Introduction to microelectronic fabrication PDF

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0506SOLUTIONS MANUALtoINTRODUCTION TOMICROELECTRONICFABRICATIONSECOND EDITIONbyRICHARD C. JAEGERCHAPTER 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 Calcu...

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0506

SOLUTIONS MANUAL to

INTRODUCTION TO MICROELECTRONIC FABRICATION SECOND EDITION by RICHARD C. JAEGER

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© 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.

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© 2002 Prentice Hall

Introduction to Microelectronic Fabrication – Second Edition 1.2

1.3

(a) A =

d2/4

d (mm) A (mm2)

25 49 1

(b) n =

(450)2/(4)(12) = 159043

(a) n =

(300)2/(4)(202) = 177

50 196 0

75 442 0

100 785 0

125 1230 0

150 1770 0

200 3140 0

(b) n =

300 7070 0

450 159000

(450)2/(4)(252) = 254

(b) n = 148

1.4

B 19.97 x 10 0.1977 2020 1960

1.5

N 1 0 2 7x 1 0

0.1505 2020 1970

1.45 x 1013 bi ts

9

3 4.4 x 1 0 tr a n s is tos

1.6

lo g B 2B 1 B 1 9.9 7 x 1 00.1977 Y 1960 Y2 Y1 0.1 9 7 7 lo g 2 lo g1 0 5.0 6 y e a r 1.5 2 y e a r s b Y2 Y1 a Y2 Y1 0.1 9 7 7 0.1 9 7 7 1.7

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© 2002 Prentice Hall

Introduction to Microelectronic Fabrication – Second Edition

lo g N2 N 1027 x 100. 1505 Y a Y2 Y1

1970

Y2 Y1

N1 0.1505

l o g2 2.00 years b Y2 0.1505

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Y1

lo g10 6.65 yea rs 0.1505

© 2002 Prentice Hall

Introduction to Microelectronic Fabrication – Second Edition 0.06079 2020 1970

3

1.8

m = 75 Å. 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

m

7.50 x 10

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

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© 2002 Prentice Hall

Introduction to Microelectronic Fabrication – Second Edition (a) N (b) N (c) N

5000

2

25 1 2

1 million transistors

5000

2

25 0.25

2

5000

2

2

25 0.1

16 million transistors 100 million transistors

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© 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

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© 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 m

(b)

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3 m

© 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

Contact Mask

Metal Mask

2.5 a NA

1 2F

1 193n m 2 180n m

b DF

0.6

0.6

a

NA

1 2 F

DF

0.6

NA

1 2 F

DF

0.6

N A2

0.536

4F 2

2

0.6

4 18 0n m 193n m

0.4 03 m

2.6

b

2.7

Fm in

2

1 2 0.25 m

1

2

NA

0.6 0.5

NA2

193nm 2

0.5 m 12

= 0.5 m = 500 nm 0.3 m

1 2 0.25 m

0.6

0.25 m 0.5 2

= 250 nm 0.6 m

96.5 nm or Fm in

0.1 m

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© 2002 Prentice Hall

Introduction to Microelectronic Fabrication – Second Edition

2.8

Fm in

2

13nm 2

6.5 nm or Fm in

0.0065 m

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© 2002 Prentice Hall

Introduction to Microelectronic Fabrication – Second Edition

CHAPTER 3 3.1

Using Fig. 3.6 with 100 nm = 0.1 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 m takes 0.45 hours or 27 minutes. The second 0.4 m takes (1.5-0.45) hours or 63 minutes. The third 0.4 m takes (3.2-1.5) hours or 102 minutes.

