# 42581030 20980796 solutions fundamentals of semiconductor fabrication

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Solutions Manual to Accompany FUNDAMETALS OF SEMICONDUCTOR FABRICATION G. S. May Motorola Foundation Professor School of Electrical & Computer Engineering Georgi a Institute of Technology Atlanta, GA, USA S. M. SZE UMC Chair Professor National Chiao Tung University National Nano Device Laborato ries Hsinchu, Taiwan 1

John Wiley and Sons, Inc New York. Chicester / Weinheim / Brisband / Singapore / Toronto 2

Contents Ch.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 Integrat ion ---------------------------------------------------------------------40 Ch.1 0 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 = 0.31017(1- x)-0.7 = 31016/(1 - l/50)0.7 x l (cm) CS (cm-3) 0 0 31016 0.2 10 3.51016 0.4 20 4.281016 0.6 30 5.681016 0.8 40 1.071017 0.9 45 1.51017 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 as r= 3 a 8 3 5.43 = 1.175 8 so r = (b) The numbers of Si atom in its diamond structure are 8. So the density of sil icon atoms is n= 8 8 = = 5.0 10 22 atoms/cm 3 3 3 a (5.43) (c) The density of Si is = M / 6.02 10 23 1/ n N (10 D 16 cm ) -3 = 28.09 5 10 22 6.02 10 23 g / cm 3 = 2.33 g / cm3. 1

3. k0 = 0.8 fo bo on in silicon M / M0 = 0.5 The density of Si is 2.33 g / cm3. The accepto concent ation fo = 0.01 cm is 91018 cm-3. The doping concent ation C S is given by C s = k 0 C 0 (1 Therefore C0 = 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 that 10.8g/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 gi ven as nb = number of boron atoms volume of silicon wafer 5.41 10 3 g / 10.81g 6.02 10 2 3 = 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 avera ge distance between two boron atoms. Then r= 3V = 2.9 10 7 cm . 4

5. The cross-sectional area of the seed is 0.55 2 = 0.24 cm 2 The maximum weig that can be su orted by the seed equals the roduct of the critical yield stren gth and the seeds 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 3 2 6. We have M Cs / C0 = k0 1 M 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 0 ( 2.33g/cm ) l = 480000 g 2

l = 656 cm = 6.56 m.

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 u nity, so the solubility of B in Si under solid phase is smaller than that of the melt. Therefore, the excess B atoms will be thrown off into the melt, then the concentration of B in the melt will be increased. The tail end of the crystal is the last to solidify. Therefore, the concentration of B in the tail end of grow n crystal will be higher than that of seed end. 8. The reason is that the solubi lity in the melt is proportional to the temperature, and the temperature is high er 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 segre gation coefficient of Ga in Si is 8 10 3 From Eq. 18 C s / C 0 = 1 (1 k )e kx / L We have x= = L 1 k ln k 1 C s / C0

10. We have from Eq.18 Cs = C0[1 (1 ke ) exp( ke x / L)] 4

2 1 8 10 3 ln

8 10 3

1 5 1015 / 5 1016

= 250 ln(1.102) = 24 cm.

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 conventionally doped silicon, the resistivity varies from 120 -cm to 155 -cm. The co esponding doping concent ation va ies f om 2.51013 to 41013 cm-3. The efo e the ange of b eakdown voltages of p+ - n junctions is given by VB s Ec 2 2q ( 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 = 1.3 10 17 / N B = 2.9 10 17 / 3 10 13 ( 1%) = 9570 to 9762 V . V B = 9762 9570 = 192 V

/ 9570 = 1% . M s w ight of GaAs at Tb C m C l s = = = M l w ight of liquid at Tb C s C m l

5

Th r for , th fraction of liquid r main d f can b

V B

2 12. W

hav

obtain d as following

V B / 7250 = 30% 2 For th n utron irradiat d silicon, = 148 ncent ation is 31013 ( 1%). The ange of b eakdown voltage is

1.5 -cm.

f = Ml l 30 = = 0.65 . M s + M l s + l 16 + 30 13. From th Fig.2.11, w find th vapor pr ssur of As is much high r than that of th Ga. Th r for , th As cont nt will b lost wh n th t mp ratur is incr as d. Thus th composition of liquid GaAs always b com s gallium rich. 88.8 (T / 300)

= 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 0 173 K at 1200 0 C = 1473 K .

