Solutions Manual to Accompany

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

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

Contents Ch. Introduction----------------------------------------------------------------------------- N/A Ch. Crystal Growth ------------------------------------------------------------------------ Ch.3 Silicon Oxidation ---------------------------------------------------------------------- 8 Ch.4 Photolithography------------------------------------------------------------------------ Ch.5 Etching -----------------------------------------------------------------------------------5 Ch.6 Diffusion ---------------------------------------------------------------------------------8 Ch.7 Ion Implantation ------------------------------------------------------------------------6 Ch.8 Film Deposition -------------------------------------------------------------------------3 Ch.9 Process Integration ---------------------------------------------------------------------4 Ch. IC Manufacturing----------------------------------------------------------------------65 Ch. Future Trends and Challenges--------------------------------------------------------78

CHAPTER. C 7 cm -3 k (As in Si).3 C S k C ( - M/M ) k-.3 7 (- x) -.7 3 6 /( - l/5).7 x..4.6.8.9 l (cm) 3 4 45 C S (cm-3) 3 6 3.5 6 4.8 6 5.68 6.7 7.5 7 ND ( 6 cm -3 ) 6 4 8 6 4 3 4 5 l (cm). (a) The radius of a silicon atom can be expressed as r so 3 8 r a 3 5.43.75Å 8 (b) The numbers of Si atom in its diamond structure are 8. So the density of silicon atoms is 8 n a 8 (5.43Å) (c) The density of Si is 3 5. atoms/cm 3 3 M / 6. 8.9 5 3 3 ρ g / cm.33 g / cm 3. 3 / n 6.

3. k.8 for boron in silicon M / M.5 The density of Si is.33 g / cm 3. The acceptor concentration for ρ. Ω cm is 9 8 cm -3. The doping concentration C S is given by C s k C ( M M ) k Therefore C k Cs M ( ) M k 8 9.8(.5). 9.8 8 cm 3 The amount of boron required for a kg charge is, 9.8.338 8 4. boron atoms So that 4. atoms.8g/mole.75g boron. 3 6. atoms/mole 4. (a) The molecular weight of boron is.8. The boron concentration can be given as n b number of boron atoms volume of silicon wafer 3 5.4 g /.8g 6.. 3.4. 8 3 9.78 atoms/cm 3 (b) The average occupied volume of everyone boron atoms in the wafer is

V n b 9.78 cm 3 8 We assume the volume is a sphere, so the radius of the sphere ( r ) is the average distance between two boron atoms. Then 3 V r.9 7 cm. 4π 5. The cross-sectional area of the seed is.55 π.4 cm The maximum weight that can be supported by the seed equals the product of the critical yield strength and the seed s cross-sectional area: 6 5 ( ).4 4.8 g 48 kg The corresponding weight of a -mm-diameter ingot with length l is 3. (.33g/cm ) π l 48 g l 656 cm 6.56 m. 6. We have C / C s k M k M Fractional..4.6.8. solidified C s /C.5.6.8..3 3

Cs/Co.....4.6.8 Fraction Solidified 7. The segregation coefficient of boron in silicon is.7. 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 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 grown crystal will be higher than that of seed-end. 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 8-3 From Eq. 8 C s / C ( k) e kx / L We have x L k k ln C s / C 3 8 ln -3 8 5 5 / 5 5 ln(.) 4 cm. 6. We have from Eq.8 C s C [ ( k e )exp( k e x/ L)] 4

So the ratio Cs / C [ ( k e )exp( k e x/ L)] (.3) exp(.3 ).5 at x / L.38 at x/l.. For the conventionally-doped silicon, the resistivity varies from Ω-cm to 55 Ω-cm. The corresponding doping concentration varies from.5 3 to 4 3 cm -3. Therefore the range of breakdown voltages of p + - n junctions is given by V B ε se q c ( N B ) 5.5 (3 ) 9.6 ( N B ).9 7 / N B 75 to6 V V B 6 75 435 V V B / 75 ± 3% For the neutron irradiated silicon, ρ 48 ±.5 Ω-cm. The doping concentration is 3 3 (±%). The range of breakdown voltage is V B 7.3 / N B 957 to 976 V..9 7 / 3 3 ( ± %) V B 976 957 9 V V B / 957 ± %.. We have M M s l weight of GaAs at T weight of liquid at T b b C C m s C C l m s l Therefore, the fraction of liquid remained f can be obtained as following 5

f M s M l + M l l s + l 3 6 + 3. 65. 3. From the Fig.., we find the vapor pressure of As is much higher than that of the Ga. Therefore, the As content will be lost when the temperature is increased. Thus the composition of liquid GaAs always becomes gallium rich. 88.8 4. n s N exp( Es / kt ) 5 exp(.3 ev / kt) 5 exp ( T / 3).3 6 cm 3 at 7 C 3 K 3 6.7 cm at 9 C 73 K 4 3 6.7 cm at C 473 K. ` 5. n NN exp( E / kt ) f f 5 7 e 3.8eV / kt e.ev / kt 7.7 4 e 94.7 /( T / 3) 7 5.7 at 7 o C 3 K.4 4 at 9 o C 73 K. 6. 37 4 48 chips In terms of litho-stepper considerations, there are 5 µm space tolerance between the mask boundary of two dice. We divide the wafer into four symmetrical parts for convenient dicing, and discard the perimeter parts of the wafer. Usually the quality of the perimeter parts is the worst due to the edge effects. 6

7

CHAPTER 3. From Eq. (with τ) x +Ax Bt From Figs. 3.6 and 3.7, we obtain B/A.5 µm /hr, B.47 µm /hr, therefore A.3 µm. The time required to grow.45µm oxide is t (x + Ax) (.45 B.47 +.3.45).7 hr 44 min.. After a window is opened in the oxide for a second oxidation, the rate constants are B. µm /hr, A.6 µm (B/A 6 - µm /hr). If the initial oxide thickness is nm. µm for dry oxidation, the value ofτcan be obtained as followed: or (.) +.66(.). ( +τ) τ.37 hr. For an oxidation time of min (/3 hr), the oxide thickness in the window area is x +.66x.(.333+.37).7 or x.35 µm 35 nm (gate oxide). For the field oxide with an original thickness.45 µm, the effectiveτis given by τ ( x + Ax) (.45 +.66.45) 7.7 hr. B. x +.66x.(.333+7.7).853 or x.453 µm (an increase of.3µm only for the field oxide). 3. x + Ax B ( t + τ ) A ( x + ) A 4 B( t + τ ) ( x + A ) A B + ( t + τ ) 4B 8