3.3

d Xo dt

D D No D No 1 o r Xo d Xo t D M ks M X o ks Integrating and rearranging where is an integration constant yields: 2

t

Xo

B Assuming

= 0 at Xo = Xi:

M 2DN o

2D No M

Xo

A=

X 2i B

M No k s

2D ks

M DN o

=

M D No

Xi B/ A

Problems 3.4 through 3.10 evaluate the following equations with spreadsheets. Xo

0.5 A 1

4B t A2

1

X2i B X i B A

t

X2o B

Xo B A

3.4 T 1150 1150 1150

3.5

B/A 5.322 5.322 5.322

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) Silicon - Wet Oxygen

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© 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.

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© 2002 Prentice Hall

Introduction to Microelectronic Fabrication – Second Edition 3.5

(b) T 1000

3.6

B/A 4.478E-02

Silicon - Dry Oxygen B Xi tau 0.025 6.182E-01 1.042E-02 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

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

Silicon - Wet Oxygen B Xi tau 4.123E-01 0 0 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)

Silicon - Dry Oxygen

T B/A B A Xi tau t Xo ( m) 1100 0.169 0.024 0.140 0.025

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© 2002 Prentice Hall

Introduction to Microelectronic Fabrication – Second Edition 0.174 0.500 0.074

Silicon - Wet Oxygen

T B/A B A Xi tau t Xo ( m) 1100 2.895 0.529 0.183 0.074 0.036 2.000 0.950

Silicon - Dry Oxygen

T B/A B A Xi tau t Xo ( m) 1100 0.169 0.024 0.140 0.950 43.931

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© 2002 Prentice Hall

Introduction to Microelectronic Fabrication – Second Edition 0.500 0.956

(b)

Silicon - Dry Oxygen

T B/A B A Xi tau t Xo ( m) 1100 0.284 0.024 0.083 0.025 0.115 0.500 0.086

Silicon - Wet Oxygen

T B/A B A Xi tau

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© 2002 Prentice Hall

Introduction to Microelectronic Fabrication – Second Edition t Xo ( m) 1100 4.865 0.529 0.109

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© 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

Silicon - Dry Oxygen A 0.233 Silicon - Wet Oxygen A 0.183

Xi 0.025

tau 0.618

t 1.000

Xo ( m) 0.058

Xi 0.058

tau 0.026

t 5.000

Xo ( m) 1.542

(b) From Fig. 3.6, 1 hr at 1000 oC in dry oxygen produces approximately 0.053 m oxide, and 5 hours at 1100 oC in wet oxygen produces a 1.5 m thick oxide. The 0.053- 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

Silicon - Dry Oxygen A 0.083 Silicon - Wet Oxygen A 0.109

Xi 0.025

tau 0.115

t 1.000

Xo ( m) 0.126

Xi 0.126

tau 0.056

t 5.000

Xo ( m) 1.582

(b) From Fig. 3.7, 1 hr at 1100 oC in dry oxygen produces approximately 0.12- m oxide, and 5 hours at 1100 oC in wet oxygen produces a 1.5 m thick oxide. The 0.12- 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 silicon yields (B/A) = 2.895 m/hr and B = 0.529 m2/hr. In the oxidized region, the initial oxide Xi = 2 0.2 m which gives X i B X i / B/ A = 0.144 hrs. The time required to reach a thickness of 0.5 m = 0.52/0.53 + 0.5/2.9 - 0.144 = 0.50 hrs. In the unoxidized region, 0.5 hours oxidation yields Xo

0.5 0.183

1

4 0.53 0.5 / 0.183

2

1

0.43

m

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

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© 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 silicon yields (B/A) = 2.895 m/hr, B = 0.529 m2/hr and A = 0.183 m. In the unoxidized region, we desire Xo = 1 m which gives t X 2o B X o / B/ A = 2.24 hrs. In the oxidized region, the initial oxide Xi = 1 m which gives X 2i B X i / B/ A = 2.24 hrs. The final thickness in the oxidized region is Xo

0.5 0.183 m

1 m

1 4

2.895 4.472 0.183

1

1.45 m

1.45 m

1 m

Original Final Note that this result is almost independent of the temperature chosen. The total growth in the oxidized area ranges from 1.49 m at 1000 oC to 1.43 m at 1200 oC. The 1- m region will appear carnation pink in color, and the 1.45- m region will appear violet.

