.7 for the 15, 20 and 30 cm diameter wafers.
8.12
For both wafers, Yx = (1 + 10Ax/2)-2 = 1/(1 + 5Ax)2. For the 100 mm wafer, N1 =
π (50- A 1 )2/A1 , and the cost per die is $150/N1Y1.
For the 150 mm wafer, N2 = π (75- A 2 )2/A2 , and the cost per die is $250/N2Y2.
For equal cost, $150/N1Y1 = $250/N2Y2. For a given A1, we can find the
corresponding A2. Thus we need to choose some cost, say $1/die. Setting $150/N 1Y1
= $1 and $250/N2Y2 = $1 and solving for the areas gives A1 = 14.8 mm2 and A2 =
17.8 mm2.
8.13
There are now 10 good dice out of 60 sites for a yield of 17%. Two good dice exist
out of with 24 sites with 4 times the area for a yield of 8.3%.
∞
8.14
Y = ∫ exp(−DA ) f (D) dD . After substitution of our f(D) and changing variables
0
with Z = D/Do, the integral becomes
2
2
2
Y1 =
exp[− Z(D o A )] exp -4(Z -1) dZ
∫
π0
This can be integrated numerically, and the results are compared below with Y2 = [1exp(-DoA)/DoA]2.
[
DoA
Y1
Y2
1
0.39
0.40
- 57 -
]
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
2
3
4
5
0.17
0.084
0.046
0.028
0.19
0.10
0.060
0.040
- 58 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
6
7
8
9
10
8.15
0.018 0.028
0.012 0.020
0.0092 0.016
0.0071 0.012
0.0056 0.010
We need to evaluate Pk = [n!/k!(n-k)!]Nn(N-1)n-k for N = 120 and k = {0,1,2,3,4,5}.
Let λ = n/N. For N = 120, n = 120, λ = 1. Pk may be approximated by
Pk = [λ n exp(-λ )]/k! = exp(-1)/k!.
# of
defects
0
1
2
3
4
5
8.16
N x Pk
44
44
22
7
2
0
 D o A  −5
(a) 0.70 = 1 +
 yields DoA = 0.370. Do = 0.370/1.5 = 0.247 defects/cm2.

5 
 Do A  −5
(b) 0.80 = 1 +
 yields DoA = 0.228. Do = 0.228/1.5 = 0.152 defects/cm2.

5 
 D o A  −5
(c) 0.90 = 1 +
 yields DoA = 0.107. Do = 0.107/1.5 = 0.071 defects/cm2.

5 
8.17
 D o A  −6
(a) 0.75 = 1 +
 yields DoA = 0.295. Do = 0.295/4 = 0.0737 defects/cm2.