when t >> τ, t >> then, x Bt similarly, when t >> τ, t >> B then, x ( t + τ ) A, A 4B, A 4B 4. At 98 (53K) and atm, B 8.5-3 µm /hr, B/A 4 - µm /hr (from Figs. 3.6 and 3.7). Since A D/k, B/A kc /C, C 5. 6 molecules/cm 3 and C. cm -3, the diffusion coefficient is given by Ak A D 8.5 3.79 µ m 3 4.79-9 B A. 5. cm C C / hr / s. B C C 6 µ m / hr 5. For impurity in the oxidation process of silicon, segregatio n coefficeint equilibrium concentration of impurity in silicon. equilibrium concentration of impurity in SiO 3 3 6. κ.6. 3 5 5 7. The SUPREM input file for the first oxidation step is: TITLE Problem 3-7a COMMENT Initialize silicon substrate INITIALIZE <> Silicon COMMENT Oxidize the wafer for 6 minutes at C in dry O DIFFUSION Time6 Temperature DryO PRINT Layers STOP End Problem 3-7a The result of the PRINT Layers command is: 9

layer material type thickness dx dxmin top bottom no. (microns) (microns) node node OXIDE.88.. 798 84 SILICON.95.. 85 This indicates an oxide thickness of.88 µm, which means.44 *.88 µm.479 µm of silicon has been consumed during the process (i.e., the Si/SiO interface is.479 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 re-oxidized in wet O. For this half, we use the following SUPREM input file: TITLE Problem 3-7b COMMENT Initialize silicon substrate INITIALIZE <> Silicon COMMENT Oxidize the wafer for 6 minutes at C in dry O DIFFUSION Time6 Temperature DryO COMMENT Remove the oxide ETCH Oxide All COMMENT Oxidize the wafer for 3 minutes at C in wet O DIFFUSION Time3 Temperature WetO PRINT Layers STOP End Problem 3-7b The result of this PRINT Layers command is: layer material type thickness dx dxmin top bottom no. (microns) (microns) node node OXIDE.9.. 83 84 SILICON.853.. 85 This indicates a final oxide thickness of.9 µm on the etched side, which means an additional.44

*.9 µm.8 µm of silicon has been consumed during the process. The total distance from the Si/SiO interface to the original Si surface on this side is therefore.479 µm +.8 µm.487 µm. The unetched side is simulated using the SUPREM input file: TITLE Problem 3-7c COMMENT Initialize silicon substrate INITIALIZE <> Silicon COMMENT Oxidize the wafer for 6 minutes at C in dry O DIFFUSION Time6 Temperature DryO COMMENT Oxidize the wafer for 3 minutes at C in wet O DIFFUSION Time3 Temperature WetO PRINT Layers STOP End Problem 3-7c The result of this PRINT Layers command is: layer material type thickness dx dxmin top bottom no. (microns) (microns) node node OXIDE.897.. 798 8 SILICON.875.. 83 On this side, a total of.44 *.897 µm.75 µm of Si is consumed. The total distance from the Si/SiO interface to the original Si surface on this side is.44 *.987 µm.75 µm. The figure below is a graphical representation of the final structure.

The step heights on the surface and in the substrate are.88 µm and. µm, respectively.

CHAPTER 4. With reference to Fig. for class clean room we have a total of 35 particles/m 3 with particle sizes.5 µm 35 735 particles/m with particle sizes. µm 4.5 35 57 particles/m with particle sizes. µm Therefore, (a) 35-735 765 particles/m 3 between.5 and µm (b) 735-57 578 particles/m 3 between and µm (c) 57 particles/m 3 above µm.. 9 D A Y Π e n A 5 mm.5 cm 4(..5) 4(.5.5) (.5). Y e e e e 3.%. 3. The available exposure energy in an hour is.3 mw /cm 36 s 8 mj/cm For positive resist, the throughput is 8 4 7 wafers/hr For negative resist, the throughput is 8 wafers/hr. 9 4. (a) The resolution of a projection system is given by l m k λ.93μm.6 NA.65.78 µm λ.93µ m DOF k.5.8 µm ( NA) (.65) (b) We can increase NA to improve the resolution. We can adopt resolution enhancement techniques (RET) such as optical proximity correction (OPC) and phase-shifting Masks (PSM). We can also develop new resists that provide lower k and higher k for better 3

resolution and depth of focus. (c) PSM technique changes k to improve resolution. 5. (a) Using resists with high γ value can result in a more vertical profile but throughput decreases. (b) Conventional resists can not be used in deep UV lithography process because these resists have high absorption and require high dose to be exposed in deep UV. This raises the concern of damage to stepper lens, lower exposure speed and reduced throughput. 6. (a) A shaped beam system enables the size and shape of the beam to be varied, thereby minimizing the number of flashes required for exposing a given area to be patterned. Therefore, a shaped beam can save time and increase throughput compared to a Gaussian beam. (b) We can make alignment marks on wafers using e-beam and etch the exposed marks. We can then use them to do alignment with e-beam radiation and obtain the signal from these marks for wafer alignment. X-ray lithography is a proximity printing lithography. Its accuracy requirement is very high, therefore alignment is difficult. (c) X-ray lithography using synchrotron radiation has a high exposure flux so X-ray has better throughput than e-beam. 7. (a) To avoid the mask damage problem associated with shadow printing, projection printing exposure tools have been developed to project an image from the mask. With a : projection printing system is much more difficult to produce defect-free masks than it is with a 5: 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 through the reticle. So we can not build a step-and-scan X-ray lithography system. 8. All of the above values can be entered from the Parameters menu or by clicking on the appropriate 4

icon on the toolbar. The resulting resist profile is shown in the figure below. In comparison to Example 3, we see that no resist feature is printed under the modified process conditions. The combination of the long pre-bake time and low exposure dose prevents the feature from being defined. 5

CHAPTER 5. As shown in the figure, the profile for each case is a segment of a circle with origin at the initial mask-film edge. As overetching proceeds the radius of curvature increases so that the profile tends to a vertical line.. (a) sec.6 /6. µm..() plane.6/6 /6.5 µm..() plane.6/ /6. µm.() plane W b W l.5.. µm (b) 4 sec.6 4/6.4 µm.()plane.6/6 4/6.5 µm. () plane.6/ 4/6.4 µm..() plane W b W l.5.4.93 µm (c) 6 sec.6.6 µm.()plane.6/6.375 µm. () plane.6/.6 µm..() plane W b W l.5. 6.65 µm. 3. Using the data in Prob., the etched pattern profiles on <>-Si are shown in below. (a) sec l. µm, W W b.5 µm 6

