Solutions Manual to AccompanyFUNDAMETALS OF SEMICONDUCTOR FABRICATION

G. S. MayMotorola Foundation Professor School of Electrical & Computer Engineering Georgia Institute of Technology Atlanta, GA, USA

S. M. SZEUMC Chair Professor National Chiao Tung University National Nano Device Laboratories Hsinchu, Taiwan

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John Wiley and Sons, Inc New York. Chicester / Weinheim / Brisband / Singapore / Toronto

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ContentsCh.1 Introduction ----------------------------------------------------------------------------- N/A Ch.2 Crystal Growth ------------------------------------------------------------------------ 1 Ch.3 Silicon Oxidation ---------------------------------------------------------------------- 8 Ch.4 Photolithography------------------------------------------------------------------------12 Ch.5 Etching -----------------------------------------------------------------------------------15 Ch.6 Diffusion ---------------------------------------------------------------------------------18 Ch.7 Ion Implantation ------------------------------------------------------------------------26 Ch.8 Film Deposition -------------------------------------------------------------------------32 Ch.9 Process Integration ---------------------------------------------------------------------40 Ch.10 IC Manufacturing----------------------------------------------------------------------65 Ch.11 Future Trends and Challenges--------------------------------------------------------78

0

CHAPTER 2

1. C0 = 1017 cm-3 k0(As in Si) = 0.3 CS= k0C0(1 - M/M0)k0-1 = 0.3×1017(1- x)-0.7 = 3×1016/(1 - l/50)0.7 x l (cm) CS (cm-3) 0 0 3×1016 0.2 10 3.5×1016 0.4 20 4.28×1016 0.6 30 5.68×1016 0.8 40 1.07×1017 0.9 45 1.5×1017

16 14 12 10 8 6 4 2 0 0 10 20 30 40 50 l ( cm )

2. (a) The radius of a silicon atom can be expressed asr= 3 a 8 3 × 5.43 = 1.175Å 8

so r =

(b) The numbers of Si atom in its diamond structure are 8. So the density of silicon atoms is

n=

8 8 = = 5.0 × 10 22 atoms/cm 3 3 3 a (5.43Å)

(c) The density of Si is

ρ =

M / 6.02 × 10 23 1/ n

N (10 D

16

cm )

-3

=

28.09 × 5 × 10 22 6.02 × 1023

g / cm 3 = 2.33 g / cm3.

1

3. k0 = 0.8 fo boon in silicon M / M0 = 0.5 The density of Si is 2.33 g / cm3. The accepto concentation fo ρ = 0.01 Ω–cm is 9×1018 cm-3. The doping concentation CS is given by C s = k 0 C 0 (1 − ThereforeC0 = Cs 9 × 1018 = M k 0 −1 0.8(1 − 0.5) − 0.2 ) k 0 (1 − M0

M k0 −1 ) M0

= 9.8 × 1018 cm − 3 The amount of boron required for a 10 kg charge is

10,000 × 9.8 × 1018 = 4.2 × 10 22 boron atoms 2.338 So that10.8g/mole × 4.2 × 10 22 atoms = 0.75g boron . 6.02 × 10 23 atoms/mole

4. (a) The molecular weight of boron is 10.81. The boron concentration can be given asnb = number of boron atoms volume of silicon wafer 5.41 × 10 −3 g / 10.81g × 6.02 × 10 23 = 10.0 2 × 3.14 × 0.1 = 9.78 × 1018 atoms/cm 3

(b) The average occupied volume of everyone boron atoms in the wafer is

2

V =

1 1 = cm 3 18 nb 9.78 × 10

We assume the volume is a sphere, so the radius of the sphere ( r ) is the average distance between two boron atoms. Thenr= 3V = 2.9 × 10 −7 cm . 4π

5. The cross-sectional area of the seed is 0.55 2 π = 0.24 cm 2 The maximum weighthat can be su

orted by the seed equals the

roduct of the critical yield stren

gth and the seed’s cross-sectional area: (2 × 10 6 ) × 0.24 = 4.8 × 10 5 g = 480 kg The corres

onding weight of a 200-mm-diameter ingot with length l is

2

20.0 ( 2.33g/cm )π l = 480000 g 2 ∴ l = 656 cm = 6.56 m.3

2

6. We have

M Cs / C0 = k0 1− M 0 Fractional 0 solidified 0.2

k0 −1

0.4

0.6 0.8 1.0

Cs /C0

0.05

0.06

0.08

0.12

0.23

∞

3

Cs/Co

0.21 0.11 0.01 0 0.2 0.4 0.6 0.8 1 Fraction Solidified

7. The segregation coefficient of boron in silicon is 0.72. It is smaller than unity, so the solubility of B in Si under solid phase is smaller than that of the melt. Therefore, the excess B atoms will be thrownoff into the melt, then the concentration of B in the melt will be increased. The tailend of the crystal is the last to solidify. Therefore, the concentration of B in the tailend of grown crystal will be higher than that of seedend. 8. The reason is that the solubility in the melt is proportional to the temperature, and the temperature is higher in the center part than at the perimeter. Therefore, the solubility is higher in the center part, causing a higher impurity concentration there. 9. The segregation coefficient of Ga in Si is 8 ×103 From Eq. 18C s / C 0 = 1 − (1 − k )e − kx / L We havex= = L 1− k ln k 1 − C s / C0

2 1 − 8 × 10 − 3 ln 8 × 10 3 1 − 5 × 1015 / 5 × 1016 = 250 ln(1.102) = 24 cm.10. We have from Eq.18

Cs = C0[1 − (1 − ke ) exp( ke x / L)] −4

− So the ratio Cs / C0 = [1 − (1 − ke ) exp( ke x / L)]= 1 − (1 − 0.3) • exp(−0.3 × 1) = 0.52

at x / L = 1

= 0.38

at x/L = 2.

11. For the conventionallydoped silicon, the resistivity varies from 120 Ω-cm to 155 Ω-cm. The coesponding doping concentation vaies fom 2.5×1013 to 4×1013 cm-3. Theefoe the ange of beakdown voltages of p+ - n junctions is given by VB ≅ε s Ec 22q

( N B ) −1

=

1.05 × 10 −12 × (3 × 10 5 ) 2 ( N B ) −1 = 2.9 × 1017 / N B = 7250 to 11600 V −19 2 × 1.6 ×

∆V B = 11600 − 7250 = 4350 V

∆V ∴ B / 7250 = ±30% 2 For th nutron irradiatd silicon, ρ = 148 1.5 Ω-cm. Tncentation is 3×1013 (1%). The ange of beakdown voltage isV B = 1.3 × 10 17 / N B = 2.9 × 10 17 / 3 × 10 13 ( 1%) = 9570 to 9762 V .∆V B = 9762 − 9570 = 192 V

∆V ∴ B 2 12. W hav / 9570 = ±1% .

M s wight of GaAs at Tb C m − C l s = = = M l wight of liquid at Tb C s − C m lThrfor, th fraction of liquid rmaind f can b obtaind as following5

f =

Ml l 30 = ≈ = 0.65 . M s + M l s + l 16 + 30

13. From th Fig.2.11, w find th vapor prssur of As is much highr than that of th Ga. Thrfor, th As contnt will b lost whn th tmpratur is incrasd. Thus th composition of liquid GaAs always bcoms gallium rich. − 88.8 (T / 300)

14. n s = N xp(− E s / kT ) = 5 × 10 22 xp(2.3 V / kT ) = 5 × 10 22 xp = 1.23 × 10 −16 cm −3 ≈ 0 at 27 0 C = 300 K = 6.7 × 1012 cm −3 = 6.7 × 1014 cm −3 at 900 0 C173 K at 1200 0 C = 1473 K .

15. n f = NN ` xp(− E f / 2kT ) = 5 × 10 22 × 1 × 10 27 −3.8V / kT × −1.1V / 2 kT = 77 × 10 24 × −94.7 /(T / 300 ) = 5.27 × 10 −17 at 27oC = 300 K =2.14×1014 at 900oC = 1173 K. 16. 37 × 4 = 148 chips In trms of lithostppr considrations, thr ar 500 µm spac tolranc btwn th mask boundary of two dic. W divid th wafr into four symmtrical parts for convnint dicing, and discard th primtr parts of th wafr. Usually th quality of th primtr parts is th worst du to th dg ffcts.6

7

CHAPTER 3

1. From Eq. 11 (with τ=0) x2+Ax = B From Figs. 3.6 and 3.7, we ob

ain B/A =1.5 µm /

hr, B=0.47 µm 2/hr, herefore A= 0.31 µm. The

ime required

o grow 0.45µm oxide is

1 2 1 (x + Ax) = (0.45 2 + 0.31 × 0.45) = 0.72 hr = 44 min . B 0.47=

2. Afer a window is opened in

he oxide for a second oxida

ion,

he ra

e cons

a

ns are B = 0.01 µm 2/hr, A= 0.116 µm (B/A = 6 ×10-2 µm /hr). If

he ini

ial oxide

hick

ness is 20 nm = 0.02 µm for dry oxidaion,

he value ofτcan be ob

ained as followed:

(0.02)2 + 0.166(0.02) = 0.01 (0 +τ) orτ= 0.372 hr.

