introduction to microelectronic fabrication

0506

SOLUTIONS MANUAL

to

INTRODUCTION TO

MICROELECTRONIC

FABRICATIONSECOND EDITION

by

RICHARD C. JAEGER

- 1 - © 2002 Prentice Hall

CHAPTER 1

1.1

Answering machineAlarm clockAutomatic doorAutomatic lightsATMAutomobile:

Engine controllerTemp. controlABSElectronic dash

Automotive tune-up equip.Bar code scannerBattery chargerCalculatorCamcorderCarbon monoxide detectorCash registerCellular phoneCopierCordless phoneDepth finderDigital watchDigital scaleDigital thermometerDigital ThermostatElectric guitarElectronic door bellElectronic gas pumpExercise machineFax machineFish finderGarage door openerGPSHearing aidInkjet & Laser PrintersLight dimmerMusical greeting cardsKeyboard synthesizerKeyless entry systemLaboratory instrumentsModel airplanesMicrowave ovenMusical tunerPagersPersonal computerPersonal planner/organizer

Radar detectorRadioSatellite receiver/decoderSecurity systemsSmoke detectorStereo system

AmplifierCD playerReceiverTape player

Stud sensorTelephoneTraffic light controllerTV & remote controlVariable speed appliances

BlenderDrillMixerFood processorFan

Vending machinesVideo gamesWorkstations

Electromechanical Appliances*

Air conditioningClothes washerClothes dryerDish washerElectrical timerThermostatIronOvenRefrigeratorStoveToasterVacuum 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.


Introduction to Microelectronic Fabrication – Second Edition

1.2 (a) A = π d2/4

d (mm) 25 50 75 100 125 150 200 300 450A (mm2)

491

1960

4420

7850

12300

17700

31400

70700

159000

(b) n = π (450)2/(4)(12) = 159043 (b) n = π (450)2/(4)(252) = 254

1.3 (a) n = π (300)2/(4)(202) = 177

(b) n = 148

1.4 B = 19.97 x 100.1977 2020−1960( ) = 1.45 x 1013 bits

1.5 N = 1027 x 100.1505 2020−1970( ) = 34.4 x 109 tra n s is tors

1.6

B = 1 9.97 x 1 00.1977 Y −1960( ) Y2 − Y1 =log B2

B1

0.1 9 7 7

a( ) Y2 − Y1 =log2( )

0.1 9 7 7= 1.5 2 y e a r s b( ) Y2 − Y1 =

log1 0( )0.1 9 7 7

= 5.0 6 y e a r s

1.7



N = 1027 x 100.1505 Y−1970( ) Y2 − Y1 =lo g N2

N1

0.1505

a( ) Y2 − Y1 =log2( )

0.1505= 2.00 years b( ) Y2 − Y1 =

log10( )0.1505

= 6.65 years



1.8 F = 8.214 x 10 −0.06079 2020 −1970( )µm = 7.50 x 10 −3 µm = 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! IRMS = (150 MW)/(220 VRMS) = 685 kA

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

(b) L = (25mm)(18mm/0.2mm) = 2.25 m !!

1.11 Two Possibilities

276 Dice 277 Dice

1.12 (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



(a) N = 5000 2 25 12( )=1 million transistors

(b) N = 5000 2 25 0.25 2( )=16 million transistors

(c) N = 5000 2 25 0.12( )=100 million transistors



1.14 Thermal oxidationn+ diffusion mask Mask 1Oxide etchn+ diffusion and oxidationContact opening mask Mask 2Oxide etchMetal depositionMetal etch mask Mask 3Metallization etch

1.15

C

B

E

n+p

n+



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

2.3SiO2SiO2

3 µm3 µm(a) (b)



2.4 1) Negative resist – n+ mask2) Negative resist – Contact mask3) Positive resist – Metallization mask

n+ Mask Contact Mask

Metal Mask

2.5

a( ) N A =12

λF

=12

193nm180nm

= 0.536

b( ) DF = 0.6λ

NA2 = 0.64F2

λ= 0.6

4 180nm( )2

193nm= 0.403 µm

2.6

a( ) NA = 12

λF

1 = 12

λ0.25µm

λ = 0.5 µm = 500 nm

DF = 0.6 λNA2 = 0.6 0.5µm

12 = 0.3 µm

b( ) NA = 12

λF

0.5 = 12

λ0.25µm

λ = 250 nm

DF = 0.6 λNA2 = 0.6 0.25µm

0.52 = 0.6 µm

2.7 Fmin ≅

λ2

=193nm

2= 96.5 nm or Fmin ≅ 0.1 µm



2.8 Fmin ≅

λ2

=13nm

2= 6.5 nm or Fmin ≅ 0.0065 µm



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

dXo

dt=

DNo

M

1

X o +Dk s

or X o +Dk s

dXo =

DNo

Mt

Integrating and rearranging where α is an integration constant yields:

t = Xo

2 M2DNo

+ X o

MNoks

+

MαDNo

B =

2D No

M A=

2Dk s

τ =MαD No

Assuming τ = 0 at Xo = Xi:

X i2

B+

X i

B/A( ) = τ

Problems 3.4 through 3.10 evaluate the following equations with spreadsheets.

X o = 0.5A 1 +

4BA2 t + τ( ) − 1

τ = X i

2 B + X i B A( ) t = Xo2 B + X o B A( )− τ

3.4<100> Silicon - Wet Oxygen

T B/A B Xi tau Xo t (hrs)1150 5.322 0.667 0 0.000 1 1.6881150 5.322 0.667 1 1.688 2 4.6871150 5.322 0.667 2 6.375 3 6.687

3.5 (a)<100> Silicon - Wet Oxygen



T B/A B Xi tau Xo t (hrs)850 6.116E-02 1.219E-01 0 0 0.01 1.643E-01

0.164 hours represents only 9.86 minutes and is too short a time for good control.