6

15. n f = NN ` xp( E f / 2kT ) = 5 10 22 1 10 27 7 10 24 94.7 /(T / 300 ) = 5.27 10 17 at 27oC = . 16. 37 4 = 148 chips In t rms of litho st pp r consid spac tol ranc b tw n th mask boundary of two dic . W four symm trical parts for conv ni nt dicing, and discard th waf r. Usually th quality of th p rim t r parts is dg ff cts.

3.8 V / kT 1.1 V / 2 kT = 300 K =2.141014 at 900oC = 1173 rations, th r ar 500 m divid th waf r into th p rim t r parts of th worst du to th

14. n s = N xp( E s / kT ) = 5 10 22 xp(2.3

V / kT ) = 5 10 22 xp

7

CHAPTER 3 1. From Eq. 11 (with =0) x2+Ax = B From Figs. 3.6 and 3.7, we ob ain B/A =1.5 m / hr, B=0.47 m 2/hr, herefore A= 0.31 m. The ime required o grow 0.45m oxide is 1 2 1 (x + Ax) = (0.45 2 + 0.31 0.45) = 0.72 hr = 44 min . B 0.47 =

2. Af er a window is opened in he oxide for a second oxida ion, he ra e cons a n s are B = 0.01 m 2/hr, A= 0.116 m (B/A = 6 10-2 m /hr). If he ini ial oxide hick ness is 20 nm = 0.02 m for dry oxida ion, he value ofcan be ob ained as followed: (0.02)2 + 0.166(0.02) = 0.01 (0 +) or = 0.372 hr.

hr. ( x + Ax) = B 0.01

x2+ 0.166x = 0.01(0.333+27.72) = 0.28053 or x = 0.4530 m (an increase of 0.003m on ly for he field oxide). 3. x2 + Ax = B ( + ) (x + A 2 A2 ) = B (t + ) 2 4 (x + A2 8 A 2 ) =B + ( + )

2

4B

For an oxida ion ime of 20 min (=1/3 a is x2+ 0.166x = 0.01(0.333+0.372) = For he field oxide wi h an original = 1 2 1 (0.45 2 + 0.166 0.45) = 27.72

hr), he oxide hickness in he window are 0.007 or x = 0.0350 m = 35 nm (ga e oxide). hickness 0.45 m, he effec iveis given by

when >> , B ( + ) A A2 , 4B A2 , 4B

>> hen, x2 = B similarly, when

>> ,

>> hen, x =

4. A 980(=1253K) and 1 a m, B = 8.510-3 m 2/hr, B/A = 410-2 m /hr (from Figs. 3.6 an d 3.7). Since A 2D/k , B/A = kC0/C1, C0 = 5.21016 molecules/cm3 and C1 = 2.21022 cm -3 , he diffusion coefficien is given by Ak A B C1 = 2 2 A C0 B C1 = 2 C 0 D= =

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. = 7. The SUPREM input file for the first oxidation step is: TITLE COMMENT INITIALIZE COMMENT DIFFUSION PRINT STOP Problem 3 7a Initialize si licon substrate Silicon Oxidize the wafer for 60 minutes at 1100 C in dry O2 Time=60 Temperature=1100 DryO2 Layers End Problem 3 7a The result of the PRINT Layers command is: 9

8.5

10 3 2.2

10 22 m 2 / hr 2 5.2

1016 = 1.79

10 3 m 2 / hr = 4.79

10 9 cm 2

layer (microns) 2 1 material type (microns) OXIDE SILICON thic ness 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 thic ness of 0.1088 m, which means 0.44 * 0.1088 m = 0.047 9 m of silicon has been consumed during the process (i.e., the Si/SiO2 interface is 0.0479 m below the original Si surface). The figure below is a graphical repre sentation. For the half of the wafer, the oxide is removed, and the wafer is re oxidized in wet O2. For this half, we use the following SUPREM input file: TITLE COMMENT INITIALIZE COMMENT DIFFUSION COMMENT ETCH Problem 3 7b Initialize silicon substrate Silicon Oxidize the wafer for 60 minutes at 1100 C in dr y O2 Time=60 Temperature=1100 DryO2 Remove the oxide Oxide All COMMENT PRINT STOP Oxidize the wafer for 30 minutes at 1000 C in wet O2 Layers End Problem 3 7b DIFFUSION 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

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