For an oxidaion

ime of 20 min (=1/3 hr),

he oxide

hickness in

he window are

a is x2+ 0.166x = 0.01(0.333+0.372) = 0.007 or x = 0.0350 µm = 35 nm (gae oxide).

For he field oxide wi

h an original

hickness 0.45 µm,

he effec

iveτis given by

τ=1 2 1 (0.45 2 + 0.166 × 0.45) = 27.72 hr. ( x + Ax) = B 0.01

x2+ 0.166x = 0.01(0.333+27.72) = 0.28053 or x = 0.4530 µm (an increase of 0.003µm only for

he field oxide). 3. x2 + Ax = B (

+ τ )

(x + A 2 A2 ) − = B (t + τ ) 2 4

(x +

A2 A 2 ) =B + ( + τ ) 2 4B

8

when >> τ ,

>>

hen, x2 = B

similarly, when

>> τ ,

>>

hen, x =

B ( + τ ) A

A2 , 4B

A2 , 4B

4. A 980(=1253K) and 1 a

m, B = 8.5×10-3 µm 2/hr, B/A = 4×10-2 µm /hr (from Figs. 3.6 an

d 3.7). Since A ≡2D/k , B/A = kC0/C1, C0 = 5.2×1016 molecules/cm3 and C1 = 2.2×1022 cm-3 ,

he diffusion coefficien

is given by

Ak A B C1 = ⋅ 2 2 A C0 B C1 = 2 C 0

D= =

8.5 10 −3 2.2 10 22 µm 2 / hr 2 5.2 1016 = 1.79 10 3 µm 2 / hr = 4.79 10 9 cm 2 /5. For impurity in the oxidation process of silicon, segregation coefficeint = 3 1011 3 = = 0.006 . 13 500 5 10 equilibrium concentration of impurity in silicon . equilibrium concentration of impurity in SiO 2

6.

κ=

7. The SUPREM input file for the first oxidation step is:TITLE COMMENT INITIALIZE COMMENT DIFFUSION PRINT STOP Problem 37a Initialize silicon substrate <100> Silicon Oxidize the wafer for 60 minutes at 1100 C in dry O2 Time=60 Temperature=1100 DryO2 Layers End Problem 37aThe result of the PRINT Layers command is:9

layer (microns) 2 1

material type (microns) OXIDE SILICON

thicness node node

dx

dxmin

top 798 805

bottom no. 804 1000

0.1088 0.0100 0.0010 1.9521 0.0100 0.0010

This indicates an oxide thicness of 0.1088 µm, which means 0.44 * 0.1088 µm = 0.047

9 µm of silicon has been consumed during the process (i.e., the Si/SiO2 interface is 0.0479 µm below the original Si surface). The figure below is a graphical representation.

For the half of the wafer, the oxide is removed, and the wafer is reoxidized in wet O2. For this half, we use the following SUPREM input file:TITLE COMMENT INITIALIZE COMMENT DIFFUSION COMMENT ETCH Problem 37b Initialize silicon substrate <100> Silicon Oxidize the wafer for 60 minutes at 1100 C in dry O2 Time=60 Temperature=1100 DryO2 Remove the oxide Oxide All

COMMENTPRINT STOP

Oxidize the wafer for 30 minutes at 1000 C in wet O2Layers End Problem 37bDIFFUSION Time=30 Temperature=1000 WetO2

The result of this PRINT Layers command is:layer no. 2 1 material type OXIDE SILICON thic

ness (microns) dx dxmin top botto

m (microns) node node 803 815 814 1000

0.2291 0.0100 0.0010 1.8513 0.0100 0.0010

This indicates a final oxide thicness of 0.2291 µm on the etched side, which mean

s an additional 0.4410

* 0.2291 µm = 0.1008 µm of silicon has been consumed during the process. The total distance from the Si/SiO2 interface to the original Si surface on this side is therefore 0.0479 µm + 0.1008 µm = 0.1487 µm. The unetched side is simulated using the SUPREM input file:

TITLE COMMENT INITIALIZE COMMENT DIFFUSION

Problem 37c Initialize silicon substrate <100> Silicon Oxidize the wafer for 60 minutes at 1100 C in dry O2 Time=60 Temperature=1100 DryO2

COMMENTPRINT STOP

Oxidize the wafer for 30 minutes at 1000 C in wet O2Layers End Problem 37cDIFFUSION Time=30 Temperature=1000 WetO2

The result of this PRINT Layers command is:layer no. 2 1 material type thic

ness (microns) dx dxmin top bottom (microns) no

de node 798 813 812 1000

OXIDE SILICON

0.2897 0.0100 0.0010 1.8725 0.0100 0.0010

On this side, a total of 0.44 * 0.2897 µm = 0.1275 µm of Si is consumed. The total distance from the Si/SiO2 interface to the original Si surface on this side is 0.44 * 0.2987 µm = 0.1275 µm. The figure below is a graphical representation of the final structure.

11

The step heights on the surface and in the substrate are 0.0818 µm and 0.0212 µm, respectively.

12

CHAPTER 4

1. With reference to Fig. 2 for class 100 clean room we have a total of 3500 particles/m3 with particle sizes ≥ 0.5 µm 21 3500 = 735 particles/m2 with particle sizes ≥ 1.0 µm 100 4.5 3500 = 157 particles/m2 with particle sizes ≥ 2.0 µm 100 Therefore, (a) 3500735 = 2765 particles/m3 between 0.5 and 1 µm (b) 735157 = 578 particles/m3 between 1 and 2 µm (c) 157 particles/m3 above 2 µm. 2.Y = Π e − D1 An =1

9

A = 50 mm2 = 0.5 cm2 Y = e −4( 0.10.5) e −4 ( 0.250.5) e −1(10.5) = e −1.2 = 30.1% .3. The available exposure energy in an hour is 0.3 mW2/cm2 3600 s =1080 mJ/cm2 For positive resist, the throughput is 1080 = 7 wafers/hr 140 For negative resist, the throughput is 1080 = 120 wafers/hr . 9 4. (a) The resolution of a projection system is given byl m =

1

λNA

= 0 .6 ×

0.193μ m = 0.178 µm 0.65

DOF = k 2

0.193µm = 0.228 µm = 0 .5 2 ( NA) (0.65) 2

λ

(b)

We can increase NA to improve the resoution. We can adopt resoution enhancement techniques (RET) such as optica proximity correction (OPC) and phase-shifting Masks (PSM). We can aso deveop new resists that provide ower k1 and higher k2 for better13

resoution and depth of focus. (c) PSM technique changes k1 to improve resoution. 5. (a) Using resists with high γ value can result in a more vertical profile but throuhput decreases. (b) Conventional resists can not be used in deep UV lithoraphy process because these resists have hih absorption and require hih dose to be exposed in deep UV. This raises the concern of damae to stepper lens, lower exposure speed and reduced throuhput.6. (a)

A shaped beam system enables the size and shape of the beam to be varied, thereby minimizin the number of flashes required for exposin a iven area to be patterned. Therefore, a shaped beam can save time and increase throuhput compared to a Gaussian beam.

(b)

We can make alinment marks on wafers usin e-beam and etch the exposed marks. We can then use them to do alinment with e-beam radiation and obtain the sinal from these marks for wafer alinment. X-ray lithoraphy is a proximity printin lithoraphy. Its accuracy requirement is very hih, therefore alinment is difficult.

(c) X-ray lithoraphy usin synchrotron radiation has a hih exposure flux so X-ray has better throuhput than e-beam.7. (a) To avoid the mask damae problem associated with shadow printin, projection printin exposure tools have been developed to project an imae from the mask. With a 1:1 projection printin system is much more difficult to produce defect-free masks than it is with a 5:1 reduction step-and-repeat system. (b) It is not possible. The main reason is that X-rays cannot be focused by an optical lens. When it is throuh the reticle. So we can not build a step-and-scan X-ray lithoraphy system.8. All of the above values can be entered from the Parameters menu or by clickin on the appropriate14

icon on the toolbar. The resultin resist profile is shown in the fiure below.In comparison to Example 3, we see that no resist feature is printed under the modified process conditions. The combination of the lon pre-bake time and low exposure dose prevents the feature from bein defined.15

CHAPTER 5

1.

As shown in the fiure, the profile for each case is a sement of a circle with oriin at the initial mask-film ede. As overetchin proceeds the radius of curvature increases so that the profile tends to a vertical line.

2.