3.5 (b)<100> Silicon - Dry Oxygen

T B/A B Xi tau Xo t (hrs)1000 4.478E-02 1.042E-02 0.025 6.182E-01 0.01 ---

Can't grow 0.01 um (< 0.025 um)

3.6 (a) Slightly over six hours

(b)<100> Silicon - Wet Oxygen

T B/A B Xi tau Xo t (hrs)1150 5.322 0.667 0.000 0.000 2.000 6.375

3.7 (a) Approximately 3 hours in wet oxygen (b) Over 70 hours in dry oxygen

(c)<100> Silicon - Wet Oxygen

T B/A B Xi tau Xo t (hrs)1050 1.504E+00 4.123E-01 0 0 1 3.090

<100> Silicon - Dry OxygenT B/A B Xi tau Xo t (hrs)

1050 8.920E-02 1.592E-02 0.025 3.195E-01 1 73.71

3.8 (a)

<100> Silicon - Dry Oxygen

TB/ABAXitaut

Xo (µ m)

11000.1690.0240.1400.025



0.1740.5000.074

<100> Silicon - Wet Oxygen

TB/ABAXitaut

Xo (µ m)

11002.8950.5290.1830.0740.0362.0000.950


TB/ABAXitaut

Xo (µ m)

11000.1690.0240.1400.95043.931



0.5000.956

(b)


TB/ABAXitaut

Xo (µ m)

11000.2840.0240.0830.0250.1150.5000.086

<111> Silicon - Wet Oxygen

TB/ABAXitau



tXo (µ m)

11004.8650.5290.109



3.9 (a)<100> Silicon - Dry Oxygen

T B/A B A Xi tau t Xo (µ m)1000 4.478E-02 1.042E-02 0.233 0.025 0.618 1.000 0.058

<100> Silicon - Wet OxygenT B/A B A Xi tau t Xo (µ m)

1100 2.895 0.529 0.183 0.058 0.026 5.000 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)<111> Silicon - Dry Oxygen

T B/A B A Xi tau t Xo (µ m)1100 0.284 0.024 0.083 0.025 0.115 1.000 0.126

<111> Silicon - Wet OxygenT B/A B A Xi tau t Xo (µ m)

1100 4.865 0.529 0.109 0.126 0.056 5.000 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 = 0.2 µ m which gives τ = X i

2 B + X i / B/ A( ) = 0.144 hrs. The time required to reach a thickness of 0.5 µ m = 0.52/0.53 + 0.5/2.9 - 0.144 = 0.50 hrs. In the unoxidized region, 0.5 hours oxidation yields

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

This result can also be obtained using Fig. 3.6 in a manner similar to the solution of Problem 3.13.

20 nm

50 nm43 nm

Original Final



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 desire Xo = 1 µ m which gives t = Xo

2 B + Xo / B/ A( ) = 2.24 hrs. In the oxidized

region, the initial oxide Xi = 1 µ m which gives τ = X i2 B + X i / B/ A( ) = 2.24 hrs.

The final thickness in the oxidized region is

X o = 0.5 0.183µm( ) 1 + 4

2.8950.183

4.472( ) −1

= 1.45 µm

1.45 µm

Original Final

1 µm1 µm

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.

(100) Silicon - Wet OxygenT B/A B Xi tau Xo t (hrs)

1100 2.895 0.529 0.400 0.441 1.400 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.



3.16 Using Figure 3.10: A 20-hour phosphorus diffusion at 1200 oC requires a minimum of 3 µ m of oxide as a barrier layer.

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

3.18 2Xox = kλ /n = 0.57k/1.46 = 0.39k µ m yielding 0.39, 0.78, 1.17 and 1.56 µ m.

3.19 Computer program – Implement oxidation equations.

3.20 Computer program – Implement oxidation equations.



3.21(a) TITLE PROBLEM 3.21INITIALIZE <100> SILICON, BORON CONCENTRATION=1E15

THICKNESS=5.0 XDX=0 DX=0.02 SPACE=200DIFFUSION TEMP=1100 TIME=30 DRY02DIFFUSION TEMP=1100 TIME=120 WET02DIFFUSION TEMP=1100 TIME=30 DRY02PRINT LAYERSPLOT CHEMICAL NET LP.PLOTSTOP

(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 PROBLEM 3.22INITIALIZE <111> SILICON, BORON CONCENTRATION=3E15

THICKNESS=5.0 XDX=0 DX=0.02 SPACE=200DIFFUSION TEMP=1100 TIME=30 DRY02DIFFUSION TEMP=1100 TIME=120 WET02DIFFUSION TEMP=1100 TIME=30 DRY02PRINT LAYERSPLOT CHEMICAL NET LP.PLOTSTOP

XO = 0.96 µ m. Boron is slightly depleted at the silicon surface. Problem 3.8 yielded 0.99 µ m.

3.23 TITLE PROBLEM 3.23INITIALIZE <100> SILICON, BORON CONCENTRATION=2.7E15

THICKNESS=5.0 XDX=0 DX=0.02 SPACE=200DIFFUSION TEMP=1150 TIME=408.7 WET02PRINT LAYERSPLOT CHEMICAL BORON LP.PLOTSTOP

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



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 PROBLEM 3.24INITIALIZE <100> SILICON, BORON CONCENTRATION=5E15

THICKNESS=5.0 XDX=0 DX=0.02 SPACE=200DIFFUSION TEMP=1050 TIME=197.2 WET02PRINT LAYERSPLOT CHEMICAL BORON LP.PLOTSTOP

XO = 1.0 µ m.

For dry oxidation:

DIFFUSION TEMP=1050 TIME=4419 DRY02

For phosphorus, change the second statement to:

INITIALIZE <100> SILICON, PHOSPHORUS CONCENTRATION=5E15

3.25 TITLE PROBLEM 3.25 Region 1INITIALIZE <100> SILICON, THICKNESS=5.0 XDX=0 DX=0.02

SPACE=200DIFFUSION TEMP=1100 TIME=141.5 WET02PRINT LAYERSPLOT CHEMICAL NET LP.PLOTSTOP

XOX = 1.0 µ m.

TITLE PROBLEM 3.25 Region 2INITIALIZE <100> SILICON, THICKNESS=5.0 XDX=0 DX=0.02

SPACE=200DIFFUSION TEMP=1100 TIME=141.5 WET02DIFFUSION TEMP=1100 TIME=141.5 WET02PRINT LAYERSPLOT CHEMICAL NET LP.PLOTSTOP

XO = 1.44 µ m.



If the oxidation times are changed to 134.2 min., the oxide thicknesses are 0.97 µ m and 1.40 µ m.



CHAPTER 4

4.1 (a) 1015 = 5x1018 exp −

x j

2 Dt

2

where Dt=108cm2

x j = 2.92 2 Dt( )= 5.8 µm

(b) x j = 2 D t( ) erfc−1 1015

5x1018

and xj = 5.3 µm

(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 x j = 5.3 µ m.

(d)

0 1 2 3 4 5 6 7 81014

1015

1016

1017

1018

1019

1020

Distance From Surface (um)

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( )2[ ] = 4.0 x 1015. Solving for Dt with xj = 4 x



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.