(b) 4 sec l.5 µm, W W b.5 µm (c) 6 sec l.375 µm W W b.5 µm. 4. If we protect the IC chip areas (e.g. with Si 3 N 4 layer) and etch the wafer from the top, the width of the bottom surface is W W + l + 65 884 µm The fraction of surface area that is lost is ( W W ) / W %(884 - ) /884 % 7.8 % In terms of the wafer area, we have lost 7.8 % π (5/ ) 7 cm Another method is to define masking areas on the backside and etch from the back. The width of each square mask centered with respect of IC chip is given by W W l 65 6 µm Using this method, the fraction of the top surface area that is lost can be negligibly small. 5. Pa 7.5 m Torr PV nrt 7.5 /76-3 n/v.8 73 n/v 4.4-7 mole/liter 4.4-7 6. 3 /.7 4 cm -3 mean free path λ 5 3 / P cm 5-3 / 7.5.6649 cm 6649 µm 5Pa 8 m Torr PV nrt 7

8/ 76-3 n/v.8 73 n/v 6.63-5 mole/liter 6.63-5 6. 3 / 4 6 cm -3 mean-free-path λ 5 3 / P cm 5-3 /8.44 cm 44 µm. Ea 6. Si Etch Rate (nm/min).86-3 RT n T e F.48.86-3 3 5.987 98 ( 98) e 3 4.7 nm/min. 3.76 7. SiO Etch Rate (nm/min).64-3 3 5.987 98 ( 98) e 5.6 nm/min 3 5.6 Etch selectivity of SiO over Si. 5 4.7.64 ( 3.76+.48).987 98 Or etch rate (SiO )/etch rate (Si) e. 5..86 8. A three step process is required for polysilicon gate etching. Step is a nonselective etch process that is used to remove any native oxide on the polysilicon surface. Step 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 %, the polysilicon may be etched % longer or for an equivalent thickness of 4 nm. The selectivity is therefore 4 nm/ nm 4.. Assuming a 3% overetching, and that the selectivity of Al over the photoresist maintains 3. The minimum photoresist thickness required is 8

(+ 3%) µm/3.433 µm 433.3 nm.. ω e qb m e.6 B 9. 9 9 π.45 3 B 8.75 - (tesla) 875 (gauss).. Traditional RIE generates low-density plasma ( 9 cm -3 ) with high ion energy. ECR and ICP generate high-density plasma ( to cm -3 ) with low ion energy. Advantages of ECR and ICP are low etch damage, low microloading, low aspect-ratio dependent etching effect, and simple chemistry. However, ECR and ICP systems are more complicated than traditional RIE systems. 3. The corrosion reaction requires the presence of moisture to proceed. Therefore, the first line of defense in controlling corrosion is controlling humidity. Low humidity is essential,. especially if copper containing alloys are being etched. Second is to remove as much chlorine as possible from the wafers before the wafers are exposed to air. Finally, gases such as CF 4 and SF 6 can be used for fluorine/chlorine exchange reactions and polymeric encapsulation. Thus, Al-Cl bonds are replaced by Al-F bonds. Whereas Al-Cl bonds will react with ambient moisture and start the corrosion process, Al-F bonds are very stable and do not react. Furthermore, fluorine will not catalyze any corrosion reactions. 9

CHAPTER 6. E a (boron) 3.46 ev, D.76 cm /sec Ea 3.46 5 From Eq. 6, D D exp( ).76 exp 4.4 cm /s 5 kt 8.64 3 L Dt 4.4 8.73 5 6 cm From Eq. 9, x x C ( x) Cserfc( ).8 erfc 6 L 5.46 3 If x, C().8 atoms /cm ; x.5-4, C(5-6 ) 3.6 9 atoms/cm 3 ; x.75-4, C(7.5-6) 9.4 8 atoms/cm 3 ; x. -4, C( -5 ).8 8 atoms/cm 3 ; x.5-4, C(.5-5 ).8 6 atoms/cm 3. The x j Dt (erfc - C C sub s ).5µ m Total amount of dopant introduced Q(t) C s L π 5.54 4 atoms/cm. Ea 3.46 4. D D exp.76 exp 4.96 cm /s 5 kt 8.64 33 From Eq. 5, C S S C(, t).34 9 atoms/cm 3 πdt x 9 x ( x) C erfc.34 erfc 5 L.673 C S If x, C().34 9 atoms/cm 3 ; x. -4, C( -5 ).4 9 atoms/cm 3 ; x. -4, C( -5 ) 6.79 8 atoms/cm 3 ; x.3-4, C(3-5 ).65 8

atoms/cm 3 ; x.4-4, C(4-5 ) 9.37 7 atoms/cm 3 ; x.5-4, C(5-5 ).87 7 atoms/cm 3 ; x.6-4, C(6-5 ) 3.5 6 atoms/cm 3 ; x.7-4, C(7-5 ) 7.3 5 atoms/cm 3 ; x.8-4, C(8-5 ) 5.6 4 atoms/cm 3. The S x j 4 Dt ln.7 µ m. C πdt B 8 5 8 3. exp 3 4.3 t t 573 s 6 min For the constant-total-dopant diffusion case, Eq. 5 gives C S S πdt 8 3 3 S π.3 573 3.4 atoms/cm. 4. The process is called the ramping of a diffusion furnace. For the ramp-down situation, the furnace temperature T is given by T T - rt where T is the initial temperature and r is the linear ramp rate. The effective Dt product during a ramp-down time of t is given by t ( Dt) eff D( t) dt In a typical diffusion process, ramping is carried out until the diffusivity is negligibly small. Thus the upper limit t can be taken as infinity:

T T rt T rt ( + T +...) and D D E E rt re exp a a a a D exp (...) D (exp )(exp...) D( T ) exp kt + + kt T kt kt E t re kt a t where D(T ) is the diffusion coefficient at T. Substituting the above equation into the expression for the effective Dt product gives ( Dt ) eff reat D( T ) exp dt D( T kt ) kt re a Thus the ramp-down process results in an effective additional time equal to kt /re a at the initial diffusion temperature T. For phosphorus diffusion in silicon at C, we have from Fig. 6.4: D(T ) D (73 K) -4 cm /s 73 773 r.47k / s 6 E a 3.66 ev Therefore, the effective diffusion time for the ramp-down process is kt re a.38 3 (73) 9.47(3.66.6 9s.5 min. ) 5. For low-concentration drive-in diffusion, the diffusion is given by Gaussian distribution. The surface concentration is then C(, t) S πdt S Ea exp πd t kt

dc dt 3 / S Ea t exp D kt. 5 π C t or dc C dt. 5 t which means % change in diffusion time will induce.5% change in surface concentration. dc dt S E exp π D t E C a a kt kt E kt a or dc C E kt 9 dt 3.6.6 dt 3 T.38 73 T a dt 6.9 T which means % change in diffusion temperature will cause 6.9% change in surface concentration. 6. At C, n i 6 8 cm -3. Therefore, the doping profile for a surface concentration of 4 8 cm -3 is given by the intrinsic diffusion process: C( x, t) Cserfc x Dt where C s 4 8 cm -3, t 3 hr 8 s, and D 5x -4 cm /s. The diffusion length is then Dt.3 5 cm.3µ m The distribution of arsenic is 8 x C ( x) 4 erfc 5 4.64 The junction depth can be obtained as follows 5 x j 8 4 erfc 5 4.64 x j. -4 cm. µm. 3