6 
 Do A  −6
(b) 0.85 = 1 +
 yields DoA = 0.165. Do = 0.295/4 = 0.0412 defects/cm2.

6 
- 59 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
CHAPTER 9
9.1
10 nm = 10-8 m = 10-6 cm = 100Å. V = (5 x 106 V/cm) x Tox = 5 x 106 x 10-6 = 5 volts.
9.2
(a) The built-in potential = 0.56 + 0.0258 ln (1015/1010) = 0.857 V. The total voltage
across the junction is (3 + 2 + 0.857) = 5.86 volts. Dividing by the background
concentration yields a value of 5.86 x 10-15 V-cm3. From Fig. 9.4, the depletion layer
width will be 3.0 µ m. The minimum line spacing must be twice this value or 6.0
µ m.
Using Eq. 9.3 directly, 2Wd = 2
2(11.7 )(8.854 x10 −14 )(5.86 )
1.602 x10 −19 (10 15 )
= 5.51 µm .
(b) The built-in potential = 0.56 + 0.0258 ln (3x1016/1010) = 0.945 V.
2Wd = 2
9.3
2(11.7 )(8.854x10−14 )(5.95)
1.602x10−19 (3x1016 )
= 1.01 µm
The built-in potential is Φ bi = 0.56 + 0.0258 ln (3x1016/1010) = 0.945 V. For NB = 3
x1016 /cm3, Eqn. 9.3 becomes W = 2.08 x 10-5 VA + Φbi . At the source side, the
depletion layer has only the built-in potential across the junction, and Wd = 0.202
µ m. At punch-through, the depletion-layer width at the drain will be (1-0.202) =
0.798 µ m. Finding VA for Wd = 0.798 µ m yields an applied voltage of 13.8 volts.
9.4
- 60 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
4
3
2
1
0
-1
-2
-3
-4 14
10
10 15
10 16
10 17
3
Doping Concentration (#/cm )
- 61 -
10 18
10 19
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
9.5
An NMOS transistor is an enhancement-mode device if VTN > 0. Fig. 9.2 is for
polysilicon gate devices with 10-nm gate oxides. From this graph, VTN > 0 for N > 2
x 1016/cm3. A more exact estimate can be obtained from Eq. (9.2):
1.12
VTN = −
+ Φ F + 2(11.7 )(8.854x10−14 )(1.602x10−19 ) NB Φ F /CO for
2
 NB 
3.9(8.854x10 −14 )
Φ F = 0.0258 ln 10  and CO =
= 3.45 x 10-7 F /cm 2
 10 
10 −6
Using a spreadsheet or solver yields NB = 2.9 x 1016/cm3.
9.6
From Fig. P9.6, NB = 2 x 1015/cm3. Using Eq. 9.2, the threshold voltage with no
implant would be VTN ≅ 0 V. The shift caused by the implant is ∆ VTN = qQi/CO, so
we need the implanted dose which is given by Q = 2π N P ∆R P with the implant
peak at the surface. The characteristics of the implant are found by subtracting the
background concentration from the profile in Fig. P9.6. The peak concentration of
the implant is 2 x 1016/cm3 at x = 0, and the implanted profile drops to 4 x 1015 at x =
0.25 µ m. The projected range is found from 4 x 10 15 = 2 x 1016 exp-(0.25)2/2∆ Rp2,
16
−5
and the projected range is 0.139 µ m. The dose Q = 2π (2x10 ) (1.39 x10 ) =
6.99 x 1011/cm2. The oxide capacitance is 1.73 x 10-7 F/cm2 for a 20 nm gate oxide.
∆ VTN = (1.602 x10-19)(6.99x1011)/(1.73x10-7) = 0.647 V. So the resulting threshold
voltage is 1.61V.
9.7
For the rectangular distribution N(x) = Ni for 0 ≤ x ≤ xi:
∞
∞
M1 = ∫ N(x) dx = Ni x i & M2 = ∫ x N( x) dx =
0
0
Ni x2i
2
2
For the Gaussian distribution N(x) = Np exp(-x /2∆ Rp2), and
π
2
M1 =
N ∆R P a nd M2 = N P ∆R P
2 P
Equating moments yields the specified results.
9.8
Scaling factor α = 1/0.25 = 4. The circuit density increases by α 2 or 16 times more
circuits/cm2. PDP ∝ 1/ α 3, so the power-delay product is reduced (improved) by a
factor of 64.
- 62 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
K ε
*
I D = µ n  O O
XO


 α 
9.9
 W
 
 α
VD S 

V
−
V
−
 VD S = α ID
G
S
TN
 L 
2 
 
 α
ID increases by the scale factor α .
P* = α ID VDD = α P - Power/circuit increases by the scale factor.
P * αP
3 P
* = A = α
A
A
α2
-
The power density increases by the cube of the scale factor
which is very bad!
9.10
The non-implanted device will have a threshold given by
1.12
−14
−19
16
+ Φ F + 2(11.7 )(8.854 x10 )(1.602 x10 )(3x10 ) Φ F /CO = 0.721V
2
A
 3x1016 
3.9(8.854 x10 −14 )
-8
2
for ΦF = 0.0258 ln

