Introduction to Microelectronic Fabrication

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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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 1 0

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

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Introduction to Microelectronic Fabrication – Second Edition

  lo g N2   N1  0.1505( Y−1970) N = 1027 x 1 0 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

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Introduction to Microelectronic Fabrication – Second Edition −0.06079 ( 2020 −1970 )

−3

1.8

F = 8.214 x 10

µm = 7.50 x 10 µm = 75 Å.

1.9

(3 x 108 tubes)(0.5 W/tube) = 150 MW!

1.10

(a) L = (25mm)(18mm/0.5mm) = 0.90 m !

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.

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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Introduction to Microelectronic Fabrication – Second Edition (a) N = 5000

2

(b) N = 5000

2

(c) N = 5000

2

25 (12 ) = 1 million transistors

25 (0.25 2 )= 16 million transistors

25 (0.1

2

) = 100 million transistors

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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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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)

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3 µm



2.4

1) Negative resist – n+ mask 2) Negative resist – Contact mask 3) Positive resist – Metallization mask n+ Mask

2.5

(a ) N A =

2.6

(a ) NA =

Metal Mask

1 λ 1  193n m =   = 0.536 2 F 2  180n m

(b ) DF = 0.6

Contact Mask

λ 4F 2 4(180n m) = 0.6 = 0.6 = 0.403 µm 2 NA λ 193n m 2

 1λ 1 λ  λ = 0.5 µm = 500 nm 1 =  2F 2  0.25µm 

λ 0.5µm DF = 0.6 = 0.6 = 0.3 µm 2 NA 12  1λ 1 λ  λ = 250 nm 0.5 =  (b ) NA = 2F 2  0.25µm  λ 0.25µm DF = 0.6 = 0.6 = 0.6 µm 2 NA 0.5 2

2.7

λ 193nm Fmin ≅ = = 96.5 nm or Fm in ≅ 0.1 µm 2 2

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2.8

λ 13nm Fmin ≅ = = 6.5 nm or Fm in ≅ 0.0065 µm 2 2

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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  o + D  d Xo = D No t =  or X dt  M  X + D ks M  o k s 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 2D Mα A= τ = M ks 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

τ = X i B + X i (B A ) 2

t = Xo B + X o (B A ) − τ 2

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

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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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(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

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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


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

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Introduction to Microelectronic Fabrication – Second Edition 0.500 0.956

(b)


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

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

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(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 τ = X i B + X i /(B/ A ) = 0.144 hrs. The time required to reach 0.2 µ m which gives 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

[

]

X o = 0.5 (0.183 ) 1 + 4 (0.53 )(0.5 )/(0.183 ) −1 = 0.43 µm 2

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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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 t = X2o B + Xo /(B/ A ) = 2.24 hrs. In the oxidized desire Xo = 1 µ m which gives τ = X 2i B + X i /(B/ A ) = 2.24 hrs. region, the initial oxide Xi = 1 µ m which gives 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.

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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.

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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

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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 -


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.

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CHAPTER 4  xj  2  where D t= 108 cm2 (a) 10 = 5x10 exp −  2 D t 15

4.1

18

(

)

x = 2.92 2 Dt = 5.8 µm j

 1015  −1   x = 2 D t erfc (b) j 18  and x j = 5.3 µm  5 x10  (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

[

]

1015 /cm3. So: 5 x1018 exp −(x j / 2 Dt ) = 4.0 x 1015. Solving for Dt with xj = 4 x

2

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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 2 Dt = boron profile, N(x) = 5 x 1018 exp -(x2/4Dt). Setting N(4µ m) = NB yields 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 -


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 Q = 2NO Dt π = 6.93 x 1014/cm2. This dose is almost x 10-15 cm2/sec for boron. 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 11 2 = NB  3x1 016  → D t= 1 .1 8 9 x 1 0c m 2 ln   NO 2 ln 1 021 

(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 17 3 and ni = 9.07 x 10 /cm .  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.

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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 -


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 -



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 -


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. 22(2 ) − 2(1.63 ) 20 (2) − 2 (1.63) 6 − 2 (1.63 ) N=3 +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 22(2 ) − 2(2.45) 20(2 ) − 2(2.45) 6 − 2 (2.45) N=3 +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

and

0

(Dt ) = D(To )(kT o E AR ) for large enough to. eff 2

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 dT =2 exp(−E A kT )(E A 2kT 2 ) t ln (N O NB ) j

dx j x S Tj = xj

 T  EA   =  dT  kT

For boron or phosphorus, EA = 3.69 eV: dx j  10  3.69 xj ST = = 31.2 a n d = 31.2   = 0.227  1373 8.62x10−5 (1373) xj 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 -


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. The dose Q = N O πDt = 5x10 π(1.676 x10 ) = 1.147 x10 /cm . For the preN O Dt = Q π 2 = 1.017x1013 / cm2 . This is a low dose, so let us deposition step, 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. −8

16

13

3

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 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 /❏.

For

NO

=

2

x

1018/cm3,

( )

( )

( )

( )

(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 -



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 = 2.45 (1.1x10 20 )(1.21x1015 )(900)/ 4x1018 = 0.134 µm j

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) − =− 2 D D dx 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    s D 2 Dt   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 -


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 -



$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 -



4.22

V=

IRS ln(2) ln(2 ) = (1 0−5 )(3 0 0) = 0.6 6 2 m V π π

- 34 -



(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 -



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 in µm     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 -



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 -5 Dt = 4.09 x 10 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 -



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 -6 Dt = 4.24 x 10 cm. For the final layer, the dose in silicon is Q Si = NP πDt = 14 2 4.5 x 10 /cm . 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 in µm   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 x 3.39x10 exp −  Then: j = 0.80 µm which agrees  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



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 -



5.11

mv2 2E E= 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. v=

2(5keV)(1.602x10 −19 J/ eV) −26

1.809x10

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 -



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



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 x 5x1 0 e xp−  j1 = 0.0 53 µm  0.0 7 2      x − 0.5 5 2  j2 18   = 1 016 yie ld s x 5x10 e xp−  j2 = 0.87 µm  0.0 9 2    These two junction values agree well with the graph of the profile.