(c) Q = No πDt = 5x1018 π 5.61x10−9( )= 6.64 x 1014 /cm2

(d) Assume a solid-solubility limited constant source predeposition with Q =

2No Dt / π . Try T = 1000 oC. No = 1 x 1021/cm3 and D = 10.5 exp (-3.69/8.617 x 10-5 x 1273) = 2.58 x 10-14. Solving for Dt yields Dt = 3.46 x 10-13 and t = 13.4 sec which is a too short to control. Try T = 900 oC. No = 5.5 x 1020/cm3 and D = 1.47 x 10-15. Solving for t yields 13.0 minutes which is short but probably usable.

4.4 (a) An 1 ohm-cm n-type wafer has a doping NB = 4 x 1015/cm3 from Fig. 4.8. For the boron profile, N(x) = 5 x 1018 exp -(x2/4Dt). Setting N(4µ m) = NB yields 2 Dt = 1.5 x 10-4 cm. For phosphorus at 950 oC, Ns = 7 x 1020/cm3, and D = 6.53 x 10-15

cm2/sec. Using t = 1800 sec yields 2 Dt = 6.86 x 10-6 cm. The junction occurs for:

4x1015+7x1020erfc (xj/6.86x10-6) = 5x1018exp [-(xj/1.5x10-4)2]

This equation can be solved approximately by realizing that the boron profile is almost constant near the surface. Thus, 7 x 1020 erfc (xj/6.86 x 10-6) ≈ 5 x 1018. Solving for xj yields a junction depth of 0.154 µ m. Checking the boron profile at this depth yields N = 4.95 x 1018 /cm3 so that the approximation is justified.

(b) Working iteratively with Fig. 4.21, one finds that the phosphorus and boron profiles each have a value of approximately 4 x 1018/cm3 at a depth of 0.75 µ m which is the junction depth.

(c) Using Fig. 4.12, we find that the 30 min curve reaches 5 x 1018/cm3 at a depth of slightly over 0.7 µ m.

(d) From Prob. 4.3, Dt = 1.14 x 10-12 cm2 for the predeposition step, and Dt = 5.61 x 10-9 cm2 for the drive-in step. The total is Dt = 5.61 x 10-9 cm2. The Dt product for the phosphorus step is 1.18 x 10-11 cm2, which is much smaller than the total Dt product for the boron step. Thus, the assumption is justified.

4.5 (a) From Fig. 4.8, a 5 ohm-cm n-type wafer corresponds to NB = 9 x 1014/cm3, and Rs

xj = 7500 ohm-µ m. A p-type Gaussian diffusion gives NS = 5 x 1016/cm3. So 9 x 1014 = 5 x 1016 exp -(7.5 x 10-4/ 2 Dt ). Solving for Dt yields = 3.5 x 10-8 cm2.

Using Fig. 4.5 to find an appropriate temperature: at 1100 oC, D is of the order of 10-13cm2/sec which gives a time over 25 hours - so we will try 1150 oC. For Do = 10.5



and EA = 3.69 eV, D = 9.05 x 10-13cm2/sec at T = 1423 K, yielding t = 3.87 x 104 sec = 10.7 hours. The diffusion schedule would be 1150 oC for 10.7 hours. A similar calculation using 1100 oC yields t = 32.1 hours which is a little long.

(b) As found above, NS = 5 x 1016/cm3

(c) Q = NS πDt = 1.66 x 1013/cm2

(d) Using 900 oC for 15 minutes (about as short as can be controlled), yields D = 1.49 x 10-15 cm2/sec for boron. Q = 2NO Dt π = 6.93 x 1014/cm2. This dose is almost two orders of magnitude too high. It is very difficult to get a low enough dose by direct diffusion.

4.6 (a) At 1000 oC, Fig. 4.6 indicates the arsenic surface concentration will be 1021/cm3. The Dt product can be found from Eq. 4.10:

D t =x j

2 lnNB

NO

=2µm

2ln3x1 016

1 021

→ D t=1 .1 8 9 x 1 011c m2

(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 .

n i

2 = 1.0 8x1 031T 3 e xp−EG

k T

= 1.0 8x1 031 1 2 7 3( )3 e xp −

1.1 28.6 2x1 0−5 x1 2 7 3

and ni = 9.07 x 1017/cm3.

Dt =

0.2x10−4 cm2.29

29.07x1017

1021

= 6.92 x 10−14cm2

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.



100 µm

90 µm30 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.



(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)3λ 2λ

3λ2λ

7λ 3λ2λ

(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 .



2λ

2λ

3λ3λ

3λ

19λ

(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.



(d) Assume a Gaussian profile. Using Fig. 4.10(b) with N/NO = 1016/1019 = 10-3, we find the ratio of lateral to vertical diffusion to be 2.2/2.7, and the lateral diffusion = 2 µ m (2.2/2.7) = 1.63 µ m.

N = 3

22 2( ) − 2 1.63( )2 2( ) + 2 1.63( ) + 2

20 2( ) − 2 1.63( )2 2( )+ 2 1.63( ) + 4

6 − 2 1.63( )2 2( )+ 2 1.63( ) + 8(0.56) + 2 0.65( )N =

34.2 .

(e) In this case, the lateral diffusion = 3 µ m(2.2/2.7) = 2.45 µ m, and

N = 3

22 2( ) − 2 2.45( )2 2( ) + 2 2.45( ) + 2

20 2( ) − 2 2.45( )2 2( ) + 2 2.45( ) + 4

6 − 2 2.45( )2 2( )+ 2 2.45( ) + 8(0.56) + 2 0.65( )N =

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

Dt( )eff = Do exp −EA kTo( ) exp − EARt kTo

2( )dt0

t o

∫and

Dt( )eff = D To( ) kTo2 E AR( ) for large enough to.

4.12

x j = 2 Do exp −EA kT( ) t ln NO NB( ) = 2exp − EA kT( ) t ln NO NB( )dx j dT =2exp −EA kT( )EA 2kT 2( ) t ln NO NB( )

ST

x j =dx j

x j

TdT

=

EA

kT

For boron or phosphorus, EA = 3.69 eV:

ST

x j =3.69

8.62x10−5 1373( ) = 31.2 and dx j

x j

= 31.210

1373

= 0.227

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



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 = NO πDt = 5x1016 π1.676 x10 −8( )=1.147 x1013 /cm 3 . For the pre-

deposition step, NO Dt = Q π 2 = 1.017x1013 /cm2 . This is a low dose, so let us try 900 oC, the lowest temperature in Fig. 4.6. From Fig. 4.6, NO = 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 ion-implantation step (discussed in the next chapter). Another possibility is a liquid “spin-on” doping source with a fixed concentration.