and
C
=
=
6.91
x
10
F
/cm
O
10 
−6
 10

5x10
VTN = −
threshold voltage shift of 3.72 volts is required to achieve a -3-V threshold. The
required dose Q = CO∆ VTN/q = 1.61 x 1012/cm2.
9.11
In Fig. 1.8, reverse all the n- and p-type regions as well as the arsenic and boron
implantations.
9.12
A = 110 λ 2 versus 168 λ
2
- 63 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
22 α 11
= λ
20 α 10
= λ
- 64 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
9.13
(a) The well doping is almost constant and equal to the surface concentration in the
depletion-region beneath the gate of the MOSFET. So the substrate doping for the
NMOS device is 3 x 1015/cm3 and 3 x 1016/cm3 for the PMOS device. VTN is
calculated using Eqns. 9.2 with CO = 2.30 x 10-7 F/cm2 for a 15 nm oxide thickness.
 3x1015 
 = 0.325 V
Φ F = 0.0258ln
 1010 
VTN = −0.56 + 0.325 + 2(11.7)(8.854x10 −14 )(1.602x10−19 )(3x1015 )(0.325)/ CO = −0.16 V
 3x1016 
 = 0.385 V
Φ F = 0.0258ln
 1010 
VTP = −0.56 − 0.385 − 2(11.7)(8.854x10 −14 )(1.602x10−19 )(3x1016 )(0.385 ) /CO = −1.21V
(b) The NMOS device requires a +1.16-V shift, and the PMOS device requires a
+0.21-V shift. The required dose is given approximately by Q = C O∆ VT/q. The dose
for the NMOS device is 1.67 x 1012/cm2, whereas it is 3.02 x 1011/cm2 for the PMOS
device. Note that both shifts are positive and would utilize boron implantations.
9.14
0V
8V
0V
8V
p+
n+
3 x 1015 /cm3
n
5 x 1016 /cm3
p-well
p+n junction with 8 V bias:
Φbi = 0.56 + 0.0258 ln(3x1015 1010 ) = 0.885V
Wd =
2(11.7 )(8.854x10 −14 )(8.885 )
1.602x10 −19 (3x1015 )
= 1.96 µm
n+p junction with 8 V bias:
Φbi = 0.56 + 0.0258 ln(5x10
Wd =
16
10
10
) = 0.958V
2(11.7 )(8.854x10 −14 )(8.958 )
1.602x10 −19 (5x1016 )
= 0.481 µm
Two-sided formula for the well-substrate junction:
 3x1015 • 5x1016 
kT  N AN D 


 = 0.723V
Φ bi =
ln 2  = 0.0258ln
20
q  ni 

10

- 65 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
Wd =
2K S ε o (V A + Φbi )  N A ND 


q
 N A + ND 
Wd =
2(11.7 )(8.854x10 −14 )(8.732 )  3x1015 • 5x10 16 

 = 2.00 µm
−19
16
1.602x10

5.3x10

The minimum spacing is W = 1.96 + 0.48 + 2.00 = 4.44 µ m with no safety margin or
alignment tolerances considered.
9.15
(a) 10 Ω -cm ν (n-type) material has a doping of 4.2 x 1014/cm3 based upon Fig. 4.8.
For a junction depth of 2 µ m, we have
  2x10−4  2 
14
16
  → Dt = 3.154x10−9 cm2
4.2 x10 = 10 exp − 
  2 Dt  
For boron at 1075 oC, D = 10.5 exp[-3.69/(8.617 x 10-5 x 1348)] = 1.678 x 10-13
cm2/sec, and the drive-in time t = 5.22 hours. The dose Q = NO πDt = 9.95 x
1011/cm2. From Fig. 4.10(b) with N/NO = 0.042, X Y = 1.3/1.8 = 0.722. Lateral
diffusion = 1.4 µ m.
(b) For this case, we have
4.2 x10
14
  1.5x10−4  2 
  → Dt = 1.177x10 −9 cm2
= 5x10 exp −