7. At 9 C, n i 8 cm -3. For a surface concentration of 4 8 cm -3, given by the extrinsic diffusion process D D e E a kt n n i 45.8e 9 4.5.6 3.38 73 4 8 8 3.77 6 cm /s x j.6 Dt.6 3.77 6 8 3.3 6 cm 3.3 nm. 8. Intrinsic diffusion is for dopant concentration lower than the intrinsic carrier concentration n i at the diffusion temperature. Extrinsic diffusion is for dopant concentration higher than n i. 9. The SUPREM input file for this problem is: TITLE Problem 6-9 COMMENT Initialize silicon substrate INITIALIZE Thickness 5 <> Silicon Phosphor Concentratione6 COMMENT Diffuse Boron DIFFUSIONT ime5 Temperature85 Boron Solidsol COMMENT Perform Drive-In DIFFUSIONT ime36 Temperature75 PRINT Layers Active Concentration Phosphorus Boron Net PLOT Active Net Cmine STOP End Problem 6-9 Note that the thickness of the structure has been increased to 5 µm (in the INITIALIZE statement) to accommodate an anticipated deeper junction. The resulting plot is shown below. The junction depth occurs at approximately 3.48 µm. 4

. The SUPREM input file for this problem is: TITLE Problem 6- COMMENT Initialize silicon substrate INITIALIZE Thickness 5 <> Silicon Phosphor Concentratione6 COMMENT Boron Predep DIFFUSION ime5 Temperature85 Boron Solidsol COMMENT Boron Drive-In DIFFUSION Time36 Temperature75 COMMENT Phosphorus Predep DIFFUSION Time3 Temperature85 Phosphor Solidsol COMMENT Phosphorus Drive-In DIFFUSION Time3 Temperature PRINT Layers Active Concentration Phosphorus Boron Net PLOT Active Net Cmine3 5

STOP End Problem 6- The resulting plot is shown below. There are pn junctions formed. The junction depths occur at approximately.45 and 3.49 µm, respectively. 6

CHAPTER 7. The ion dose per unit area is N A It q A 6 9.6 π ( ) 5 6.38 ions/cm From Eq. and Example, the peak ion concentration is at x R p. Figure 7.6 indicates the σ p is nm. Therefore, the ion concentration is S σ p π.38 4.74 7 π 7 cm 3.. From Fig. 7.6, the R p 3 nm, and σ p 6 nm. The peak concentration is S 7 σ p π 6 5.9 π cm 3 From Eq., 5 ( x j R p ).9 exp σ p x j.53 µm. 3. Dose per unit area Q C VT q q 4 3.9 8.85 8.6 cm 8 9 5.6 From Fig. 7.6 and Example, the peak concentration occurs at 4 nm from the surface. Also, it is at (4-5) 5 nm from the Si-SiO interface. 7

4. The total implanted dose is integrated from Eq. Q T S ( x R p ) S R p S S exp dx + erfc( ) [ erfc(.3)].9989 σ p π σ p σ p The total dose in silicon is as follows (d 5 nm): Q Si ( ) S x R p S R p d S S exp erfc( ) [ erfc(.87)].998 dx + σ p π σ p σ d p the ratio of dose in the silicon Q Si /Q T 99.6%. 5. The projected range is 5 nm (see Fig. 7.6). The average nuclear energy loss over the range is 6 ev/nm (Fig. 7.5). 6.5 5 ev (energy loss of boron ion per each lattice plane) the damage volume V D π (.5 nm) (5 nm) 3-8 cm 3 total damage layer 5/.5 6 displaced atom for one layer 5/5 damage density 6/V D cm -3 /5..4%. 6. The higher the temperature, the faster defects anneal out. Also, the solubility of electrically active dopant atoms increases with temperature. 7. V t V Q C ox where Q is the additional charge added just below the oxide-semiconductor surface by ion implantation. C OX is a parallel-plate capacitance per unit area ε s given by Cox d (d is the oxide thickness, ε s is the permittivity of the semiconductor) 8

4 V 3.9 8.85 F/cm 7 C Q V t Cox 8.63 6.4 cm cm 8.63.6 7 9 5.4 ions/cm Total implant dose 5.4 45%. 3 ions/cm. 8. The discussion should mention much of Section 7.3. Diffusion from a surface film avoids problems of channeling. Tilted beams cannot be used because of shadowing problems. If low energy implantation is used, perhaps with preamorphization by silicon, then to keep the junctions shallow, RTA is also necessary. 9. From Eq. S d S.4.6 erfc.84. The effectiveness of the photoresist mask is only 6%. S d S.6 erfc.3. The effectiveness of the photoresist mask is 97.7%.. T -u 5 e π u u 3. d R p + 4.7σ.53 + 4.7.93.97 µm p 9

. The SUPREM input file for this problem is: TITLE Problem 7- COMMENT Initialize silicon substrate INITIALIZE <> Silicon Phosphor Concentratione4 COMMENT Implant Boron IMPLANT Boron Energy3 Dosee3 PRINT Layers Active Concentration Phosphorus Boron Net PLOT Active Net Cmine STOP End Problem 7- The resulting doping profile is shown in the figure below. Examining this figure and the SUPREM output file gives: (a) The peak of the implanted boron occurs at a depth of. µm. (b) The boron concentration at the peak is 8.59e7 cm -3. (c) The junction depth is.449 µm. 3

. The SUPREM input parameters that must be determined are the dose and implant energy. The dose can be determined from Eq. in Chapter 6 as Q( t).3c s Dt where C s can be read directly from the SUPREM output file for Example 3 in Chapter 6 as 4.6e9 cm -3, D.3e 6 cm /s for boron at 85 o C (see Figure 6.4), and t 9 s. Substituting these numbers into the above expression gives a dose of.36e3 cm -. The implant energy required can be approximated by matching the diffused and implanted concentration profiles at the surface (x ) and at the junction and using Eq. to solve for R p and σ p simultaneously. Note from the SUPREM output file corresponding to Example 3 in Chapter 6 that the junction occurs at x j.555 µm, at which point the doping concentration is 6 cm -3. As stated before, the surface concentration is 4.6e9 cm -3. These equations cannot be solved analytically, but after several iterations, the approximate values of R p and σ p required are found to be. µm and.8 µm, respectively. These values correspond to an implant energy of 5 kev (extrapolating from Figure 7.6a). The require SUPREM input file is therefore: TITLE Problem 7- COMMENT Initialize silicon substrate INITIALIZE <> Silicon Phosphor Concentratione6 COMMENT Implant Boron IMPLANT Boron Energy5 Dose.36e3 PRINT Layers Active Concentration Phosphorus Boron Net PLOT Active Net Cmine STOP End Problem 7- The resulting doping profile appears in the figure below. 3