(a) 20 sec 0.6 × 20/60 = 0.2 µm…..(100) plane 0.6/16 × 20/60 = 0.0125 µm……..(110) plane 0.6/0 × 20/60 = 0.002 µm…….(111) planeWb = W0 − 2l = 1.5 − 2 0.2 = 1.22 µm(b) 40 sec 0.6 40/60 = 0.4 µm….(100)plane 0.6/16 40/60 = 0.025 µm…. (110) plane 0.6/100 40/60 = 0.004 µm…..(111) planeWb = W0 − 2l = 1.5 − 2 0.4 = 0.93 µm(c)

60 sec 0.6 1 = 0.6 µm….(100)plane0.6/16 1 = 0.0375 µm…. (110) plane 0.6/100 1= 0.006 µm…..(111) planeWb = W0 − 2l = 1.5 − 2 0.6 = 0.65 µm.3. Using the data in Prob. 2, the etched pattern profiles on <100>Si are shown in below. (a) 20 sec l = 0.012 µm, W0 = Wb = 1.5 µm16

(b) 40 sec

l = 0.025 µm, W0 = Wb = 1.5 µm

(c) 60 sec l = 0.0375 µm W0 = Wb = 1.5 µm.

4. If we protect the IC chip areas (e.g. with Si3N4 layer) and etch the wafer from the top, the width of the bottom surface is W = W1 + 2l = 1000 + 2 625 = 1884 µm The fraction of surface area that is lost is (W 2 − W12 ) / W 2 100%=(1884210002) /18842 100% = 71.8 % In terms of the wafer area, we have lost 71.8 % π (15 / 2) 2 =127 cm2 Another method is to define masking areas on the backside and etch from the back. The width of each square mask centered with res

ect of IC chi

is g

iven by W = W1 − 2l = 1000 − 2 625 = 116 µm Using this method, the fraction of the top surface area that is lost can be negligibly small.

5. 1 Pa = 7.52 m Torr PV = nRT 7.52 /760 103 = n/V 0.082 273 n/V = 4.42 107 mole/liter = 4.42 107 6.02 1023/1000 =2.7 1014 cm3 mean–free–pathλ = 5 × 10 −3 / P cm = 5 103 1000/ 7.52 = 0.6649 cm = 6649 µm150Pa = 1128 m Torr PV = nRT17

1128/ 760 103 = n/V 0.082 273 n/V = 6.63 105 mole/liter = 6.63 1056.021023/100= 4 1016 cm3 meanfreepathλ = 5 × 10 −3 / P cm = 5 103 1000/1128 = 0.0044 cm = 44 µm.6. Si Etch Rate (nm/min) = 2.86 1013 n F T1 2e− Ea

RT

= 2.86 1013310 (298)15

1

2e−2.48103 1.987 298= 224.7 nm/min.

7. SiO2 Etch Rate (nm/min) = 0.614 10 Etch selectivity of SiO2 over Si =13310 (298)15

1

2e−3.76103 1.987298= 5.6 nm/min

5.6 = 0.025 224.7 0.614 ( −3.76+ 2.48)1.987298 e = 0.025 . 2.86Or etch rate (SiO2)/etch rate (Si) =

8. A three–step process is required for polysilicon gate etching. Step 1 is a nonselective etch process that is used to remove any native oxide on the polysilicon surface. Step 2 is a high polysilicon etch rate process which etches polysilicon with an anisotropic etch profile. Step 3 is a highly selective polysilicon to oxide process which usually has a low polysilicon etch rate. 9. If the etch rate can be controlled to within 10 %, the polysilicon may be etched 10 % longer or for an equivalent thic

ness of 40 nm. The selectivity is therefore 40 nm/1 nm =

40. 10. Assuming a 30% overetching, and that the selectivity of Al over the photoresist maintains 3. The minimum photoresist thic

ness required is

18

(1+ 30%) 1 µm/3 = 0.433 µm = 433.3 nm. 11.ωe =qB me9

1.6 × 10 −19 × B 2π × 2.45 × 10 = 9.1 × 10 −31 B = 8.75 × 102(tesla) = 875 (gauss). 12. Traal RIE generates lodensity lasma (109 cm3) ith high ion energy. ECR and ICP generate highdensity lasma (1011 to 1012 cm3) ith lo ion energy. Advantages of ECR and ICP are lo etch damage, lo microloading, lo asectratio deendent etching effect, and simle chemistry. Hoever, ECR and ICP systems are more comlicated than traditional RIE systems. 13. The corrosion reaction requires the resence of moisture to roceed. Therefore, the first line of defense in controlling corrosion is controlling humidity. Lo humidity is essential,. esecially if coer containing alloys are being etched. Second is to remove as much chlorine as ossible from the afers before the afers are exosed to air. Finally, gases such as CF4 and SF6 can be used for fluorine/chlorine exchange reactions and olymeric encasulation. Thus, AlCl bonds are relaced by AlF bonds. Whereas AlCl bonds ill react ith ambient moisture and start the corrosion rocess , AlF bonds are very stable and do not react. Furthermore, fluorine ill not catalyze any corrosion reactions.

19

CHAPTER 6

1. Ea(boron) = 3.46 eV, D0 = 0.76 cm2/sec From Eq. 6,D = D0 ex(− Ea − 3.46 −15 2 ) = 0.76 ex = 4.142 × 10 cm /s −5 kT 8.614 × 10 × 1223 L = Dt = 4.142 × 10 −15 × 1800 = 2.73 × 10 −6 cm

From Eq. 9, C ( x ) = C s erfc(

x x ) = 1.8 × 10 20 erfc −6 2L 5.46 × 10

If x = 0, C (0) = 1.8 × 10 20 atoms /cm 3 ; x = 0.05 ×104, C(5× 106) = 3.6 × 1019 atoms/cm3; x = 0.075 ×104 , C(7.5×106) = 9.4 ×1018 atoms/cm3; x = 0.1×104, C(105) = 1.8 × 1018 atoms/cm3;x = 0.15× 104, C(1.5×105) = 1.8× 1016 atoms/cm3.The x j = 2 Dt (erfc 1C sub ) = 0.15µm Cs

Total amount of doant introduced = Q(t) = 2C s L = 5.54 × 1014 atoms/cm2.

π

− 3.46 − Ea −14 2 2. D = D0 ex = 0.76 ex = 4.96 × 10 cm /s −5 kT 8.614 × 1, C S = C (0, t ) =S

πDt

= 2.342 × 1019 atoms/cm 3

x x C ( x) = C S erfc = 2.342 × 1019 erfc −5 2L 2.673 × 10

If x = 0, C(0) = 2.342 × 1019 atoms/cm3; x = 0.1×104, C(105) = 1.41×1019 atoms/cm3;x = 0.2×104, C(2×105) = 6.79×1018 atoms/cm3; x = 0.3×104, C(3×105) = 2.65×101820

atoms/cm3;x = 0.4×104, C(4×105) = 9.37×1017 atoms/cm3; x = 0.5×104, C(5×105) = 1.87×1017atoms/cm3;x = 0.6×104, C(6×105) = 3.51×1016 atoms/cm3; x = 0.7×104, C(7×105) = 7.03×1015atoms/cm3;x = 0.8×104, C(8×105) = 5.62×1014 atoms/cm3.The x j = 4 Dt ln

S C B πDt

= 0.72 µm .

10 −8 3. 1 × 1015 = 1 × 1018 ex 4 × 2.3 × 10 −13 t t = 1573 s = 26 min For the constanttotaldoant diffusion case, Eq. 15 gives C S =S = 1 × 1018 π × 2.3 × 10 −13 × 1573 = 3.4 × 1013 atoms/cm 2 .

S

πDt

4. The rocess is called the raming of a diffusion furnace. For the ramdon situation, the furnace temerature T is given by T = T0 rt here T0 is the initial temerature and r is the linear ram rate. The effective Dt roduct during a ramdon time of t1 is given by( Dt ) eff = ∫t1 0

D (t )dt

In a tyical diffusion rocess, raming is carried out until the diffusivity is negligibly small. Thus the uer limit t1 can be taken as infinity:21

1T

=

1T0 − rt

≈

1T0

(1 +

rt T0

+ ...)

and− Ea = D ex − E a (1 + rt + ...) = D (ex − E a )(ex − rE a t ...) ≈ D (T ) ex − rE = D0 ex 0 0 0 kT 2 2 kT T0 kT0 kT0 kT0 0here D(T0) is the diffusion coefficient at T0. Substituting the above equation into the exression for the effective Dt roduct gives( Dt ) eff ≈ ∫ D (T0 ) ex0

∞

− rE a t kT02

dt = D (T0 )

kT0 rE a

2

Thus the ramdon rocess results in an effective additional time equal to kT02/rEa at the initial diffusion temerature T0. For hoshorus diffusion in silicon at 1000°C, e have from Fig. 6.4:D(T0) = D (1273 K) = 2× 1014 cm2/s r=1273 − 773 20 × 60

= 0.417 K / s

Ea = 3.66 eV

Therefore, the effective diffusion time for the ramdon rocess iskT2 0 a

rE

=

1.38 × 10

−23

(1273)

2 −19

0.417( 3.66 × 1.6 × 10

)

= 91s ≈ 1.5 min .