2
Dt
 

16
D for phosphorus is the same as D for boron, and the drive-in time t = 1.95 hours.
The dose Q = NO πDt = 3.04 x 1012/cm2. From Fig. 4.10(b) with N/NO = 0.0084,
X Y = 2.25/2.75 = 0.818. Lateral diffusion = 1.23 µ m.
(c) The total diffusion time is 7.17 hours or 2.58 x 104 sec. For arsenic at 1075 oC, D
= 0.32 exp[-3.56/(8.617 x 10-5 x 1348)] = 1.566 x 10-14 cm2/sec. The out diffusion is
modeled approximately by Eq. (6.31). The out diffusion boundary is given by
 x − 3x1 0−4    x − 3x1 0−4 
1 020 
14
  → 
 = −3.7 8 9
4.2 x1 0 =
1 + e r f
2 
 2 D t    2 D t 
which yields x – 3 µ m = 1.52 µ m of out diffusion.
- 66 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
9.16
Contacts
Thin Oxide
Metal
9.17
14 λ
24 λ
The area is reduced from 416 λ 2 to 336 λ 2, a 19% improvement, but the gate
overlap capacitance is significantly larger. The overlap area is now 72 λ 2 versus 44
λ 2 in the original layout.
- 67 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
9.18
24 λ
24 λ
All Levels Aligned to First level (Diffusion)
Lateral Diffusion = 2µm = 1 λ
2 λ x 2 λ Contacts
22 λ
22 λ
Thin Oxide Aligned to Diffusion
Contacts and Metal Aligned to Thin oxide
Lateral Diffusion = 2µm = 1 λ
- 68 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
9.19
+VDD
VO
A
B
C
9.20
Area = (12 λ )(44 λ ) = 528 λ
2
--- Active gate area = 2(10 λ )(2 λ ) = 40 λ
- 69 -
2
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
Well Boundary
12 λ
16 λ
44 λ
9.21
Area = (12 λ )(32 λ ) = 384 λ
2
--- Active gate area = 2(10 λ )(2 λ ) = 40 λ
2
Well Boundary
12 λ
4 λ
32 λ
9.22
Approximately (5 x 1022) x (0.25 x 10-4) = 1.25 x 1018 silicon atoms/cm2 are in the
layer to be formed. We need to implant two oxygen atoms per silicon atom or a total
dose of 2.5 x 1018 oxygen atoms/cm2. Five 200-mm wafers have a total area of 1571
cm2, so a total number of 3.93 x 1021 oxygen atoms is required per hour. The beam
current will be (3.93 x 1021/hr)(1.602x10-19C)/(3600sec/hr) = 175 mA, and the power
in the 4-MeV beam is 699 kW. The wafers will be destroyed!
9.23
(a)
VDE − VCD =
 L

IR S   L
2∆Y
  + ∆Y −  − ∆Y  = IR S
 2

W  2
W
- 70 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
and
VAC = IR S
VDE − VCD
2
= ∆Y
VAC
L
L
W
a n d ∆Y =
L  VDE − VCD 


2 
VAC 
(b)
VIJ − VJH =
 L

IR S  L
2∆X
 + ∆X −  − ∆X  = IR S
 2

W  2
W
and
VGH = IR S
VIJ − VJH
2
= ∆X
VIH
L
L
W
a n d ∆X =
- 71 -
L  VIJ − V JH 
2  VIH 
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
CHAPTER 10
10.1
Current gain can be estimated using Eq. (10.1). For constant doping levels, the
Gummel numbers in the emitter and base are given respectively by:
GE = NELE/DE = (1020/cm3)(2 x 10-3 cm)/(5 cm2/sec) = 4 x 1016 sec/cm4
GB = NBWB/DB = (1018/cm3)(4 x 10-4 cm)/(20 cm2/sec) = 2 x 1013 sec/cm4
β
10.2
-1
= (2 x 1013)/(4 x 1016) + 42/1(502) & β = 145.
The base profile is N(x) = 3 x 1018 exp [-(x2/4Dt)] which must equal 1015 at xj = 4
µ m. This gives Dt = 7.07 x 10-5 cm. To simplify the problem, neglect the
depletion layer regions in the base. Then,
4.0µm
4.0x10 −4
  x 2
G B = ∫ [N(x )/D B ]dx = 1.5x10
exp − 
 dx
∫