3

CHAPTER 8. ν av vf f v v dv dv 8kT πm Where f ν 4 π M kt 3 / Mν ν exp kt M: Molecular mass k: Boltzmann constant.38-3 J/k T: The absolute temperature ν: Speed of molecular So that ν av 4 468 π 3.38 3 7 9.67 m/sec 4.68 cm/sec..66. λ cm P( in Pa).66.66 3 P 4.4 Pa. λ 5 3. For close-packing arrange, there are 3 pie shaped sections in the equilateral triangle. Each section corresponds to /6 of an atom. Therefore N s number of atoms contained in the triangle area of the triangle 3d 3(4.68 8 5.7 4 atoms/cm. ) 3 6 3 d d 33

dd 4. (a) The pressure at 97 C (43K) is.9 - Pa for Ga and 3 Pa for As. The arrival rate is given by the product of the impringement rate and A/πL : Arrival rate.64 P MT A πl.64.9 5 69.7 43 π.9 5 Ga molecules/cm s The growth rate is determined by the Ga arrival rate and is given by (.9 5 ).8/(6 4 ) 3.5 Å/s 8 Å/min. (b) The pressure at 7ºC for tin is.66-6 Pa. The molecular weight is 8.69. Therefore the arrival rate is.64 6.66 5 8.69 973 π.8 molecular/cm s If Sn atoms are fully incorporated and active in the Ga sublattice of GaAs, we have an electron concentration of.8.9 4.4.74 7-3 cm 5. 5. The x value is about.5, which is obtained from Fig. 8.7. 34

6. The lattice constants for InAs, GaAs, Si and Ge are 6.5, 5.65,5.43, and 5.65 Å, respectively. Therefore, the f value for InAs-GaAs system is f ( 5.65 6.5) 6.5.66 And for Ge-Si system is f ( 5.43 5.65) 5.65.39. 7. (a) For SiN x H y Si. N x x.83 y atomic % H +.83 + y y.46 The empirical formula is SiN.83 H.46. (b) ρ 5 8 e -33.3. Ω-cm As the Si/N ratio increases, the resistivity decreases exponentially. 8. Set Ta O 5 thickness 3t, ε 5 then SiO thickness t, ε 3.9 Si 3 N 4 thickness t, ε 3 7.6, area A εε A 3t CTa O5 C C ONO ONO t ε ε t t + + ε ε A ε ε A A 3 ε ε 3ε A ( ε + ε )t CTa O ε 5 CONO 3ε ε 3 3 ( ε + ε ) 5( 3.9 + 7.6) 3 3 3.9 7.6 5.37. 35

9. Set BST thickness 3t, ε 5, area A SiO thickness t, ε 3.9, area A Si 3 N 4 thickness t, ε 3 7.6, area A then ε ε A ε ε 3ε A 3t ( ε + ε )t A A.93. 3. Let Ta O 5 thickness 3t, ε 5 SiO thickness t, ε 3.9 Si 3 N 4 thickness t, ε 3 7.6 area A then εε A ε ε A 3t d 3ε t d.468t. ε 36

. The deposition rate can be expressed as r r exp (-E a /kt) where E a.6 ev for silane-oxygen reaction. Therefore for T 698 K r( T ) exp.6 r( T ) kt kt.6 3 3 ln.59 698 T 4 T 3 K 757.. We can use energy-enhanced CVD methods such as using a focused energy source or UV lamp. Another method is to use boron doped P-glass which will reflow at temperatures less than 9. 3. Moderately low temperatures are usually used for polysilicon deposition, and silane decomposition occurs at lower temperatures than that for chloride reactions. In addition, silane is used for better coverage over amorphous materials such SiO. 4. There are two reasons. One is to minimize the thermal budget of the wafer, reducing dopant diffusion and material degradation. In addition, fewer gas phase reactions occur at lower temperatures, resulting in smoother and better adhering films. Another reason is that the polysilicon will have small grains. The finer grains are easier to mask and etch to give smooth and uniform edges. However, for temperatures less than 575 ºC the deposition rate is too low. 5. The flat-band voltage shift is V FB.5 V ~ ε ox Number of fixed oxide charge is Q ot C 4 3.9 8.85 8 6.9 8 C F/cm d 5. 37

.5C q.5 6.9 9.6 8. To remove these charges, a 45 heat treatment in hydrogen for about 3 minutes is required. cm - 6. /.5 8 sqs. Therefore, the resistance of the metal line is 5 5 4 Ω. 7. For TiSi 3.37 7.nm For CoSi 3 3.56 6.8nm. 8. For TiSi : Advantage: low resistivity It can reduce native-oxide layers TiSi on the gate electrode is more resistant to high-field-induced hot-electron degradation. Disadvantage: bridging effect occurs. Larger Si consumption during formation of TiSi Less thermal stability For CoSi : Advantage: low resistivity High temperature stability No bridging effect A selective chemical etch exits Low shear forces Disadvantage: not a good candidate for polycides 38

9. (a) L 6 R ρ.67 4 A.8.3 3. Ω 3 4 4 εa εtl 3. 9 8. 85. 3 C. 9 4 d S. 36 4 4 6 3 F RC 3. 5.9 5.93 ns (b) L 6 R ρ.7 4 A.8.3 3 4 4 ε εtl.8 8.85.3 C 4 d S.36 3 3 RC..4 ns Ω. 4 A 3.4 (c) We can decrease the RC delay by 55%. Ratio 93.45. F. (a) L 6 3 R ρ.67 3. Ω 4 4 A.8.3 4 4 εa εtl 3.9 8.85.3 3 3 C 8.7 F 4 d S.36 RC 3. 3 8.7-3.8 ns.. (b) R ρ L A.7.8.3 6 3 4 4 Ω 4 εa εtl.8 8.85.3 C 4 d S.36 RC 3 RC 3. 8.7 3 3 8.7.5 ns 3.5 ns. 4 3 6.3 3 F. (a) The aluminum runner can be considered as two segments connected in series: % (or.4 mm) of the length is half thickness (.5 µm) and the remaining.6 mm is full thickness (µm). The total resistance is 6.6.4 R ρ + + 4 4 4 A (.5 A 3 4 ) 39