5. For loconcentration drivein diffusion, the diffusion is given by Gaussian distribution. The surface concentration is thenC (0, t ) = S S E = ex a πDt πD0 t 2kT 22

dC S E − t 3 / 2 = ex a dt πD 0 2kT 2 C = −0.5 × t

or

dC C

= −0.5 ×

dt thich means 1% change in diffusion time ill induce 0.5% change in surface concentration.dC S E − E a ex a = 2 dT πD0 t 2kT 2kT E = −C a 2 2kT or

dC C

=

−E

a

2 kT

×

dT T

=

− 3.6 × 1.6 × 10 −19 2 × 1.38 × 10− 23

× 1273

×

dT T

= −16.9 ×

dT Thich means 1% change in diffusion temerature ill cause 16.9% change in surface concentration.

6. At 1100°C, ni = 6×1018 cm3. Therefore, the doing rofile for a surface concentration of 4 × 1018 cm3 is given by the “intrinsic” diffusion rocess: x C ( x, t ) = C s erfc 2 Dt here Cs = 4× 1018 cm3, t = 3 hr = 10800 s, and D = 5x1014 cm2/s. The diffusion length is thenDt = 2.32 × 10 −5 cm = 0.232µm x The distribution of arsenic is C ( x) = 4 × 1018 erfc .64 × 10

The junction deth can be obtained as follos

xj 1015 = 4 × 1018 erfc 4.64 × 10 −5

xj = 1.2× 104 cm = 1.2 µm.23

7. At 900°C, ni = 2× 1018 cm3. For a surface concentration of 4×1018 cm3, given by the “extrinsic” diffusion rocessD = D0 e− Ea kT

4 × 1018 n − 23 × = 45.8e 1.38×10 ×1173 × = 3.77 × 10 −16 cm 2 /s ni 2 × 1018

−4.05×1.6×10 −19

x j = 1.6 Dt = 1.6 3.77 × 10 −16 × 10800 = 3.23 × 10 −6 cm = 32.3 nm .

8. Intrinsic diffusion is for doant concentration loer than the intrinsic carrier concentration ni at the diffusion temerature. Extrinsic diffusion is for doant concentration higher than ni.9. The SUPREM inut file for this roblem is:TITLE COMMENT INITIALIZE COMMENT DIFFUSIONT COMMENT DIFFUSIONT PRINT PLOT STOP Problem 69 Initialize silicon substrate Thickness = 5 <100> Silicon Phoshor Concentration=1e16 Diffuse Boron ime=15 Temerature=850 Boron Solidsol Perform DriveIn ime=360 Temerature=1175 Active Net Cmin=1e12 End Problem 69 Layers Active Concentration Phoshorus Boron NetNote that the thickness of the structure has been increased to 5 µm (in the INITIALIZE statement) to accommodate an anticiated deeer junction. The resulting lot is shon belo. The junction deth occurs at aroximately 3.48 µm.24

10. The SUPREM inut file for this roblem is:TITLE COMMENT INITIALIZE COMMENT DIFFUSION COMMENT DIFFUSION COMMENT DIFFUSION COMMENT DIFFUSION PRINT PLOT Problem 610 Initialize silicon substrate Thickness = 5 <100> Silicon Phoshor Concentration=1e16 Boron Prede ime=15 Temerature=850 Boron Solidsol Boron DriveIn Time=360 Temerature=1175 Phoshorus Prede Time=30 Temerature=850 Phoshor Solidsol Phoshorus DriveIn Time=30 Temerature=1000 Active Net Cmin=1e13 Layers Active Concentration Phoshorus Boron Net25

STOP

End Problem 610The resulting lot is shon belo. There are 2 n junctions formed. The junction deths occur at aroximately 0.45 and 3.49 µm, resectively.26

CHAPTER 7

1. The ion dose er unit area isIt 10 × 10 −6 × 5 × 60 N q 1.6 × 10 −19 = = = 2.38 × 1012 ions/cm 2 10 2 A A π ×( ) 2

From Eq. 1 and Examle 1, the eak ion concentration is at x = R. Figure 7.6 indicates the σp i 20 nm. Therefore, the ion concentration iS

σ p 2π2.

=

2.38 × 1012 20 × 10−7

2π

= 4.74 × 1017 cm −3 .

From Fig. 7.6, the R = 230 nm, and σp = 62 nm. The peak concentration iS

σ p 2πFrom Eq. 1,

=

2 × 1015 62 × 10−7

2π

= 1.29 × 10 20 cm −3

− (x j − R )2 1015 = 1.29 × 10 20 ex 2 2σ p xj = 0.53 µm.

3. Doe per unit area =Q C 0 ∆VT 3.9 × 8.85 × 10 −14 × 1 = = = 8.6 × 1011 cm − 2 q q 250 × 10 −8 × 1.6 × 10 −19

From Fig. 7.6 and Exampl 1, th pak concntration occurs at 140 nm from th surfac. Also, it is at (14025) = 115 nm from th SiSiO2 intrfac.27

4. Th total implantd dos is intgratd from Eq. 1QT = ∫∞ 0

S

σp

− (x − Rp )2 R p S S S ) = [2 − erfc( 2.3)] = × 1.9989 exp dx = 1 + 1

The total doe in ilicon i a follow (d = 25 nm): Q Si = ∫∞ d

S

σp

− (x − Rp )2 R p − d S S S ) = [2 − erfc(1.87)] = × 1.9918 exp dx = 1 +

the ratio of doe in the ilicon = QSi/QT = 99.6%.5. The projected range i 150 nm (ee Fig. 7.6). The average nuclear energy lo over the range i 60 eV/nm (Fig. 7.5). 60× 0.25 = 15 eV (energy lo of boron ion per each lattice plane) the damage volume = VD = π (2.5 nm)2(150 nm) = 3× 1018 cm3 total damage layer = 150/0.25 = 600 dislaced atom for one layer = 15/15 = 1 damage density = 600/VD = 2×1020 cm3 2×1020/5.02×1022 = 0.4%. 6. The higher the temerature, the faster defects anneal out. Also, the solubility of electrically active doant atoms increases ith temerature.Q1 C ox

7.

∆V t = 1 V =

whr Q1 is th additional charg addd just blow th oxidsmiconductor surfac by ion implantation. COX is a paralllplat capacitanc pr unit ara givn by C ox =

εs

d (d is th oxid thicknss, ε s is th prmittivity of th smiconductor)28

1V × 3.9 × 8.85 × 10 −14 F/cm C Q1 = ∆Vt C ox = = 8.63 × 10 −7 −6 cm 2 0.4 × 10 cm 8.63 × 10012 ions/cm2 −19 1.6 × 10 Total implant dos = 5.4 × 1012 = 1.2 × 1013 ions/cm2. 45%8.

Th discussion should mntion much of Sction 7.3. Diffusion from a surfac film avoids problms of channling. Tiltd bams cannot b usd bcaus of shadowing problms. If low nrgy implantation is usd, prhaps with pramorphization by silicon, thn to kp th junctions shallow, RTA is also ncssary.9.

From Eq.11

Sd 1 0.4 − 0.6 = rfc = 0.84 S 2 0.2 2 Th ffctivnss of th photorsist mask is only 16%.Sd 1 1 − 0.6 = rfc = 0.023 S 2 0.2 2 Th ffctivnss of th photorsist mask is 97.7%.10. u T= = 10 −5 u 2 π 1∴u = 3.02d = R + 4.27 σ p = 0.53 + 4.27 × 0.093 = 0.927 µm2

29

11. The SUPREM input file for thi problem i:TITLE Problem 711 COMMENT Initialize ilicon ubtrate INITIALIZE <100> Silicon Phophor Concentration=1e14 COMMENT Implant Boron IMPLANT Boron Energy=30 Doe=1e13 PRINT Layer Active Concentration Phophoru Boron Net PLOT Active Net Cmin=1e11 STOP End Problem 711The reulting doping profile i hown in the figure below.Examining thi figure and the SUPREM output file give: (a) The peak of the implanted boron occur at a depth of 0.11 µm. (b) The boron concentration at the peak i 8.59e17 cm3. (c) The junction depth i 0.4492 µm.30

12. The SUPREM input parameter that mut be determined are the doe and implant energy. The doe can be determined from Eq. 11 in Chapter 6 aQ(t ) ≅ 1.13C Dtwhere C can be read directly from the SUPREM output file for Example 3 in Chapter 6 a 4.6e19 cm3, D ≈ 2.3e − 16 cm2/ for boron at 850 oC (ee Figure 6.4), and t = 900 . Subtituting thee number into the above expreion give a doe of 2.36e13 cm2. The implant energy required can be approximated by matching the diffued and implanted concentration profile at the urface (x = 0) and at the junction and uing Eq. 1 to olve for Rp and σp imultaneouly. Note from the SUPREM output file correponding to Example 3 in Chapter 6 that the junction occur at xj = 0.0555 µm, at which point the doping concentration i 1016 cm3. A tated before, the urface concentration i 4.6e19 cm3. Thee equation cannot be olved analytically, but after everal iteration, the approximate value of Rp and σp required are found to be 0.011 µm and 0.008 µm, repectively. Thee value correpond to an implant energy of 5 keV (extrapolating from Figure 7.6a). The require SUPREM input file i therefore:TITLE COMMENT INITIALIZE COMMENT IMPLANT PRINT PLOT STOP Problem 712 Initialize ilicon ubtrate <100> Silicon Phophor Concentration=1e16 Implant Boron Boron Energy=5 Doe=2.36e13 Layer Active Concentration Phophoru Boron Net Active Net Cmin=1e11 End Problem 712The reulting doping profile appear in the figure below.31

32

CHAPTER 8

1.