2
Dt
 
−4

1.5 µm
1. 5x10
17
2.83
G B = (2.12x10
13
11
G B = 2.51x10
2
) ∫ exp(−z )dz
for
z=
1 .06
and
x
-4
1.41x10
13
G
5x10
β≅ E =
= 199
G B 2.51x1011
Since the depletion-layer intrusion into the base will reduce GB, one would expect β
to be even larger.
10.3
From Figs. 9.3 and 10.7, the maximum breakdown voltage can reach approximately
90 volts for a doping concentration of approximately 5 x 10 15/cm3. This assumes a
deep base diffusion with large radius of curvature.
10.4
(a) From Fig. 10.5, NOB ≅ 4 x 1017/cm3. (b) For NOB = 2 x 1017/cm3, the breakdown
voltage is 7 V.
10.5
For a doping concentration of 4 x 1015/cm3, Φ bi = 0.56V + (0.0258V) ln (4 x
1015/1010) = 0.893 V.
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© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
 4 x1015 
 = 0.893 V
Φ bi = 0.56 + (0.0258) ln
10
 10 
Wd =
2(11.7 )(8.854x10 −14 )(40.89)
1.602x10−19 (4x1015 )
= 3.64 µm
The step junction result is somewhat greater than the 3.1 µ m found in Ex. 10.3.
- 73 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
10.6
From Table 4.2 on page 81, Q = 0.55 NO xj, and NO = 1.56 x1017/RS xj which yield Q
= 8.64 x 1015/cm3 for a 10 ohm/square sheet resistance. For the heavy doping limit in
the equations in Prob. 4.24, the minority carrier diffusion constant approaches a value
of approximately (92)(0.0258) = 2.4 cm2/sec (with Dn/µ n = kT/q). The Gummel
number can be approximated by GE ≅ Q/D = 3.6 x 1015 sec/cm4.
10.7
(a) For S ≤ 0, the collector-base and emitter-base junctions are approximately the
same. From the drawing, the emitter junction depth is approximately 2 µ m. From
Fig. 10.5, the breakdown voltage is approximately 8 volts for a junction depth of 2
µ m and a doping concentration of 1018/cm3.
(b) For S = 3 µ m, breakdown will occur when the depletion layer in the lightlydoped collector region reaches the n+ contact. This is equivalent to XBL-XBC = 3 µ m
and xJC = 5 µ m in Fig. 10.7. Using Fig. 4.8, a one ohm-cm substrate corresponds to
NC = 4 x 1015/cm3. The breakdown voltage will be limited by the 3-µ m punchthrough to approximately 50 volts
(c) For S = 5 µ m, XBL-XBC = 5 µ m. Figure 10.7 indicates that the breakdown
voltage is still limited to approximately 50 volts by avalanche breakdown with the 5µ m radius of curvature of the junction.
10.8
Breakdown will occur where the doping is the heaviest on the two sides of the
junction formed by the intersection of the phosphorus implantation with the boron
diffusion. Using the data given for the profiles, breakdown should occur at a depth of
approximately 0.225 µ m where the boron doping is approximately 3.71 x 10 18/cm3.
From Fig. 10.4, we see that the total space charge region width is very narrow at these
doping levels. The diffused boron profile changes slowly near the shallow junction,
so the intersection of the implant and the boron diffusion can be approximated by a
Gaussian profile intersecting a uniform background concentration of 3.71 x 1018/cm3.
Using Figs. 9.3 and 10.7 and guessing an effective junction radius of ∆ RP yields a
breakdown voltage of 2 volts or less.
10.9
(a) The common-base current gain α = IC/IE = AC/AE where AE is the total emitter
surface area and AC is the area of the emitter surface facing the collector. α = 4(4λ )
(2λ )/[4(4λ )(2λ ) + (4λ )2] = 2/3; The common-emitter current gain β is related to
α by β = α /(1-α ) = 2. (Note that this analysis has neglected all corner effects.)
(b) In general, α = 2d(L+W)/[2d(L+W) + LW] where L, W and d are the length,
width and depth of the emitter. β is given by β = 2d(L+W)/LW = 2d[(1/W)+(1/L)].
β will be maximum when L and W are as small as possible. Setting L and W to a
minimum feature size of 2λ , β = 2d/λ .
(c) Assuming one side is a minimum feature size, the square is optimum.
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© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
(d) For the circle, α = π (4λ )(2λ )/[π (4λ )(2λ ) + π (2λ )2] = 2/3 and β = 2, the
same as the square emitter.
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© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
10.10 (a) N = 1016/cm3 at xj = 15 µ m. This is a deep boron diffusion, so we will try a
relatively high temperature, say 1150 oC. NO will be 4 x 1020/cm3 at this temperature
from Fig. 4.6. At T = 1423 K, the diffusion coefficient for boron is D = 10.5 exp(3.69/kT) = 8.86 x 10-13 cm2/sec. To find t: 1016 = 4x1020 erfc (15 x 10-4/2 Dt ) or (15
x 10-4/2 Dt ) = erfc-1 (2.5 x 10-5) = 2.965 using MATLAB®. Dt = 2.53 x 10-4 cm.
This yields t = 7.22 x 104 sec or 20.1 hours which is reasonable. So the diffusion
schedule would be 20.1 hours at 1150 oC.
(b) Let XD be the depth of the down diffusion, and XU = 15-XD is the amount of up