7 Ω. The limiting current I is given by the maximum allowed current density times cross-sectional area of the thinner conductor sections: I 5 5 A/cm ( -4.5-4 ).5-3 A.5 ma. The voltage drop across the whole conductor is then..5 µm V RI 7Ω.5 3 A.8V..5 µm Cu Al 4 nm 6 nm h: height, W : width, t : thickness, assume that the resistivities of the cladding layer and TiN are much larger than R Al ρ R Cu ρ Al Cu h W When R Al R Cu.7 (.5 ρ A.).5 and ρ.7 h W (.5 t) (.5 t).7.7 Then.4.5 (.5 t) t.73 µm 73 nm. Cu 4

CHAPTER 9. Each U-shape section (refer to the figure) has an area of 5 µm 8 µm 4 µm. Therefore, there are (5) / 4 3.5 U-shaped section. Each section contains long lines with 48 squares each, 4 corner squares, bottom square, and half squares at the top. Therefore the resistance for each section is kω / (48 + 4.65 +) 5.6 kω The maximum resistance is then 3.5 5.6 7.8 8 Ω 78 MΩ. The area required on the chip is 4

C d A ε ox 7 ( 3 )(5 ) 5 4.35 4 3.9 8.85 cm 4.35 3 µm 66 66 µm Refer to Fig.9.4a and using negative photoresist of all levels (a) Ion implantation mask (for p + implantation and gate oxide) (b) Contact windows ( µm) (c) Metallization mask (using Al to form ohmic contact in the contact window and form the MOS capacitor). Because of the registration errors, an additional µm is incorporated in all critical dimensions. 4

3. If the space between lines is µm, then there is 4 µm for each turn (i.e., n, for one turn). Assume there are n turns, from Eq.6, L µ n r. -6 n r, where r can be replaced by n. Then, we can obtain that n is 3. 43

4. (a) Metal, (b) contact hole, (c) Metal. (a) Metal, (b) contact hole, (c) Metal. 44

5. The circuit diagram and device cross-section of a clamped transistor are shown in (a) and (b), respectively. 6. (a) The undoped polysilicon is used for isolation. (b) The polysilicon is used as a solid-phase diffusion source to form the extrinsic base region and the base electrode. (c) The polysilicon is used as a solid-phase diffusion source to form the emitter region and the emitter electrode. 7. (a) For 3 kev boron, R p nm and R p 34 nm. Assuming that R p and R p for 45

boron are the same in Si and SiO the peak concentration is given by S π R p 8 π (34 7 ) 9.4 6 cm 3 The amount of boron ions in the silicon is Q q d S R erfc p d R p 8 7.88 S π R erfc cm p exp ( x R ) p R p 75 34 dx Assume that the implanted boron ions form a negative sheet charge near the Si-SiO interface, then 9 Q.6 (7.88 ) V T q / C ox.9 V 4 7 q 3.9 8.85 /( 5 ) (b) For 8 kev arsenic implantation, R p 49 nm and R p 8 nm. The peak arsenic concentration is S π R p 6 π (8 7. ) cm 3. 46

8. (a) Because ()-oriented silicon has lower (~ one tenth) interface-trapped charge and a lower fixed oxide charge. (b) If the field oxide is too thin, it may not provide a large enough threshold voltage for adequate isolation between neighboring MOSFETs. (c) The typical sheet resistance of heavily doped polysilicon gate is to 3 Ω /, which is adequate for MOSFETs with gate lengths larger than 3 µm. For shorter gates, the sheet resistance of polysilicon is too high and will cause large RC 47

delays. We can use refractory metals (e.g., Mo) or silicides as the gate material to reduce the sheet resistance to about Ω /. (d) A self-aligned gate can be obtained by first defining the MOS gate structure, then using the gate electrode as a mask for the source/drain implantation. The self-aligned gate can minimize parasitic capacitance caused by the source/drain regions extending underneath the gate electrode (due to diffusion or misalignment). (e) P-glass can be used for insulation between conducting layers, for diffusion and ion implantation masks, and for passivation to protect devices from impurities, moisture, and scratches. 9. The lower insulator has a dielectric constant ε /ε 4 and a thickness d nm The upper insulator has a dielectric constant ε /ε and a thickness d nm. Upon application of a positive voltage V G to the external gate, electric field E and E are established in the d and d respectively. We have, from Gauss law, that ε E ε E +Q and V G E d + E d where Q is the stored charge on the floating gate. From these above two equations, we obtain E d + d V G ( ε / ε ) ε + ε ( / d ) + Q d 48

J 7 7 Q σ E +..6 4 4 + 4 8.85 + 5 Q (a) If the stored charge does not reduce E by a significant amount (i.e.,. >>.6 5 Q, Q we can write 6 (.5 ) t 8 σedt'. t. 5 8 Q 5 V T.565 V C 4 7 ( 8.85 )/( ) 5 (b) when t, J we have Q. /.6 8.84-7 C. C Then 7 Q 8.84 V T 9.98 V. 5 C / 4 ( 8.85 ). 49

5

. The oxide capacitance per unit area is given by C ox ε d SiO 3.5 7 F/cm and the maximum current supplied by the device is I DS W 5µ m 7 µ Cox ( VG VT ) 3.5 ( VG VT ) 5 ma L.5µ m 5

and the maximum allowable wire resistance is. V/5 ma, or Ω. Then, the length of the wire must be 8 R Area Ω cm L.74 cm 8 ρ.7 Ω cm or 74 µm. This is a long distance compared to most device spacing. When driving signals between widely spaced logic blocks however, minimum feature sized lines would not be appropriate.. 5

3. To solve the short-channel effect of devices. 4. The device performance will be degraded from the boron penetration. There are 53

methods to reduce this effect: () using rapid thermal annealing to reduce the time at high temperatures, consequently reduces the diffusion of boron, () using nitrided oxide to suppress the boron penetration, since boron can easily combine with nitrogen and becomes less mobile, (3) making a multi-layer of polysilicon to trap the boron atoms at the interface of each layer. 5. Total capacitance of the stacked gate structure is : ε C d ε d ε d ε + d 7.5 5 7.5 5 +. 3.9. d d 3.9..84 nm. 6. Disadvantages of LOCOS: () high temperature and long oxidation time cause V T shift, () bird s beak, (3) not a planar surface, (4) exhibits oxide thinning effect. Advantages of shallow trench isolation: () planar surface, () no high temperature processing and long oxidation time, (3) no oxide thinning effect, (4) no bird s beak. 7. For isolation between the metal and the substrate. 8. GaAs lacks of high-quality insulating film. 9. Answers will vary. If we ignore the contributions of isolation region processing, the structure can be simulated using four SUPREM input decks. The first deck simulates 54

processing in the active region of the device, up to the point of the isolation oxidation. The second deck starts with the results from the first deck and completes all processing in the active regions. This allows the doping profile through the emitter to be plotted (for part b). The third deck is similar to the second, except it eliminates the emitter implant and facilitates plotting of the doping profile through the base region (for part a). The final deck is also similar to the second, except that it eliminates the base implant and facilitates plotting the doping profile through the collector region (for part c). The complete process sequence is as follows: ) Begin with a high-resistivity, <>, p-type silicon substrate. ) Grow a µm SiO layer. 3) Remove the oxide in the areas where the buried layers are to be placed. 4) Implant antimony at a dose of e5 cm -. Drive in the buried layer for 5 hours at 5 o C. 5) Etch the silicon dioxide from the surface. 6) Grow a.6 µm arsenic-doped epitaxial layer. 7) Grow a 4 Å pad oxide. 8) Deposit 8 Å of silicon nitride. 9) Etch the oxide and nitride from the isolation regions. ) Etch the silicon halfway through the epi-layer. 55