ν av =

∫ ∫

∞

0 ∞ 0

vf v dv f v dv

=

8kT πM3/ 2

4 M Where fν = π 2kT

ν 2 ex − Mν 2 2kT

M: Molecular mass k: Boltzma costat = 1.38×1023 J/k T: The absolute temerature ν: Seed of molecular So thatν av =2. λ = 0.66 cm P( in Pa ) 0.66

2

π

2 × 1.38 × 10 −23 × 300 = 468 m/sec = 4.68 × 10 4 cm/sec . 29 × 1.67 × 10 − 27

∴P = 3.

λ

=

0.66 = 4.4 × 10 −3 Pa . 150

For closeackig arrage, there are 3 ie shaed sectios i the equilateral triagle. Each sectio corresods to 1/6 of a atom. Therefore1 3× umber of atoms cotaied i the triagle 6 = Ns = area of the triagle 1 3 d× d 2 2 =2 3d 2=

2 3 (4.68 × 10 −8 ) 2

= 5.27 × 1014 atoms/cm 2 .

33

d d

4.

(a) The ressure at 970°C (=1243K) is 2.9×101 Pa for Ga ad 13 Pa for As2. The arrival rate is give by the roduct of the imrigemet rate ad A/πL2 : P A Arrival rate = 2.64×1020 2 MT πL 2.9 × 10 −1 5 = 2.64×1020 2 69.72 × 1243 π × 12

= 2.9×1015 Ga molecules/cm2 –s The growth rate is determied by the Ga arrival rate ad is give by (2.9×1015)×2.8/(6×1014) = 13.5 Å/s = 810 Å/mi . (b) The ressure at 700ºC for ti is 2.66×106 Pa. The molecular weight is 118.69. Therefore the arrival rate is 2.66 × 10 −6 5 10 2 2.64 × 10 20 118.69 × 973 π × 12 2 = 2.28 × 10 molecular

If Sn atoms are fully incorporated and active in the Ga sublattice of GaAs, we have an electron concentration of 2.28 1010 2.9 1015 4.42 10 22 2 = 1.74 1017 cm 3 . 5. The x value is about 0.25, which is obtained from Fig. 8.7.

34

6. The lattice constants for InAs, GaAs, Si and Ge are 6.05, 5.65,5.43, and 5.65 Å, respectively. Therefore, the f value for InAsGaAs system isf = (5.65 − 6.05) 6.05 = −0.066

And for GeSi system isf = (5.43 − 5.65) 5.65 = −0.39 .

7. (a) For SiNxHySi 1 = = 1.2 N x

∴ x = 0.83

atomic % H =

100 y = 20 1 + 0.83 + y

∴ y = 0.46

The empirical formula is SiN0.83H0.46.(b) ρ= 5× 1028e-33.3×1.2 = 2× 1011 Ω-cm As the Si/N atio inceases, the esistivity deceases exponentially.

8.

Set Ta2O5 thickness = 3t, ε1 = 25 SiO2 thicknss = t, ε2 = 3.9 Si3N4 thicknss = t, ε3 = 7.6, ara = A thn C Ta 2O5 = 1 C ONO C ONO C Ta 2 O5 C ONO =ε 1ε 0 A3t t + t + t

ε 2ε 0 A ε 3ε 0 A ε 2ε 0 A εεε A = 2 3 0 (ε 2 + 2ε 3 )t=

ε 1 (ε 2 + 2ε 3 ) 25(3.9 + 2 × 7.6) = = 5.37 . 3ε 2 ε 3 3 × 3.9 × 7.6

35

9.

St BST thicknss = 3t, ε1 = 500, ara = A1 SiO2 thicknss = t, ε2 = 3.9, ara = A2 Si3N4 thicknss = t, ε3 = 7.6, ara = A2 thnε 1ε 0 A13t

=

ε 2 ε 3ε 0 A2 (ε 2 + 2ε 3 )t

A1 = 0.0093. A2

10. Lt Ta2O5 thicknss = 3t, ε1 = 25 SiO2 thicknss = t, ε2 = 3.9 Si3N4 thicknss = t, ε3 = 7.6 ara = A thnε 1ε 0 A3t d=

=

ε 2ε 0 Ad = 0.468t.

3ε 2 t

ε1

36

11. Th dposition rat can b xprssd as r = r0 xp (Ea/kT) whr Ea = 0.6 V for silanoxygn raction. Thrfor for T1 = 698 K 1 r (T2 ) 1 = 2 = xp 0.6 kT − kT r (T1 ) 2 1 ln 2 = 0.6 300 304 T2 =1030 K= 757 . 12. We can use energy-enhanced CVD me

hods such as using a fo

cused energy source or UV lamp. Anoher me

hod is

o use boron doped

-glass whi

ch will reflow a empera

ures less

han 900 . 13. Modera

ely low

empera

ures a

re usually used for polysilicon deposiion, and silane decomposi

ion occurs a

l

ower empera

ures

han

ha for chloride reac

ions. In addi

ion, silane is used

for be

er coverage over amorphous maerials such SiO2. 14. There are

wo reason

s. One is o minimize

he

hermal budge

of

he wafer, reducing dopan

diffusion

and maerial degrada

ion. In addi

ion, fewer gas phase reac

ions occur a

lower

empera

ures, resul

ing in smoo

her and be

er adhering films. Ano

her reason i

s ha he polysilicon will have small grains. The finer grains are easier

o ma

sk and ech

o give smoo

h and uniform edges. However, for

empera

ures less

ha

n 575 C he deposiion rae is oo low. 15. The fla-band volage shif isQo C0

∆VFB = 0.5 V ~

C0 =

ε oxd

=

3.9 × 8.85 × 10 −14 = 6.9 × 10 −8 F/cm − 2 . 500 × 10 −8

∴ Numbr of fixd oxid charg is37

0.5C 0 0.5 × 6.9 × 10 −8 = = 2.1 × 1011 cm 2 −19 q 1.6 × 10To rmov ths chargs, a 450 hea reamen in hydrogen for abou 30 minues is required. 16. 20/0.25 = 80 sqs. Therefore,

he resis

ance of

he me

al line is

5×50 = 400 Ω .

17.

Fo TiSi2 30 × 2.37 = 71.1nm Fo CoSi2 30 × 3.56 = 106.8nm.18.

Fo TiSi2: Advantage: low esistivity It can educe native-oxide layes TiSi2 on the gate electode is moe esistant to high-field-induced hot-electon degadation. Disadvantage: bidging effect occus. Lage Si consumption duing fomation of TiSi2 Less themal stability Fo CoSi2: Advantage: low esistivity High tempeatue stability No bidging effect A selective chemical etch exits Low shea foces Disadvantage: not a good candidate fo polycides38

19. (a) R = ρεAd

L 1 = 2.67 × 10 − 6 × = 3.2 × 10 3 Ω −4 −4 A 0.28 × 10 × 0.3 × 10=

C=

εTLS

=

3.9 × 8.85 × 10 −14 × 0.3 × 10 −4 × 1 × 10 4 × 10 −6 = 2.9 × 10 −13 F 0.36 × 10 − 4

RC = 3.2 × 10 5 × 2.9 × 10 −15 = 0.93 ns L 1 = 1.7 × 10 − 6 × = 2 × 10 3 Ω −4 −4 A 0.28 × 10

(b) R = ρ C=

2.8 × 8.85 × 10 −14 × 0.3 × 10 −4 × 1 = 2.1 × 10 −13 F −4 d S 0.36 × 10 3 −13 RC = 2 × 10 × = = 0.42 = 0.45. 093

εA

εTL

(c) W can dcras th RC dlay by 55%. Ratio =20. (a)

R=ρ

1 L = 2.67 × 10 − 6 × = 3.2 × 10 3 Ω 4 4 A 0.28 × 10 × 0.3 × 10

3.9 × 8.85 × 10 −14 × 0.3 × 10 −4 × 1 × 3 = 8.7 × 10 −13 F C= = = 4 d S 0.36 × 10 RC = 3.2 × = 2.8 s.εA