diffusion. We need XD + XU = 15 µ m where 1016 = 4x1020 erfc (XD/2 Dt ), and
(see Eq. 6.31) 1016 = 5x1017[1 + erf (XD-15x10-4)/2 Dt ]. The first equation yields
XD = 5.93 Dt . The second equation gives XD-15x10-4 = -3.29 Dt . Combining
these two results gives Dt = 1.63 x 10-4, and yields t = 2.99 x 104 sec = 8.3 hrs.
10.11 For a one-sided step junction (see Eqs. 9.3), VA + Φ bi = Wd2qNB/2KSε o. With Wd =
5 µ m and NB = 1015/cm3, Φ bi = 0.56 + 0.0258 ln (1015/1010) = 0.86 V resulting in VA
= 18.5 volts. Figure 10.7 predicts a breakdown voltage of approximately 60 volts.
The step junction formula ignores depletion-region penetration into the base as well
as non-uniform doping of the epitaxial layer caused by up diffusion from the buried
layer.
10.12 Using Fig. 10.7, the xjc = 5 µ m and XBL-XBC = 3 µ m curves coincide for NB = 3 x
1015/cm3 corresponding to a breakdown voltage of approximately 50 V.
10.13 In the resistor, there are 3 long legs, 2 shorter legs, 4 vertical links, 8 corners and 2
contacts. The total number of squares is
N = 3(22/2) + 2(20/2) + 4(3/2) + 8(0.56) + 2(0.35) = 64.2 ❏
Using mid-range values as nominal:
(a) 3 kΩ /❏ yields 193 kΩ
(b) 150 Ω /❏ yields 9.63 kΩ
(c) 12.5 Ω /❏ yields 803 Ω
10.14 (a)
(b)
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© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
B
E
C
B
E
C
p+
n+
n+
n+
p+
p+
p-well
n-well
n-type substrate
p-type substrate
- 77 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
10.15
B
C
E
Final Emitter
Emitter Mask
Base Mask
n + contact mask
Final n +
Final Base
Final Isolation
Isolation Mask Boundary
The emitter is aligned to the base; contacts are aligned to the emitter; metal is aligned
to the contacts. The vertical height has been expanded over minimum size to
accommodate a second base contact.
10.16 Isolation regions are not shown. The Schottky diode is bounded by the p-type ring
and base.
B
p
C
E
p
n+
n+
n-type substrate
n+ buried layer
- 78 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
10.17 1. trench etch
2. selective oxidation
3. n+ ion-implantation (I/I)
4. p+ poly pattern (N/C)
5. p I/I (N/C)
6. n+ poly pattern (N/C)
7. contact windows
8. metal
9. passivation
(and probably several more for multilevel metal)
An abbreviated process flow: Etch trench, implant p+, thermal oxidation, poly
deposition, poly etch, selective oxidation, n+ I/I & drive-in, p+ poly deposition and
oxidation, p+ poly etch, thermal diffusion & oxide growth, p-I/I, n+ poly deposition
and definition, oxide deposition, contact windows, metal deposition, metal etching,
passivation layer deposition and etching.
10.18 The base intersects the collector at a depth of approximately 275 nm and a doping of
6 x 1016/cm3. From Fig. 10.4 and using the 1-µ m junction depth, we see that the
depletion layer will extend on the order of 0.1 µ m into the collector. At that point
the doping is 1-2 x 1017/cm3. Based upon Fig. 10.7, the breakdown voltage will be 45 volts. The emitter intersects the base at a doping of approximately 6 x 10 18/cm3.
Using the 1-µ m curve in Fig. 10.5, the breakdown voltage will be less than 3 V. We
can get a punch-through estimate from Fig. 10.4 using a background concentration of
1017/cm3, a base width of 270-170 = 100 nm, and xj = 1 µ m. The depletion layer x1
on the base side will reach 100 nm for a total voltage of 4 V. Assuming a built-in
potential of 1 V, the base will punch through at approximately 3 V. A smaller
junction depth and radius of curvature will reduce the breakdown and punch through
voltages below these estimates. In addition, based upon the results in Problem 10.19,
the base will punch through when x1 reaches 83 nm.
10.19 For the emitter side, use a one-sided step junction with NB = 5 x 1018/cm3: Φ bi = 0.56
+ 0.0258 ln (5x1018/1010) = 1.08 V. Wd = [2(11.7)(8.854x10-14)(1.08)/(1.602x10-19)
(5x1018)]0.5 ≅ 17 nm. On the collector side, we can estimate the space charge layer
intrusion using Fig. 10.4 with some assumptions. For Φ bi = 0.85 V, a doping of
approximately 1017/cm3 near the junction, and a 1-µ m radius, the total SCR width is
120 nm with 42 % in the base or 50 nm. The metallurgical base width is
approximately 270 – 170 = 100 nm. The neutral base with is approximately 100–17–
50 = 33 nm, a very narrow base!
10.20 (a)
 xj 
xj
 = 5x1018 at xj = 50 nm or
1021erfc
= 1.985
 2 D t
2 Dt
For arsenic at 1000 oC, D = 0.32 exp [-3.56/(8.617x10-5)(1273)] = 2.57 x 10-15
cm2/sec. Dt = 1.59 x 10-12 cm2/sec, and t = 617 sec or 10.3 min. This is a bit short.
Reducing the temperature slightly would yield a more controlled time.
- 79 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
(b)
  x 2
 xj 2
j
16