3.13

Using Fig. 3.6: At 1100 oC, 1.4 m of oxide could be grown in 4 hours. However, the wafer has 0.4 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 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 oxide. A one-hour phosphorus diffusion at 1050 oC requires 0.4 m SiO2.

3.15

Using Figure 3.10: A 15-hour boron diffusion at 1150 oC requires a minimum of approximately 0.15 m of oxide as a barrier layer.

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m of

© 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 m of oxide as a barrier layer.

3.17

Using Table 3.2: The 1- m thick oxide region will appear carnation pink in color. The 2- m thick oxide region will also appear carnation pink in color.

3.18

2Xox = k /n = 0.57k/1.46 = 0.39k

3.19

Computer program – Implement oxidation equations.

3.20

Computer program – Implement oxidation equations.

m yielding 0.39, 0.78, 1.17 and 1.56

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m.

© 2002 Prentice Hall

Introduction to Microelectronic Fabrication – Second Edition 3.21(a) TITLE INITIALIZE DIFFUSION DIFFUSION DIFFUSION PRINT PLOT STOP

PROBLEM 3.21 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

SILICON, ARSENIC CONCENTRATION=1E15

For (a) and (b), XO = 0.92 m. Problem 3.8 yielded 0.96 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 XO = 0.96 0.99 m.

3.23

PROBLEM 3.22 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

m. Boron is slightly depleted at the silicon surface. Problem 3.8 yielded

TITLE INITIALIZE DIFFUSION PRINT PLOT STOP

PROBLEM 3.23 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 m. Boron is slightly depleted at the silicon surface and approximately uniform in the oxide. Problem 3.6 yielded 2.0 m in 6.375 hours (382.4 min). The simulation requires more time to reach 2 m. SUPREM yields a

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© 2002 Prentice Hall

Introduction to Microelectronic Fabrication – Second Edition 1.93- m oxide in 382 min. The oxidation coefficients are slightly different in SUPREM. For phosphorus, change the second statement to: INITIALIZE

SILICON, PHOSPHORUS CONCENTRATION=2.7E15

The result is unchanged: XO = 2.0 m. The phosphous concentration in the oxide is much lower than for the boron doped substrate.

3.24

TITLE INITIALIZE DIFFUSION PRINT PLOT STOP XO = 1.0

PROBLEM 3.24 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

m.

For dry oxidation: DIFFUSION

TEMP=1050 TIME=4419 DRY02

For phosphorus, change the second statement to:

3.25

INITIALIZE

SILICON, PHOSPHORUS CONCENTRATION=5E15

TITLE INITIALIZE

PROBLEM 3.25 Region 1 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

m.

TITLE INITIALIZE DIFFUSION DIFFUSION PRINT PLOT STOP XO = 1.44

PROBLEM 3.25 Region 2 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

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 and 1.40 m.

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m

© 2002 Prentice Hall

Introduction to Microelectronic Fabrication – Second Edition

CHAPTER 4 4.1

15

18

(a) 10

5x10 exp

xj

2.92 2 Dt

(b) x j

2 D t erfc

-8

2

where D t= 10 cm

2 Dt 5.8

1

2

xj

m

1015 18 5 x10

a nd

x 5.3 m

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- m. Dividing by the junction depth of 5.8 m yields Rs = 47 ohms/ . For the erfc profile, use Fig. 4.16(a) yielding 320 ohm- m and 60 ohms/ with xj = 5.3 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

4.3

(a) Using Fig. 4.8, a 1 ohm-cm n-type wafer has a background concentration of 4.0 x 1015 /cm3. So: 5 x1018 exp

x j / 2 Dt

m (b) 1

2

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m

= 4.0 x 1015. Solving for Dt with xj = 4 x

© 2002 Pren...