εTL

.(b) R = ρ 1 L = 1.7 × 10 − 6 × = 2 × 10 3 Ω 4 4 A 0.28 × 10 × 0.3 × 10

C=

εAd

=

εTLS

=

2.8 × 8.85 × 10 −14 × 0.3 × 10 −4 × 1 × 3 = 6.3 × 10 −13 F 4 0.36 × 10

RC = 2 × 10 3 × 8.7 × 10 −13 = 2.5 ns RC = 3.2 × 10 3 × 8.7 × 10 −13 = 2.5 ns. 21. (a) Th a

num runnr can b considrd as two sgmnts connctd in sris: 20% (or 0.4 mm) of th lngth is half thicknss (0.5 µm) and th rmaining 1.6 mm is full thicknss (1µm). Th total rsistanc is 0.16 0.04 + −4 R = ρ 1 + 2 = 3 × 10 − 6 − 4 −4 −4 10 × (0.5 × 10 ) 10 × 139

= 72 Ω. The limiting cuent I is given by the maximum allowed cuent density times coss-sectional aea of the thinne conducto sections: I = 5×105 A/cm2× (10-4×0.5×10-4) = 2.5×10-3 A = 2.5 mA. The voltage dop acoss the whole conducto is then V = RI = 72Ω × 2.5 × 10 −3 A = 0.18V.22.0.5 µm 40 m 0.5 µmCu

=

Al60 mh: height , W : width , t : thickess, assume that the resistivities of the claddig layer ad TiN are much larger tha ρ A and ρ Cu R Al = ρ Al × RCu = ρ Cu × h ×W h ×W = = 1.7 (0.5 − 0.1) × 0.5 (0.5 − 2t ) × (0.5 − 2t )

Whe R Al = RCu The 2.7 1.7 = 0.4 × 0.5 (0.5 − 2t ) 2 t = 0.073 µm = 73 m .40

CHAPTER 9

1.

Each Ushae sectio (refer to the figure) has a area of 2500 µm × 8 µm = 2 × 104 µm2. Therefore, there are (2500)2/2 ×104 = 312.5 Ushaed sectio. Each sectio cotais 2 log lies with 1248 squares each, 4 corer squares, 1 bottom square, ad 2 half squares at the to. Therefore the resistace for each sectio is 1 kΩ / (1248×2 + 4×0.65 +2) = 2500.6 kΩ The maximum esistance is then 312.5×2500.6 = 7.81 × 108 Ω = 781 MΩ2. The aea equied on the chip is41

A=

Cd

ε ox

=

(30 × 10 −7 )(5 × 10 −12 ) = 4.35 × 10 −5 cm 2 −14 3.9 × 8.85 × 10

= 4.35 × 103 µm2 = 66 × 66 µm Rfr to Fig.9.4a and using ngativ photorsist of all lvls (a) Ion implantation mask (for p+ implantation and gat oxid) (b) Contact windows (2×10 µm) (c) Mtallization mask (using Al to form ohmic contact in th contact window and form th MOS capacitor). Bcaus of th rgistration rrors, an additional 2 µm is incorporatd in all critical dimnsions.42

3. If th spac btwn lins is 2 µm, thn thr is 4 µm for ach turn (i.., 2×n, for on turn). Assum thr ar n turns, from Eq.6, L ≈ µ0n2r ≈ 1.2 × 106n2r, whr r can b rplacd by 2 × n. Thn, w can obtain that n is 13.43

4. (a) Mtal 1, (b) contact hol, (c) Mtal 2. (a) Mtal 1,(b) contact hol,(c) Mtal 2.44

5. Th circuit diagram and dvic crosssction of a clampd transistor ar shown in (a) and (b), rspctivly.6.

(a) Th undopd polysilicon is usd for isolation. (b) Th polysilicon 1 is usd as a solidphas diffusion sourc to form th xtrinsic bas rgion and th bas lctrod. (c) Th polysilicon 2 is usd as a solidphas diffusion sourc to form th mittr rgion and th mittr lctrod.7. (a) For 30 kV boron, Rp = 100 nm and ∆Rp = 34 nm. Assuming that Rp and ∆Rp for45

boron ar th sam in Si and SiO2 th pak concntration is givn by S 2π ∆R p = 8 × 1011 2π (34 × 10 )−7

= 9.4 × 1016 cm −3

The amout of boro ios i the silico is (x − R )2 ∞ Q S ex − =∫ 2 d q 2 ∆R p 2π ∆R p = Rp − d S 2 − rfc 2 ∆R 2 p=

750 2 − rfc 2 × 340 = 7.88 × 1011 cm − 2

Assum that th implantd boron ions form a ngativ sht charg nar th SiSiO2 intrfac, thn Q 1.6 × 10 −19 × (7.88 × 1011 ) = 0.91 V ∆VT = q / C ox = q 3.9 × (25 × 10 − 7 ) (b) For 80 kV arsnic implantation, Rp = 49 nm and ∆ Rp = 18 nm. Th pak arsnic concntration is S 2π ∆R p = 1016 = 2.21 × 10 21 cm − 3 .π × (18 × 10 )−7

46

8.

(a) Because (100)orieted silico has lower (~ oe teth) iterfacetraed charge ad a lower fixed oxide charge. (b) If the field oxide is too thi, it may ot rovide a large eough threshold voltage for adequate isolatio betwee eighborig MOSFETs. (c) The tyical sheet resistace of heavily doed olysilico gate is 20 to 30 Ω /, which is adequate fo MOSFETs with gate lengths lage than 3 µm. Fo shote gates, the sheet esistance of polysilicon is too high and will cause lage RC47

delays. We can use efactoy metals (e.g., Mo) o silicides as the gate mateial to educe the sheet esistance to about 1 Ω /. (d) A self-aligned gate can be obtained by fist defining the MOS gate stuctue, then using the gate electode as a mask fo the souce/dain implantation. The self-aligned gate can minimize paasitic capacitance caused by the souce/dain egions extending undeneath the gate electode (due to diffusion o misalignment). (e) P-glass can be used fo insulation between conducting layes, fo diffusion and ion implantation masks, and fo passivation to potect devices fom impuities, moistue, and scatches. 9. The lowe insulato has a dielectic constant ε1/ε0 = 4 and a thicknss d1= 10 nm Th uppr insulator has a dilctric constant ε2/ε0 = 10 and a thicknss d2 = 100 nm. Upon application of a positiv voltag VG to th xtrnal gat, lctric fild E1 and E2 ar stablishd in th d1 and d2 rspctivly. W hav, from Gauss’ law, that ε1E1 = ε2E2 +Q and VG = E1d1 + E2d2 whr Q is th stord charg on th floating gat. From ths abov two quations, w obtainE= 1

VG Q + d1 + d 2 (ε 1 / ε 2 ) ε 1 + ε 2 (d1 / d 2 )

48

10 × 10 7 −7 + J = σE = 10 1 10 + 100 4 10

Q 5 = 0.2 − 2.26 × 10 Q 10 −14 4 + 10 100 × 8.85 × 10

(a) If the tored charge doe not reduce E1 by a ignificant amount (i.e., 0.2 >> 2.26×105 Q, we can writeQ = ∫ σE dt ' ≈ 0.2∆t = 0.2 × (0.25 × 10 − 6 ) = 5 × 10 −8 C 1t0

∆VT =

5 × 10 −8 Q = = 0.565 V C2 10 × 8.85 × 10 −14 / 100 × 10 −7

(

)(

)

(b) whn t → ∞, J → 0 w hav Q → 0.2 / 2.26 × 10 5 ≅ 8.84×107 C. Thn ∆VT = 10. 8.84 × 10 9.98 V. C2 10 × 8.85 × 10 −14 / 10 −5

(

)

49

50

11. Th oxid capacitanc pr unit ara is givn byC ox =

ε SiOd

2

= 3.5 × 10 − 7 F/cm2

and th maximum currnt supplid by th dvic is I DS ≈ 1W 1 5µm 2 µC ox (VG − VT )2 = 3.5 × 10 −7 (VG − VT ) ≈ 5 mA 2 L 2 0.5µm51

and th maximum allowabl wir rsistanc is 0.1 V/5 mA, or 20Ω. Then, the length of the wie must beL≤ R × Aeaρ

=

20Ω × 10 −8 cm 2 = 0.074 cm 2.7 × 10 −8 Ω − cm

or 740 µm. This is a log distace comared to most device sacig. Whe drivig sigals betwee widely saced logic blocks however, miimum feature sized lies would ot be aroriate.12.