 = 5x10 at xj = 100 nm or 
 = 4.605
5x10 exp − 
  2 D t 
 2 D t


18
For boron at 1000 oC, D = 10.5 exp [-3.69/(8.617x10-5)(1273)] = 2.58 x 10-14 cm2/sec.
Dt = 5.43 x 10-12 cm2/sec, and t = 210 sec or 3.51 min. This time is too short.
Both the base and emitter diffusions probably require rapid thermal processing to
achieve the small Dt products that are required for this structure.
10.21
C
E
B
p+
p
n-collector(+V)
p-sub(-V)
n+
n
n-typecollector
n+buriedlayer
p-substrate
Two vertical four-layer pnpn structures exist that can latch up if the biasing is not
properly maintained. Saturation of the pnp transistor will turn on the npn transistor in
the reverse direction. The pnp transistor will have high resistance between the
collector contact and the active collector region of the transistor.
E
B
C
n-collector (+V)
p-sub (-V)
10.22 τ = 1/2π fT = 3.18 psec.
(Cjc + Csub) < 3.18 psec/40 = 80 fF.
XC < (3.18 psec) x (2x107cm/sec) = 0.64 µ m.
- 80 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
WB < [(3.18psec)(10)(20cm2/sec)]0.5 = 250 nm.
CBE < 3.18psec/25Ω = 0.13 pF.
- 81 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
10.23 Latchup potential is related to the existence of four-layer pnpn structures. Proper
biasing must be maintained to prevent latchup.
(a) Laterally beneath the PMOS and NMOS transistors: n-well/pw/nw/p-well; pwell/nw/pw/n-epi. Vertically below the NMOS device: n+/p-well/n-iso/p-substrate
(b) Laterally across the tubs: npn-tub/ptub/n-tub/p-tub; between the NMOS and
PMOS devices: n+/p-tub/n-tub/p+ (c) Vertical npn: n+/p/n+/p-iso & p+/n+/p-iso/n-epi;
vertical pnp: p+/n+/p-iso/n-epi; PMOS device: p+/n+/p-iso/n-epi.
10.24 Approximately (5 x 1022) x (0.5 x 10-4) = 2.5 x 1018 silicon atoms/cm2 are in the layer
to be formed. We need to implant two oxygen atoms per silicon atom for a total dose
of 5 x 1018 oxygen atoms/cm2. Four 200-mm wafers have a total surface area of 1260
cm2, so a total number of 6.28 x 1021 oxygen atoms is required to be implanted per
hour. The beam current will be (6.28 x 1021/hr)(1.602 x 10-19C)/(3600sec/hr) = 280
mA, and the power in the 5-MeV beam is 1.40 MW. The wafers will be destroyed!
10.25 The CDI structure has a narrow base region and a heavily-doped collector without a
lightly-doped n-type region. Hence the collector-base breakdown voltage will be too
low for most analog applications.
10.26 (a) Six masks:
(b) Alignment
1. n+ buried layer
2. n+ isolation diffusion
3. n+ emitter diffusion
4. contact windows
5. metal layer
6. passivation layer openings
Align to 1
Align to 2
Align to 3
Align to 4
Align to 5
10.27 Based upon Fig. 4.8, the 0.25 ohm-cm p-type base region corresponds to a doping of
6 x 1016/cm3. The emitter-base junction is a 1-µ m diffusion into a uniform base
region which approximates the conditions of Fig. 10.7 for which 6 x 1016/cm3 and xJC
= 1 µ m yield VEB = 6-7 volts. The epi-substrate collector-base junction is
approximated by an n+p step junction. Using Fig. 9.3 with r j = ∞ and N = 6 x
1016/cm3 gives VBC = 20 V. However, the actual breakdown voltage will be less due
to curvature effects associated with the collector contact diffusions. For a 2-µ m epi
thickness (rj = 2 µ m), the breakdown voltage will be approximately 8 volts from Fig.
10.7.
- 82 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
CHAPTER 11
A useful triangle:
1
φ
2
11.1
11.2
11.3
54.74o
tanφ = 2
3
cos φ =
1
3
sin φ =
2
3
 1 1 1
 and that
The unit vector perpendicular to the <111> plane is v1 = 
 3 3 3
perpendicular to the <100> plane is v 2 = (1 0 0) . Thus we have c o s θ = v1 • v 2 and
 