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 -


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  (x − R )2    P  R S = q ∫ µ (N) N(x ) dx  N( x) = N P exp  − 2   2∆R  P  x j1  1270 447 µ n = 92 + 0.091 µ p = 48 +     0.076 N N 1+  1 +     1.3x1017   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


5.21


- 43 -



5.22


- 44 -



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 -


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 2 Dt =1.73 µ m. The two profiles are given by time t = 3000 sec.  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 -



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 -


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    cm3   0.081  - 48 -



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 -



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 -



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 -


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 10 −4 cm R = ρ = ( 5µΩ − cm ) 2 = 0.80 Ω 7.11 A 0.25 x10−4 cm ) (

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 -



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 -


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 D oA = {1, 2, 4}. For Y = α [1 + DoA/α ]- , numerical fitting yields an excellent fit for α = 3.1.

- 54 -



(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 -


Introduction to Microelectronic Fabrication – Second Edition α →∞

α →∞

- 56 -


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 -

]


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 -


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 -



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 -



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



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 107 F /cm 2 −6   10 10 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 2 For the Gaussian distribution N(x) = Np exp(-x /2∆ Rp2), and π 2 M1 = NP ∆R P a nd M2 = N P ∆R P 2 Equating moments yields the specified results.

9.8

Ni x2i 2

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 -



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 2 α

-

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 -



22 α 11 = λ

20 α 10 = λ

- 64 -



(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  V = −0.56 − 0.385 − 2(11.7)(8.854x10 −14 )(1.602x10−19 )(3x1016 )(0.385 ) /CO = −1.21V TP

(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 n  10    i

- 65 -



Wd = Wd =

2K S ε o (V A + Φbi )  N A ND    q  N A + ND  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 14

  1.5x10−4  2    → Dt = 1.177x10 −9 cm2 = 5x10 exp −  2 Dt    16

4.2 x10 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 D t  2  2 D t  which yields x – 3 µ m = 1.52 µ m of out diffusion.

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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 -



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 -


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



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)

 L  IR   L 2∆Y VDE − VCD = S   + ∆Y −  − ∆Y  = IR S  2  W  2 W - 70 -


Introduction to Microelectronic Fabrication – Second Edition and

(b)

and

L VAC = IR S W

VDE − VCD 2 L  V − VCD   = ∆Y a n d ∆Y =  DE VAC L 2 VAC 

 L  IR  L 2∆X VIJ − VJH = S  + ∆X −  − ∆X  = IR S  2  W  2 W L VGH = IR S W

VIJ − VJH 2 L  V − V JH  = ∆X a n d ∆X =  IJ L 2  VIH  VIH

- 71 -



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,   x 2 G B = ∫ [N(x )/D B ]dx = 1.5x10 exp −   dx ∫   2 Dt   −4  1.5 µm 1. 5x10 4.0x10 −4

4.0µm

17

G B = (2.12x10

2.83

13

x ) ∫ exp(−z )dz for z = 1.41x10 2

4

1 .06

13

G 5x10 G B = 2.51x10 and β ≅ E = = 199 G B 2.51x1011 11

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.

- 72 -


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 -



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. - 74 -


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.

- 75 -


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)

- 76 -


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 -


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 -


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!

 xj  x  = 5x1018 at xj = 50 nm or j = 1.985 1021erfc  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.

10.20 (a)

- 79 -



  x 2  xj 2 j 16    = 5x10 at xj = 100 nm or   = 4.605 5x10 exp −  (b)   2 D t   2 D t   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. 18

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 -


Introduction to Microelectronic Fabrication – Second Edition WB < [(3.18psec)(10)(20cm2/sec)]0.5 = 250 nm. CBE < 3.18psec/25Ω = 0.13 pF.

- 81 -


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 -



CHAPTER 11 A useful triangle: 1 φ 54.74o 2

11.1

11.2

11.3

1 tanφ = 2 cos φ = sin φ = 3

3

2 3

 1 1 1 The unit vector perpendicular to the <111> plane is v1 =   and that  3 3 3  c o s θ = v1 • v 2 and perpendicular to the <100> plane is v 2 = (1 0 0) . Thus we have   −1 1 o θ = cos   = 54.74  3  1 1 1 The unit vector perpendicular to the <111> plane is v1 =   and that  3 3 3   1 1  c o s θ = v1 • v 2 perpendicular to the <110> plane is v 2 =  0 . Thus we have  2 2    −1 2 o and θ = cos   = 35.3   6 10 o  W  = tan(54.74 ) → W = 14.1 µm    2  W

54.74o

10 µm

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(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+2 µm

L

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

1 µm

Y o  W  = tan(54.74 ) =    2

2

Y = 5 + 0.5 2 = 5.71 µm W = Y 2 = 8.07 µm

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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

10 µm

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  500  µm W = 2 5 +  tan(54 .74 o )  500  W = 2 5 +  = 717 µm  2 



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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. 11 2 2 −6 3 1 4 (1.7x10 kg − m /sec / m )(2x10 m ) 1  fo =   = 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 2

φ' = tan −1

3 2

1+

3 S

For S = 400, φ ’ = 54.6 , and for S = 20, φ ’ = 52.5. o

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Introduction to Microelectronic Fabrication

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