52

13. To solve the shortchael effect of devices. 14. The device erformace will be degraded from the boro eetratio. There are53

methods to reduce this effect: (1) usig raid thermal aealig to reduce the time at high temeratures, cosequetly reduces the diffusio of boro, (2) usig itrided oxide to suress the boro eetratio, sice boro ca easily combie with itroge ad becomes less mobile, (3) makig a multilayer of olysilico to tra the boro atoms at the iterface of each layer. 15. Total caacitace of the stacked gate structure is :

C=

ε1d1

×

ε2d2

ε1 ε 2 7 25 7 25 = + × + = 2.12 d 0.5 10 0.5 10 1 d2

3.9 = 2.12 d ∴d = 3.9 =1.84 nm. 2.12

16. Disadvantags of LOCOS: (1) high tmpratur and long oxidation tim caus VT shift, (2) bird’s bak, (3) not a planar surfac, (4) xhibits oxid thinning ffct. Advantags of shallow trnch isolation: (1) planar surfac, (2) no high tmpratur procssing and long oxidation tim, (3) no oxid thinning ffct, (4) no bird’s bak. 17. For isolation btwn th mtal and th substrat. 18. GaAs lacks of highquality insulating film.19. Answrs will vary. If w ignor th contributions of isolation rgion procssing, th structur can b simulatd using four SUPREM input dcks. Th first dck simulats54

procssing in th activ rgion of th dvic, up to th point of th isolation oxidation. Th scond dck starts with th rsults from th first dck and complts all procssing in th activ rgions. This allows th doping profil through th mittr to b plottd (for part b). Th third dck is similar to th scond, xcpt it liminats th mittr implant and facilitats plotting of th doping profil through th bas rgion (for part a). Th final dck is also similar to th scond, xcpt that it liminats th bas implant and facilitats plotting th doping profil through th collctor rgion (for part c). Th complt procss squnc is as follows: 1) Bgin with a highrsistivity, <100>, ptyp silicon substrat. 2) Grow a 1 µm SiO2 layr. 3) Rmov th oxid in th aras whr th burid layrs ar to b placd. 4) Implant antimony at a dos of 115 cm2. Driv in th burid layr for 5 hours at 1150 oC. 5) Etch th silicon dioxid from th surfac. 6) Grow a 1.6 µm arsnicdopd pitaxial layr. 7) Grow a 400 Å pad oxid. 8) Dposit 800 Å of silicon nitrid. 9) Etch th oxid and nitrid from th isolation rgions. 10) Etch th silicon halfway through th pilayr.55

11) Implant boron in th fild rgions with a dos of 113 cm2 at an nrgy of 50 kV. 12) Oxidiz th fild rgions to a thicknss approximatly onhalf that of th pilayr. 13) Implant th bas rgion with boron at a dos of 114 cm2 at an nrgy of 50 kV. 14) Etch th oxid from th mittr rgion. 15) Implant mittr collctor contact rgions with arsnic at a dos of 515 cm2 at an nrgy of 100 kV. 16) Drivin th arsnic and activat th bas diffusion. Th SUPREM input dcks ar as follows:TITLE COMMENT COMMENT INITIALIZE COMMENT DIFFUSION COMMENT ETCH COMMENT IMPLANT DIFFUSION DIFFUSION COMMENT ETCH COMMENT EPITAXY BJT – Deck 1 Iitial Active Regio Processig Iitialize silico substrate <100> Silico Boro Cocetratio=5e14 Grow maskig oxide for oactive regios Time=100 Temerature=1150 WetO2 Etch oxide over buried layer regios Oxide Imlat ad drivei atimoy buried layer Atimoy Dose=5e14 Eergy=120 Time=15 Temerature=1150 DryO2 Time=300 Temerature=1150 Etch the oxide Oxide Grow 1.6 µm of arseicdoed ei Temerature=1050 Time=4 Growth.Rate=0.4 Arseic Gas.Coc=5e1556

COMMENT DIFFUSION COMMENT DEPOSITION SAVEFILE STOP

Grow 400 A ad oxide Time=20 Temerature=1060 DryO2 Deosit itride to mask the field oxidatio Nitride Thickess=0.08 Structur Fileame=bjtactiveiit.str Ed BJT 1

TITLE COMMENT COMMENT INITIALIZE COMMENT DIFFUSION DIFFUSION DIFFUSION DIFFUSION DIFFUSION COMMENT ETCH ETCH ETCH COMMENT IMPLANT COMMENT ETCH COMMENT IMPLANT COMMENT DIFFUSION PRINT PLOT STOP

BJT – Deck 2 Fial Active Regio Processig for Emitter Profile Start with revious results Structur=bjtactiveiit.str Field oxide growth Time=30 Temerature=800 t.rate=10 Time=15 Temerature=1000 DryO2 Time=210 Temerature=1100 Wet02 Time=15 Temerature=1100 Etch the oxide ad itride layers Oxide Nitride Oxide Imlat boro base Boro Dose=1e14 Eergy=50 Remove oxide from emitter regio Oxide Imlat arseic emitter ad collector cotacts Arseic Dose=5e15 Eergy=100 Drivei emitter ad collector cotact regios Time=20 Temerature=1000 Layers Chemical Boro Arseic Phoshor Net Ed BJT 2 DryO2 Time=10 Temerature=1100 t.rate=30TITLE COMMENT COMMENT

BJT – Deck 3 Active Regio Processig for Base Profile Start with revious results57

INITIALIZE COMMENT DIFFUSION DIFFUSION DIFFUSION DIFFUSION DIFFUSION COMMENT ETCH ETCH ETCH COMMENT IMPLANT COMMENT ETCH COMMENT DIFFUSION PRINT PLOT STOP

Structur=bjtactiveiit.str Field oxide growth Time=30 Temerature=800 t.rate=10 Time=15 Temerature=1000 DryO2 Time=210 Temerature=1100 Wet02 Time=15 Temerature=1100 Etch the oxide ad itride layers Oxide Nitride Oxide Imlat boro base Boro Dose=1e14 Eergy=50 Remove oxide from emitter regio Oxide Drivei emitter ad collector cotact regio Time=20 Temerature=1000 Layers Chemical Boro Arseic Phoshor Net Ed BJT 3 DryO2 Time=10 Temerature=1100 t.rate=30TITLE COMMENT COMMENT INITIALIZE COMMENT DIFFUSION DIFFUSION DIFFUSION DIFFUSION DIFFUSION COMMENT ETCH ETCH ETCH COMMENT ETCH

BJT – Deck 4 Active Regio Processig for Collector Profile Start with revious results Structur=bjtactiveiit.str Field oxide growth Time=30 Temerature=800 t.rate=10 Time=15 Temerature=1000 DryO2 Time=210 Temerature=1100 Wet02 Time=15 Temerature=1100 Etch the oxide ad itride layers Oxide Nitride Oxide Remove oxide from emitter regio Oxide DryO2 Time=10 Temerature=1100 t.rate=3058

COMMENT IMPLANT COMMENT DIFFUSION PRINT PLOT STOP

Imlat arseic emitter ad collector cotacts Arseic Dose=5e15 Eergy=100 Drivei emitter ad collector cotact regios Time=20 Temerature=1000 Layers Chemical Boro Arseic Phoshor Net Ed BJT 4The resultig doig rofiles though the base, emitter, ad collector (arts a, b, ad c), resectively, are show i the followig three figures:(a)59

(b)

60

(c)

20. Aswers will vary. For the sake of simlicity, we will igore isolatiorelated rocessig. The structure ca be simulated usig four SUPREM iut decks (oe for each requested rofile). The comlete rocess sequece is as follows:61

1) 2) 3) 4) 5) 6) 7) 8) 9)

Start with a <100>, tye silico substrate. Grow a 0.9 µm SiO2 layer. Remove the oxide i the well areas. Imlat boro well at a dose of 5e14 cm2 at 50 keV. Drive i the well for 6 hours at 1150 oC. Remove oxide i PMOS source/drai regios. Imlat boro for PMOS source/drai at a dose of 1e14 cm2 at 20 keV. Drive i the PMOS source/drai regios for 2.5 hours at 1100 oC. Etch oxide i NMOS source/drai regios.10) Imlat hoshorus for NMOS source/drai at a dose of 1e14 cm2 at 20 keV. 11) Drive i the NMOS source/drai regios for 2.5 hours at 1100 oC. 12) Etch oxide i gate areas. 13) Grow 500 Å gate oxide. 14) Deosit ad atter olysilico gates. 15) Grow assivatio oxide. 16) Deosit ad atter metallizatio. The SUPREM iut decks are ad corresodig oututs as follows:TITLE COMMENT COMMENT INITIALIZE CMOS – Deck 1 PMOS source/drai Iitialize silico substrate <100> Silico Phoshorus Cocetratio=5e15 Thickess=562

COMMENT DIFFUSION COMMENT ETCH COMMENT DIFFUSION COMMENT ETCH COMMENT IMPLANT COMMENT DIFFUSION COMMENT DIFFUSION COMMENT ETCH COMMENT DEPOSITION PRINT PLOT STOP

Grow field oxide Time=120 Temerature=1100 WetO2 Etch oxide after well imlat Oxide Pwell drivei Time=900 Temerature=1150 DryO2 Etch the oxide rior to PMOS source/drai imlat Oxide PMOS source/drai imlat Boro Dose=1e14 Eergy=20 PMOS source/drai drivei Time=150 Temerature=1100 DryO2 NMOS source/drai drivei ad gate oxidatio Time=150 Temerature=1100 DryO2 Etch oxide Oxide Deosit metal Alumium Thickess=1.0 Layers Chemical Boro Phoshor Net Ed CMOS 163