4.14 For NO = 2 x 1018/cm3, NB = 1016/cm3 and xj = 2 um,

2 Dt = 2x10 −4 ln 2x1018 1016( )= 8.689 x 10-5 cm. Using Eq. 4.13,

RS = qµ pNA x( )dx0

x j

∫

−1

= 1.602x10−19 300( ) 2x1018( ) exp −x

2 Dt

2

dx

0

2x10 −4

∫

−1

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

= 1020/cm3, and xj = 2 um, 2 Dt = 2x10 −4 ln 10 20 5x1016( )= 7.254 x 10-5 cm.

R S = 1.602x10−19 100( )1020( ) exp −

x2 Dt

2

dx

0

0.5x10−4

∫

−1

= 14.5 Ω /sq.

RS = 1.602x10−19 100( )1020( ) exp −

x2 Dt

2

dx

0.5x10−4

1.0x10−4

∫

−1

= 34.9 Ω /sq.

RS = 1.602x10−19 100( )1020( ) exp −

x2 Dt

2

dx

1.0x10−4

1.5x10−4

∫

−1

= 203 Ω /sq.

R S = 1.602x10−19 100( )1020( ) exp −

x2 Dt

2

dx

1.5x10 −4

2.0x10−4

∫

−1

= 2890 Ω /sq.

(b) Putting these four sheet resistance values in parallel yields RS = 9.72 Ω /.

(c) Irvin’s curves for an n-type Gaussian layer with NB = 5 x 1016/cm3 and NO = 1020/cm3 gives RS xj = 25 Ω -µ m, and Rs = 12.5 Ω /.



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 = NODt ni ; D = 3.17exp(-3.59/kT), NO = 2.78 x1017(RS xj)-1 and Q = 0.67 NO xj. At 900 oC, NO = 1.1x1020/cm3 from Fig. 4.6, and D = 1.21 x 10-15cm2/sec.

x j = 2.45 1.1x1020( )1.21x1015( )(900)/4x1018 = 0.134 µm

and Q = 9.9x1014/cm2 (7 times larger than in Ex. 4.3).

From Ex. 4.2, NB = 3 x 1016/cm3, D = 2.96 x 10-13 cm2/sec and t = 18000 sec. After the drive-in, x j = 2 Dt ln NO NB( ) = 3.44 µm (24% greater).

4.17 Using Laplace Transforms, we get an ordinary differential equation:

d2N x,s( )dx2 −

sN x,s( )D

= −N x,t = 0−( )

D (1)

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) = NOu(t), and

N x,s( ) =

NO

sexp −x

sD

. Using transform tables,

N x,t( ) = NOerfc

x2 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 = 0-

to x = 0+ yields dN(x,s)/dx = -Q/D, and therefore A s( ) = Q/ Ds . From the transform tables,

N x,t( ) =

QπDt

exp −x2

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.



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!



4.20 $SUPREM IV -- Use Default GridINITIALIZE <100> SILICON PHOSPHORUS=3E16 WIDTH=6.5$Diffusion BarrierDEPOSITION OXIDE THICKNESS=0.5ETCH OXIDE RIGHT P1.X=4$PredepositionDIFFUSION TEMP=900 TIME=15 BORON GAS.CONC=1.2E21$Reflect about right edge to complete structureSTRUCTURE REFLECT RIGHT$Plot Boron ContoursSELECT Z=LOG10(BORON) TITLE=”Contours of Boron Concentration”PLOT.2D SCALE Y.MAX=13 Y.MIN=0FOREACH X (16 17 18 19)

CONTOUR VAL=X COLOR=2ENDCONTOUR VALE=3E16 LINE.TYP=2 COLOR=2

4.21 TITLE PROBLEM 4.21 TWO STEP DIFFUSION FROM EXAMPLE 4.3INITIALIZE <100> SILICON, PHOSPHOUS CONCENTRATION=4E15

THICKNESS=6.0 XDX=0 DX=0.015 SPACE=400DIFFUSION TEMP=900 TIME=15 BORON GAS.CONC=1.2E21PRINT LAYERSPLOT CHEMICAL BORON LP.PLOTDIFFUSION TEMP=1100 TIME=304 DRY02PRINT LAYERSPLOT CHEMICAL BORON LP.PLOTDIFFUSION TEMP=1100 TIME=76 WET02PRINT LAYERSPLOT CHEMICAL BORON LP.PLOTPLOT CHEMICAL NET LP.PLOTETCH OXIDEDIFFUSION TEMP=950 TIME=30 PHOSPHORUS SOLIDSOLPRINT LAYERSPLOT CHEMICAL PHOSPHORUS LP.PLOTPLOT CHEMICAL NET LP.PLOT CMIN=1E13STOP

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.



4.22 V =

IRS ln 2( )π

= 10−5( ) 300( ) ln 2( )π

= 0.662 m V



4.23 (a) The cylinder contains 100 ft3 x 0.001 = 0.1 ft3 of diborane. The volume of the room is 10 x 12 x 8 or 960 ft3. 0.1 ft3/960 ft3 = 1.04 x 10-4 or approximately 100 ppm.

(b) Life threatening exposure is 160 ppm for 15 min. Evacuate rapidly!

(c) Life threatening exposure is 6-15 ppm for 30 min. Evacuate immediately!

4.24 From Prob. 4.15, 2 Dt = 7.254 x 10-5 cm.

R S = qµ N x( )( )N x( )dx

0

2x10 −4

∫

−1

where N x( ) = 1020 exp −x

2 Dt

2

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.



CHAPTER 55.1 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.

(b) QSi = 4.43x1018

0.25µm

∞

∫ exp −x − 0.19µm( )2

2 0.09µm( )2

dx = 0.751( ) 2πNp∆R p

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

Q = NP

0

∞

∫ exp −x − 0.0310.015 2

2

dx with x in µm

Numerical integration with MATLAB® shows that only 98.06% of the profile is in the silicon, so that

Q = 2πNP 0.015µm( )[ ]0.9806( ) and NP = 5.42 x 1020/cm3 for Q = 2 x 1015/cm2.

Then: 5.42x1020 exp −

x j − 0.031

0.015 2

2

= 1016 yields xj = 0.10 µm which agrees

well with the graph of the profile given below.



0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.51015

1016

1017

1018

1019

1020

1021


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 RS xj = 625 ohm-µ 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 QSi = 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 RS xj = 400 ohm-µ m yields NP = 4 x 1018/cm3. At x = xj, 4 x 1018exp[-(2x10-4/2 Dt )2] = 1016, and

Dt = 4.09 x 10-5 cm. For the final layer, the dose in silicon is QSi = 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.