−1 1
o
θ = cos   = 54.74
 3
 1 1 1
 and that
The unit vector perpendicular to the <111> plane is v1 = 
 3 3 3
 1 1 
0 . Thus we have c o s θ = v1 • v 2
perpendicular to the <110> plane is v 2 = 
 2 2 
 
−1 2
o
and θ = cos   = 35.3
 6
10
o
 W  = tan(54.74 ) →
 
 2
W = 14.1 µm
W
54.74o
10 µm
- 83 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
11.4
(a) Horizontal cross section through second P2-P1 contact region
(b) Horizontal cross section through the center P2-P1 contact region
P2
(a)
P1
P1
P2
(b)
P1
P1
11.5
(2/L) = sin (54.74o)
and
L = 2.45 µ m.
n+L
2 µm
L
54.74o
n+
11.6
The isolation opening W is 8.07 µ m. This is competitive with diffused isolation
regions, but not with deep or shallow trench isolation structures. At the time this
isolation was conceived, it was also difficult to planarize the topology.
W
5 µm
Y
54.74o
Y
o
 W  = tan(54.74 ) =
 
 2
2
Y = 5 + 0.5 2 = 5.71 µm
1 µm
W = Y 2 = 8.07 µm
- 84 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
11.7
10 λ
Polysilicon
Oxide Etch 1
Oxide Etch 2
Possible process flow: Deposit sacrificial oxide layer. Fully etch oxide leaving
rectangular oxide pad (oxide etch 1). Etch oxide to thin down (oxide etch 2). Deposit
polysilicon. Etch polysilicon (polysilicon mask). Remove sacrificial oxide. The
layout assumes a 1-λ alignment tolerance and a 2-λ minimum feature size.
11.8
The volume is constant, so (P2/P1) = (T2/T1). (P2/14.7) = (300K/673K) = 6.55 psi.
11.9
The volume is constant, so (P2/P1) = (T2/T1). (P2/1) = (623K/300K) = 2.08 psi.
11.10 The minimum spacing is 717 µ m with no spacing between cavity edges at the
surface.
W
500 µm
Y
54.74o


500
 µm
W = 2 5 +

tan(54 .74 o )

500 
W = 2 5 +
 = 717 µm

2 
10 µm
- 85 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
- 86 -
© 2002 Prentice Hall
Introduction to Microelectronic Fabrication – Second Edition
11.11 The values here are approximate from scaling the photograph. For the spring arms,
W/L ≅ 22/1. Assuming the fingers are F = 2.5 µ m wide and spaced by 3F, the area
of the fingers is 1.35 x 104 µ m2, the area of the finger base is 1.80 x 104 µ m2, and
the area of the mid pieces is 5.35 x 103 µ m2. The total area is 3.685 x 104 µ m2. The
mass will be (3.685 x 104 x 2 x10-12 cm3) x (2.3 x 10-3 kg/cm3) = 1.70 x 10-10 kg.
fo =
11
2
2
−6
3
1 4 (1.7x10 kg − m /sec / m )(2x10 m ) 1 


= 138 kHz
 22
2π
1.7x10−10 kg
11.12 With infinite selectivity, the vertical and horizontal distances are related by Y/X = tan
54.74o = 2 . For a finite selectivity factor S, the <111> plane at the surface will be
etched by an amount equal to Y/S, and the intersection of the <111> plane with the
surface will move by ∆ X where ∆ X = (Y/S) cos (90o–φ ) = (Y/S) sin φ .
∆X
X
φ
φ'
Y/S
Y
90o - φ
φ = 54.74o
The new angle φ ’ is given by
tan φ' =
Y
Y
=
X + ∆X
Y
Y
+
2 S
3
2
2
φ' = tan −1
1+
3
S
For S = 400, φ ’ = 54.6o, and for S = 20, φ ’ = 52.5.
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© 2002 Prentice Hall