) Implant boron in the field regions with a dose of e3 cm - at an energy of 5 kev. ) Oxidize the field regions to a thickness approximately one-half that of the epi-layer. 3) Implant the base region with boron at a dose of e4 cm - at an energy of 5 kev. 4) Etch the oxide from the emitter region. 5) Implant emitter collector contact regions with arsenic at a dose of 5e5 cm - at an energy of kev. 6) Drive-in the arsenic and activate the base diffusion. The SUPREM input decks are as follows: TITLE BJT Deck COMMENT Initial Active Region Processing COMMENT Initialize silicon substrate INITIALIZE <> Silicon Boron Concentration5e4 COMMENT Grow masking oxide for non-active regions DIFFUSION Time Temperature5 WetO COMMENT Etch oxide over buried layer regions ETCH Oxide COMMENT Implant and drive-in antimony buried layer IMPLANT Antimony Dose5e4 Energy DIFFUSION Time5 Temperature5 DryO DIFFUSION Time3 Temperature5 COMMENT Etch the oxide ETCH Oxide COMMENT Grow.6 µm of arsenic-doped epi EPITAXY Temperature5 Time4 Growth.Rate.4 Arsenic Gas.Conc5e5 56

COMMENT Grow 4 A pad oxide DIFFUSION Time Temperature6 DryO COMMENT Deposit nitride to mask the field oxidation DEPOSITION Nitride Thickness.8 SAVEFILE Structur Filenamebjtactiveinit.str STOP End BJT TITLE BJT Deck COMMENT Final Active Region Processing for Emitter Profile COMMENT Start with previous results INITIALIZE Structurbjtactiveinit.str COMMENT Field oxide growth DIFFUSION Time3 Temperature8 t.rate DIFFUSION Time5 Temperature DryO DIFFUSION Time Temperature Wet DIFFUSION Time5 Temperature DryO DIFFUSION Time Temperature t.rate-3 COMMENT Etch the oxide and nitride layers ETCH Oxide ETCH Nitride ETCH Oxide COMMENT Implant boron base IMPLANT Boron Dosee4 Energy5 COMMENT Remove oxide from emitter region ETCH Oxide COMMENT Implant arsenic emitter and collector contacts IMPLANT Arsenic Dose5e5 Energy COMMENT Drive-in emitter and collector contact regions DIFFUSION Time Temperature PRINT Layers PLOT Chemical Boron Arsenic Phosphor Net STOP End BJT TITLE BJT Deck 3 COMMENT Active Region Processing for Base Profile COMMENT Start with previous results 57

INITIALIZE Structurbjtactiveinit.str COMMENT Field oxide growth DIFFUSION Time3 Temperature8 t.rate DIFFUSION Time5 Temperature DryO DIFFUSION Time Temperature Wet DIFFUSION Time5 Temperature DryO DIFFUSION Time Temperature t.rate-3 COMMENT Etch the oxide and nitride layers ETCH Oxide ETCH Nitride ETCH Oxide COMMENT Implant boron base IMPLANT Boron Dosee4 Energy5 COMMENT Remove oxide from emitter region ETCH Oxide COMMENT Drive-in emitter and collector contact region DIFFUSION Time Temperature PRINT Layers PLOT Chemical Boron Arsenic Phosphor Net STOP End BJT 3 TITLE BJT Deck 4 COMMENT Active Region Processing for Collector Profile COMMENT Start with previous results INITIALIZE Structurbjtactiveinit.str COMMENT Field oxide growth DIFFUSION Time3 Temperature8 t.rate DIFFUSION Time5 Temperature DryO DIFFUSION Time Temperature Wet DIFFUSION Time5 Temperature DryO DIFFUSION Time Temperature t.rate-3 COMMENT Etch the oxide and nitride layers ETCH Oxide ETCH Nitride ETCH Oxide COMMENT Remove oxide from emitter region ETCH Oxide 58

COMMENT Implant arsenic emitter and collector contacts IMPLANT Arsenic Dose5e5 Energy COMMENT Drive-in emitter and collector contact regions DIFFUSION Time Temperature PRINT Layers PLOT Chemical Boron Arsenic Phosphor Net STOP End BJT 4 The resulting doping profiles though the base, emitter, and collector (parts a, b, and c), respectively, are shown in the following three figures: (a) 59

(b) 6

(c). Answers will vary. For the sake of simplicity, we will ignore isolation-related processing. The structure can be simulated using four SUPREM input decks (one for each requested profile). The complete process sequence is as follows: 6

) Start with a <>, n-type silicon substrate. ) Grow a.9 µm SiO layer. 3) Remove the oxide in the p-well areas. 4) Implant boron well at a dose of 5e4 cm - at 5 kev. 5) Drive in the p-well for 6 hours at 5 o C. 6) Remove oxide in PMOS source/drain regions. 7) Implant boron for PMOS source/drain at a dose of e4 cm - at kev. 8) Drive in the PMOS source/drain regions for.5 hours at o C. 9) Etch oxide in NMOS source/drain regions. ) Implant phosphorus for NMOS source/drain at a dose of e4 cm - at kev. ) Drive in the NMOS source/drain regions for.5 hours at o C. ) Etch oxide in gate areas. 3) Grow 5 Å gate oxide. 4) Deposit and pattern polysilicon gates. 5) Grow passivation oxide. 6) Deposit and pattern metallization. The SUPREM input decks are and corresponding outputs as follows: TITLE CMOS Deck COMMENT PMOS source/drain COMMENT Initialize silicon substrate INITIALIZE <> Silicon Phosphorus Concentration5e5 Thickness5 6

COMMENT Grow field oxide DIFFUSION Time Temperature WetO COMMENT Etch oxide after p-well implant ETCH Oxide COMMENT P-well drive-in DIFFUSION Time9 Temperature5 DryO COMMENT Etch the oxide prior to PMOS source/drain implant ETCH Oxide COMMENT PMOS source/drain implant IMPLANT Boron Dosee4 Energy COMMENT PMOS source/drain drive-in DIFFUSION Time5 Temperature DryO COMMENT NMOS source/drain drive-in and gate oxidation DIFFUSION Time5 Temperature DryO COMMENT Etch oxide ETCH Oxide COMMENT Deposit metal DEPOSITION Aluminum Thickness. PRINT Layers PLOT Chemical Boron Phosphor Net STOP End CMOS 63