5.7 Using Irvin's curves for a p-type Gaussian layer, Fig. 4.16(d), with RS xj = 62.5 ohm-µ m yields NP = 6 x 1019/cm3. At x = xj, 6 x 1019exp[-(2.5x10-5/2 Dt )2] = 1016, and

Dt = 4.24 x 10-6 cm. For the final layer, the dose in silicon is QSi = NP πDt = 4.5 x 1014/cm2. With the pre-deposition implant peak at the surface, the implanted dose will be 2 x QSi or 9.0 x 1014/cm2. D = 10.5 exp (-3.69/kT) = 2.96 x 10-13 at T = 1373 K, and the drive-in time is t = 60.7 sec! This very short time will require rapid thermal annealing, or the drive-in temperature could be reduced.

5.8 From Fig. 5.3 for phosphorus at an energy of 20 keV, Rp = 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

Q = NP

0

∞

∫ exp −x − 0.0250.012 2

2

dx with x in µm

Numerical integration with MATLAB® shows that only 98.14% of the profile is in the silicon, so that

Q = 2πNP 0.012µm( )[ ] 0.9814( ) and NP = 3.39 x 1020/cm3.for Q = 1015/cm2.

Then: 3.39x1020 exp −

x j − 0.025

0.012 2

2

= 1016 yields xj = 0.80 µm which agrees

well with the graph of the profile given below.

0 0.05 0.1 0.15 0.2 0.251015

1016

1017

1018

1019

1020

1021




5.9 (a) Irvin's curves can be used for each half of the distribution. We have 1015 = 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.



5.11 E =

mv2

2 or v=

2Em

; Also, remember 1 Joule = 1 kg-m2/sec2.

(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( )1.809x10−26 kg

= 2.98 x 105 m/sec

(b) For (BF2)++, m = [10.811+2(18.998)](1.673 x 10-27 kg) = 8.17 x 10-26 kg.

v =

2 10keV( )1.602x10−19 J/eV( )8.165x10−26 kg

= 1.98 x 105 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 =

2 5keV( )1.602x10−19 J/eV( )2.045x10−25 kg

= 8.85 x 104 m/sec

5.12 From Fig. 5.4 we see that the concentration is largest at x = RP. Using Eqn. 5.7, N(y) = 0.5 x 1020erfc [(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.



0 0.2 0.4 0.6 0.8 1 1.21014

1015

1016

1017

1018

1019

1020

1021

Distance from Center of Opening (um)



5.13 100 keV: RP = 0.30 µ m and ∆ RP = 0.07 µ m.

200 keV: RP = 0.55 µ m and ∆ RP = 0.09 µ m.

The doses are given by Q = 2πNP ∆R P and are 8.77 x 1013/cm2 and 1.13 x 1014/cm2 for the two implants with NO = 5 x 1018/cm3.

So: N x( ) = 5x1018 exp −

x − 0.30.07 2

2

+ exp −x − 0.550.09 2

2

as plotted below.

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.

5x1018 exp −

x j1 − 0.3

0.07 2

2

= 1016 yie lds xj1 = 0.053 µm

5x1018 exp −

x j2 − 0.55

0.09 2

2

= 1016 yie lds xj2 = 0.87 µm

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



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 5-MeV 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.

RS = q µ N( ) N x( ) dxx j1

x j2

∫

−1

N x( ) = NP exp −x − R P( )2

2∆R P2

µ n = 92 +1270

1+ N1.3x1017

0.091 µ p = 48 +447

1+ N6.3x1016

0.076

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.





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 =

kT2πpd2 =

1.38x10 −23 J/K( )N − m/J( )102 cm/m( )300K( )2π 10−4 N/m2( )10 −4 m2 /cm2( )3.6x10 −8 cm( )2 = 7190 cm

or 71.9 m. (10-4 Pa)(0.0075 torr/Pa) = 7.5 x 10-7 torr

6.4

n = PkT

=10−8 Pa 1 N/m2 /Pa( )

1.38x10−23 J/K 300K( )1 N m/J( )= 2.42x1012 /m3

n = 2.42x106 molecules/cm3

6.5

P = nkT = 1000/cm3( )106cm3 /m3( )1.38x10−23 J/K( )300K( )1 N m/J( )P = 4.14x10−12 N/m2 = 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 1015) 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 + 52 , and cos φ = (80/80.16). So at the wafer center, G1 = m(12)/π ρ (802). At the wafer



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 = 1002 +102 , = 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/ d2 +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 = NS/Φ and P = (2.23x1014) 32 • 300 / (2.63 x 1020)(1.44 x 104) = 5.77 x 10-9 Pa = 4.33 x 10-11 Torr - an ultrahigh vacuum (UHV) system.

6.11 v = (Ng/N) ks hg/(ks+hg). ks = 2 x 106 exp (-1.9)(1.6 x 10-19)/(1.38 x 10-23)(1473) = 0.64 cm/sec. hg = 1 cm/sec, Ng = 3 x 1016 atoms/cm3. v = (0.64 x 1/1.64)(3 x 1016/5 x 1022) = 2.3 x 10-7 cm/sec or 0.14 µ m/min.

(b) At T = 1498 K, ks becomes 0.82 cm/sec, and v increases to 0.16 µ m/min. The change is 0.02 µ m/min, a 14% increase.

(c) Setting ks = 1 yields T = 1520 K or 1247 oC.

(d) From the SiH4 curve, ks = 0.2 µ m/min at 1000/T = 0.93 and 0.01 µ m/min at 1000/T = 1.1. EA = -1000 k (∆ ln ks/∆ (1000/T)) = -1000 x 8.617 x 10-5 x (ln(.2)-ln(.01))/(.93-1.1) = 1.52 eV.

6.12 The graph on the next page is generated with MATLAB® using equations 6.31 and 6.32. For both boron and phosphorus at 1200 oC, D = 10.5 exp (-3.69/(8.617x10-5) (1473) = 2.49 x 10-12cm2/sec. xepi = 10 µ m and vx = 0.2 µ m/min give the growth time t = 3000 sec. 2 Dt =1.73 µ m. The two profiles are given by

N1 x( ) = 1018

21+ erf

x − xepi

1.73µm

N2 x( ) =1016

2erfc

x − xepi

1.73µm

+ exp

x0.0747µm

erfc

x + xepi

1.73µm

From a blow-up of the graph, xj = 7.2 µ m.



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.



Graph for Problem 6.12.

0 2 4 6 8 10 12 14 161015

1016

1017

1018

1019


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 = 2002 + 502 = 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 = 4002 +1002 = 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 = 4002 +1502 = 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)

N(x) = 5x1019 1+ erf

x − 10.081

cm3



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)

N(x) = 1020 1+ erf

x − 20.045

cm3

Graph for Problem 6.16

0 0.5 1 1.5 2 2.5 31014

1015

1016

1017

1018

1019

1020

1021


Graph for Problem 6.17



0 0.5 1 1.5 2 2.5 31014

1015

1016

1017

1018

1019

1020

1021


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.