(a)TITLE COMMENT COMMENT INITIALIZE COMMENT DIFFUSION COMMENT ETCH COMMENT DIFFUSION CMOS – Deck 2 PMOS Gate Iitialize silico substrate <100> Silico Phoshorus Cocetratio=5e15 Thickess=5 Grow field oxide Time=120 Temerature=1100 WetO2 Etch oxide after well imlat Oxide Pwell drivei Time=900 Temerature=1150 DryO2

64

COMMENT ETCH COMMENT DIFFUSION COMMENT DIFFUSION COMMENT DEPOSITION COMMENT DIFFUSION PRINT PLOT STOP

Etch the oxide rior to PMOS source/drai imlat Oxide PMOS source/drai drivei Time=150 Temerature=1100 DryO2 NMOS source/drai drivei ad gate oxidatio Time=150 Temerature=1100 DryO2 Deosit olysilico Polysilico Thickess=0.5 Grow assivatio oxide Time=30 Temerature=1000 DryO2 Layers Chemical Boro Phoshor Net Ed CMOS 265

(b)TITLE COMMENT COMMENT INITIALIZE COMMENT DIFFUSION COMMENT ETCH COMMENT CMOS – Deck 3 NMOS source/drai Iitialize silico substrate <100> Silico Phoshorus Cocetratio=5e15 Thickess=5 Grow field oxide Time=120 Temerature=1100 WetO2 Etch oxide i well regio Oxide Pwell imlat66

IMPLANT COMMENT DIFFUSION COMMENT DIFFUSION COMMENT ETCH COMMENT IMPLANT COMMENT DIFFUSION COMMENT ETCH COMMENT DEPOSITION PRINT PLOT STOP

Boro Dose=5e14 Eergy=50 Pwell drivei Time=900 Temerature=1150 DryO2 PMOS source/drai drivei Time=150 Temerature=1100 DryO2 Etch the oxide rior to NMOS source/drai imlat Oxide NMOS source/drai imlat Phoshorus Dose=1e14 Eergy=20 NMOS source/drai drivei ad gate oxidatio Time=150 Temerature=1100 DryO2 Etch oxide Oxide Deosit metal Alumium Thickess=1.0 Layers Chemical Boro Phoshor Net Ed CMOS 367

(c)TITLE COMMENT COMMENT INITIALIZE COMMENT DIFFUSION COMMENT ETCH CMOS – Deck 4 NMOS gate Iitialize silico substrate <100> Silico Phoshorus Cocetratio=5e15 Thickess=5 Grow field oxide Time=120 Temerature=1100 WetO2 Etch oxide i well regio Oxide68

COMMENT IMPLANT COMMENT DIFFUSION COMMENT DIFFUSION COMMENT ETCH COMMENT DIFFUSION COMMENT DEPOSITION COMMENT DIFFUSION PRINT PLOT STOP

Pwell imlat Boro Dose=5e14 Eergy=50 Pwell drivei Time=900 Temerature=1150 DryO2 PMOS source/drai drivei Time=150 Temerature=1100 DryO2 Etch the oxide rior to NMOS source/drai imlat Oxide NMOS source/drai drivei ad gate oxidatio Time=150 Temerature=1100 DryO2 Deosit olysilico Polysilico Thickess=0.5 Grow assivatio oxide Time=30 Temerature=1000 DryO2 Layers Chemical Boro Phoshor Net Ed CMOS 469

(d)

70

CHAPTER 10

1. x chart: Ceter = µ = 0.75 V UCL = µ + 3σ 3(0.1) = 0.75 + = 0.845 V n 10LCL = 33σ = 0.655 V nchart: Center = = c4σ = 0.9727(0.1) = 0.0973 V UCL = + 3 σ 1 − c = 0.973 + 3(0.1)(1 − 0.9727) = 0.167 V2 4 1 2

2 LCL = 3 σ 1 − c4 = 0.028 V2. x chart: Center = x = 0.734 V UCL = x +3 c4 n= 0.846 V

LCL = x −

3 c4 n= 0.622 Vchart: Center = = 0.125 V 2 UCL = + 3 1 − c 4 = 0.215 V c4 LCL = − 3 2 1 − c4 = 0.036 V c4

3. Let ED = expoure doe, DT = develop time, BT = bake temperature ED DT BT Y (1) (2) (3) Div. Eff. ID 60 137 264 534 8 66.75 Avg.71

+ + + +

+ + + +

+ + + +

77 59 68 57 83 45 85

127 140 130 17 9 26 40

270 26 66 10 10 8 1492 20 6 6 40 0 224 4 4 4 4 4 4

23 5 1.5 1.5 10 0 5.5ED DT ED x DT BT ED x BT DT x BT ED x DT x BT

4. Let: day = block (i.e., n = 3) procee = treatment (i.e., k = 5) Then we have: A Day 1 Day 2 Day 3 yt 509 505 465 493 B 512 507 472 497 C 532 542 498 524 D 506 520 483 503 E 509 519 475 501 yi 513.6 518.6 478.6 y = 503.6

ANOVA Table: Source Average (SA) Block (SB) Treatment (ST) Reidual (SR) Sum of Square 3,804,194.4 4,750 1,737.6 210 Degree of Freedom 1 2 4 8 Mean Square 3,804,194.4 2 2,375 ( B ) 2 434.4 ( T ) 2 26.25 ( R )where: S A = nky 2 = 3,804,194.4 S B = k ∑ ( y i − y ) , DF = n – 1i =1 k n

2

S T = n∑ ( y t − y ) , DF = k – 1t =1

2

S R ∑∑ ( y ti − y i − y t + y ) , DF = (n – 1)(k – 1)t =1 i =1

k

n

2

Now:

2 2 B / R = 90.4872

2 2 T / R = 16.55A. Significance level for the null hypothei that the block are the ame i very low, ince P( F 2,8 > 90.48) ~0B. The ame i true for the hypothei that the treatment are the ame, ince: P ( F4,8 > 16.55) ~ 0

I. II.

The procee are ignificantly different. The proceing date have ignificant difference.5. From Eq. 33: Y = exp(AcD0)N = exp(NAcD0) where: Y = 0.95 N = 100,000 Ac = WL = (10e4 cm)(1e4 cm) = 1e7 cm2 => D = − ln Y = 5.13 cm2 NAc∞

6. Murphy’s Yield Itegral (Eq. 34): Y = ∫ e − Ac D f ( D)dD0

Uiform defect distributio: f ( D) = 1 / 2 D0 for 0 ≤ D ≤ 2 D0Y=2 D0

∫0

e − Ac D 1 e − Ac D dD = 2 D0 2 D0 − Ac

2 D0

0

=> Yuiform =1 − e −2 D0 Ac 2 D0 Ac

Triagular defect distributio: f ( D) = D / D02 for 0 ≤ D ≤ D0 =− D 2 + for D0 ≤ D ≤ 2D0 2 D0 D0

73

Y = ∫e0

D0

− Ac D

D dD + D02

2 D0

D0

∫e

− Ac D

D 2 − 2 + dD D D0 0 2 D0

− Ac D 1 e − AD 1 e − Ac D D (− Ac D − 1) 2 D0 + 2 e Y = 2 2 (− Ac D − 1) 0 0 − 2 D0 D0 Ac D2 D0 − Ac

D0

=> Ytriagular 1 − e − D0 Ac = D A 0 c

2

Exoetial Defect Distributio: f ( D) =− D 1 ex D D0 0 ∞

∞ − D(1 + Ac D0 ) − D(1 + Ac D0 ) 1 − Ac D − D / D0 1 1 ex Y =∫ e e dD = dD = 0 D0 0 D0 D0 0 0

∞

=> Yex oetial =1 1 + D0 Ac

7. Use Murhy’s Yield Itegral (Eq. 34): Y = ∫ e − AD f ( D)dD , where: A = 100cm2, f(D) = 100D+100.1

Y=

0.05

∫e

−100 D

(− 100 D + 10)dD

0.1

= 1000.05

−100 D −100 D ∫ De dD + 10 ∫ e dD0.05

0.1

=

− 100 −100 D ( AD − 1) 0..1 − 10 e −100 D e 4 0 05 100 10

0.1 0.05

=> Y = 0.094 %

74

CHAPTER 11

1. (a)A 1 L RC = ρ ε ox = 10 −5 × d 1 × 0.5 × 10 −8 A 1 × (1 × 10 −4 )

= 2000 × (69.03 × 10 −14 ) = 1.38 × 10 − 9 s = 1.38 ns.

(b) For a polysilicon runnrL A RC = Rsquar ε ox W d 1 = 30 − 4 69.03 × 10 −14 = 2.07 × 10 −7(

)

Thrfor th polysilicon runnr’s RC tim constant is 150 tims largr than th aluminum runnr.2. Whn w combin th logic circuits and mmory on th chip, w nd multipl supply voltags. For rliability issu, diffrnt oxid thicknsss ar ndd for diffrnt supply voltags. 3. (a) 1 1 + 1C total =

C Ta 2 O5

C nitrid + 10 = 17.3 Å 7hnc EOT3.9

= 75

25

(b) EOT = 16.7 Å.

75