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,

φ bi = 0.56V + 25.8m V( )ln 1015

1010

= 0.857 V



For VR = 0, with KS = 11.7,

C =

1.602x10−19 1015( )11.7( ) 8.854x10−14( )2 0.857( ) 25x10−8( )= 2.46 fF /µm

7.9 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.

7.11 R = ρ

LA

= 5µΩ − cm( ) 10−4 cm

0.25x10−4 cm( )2 = 0.80 Ω

7.12 (a) RS = ρ /t = 1.7 x 10-6 Ω -cm/0.5x10-4cm = 0.034 Ω /.

(b) R = RS (L/W) = 0.034Ω / x (50µ m/0.5µ m) = 3.4 Ω .

(c) Cox = (2)(8.854 x 10-14 F/cm)/10-4cm = 1.77 x 10-9 F/cm2.

C = Cox (WL) = (1.77 x 10-9 F/cm2)(50x10-4cm)(0.5x10-4cm) = 0.443 fF.

(d) RC = 3.4Ω x (0.443 x 10-15 F) = 1.5 fsec.



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 (104/100) - 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 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)]/DoA2 0.01 2 - 3 6

[1- exp(-2DoA)]/2DoA 0.05 12 - 13 30 - 31

1/(1+DoA) 0.09 23 56

8.5 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



partitioning would yield no good die for 0% yield. Poisson statistics predicts Y = exp(-4) or 1.8%. For our wafer, Y = 0.43, 0.22, 0.077 for DoA = 1, 2, 4. For Y = [1 + DoA/α ]-α , numerical fitting yields an excellent fit for α = 3.1.



8.6 (a) Do = 10 and A = 0.4 cm2. Y = [1-exp(-8)]/8 = 0.125. N = π (R-S)2/S2 = π (62.5mm - 6.3mm)2/(6.3mm)2 = 248 die, where S has been approximated by 40 mm. Thus, there will be Y x N = 31 good dice. The total cost will be $1.60 + $250/31 = $9.67/die.

(b) The area of each die becomes 1.15(20) = 23 mm2. S is approximated by A = 4.8 mm, and N = 455 dice. Yield Y = [1-exp(-4.6)]/4.6 = 0.215, and Y x N = 98 good dice. The new cost per packaged die is $1.60 + $250/98 = $4.15/die. The total cost of the two-die set is $8.30, which is more economical!

8.7 (a) Do = 5, and wafer cost = $150. A = 40 mm2, and S = 40 mm. N = 248, and Y = 0.245. Y x N = 61 dice, and C = [1.60 + 150/61] = $4.06. For two dice, A = 23 mm2, N = 455, Y = 0.391 and Y x N = 178. The total cost is 2[1.60 + 150/178] = $4.89 which is more expensive.

(b) For a single chip, C = [1.60 + 300/61] = $6.52. For two dice, C = 2[1.60 + 300/178] = $6.57.

8.8 (a) For the first process: A = 25 mm2, and N = π (50-5)2/52 = 254 dice. Y = [1 – exp (-2 x 2 x 0.25)] / (2 x 2 x 0.25) = 0.632. Y x N = 161 good dice. The cost per die is C1 = M/161 for a wafer cost of M $/wafer. For the second process: A = 12.5 mm2. 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. C2 = 1.3M/199 = M/153. The new die is only slightly more expensive even at a defect level of 10 defects/cm2. 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 lim (1 + DoA/α )-α = lim 1/(1 + DoA/α )α = exp(-DoA).



α →∞ α →∞



8.10DoA exp(-DoA) α = 5 α = 5000

1 3.68E-01 4.02E-01 3.68E-012 1.35E-01 1.86E-01 1.35E-013 4.98E-02 9.54E-02 4.98E-024 1.83E-02 5.29E-02 1.83E-025 6.74E-03 3.13E-02 6.75E-036 2.48E-03 1.94E-02 2.49E-037 9.12E-04 1.26E-02 9.16E-048 3.35E-04 8.42E-03 3.38E-049 1.23E-04 5.81E-03 1.24E-0410 4.54E-05 4.12E-03 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- A1 )2/A1 , and the cost per die is $150/N1Y1.

For the 150 mm wafer, N2 = π (75- A2 )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/N1Y1

= $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( )

0

∞

∫ dD. After substitution of our f(D) and changing variables

with Z = D/Do, the integral becomes

Y1 =

2π

exp −Z DoA( )[ ] exp 4 Z 1( )2[ ]0

2

∫ dZ

This can be integrated numerically, and the results are compared below with Y2 = [1-exp(-DoA)/DoA]2.

DoA Y1 Y2

1 0.39 0.40



2 0.17 0.19 3 0.084 0.10

4 0.046 0.060 5 0.028 0.040



6 0.018 0.028 7 0.012 0.020 8 0.0092 0.016 9 0.0071 0.012 10 0.0056 0.010

8.15 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 N x Pk

0 44 1 44 2 22 3 7 4 2 5 0

8.16 (a) 0.70 = 1 +

DoA5

−5

yields DoA = 0.370. Do = 0.370/1.5 = 0.247 defects/cm2.

(b) 0.80 = 1 +

DoA5

−5


(c) 0.90 = 1 +

DoA5

−5


8.17 (a) 0.75 = 1 +

DoA6

−6

yields DoA = 0.295. Do = 0.295/4 = 0.0737 defects/cm2.

(b) 0.85 = 1 +

DoA6

−6

yields DoA = 0.165. Do = 0.295/4 = 0.0412 defects/cm2.



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

211.7( )8.854 x10−14( )5.86( )1.602 x10−19 1015( ) = 5.51 µm .

(b) The built-in potential = 0.56 + 0.0258 ln (3x1016/1010) = 0.945 V.

2Wd = 2

211.7( )8.854x10−14( )5.95( )1.602x10−19 3x1016( ) = 1.01 µm

9.3 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



1014 1015 1016 1017 1018 1019-4

-3

-2

-1

0

1

2

3

4

Doping Concentration (#/cm 3)



9.5 An NMOS transistor is an enhancement-mode device if VTN > 0. Fig. 9.2 is for polysilicon gate devices with 10-nm gate oxides. From this graph, VTN > 0 for N > 2 x 1016/cm3. A more exact estimate can be obtained from Eq. (9.2):

VTN = −1.12

2+ Φ F + 2 11.7( ) 8.854x10−14( )1.602x10−19( ) NBΦ F /CO for

Φ F = 0.0258ln NB

1010

and CO =

3.9 8.854x10 −14( )10 −6 = 3.45 x 107F /cm2

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 NP ∆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 1015 = 2 x 1016 exp-(0.25)2/2∆ Rp2, and the projected range is 0.139 µ m. The dose Q = 2π 2x10 16( ) 1.39 x10 −5( ) = 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( )

0

∞

∫ dx = Ni x i & M2 = x N x( )0

∞

∫ dx =Ni xi

2

2

For the Gaussian distribution N(x) = Np exp(-x2/2∆ Rp2), and

M1 =

π2

NP ∆R P and M2 = NP ∆R P2

Equating moments yields the specified results.

9.8 Scaling factor α = 1/0.25 = 4. The circuit density increases by α 2 or 16 times more circuits/cm2. PDP ∝ 1/ α 3, so the power-delay product is reduced (improved) by a factor of 64.



9.9

ID* = µn

KOεO

XO

α

Wα

Lα

VGS − VTN −VDS

2

VDS = α ID

ID increases by the scale factor α .

P* = α ID VDD = α P - Power/circuit increases by the scale factor.

P*

A* =αPAα 2

= α 3 PA - 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

VTN = −1.12

2+ ΦF + 2 11.7( ) 8.854x10−14( )1.602x10−19( )3x1016( ) ΦF /CO = 0.721V

for ΦF = 0.0258 ln 3x1016

1010

and CO =

3.9 8.854x10 −14( )5x10−6 = 6.91 x 10 8F /cm2

A

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



20 α = 10 λ

22 α = 11 λ



9.13 (a) The well doping is almost constant and equal to the surface concentration in the depletion-region beneath the gate of the MOSFET. So the substrate doping for the NMOS device is 3 x 1015/cm3 and 3 x 1016/cm3 for the PMOS device. VTN is calculated using Eqns. 9.2 with CO = 2.30 x 10-7 F/cm2 for a 15 nm oxide thickness.

ΦF = 0.0258ln3x1015

1010

= 0.325 V

VTN = −0.56 + 0.325 + 2 11.7( ) 8.854x10−14( )1.602x10−19( )3x1015( )0.325( )/CO = −0.16 V

ΦF = 0.0258ln3x1016

1010

= 0.385 V

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

(b) The NMOS device requires a +1.16-V shift, and the PMOS device requires a +0.21-V shift. The required dose is given approximately by Q = CO∆ 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

p+ n+

n p-well5 x 1016/cm3 3 x 1015/cm3

0 V 8 V8 V 0 V

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 5x1016 1010( )= 0.958V

Wd =2 11.7( ) 8.854x10−14( )8.958( )

1.602x10−19 5x1016( ) = 0.481 µm

Two-sided formula for the well-substrate junction:

Φ bi =

kTq

lnNAND

ni2

= 0.0258ln

3x1015 • 5x1016

1020

= 0.723V



Wd =2K Sεo VA + Φbi( )

qNAND

NA + ND

Wd =2 11.7( ) 8.854x10 −14( )8.732( )

1.602x10−19

3x1015 • 5x1016

5.3x1016

= 2.00 µm

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

4.2x1014 = 1016 exp −

2x10−4

2 Dt

2

→ Dt = 3.154x10−9

cm2

For boron at 1075 oC, D = 10.5 exp[-3.69/(8.617 x 10-5 x 1348)] = 1.678 x 10-13

cm2/sec, and the drive-in time t = 5.22 hours. The dose Q = NO πDt = 9.95 x 1011/cm2. From Fig. 4.10(b) with N/NO = 0.042, X Y = 1.3/1.8 = 0.722. Lateral diffusion = 1.4 µ m.

(b) For this case, we have

4.2x1014 = 5x1016 exp −

1.5x10−4

2 Dt

2

→ Dt = 1.177x10 −9

cm2

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

4.2x1014 =

1020

21+ e rf

x − 3x10−4

2 Dt

→x − 3x10−4

2 Dt

= −3.789

which yields x – 3 µ m = 1.52 µ m of out diffusion.



9.16

Contacts Thin Oxide Metal

9.17

14 λ

24 λ

The area is reduced from 416 λ 2 to 336 λ 2, a 19% improvement, but the gate overlap capacitance is significantly larger. The overlap area is now 72 λ 2 versus 44 λ 2 in the original layout.



9.18

All Levels Aligned to First level (Diffusion)

24 λ

24 λ

Lateral Diffusion = 2 µm = 1 λ

Thin Oxide Aligned to Diffusion Contacts and Metal Aligned to Thin oxide

22 λ

22 λ

Lateral Diffusion = 2 µm = 1 λ

2 λ x 2 λ Contacts



9.19

VO

+VDD

A

B

C

9.20 Area = (12 λ )(44 λ ) = 528 λ 2 --- Active gate area = 2(10 λ )(2 λ ) = 40 λ 2



Well Boundary

44 λ

12 λ

16 λ

9.21 Area = (12 λ )(32 λ ) = 384 λ 2 --- Active gate area = 2(10 λ )(2 λ ) = 40 λ 2

Well Boundary

32 λ

12 λ

4 λ

9.22 Approximately (5 x 1022) x (0.25 x 10-4) = 1.25 x 1018 silicon atoms/cm2 are in the layer to be formed. We need to implant two oxygen atoms per silicon atom or a total dose of 2.5 x 1018 oxygen atoms/cm2. Five 200-mm wafers have a total area of 1571 cm2, so a total number of 3.93 x 1021 oxygen atoms is required per hour. The beam current will be (3.93 x 1021/hr)(1.602x10-19C)/(3600sec/hr) = 175 mA, and the power in the 4-MeV beam is 699 kW. The wafers will be destroyed!

9.23 (a)

VDE − VCD =

IRS

WL2

+ ∆Y

−

L2

− ∆Y

= IRS

2∆YW



and

VAC = IR S

LW

VDE − VCD

VAC

= ∆Y2L

and ∆Y =L2

VDE − VCD

VAC

(b)

VIJ − VJH =

IRS

WL2

+ ∆X

−

L2

− ∆X

= IRS

2∆XW

and

VGH = IR S

LW

VIJ − VJH

VIH

= ∆X2L

and ∆X =L2

VIJ − VJH

VIH



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 -1 = (2 x 1013)/(4 x 1016) + 42/1(502) & β = 145.

10.2 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,

GB = N x( )/DB[ ]dx1.5µm

4.0µm

∫ = 1.5x1017 exp − x2 Dt

2

dx

1.5x10−4

4.0x10−4

∫

GB = 2.12x1013( ) exp −z2( )dz1.06

2.83

∫ for z =x

1.41x10 4

GB = 2.51x1011 and β ≅ GE

GB

= 5x1013

2.51x1011 = 199

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 1015/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.



Φbi = 0.56+ 0.0258( )ln4x1015

1010

= 0.893 V

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.



10.6 From Table 4.2 on page 81, Q = 0.55 NO xj, and NO = 1.56 x1017/RS xj which yield Q = 8.64 x 1015/cm3 for a 10 ohm/square sheet resistance. For the heavy doping limit in the equations in Prob. 4.24, the minority carrier diffusion constant approaches a value of approximately (92)(0.0258) = 2.4 cm2/sec (with Dn/µ n = kT/q). The Gummel number can be approximated by GE ≅ Q/D = 3.6 x 1015 sec/cm4.

10.7 (a) For S ≤ 0, the collector-base and emitter-base junctions are approximately the same. From the drawing, the emitter junction depth is approximately 2 µ m. From Fig. 10.5, the breakdown voltage is approximately 8 volts for a junction depth of 2 µ m and a doping concentration of 1018/cm3.

(b) For S = 3 µ m, breakdown will occur when the depletion layer in the lightly-doped 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 punch-through 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 5-µ 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 1018/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.



(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.



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)



n-type substrate

p-well

p+ n+ n+

E CB

p-type substrate

n-well

n+ p+ p+

E CB



10.15

Emitter Mask

Base Maskn+ contact mask

Isolation Mask Boundary

Final Base

Final Emitter

Final n+

Final Isolation

C B E

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.

n-type substrate

pn+ n+

E CB

p

n+ buried layer



10.17 1. trench etch 6. n+ poly pattern (N/C) 2. selective oxidation 7. contact windows 3. n+ ion-implantation (I/I) 8. metal 4. p+ poly pattern (N/C) 9. passivation 5. p I/I (N/C) (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 4-5 volts. The emitter intersects the base at a doping of approximately 6 x 1018/cm3. Using the 1-µ m curve in Fig. 10.5, the breakdown voltage will be less than 3 V. We can get a punch-through estimate from Fig. 10.4 using a background concentration of 1017/cm3, a base width of 270-170 = 100 nm, and xj = 1 µ m. The depletion layer x1

on the base side will reach 100 nm for a total voltage of 4 V. Assuming a built-in potential of 1 V, the base will punch through at approximately 3 V. A smaller junction depth and radius of curvature will reduce the breakdown and punch through voltages below these estimates. In addition, based upon the results in Problem 10.19, the base will punch through when x1 reaches 83 nm.

10.19 For the emitter side, use a one-sided step junction with NB = 5 x 1018/cm3: Φ bi = 0.56 + 0.0258 ln (5x1018/1010) = 1.08 V. Wd = [2(11.7)(8.854x10-14)(1.08)/(1.602x10-19)(5x1018)]0.5 ≅ 17 nm. On the collector side, we can estimate the space charge layer intrusion using Fig. 10.4 with some assumptions. For Φ bi = 0.85 V, a doping of approximately 1017/cm3 near the junction, and a 1-µ m radius, the total SCR width is 120 nm with 42 % in the base or 50 nm. The metallurgical base width is approximately 270 – 170 = 100 nm. The neutral base with is approximately 100–17–50 = 33 nm, a very narrow base!

10.20 (a) 1021erfc

x j

2 Dt

= 5x1018 at xj = 50 nm or

x j

2 Dt= 1.985

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.



(b) 5x1018 exp −

x j

2 Dt

2

= 5x1016 at xj = 100 nm or

x j

2 Dt

2

= 4.605

For boron at 1000 oC, D = 10.5 exp [-3.69/(8.617x10-5)(1273)] = 2.58 x 10-14 cm2/sec. Dt = 5.43 x 10-12 cm2/sec, and t = 210 sec or 3.51 min. This time is too short.

Both the base and emitter diffusions probably require rapid thermal processing to achieve the small Dt products that are required for this structure.

10.21

n-type collector

p n n+

EC

n+ buried layer

p+

p-substrate

B n-collector (+V) p-sub (-V)

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.



WB < [(3.18psec)(10)(20cm2/sec)]0.5 = 250 nm.

CBE < 3.18psec/25Ω = 0.13 pF.



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; p-well/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 Align to 1 3. n+ emitter diffusion Align to 2 4. contact windows Align to 3 5. metal layer Align to 4 6. passivation layer openings 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 rj = ∞ 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.



CHAPTER 11

A useful triangle:

23

1

54.74oφ

tanφ = 2 cos φ =

13

sin φ =23

11.1 The unit vector perpendicular to the <111> plane is v1 =

13

13

13

and that

perpendicular to the <100> plane is v2 = 1 0 0( ). Thus we have c o s θ = v1 • v2 and

θ = cos−1 1

3

= 54.74o

11.2 The unit vector perpendicular to the <111> plane is v1 =

13

13

13

and that

perpendicular to the <110> plane is v2 =

12

12

0

. Thus we have c o s θ = v1 • v2

and θ = cos−1 2

6

= 35.3o

11.3

10W2

= tan54.74o( ) → W =14.1 µm

W

10 µm 54.74o



11.4 (a) Horizontal cross section through second P2-P1 contact region

(b) Horizontal cross section through the center P2-P1 contact region

P1

P2

P1

(a)

P1

P2

P1

(b)

11.5 (2/L) = sin (54.74o) and L = 2.45 µ m.

2 µm 54.74o

n+-

n+

L L

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

1 µm

54.74o5 µm Y

YW2

= tan 54.74o( )= 2

Y = 5 + 0.5 2 = 5.71 µm

W = Y 2 = 8.07 µm



11.7

Oxide Etch 1

Oxide Etch 2

Polysilicon

10 λ

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

10 µm

54.74o500 µm Y

W = 2 5 +500

tan 54.74o( )

µm

W = 2 5 +500

2

= 717 µm



11.11 The values here are approximate from scaling the photograph. For the spring arms, W/L ≅ 22/1. Assuming the fingers are F = 2.5 µ m wide and spaced by 3F, the area of the fingers is 1.35 x 104 µ m2, the area of the finger base is 1.80 x 104 µ m2, and the area of the mid pieces is 5.35 x 103 µ m2. The total area is 3.685 x 104 µ m2. The mass will be (3.685 x 104 x 2 x10-12 cm3) x (2.3 x 10-3 kg/cm3) = 1.70 x 10-10 kg.

fo =

12π

4 1.7x1011 kg − m/sec2/m2( )2x10−6 m( )1.7x10−10 kg

122

3

= 138 kHz

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 .

Y

X ∆X

Y/S

φ'

φ = 54.74o

90o - φ

φ

The new angle φ ’ is given by

tan φ' = YX + ∆X

= YY2

+ YS

32

φ' = tan−1 2

1+ 3S

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


introduction to microelectronic fabrication

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