(a) TITLE CMOS Deck COMMENT PMOS Gate COMMENT Initialize silicon substrate INITIALIZE <> Silicon Phosphorus Concentration5e5 Thickness5 COMMENT Grow field oxide DIFFUSION Time Temperature WetO COMMENT Etch oxide after p-well implant ETCH Oxide COMMENT P-well drive-in DIFFUSION Time9 Temperature5 DryO 64

COMMENT Etch the oxide prior to PMOS source/drain implant ETCH Oxide COMMENT PMOS source/drain drive-in DIFFUSION Time5 Temperature DryO COMMENT NMOS source/drain drive-in and gate oxidation DIFFUSION Time5 Temperature DryO COMMENT Deposit polysilicon DEPOSITION Polysilicon Thickness.5 COMMENT Grow passivation oxide DIFFUSION Time3 Temperature DryO PRINT Layers PLOT Chemical Boron Phosphor Net STOP End CMOS 65

(b) TITLE CMOS Deck 3 COMMENT NMOS source/drain COMMENT Initialize silicon substrate INITIALIZE <> Silicon Phosphorus Concentration5e5 Thickness5 COMMENT Grow field oxide DIFFUSION Time Temperature WetO COMMENT Etch oxide in p-well region ETCH Oxide COMMENT P-well implant 66

IMPLANT Boron Dose5e4 Energy5 COMMENT P-well drive-in DIFFUSION Time9 Temperature5 DryO COMMENT PMOS source/drain drive-in DIFFUSION Time5 Temperature DryO COMMENT Etch the oxide prior to NMOS source/drain implant ETCH Oxide COMMENT NMOS source/drain implant IMPLANT Phosphorus Dosee4 Energy COMMENT NMOS source/drain drive-in and gate oxidation DIFFUSION Time5 Temperature DryO COMMENT Etch oxide ETCH Oxide COMMENT Deposit metal DEPOSITION Aluminum Thickness. PRINT Layers PLOT Chemical Boron Phosphor Net STOP End CMOS 3 67

(c) TITLE CMOS Deck 4 COMMENT NMOS gate COMMENT Initialize silicon substrate INITIALIZE <> Silicon Phosphorus Concentration5e5 Thickness5 COMMENT Grow field oxide DIFFUSION Time Temperature WetO COMMENT Etch oxide in p-well region ETCH Oxide 68

COMMENT P-well implant IMPLANT Boron Dose5e4 Energy5 COMMENT P-well drive-in DIFFUSION Time9 Temperature5 DryO COMMENT PMOS source/drain drive-in DIFFUSION Time5 Temperature DryO COMMENT Etch the oxide prior to NMOS source/drain implant ETCH Oxide COMMENT NMOS source/drain drive-in and gate oxidation DIFFUSION Time5 Temperature DryO COMMENT Deposit polysilicon DEPOSITION Polysilicon Thickness.5 COMMENT Grow passivation oxide DIFFUSION Time3 Temperature DryO PRINT Layers PLOT Chemical Boron Phosphor Net STOP End CMOS 4 69

7 (d)

CHAPTER. x -chart: Center µ.75 V 3σ 3(.) UCL µ +.75 +.845 V n 3σ LCL 3-.655 V n s-chart: Center s c 4 σ.977(.).973 V 4 ) UCL s + 3σ c.973 + 3(.)(.977.67 V LCL s - 3σ c 4.8 V. x -chart: Center x.734 V 3s UCL x +.846 V c n 4 LCL 3s x c n 4.6 V s-chart: Center s.5 V s UCL s + 3 c4.5 V c s LCL s 3 c c 4 4 4.36 V 3. Let ED exposure dose, DT develop time, BT bake temperature ED DT BT Y () () (3) Div. Eff. ID - - - 6 37 64 534 8 66.75 Avg. 7

+ - - 77 7 7 9 4 3 ED - + - 59 4 6-4 -5 DT + + - 68 3 66 6 4.5 ED x DT - - + 57 7-6 4.5 BT + - + 83 9-4 4 ED x BT - + + 45 6-8 4 DT x BT + + + 85 4 4 4 5.5 ED x DT x BT 4. Let: days blocks (i.e., n 3) processes treatments (i.e., k 5) Then we have: A B C D E y i Day 59 5 53 56 59 53.6 Day 55 57 54 5 59 58.6 Day 3 465 47 498 483 475 478.6 y 493 497 54 53 5 y 53.6 t ANOVA Table: Source Sum of Squares Degrees of Freedom Mean Square Average (S A ) 3,84,94.4 3,84,94.4 Blocks (S B ) 4,75,375 ( ) Treatments (S T ),737.6 4 434.4 ( ) Residual (S R ) 8 6.5 ( s R ) s B s T where: S A nky 3,84,94.4 n i ( S k y y), DF n B k t ( i S n y y), DF k T k n R t i ( t S y y y + y), DF (n )(k ) ti i t Now: s B s 9.48 / R 7

s T / sr 6.55 A. Significance level for the null hypothesis that the blocks are the same is very low, since P( F 9.48) ~,8 > B. The same is true for the hypothesis that the treatments are the same, since: P (, 8 F 4 > 6.55) ~ I. The processes are significantly different. II. The processing dates have significant differences. 5. From Eq. 33: Y exp(-a c D ) N exp(-na c D ) where: Y.95 N, A c WL (e-4 cm)(e-4 cm) e-7 cm > lny D 5.3 cm - NA c 6. Murphy s Yield Integral (Eq. 34): Y e A D c f ( D ) dd Uniform defect distribution: f ( D) / D for D D Y D e Ac D dd D D e A Ac D c D > Y uniform D D Ac e A c Triangular defect distribution: f ( D) D / D for D D D + for D D D D D 73

Y D D A D D c e dd + e D D A D c D D + dd D Y D e A AD c c D ( A D ) ( A D ) c D e A A D c c D D + D c e A A D c D D > Y triangular D e D Ac A c Exponential Defect Distribution: f ( D) D D exp D Y A / D( e c D D D e dd exp D D D + A cd ) dd + Ac D D( + Ac D exp D ) > Y exp onential + D A c 7. Use Murphy s Yield Integral (Eq. 34): Y AD e f ( D) dd, where: A cm, f(d) -D+ Y. e.5 D ( D + )dd -..5 De D dd +. D.5 e dd. e D 4.5 D e ( AD )..5 > Y.94 % 74

CHAPTER. (a) L RC ρ ε A ox A d 5.5 8 4 9 ( 69.3 ).38 s.38 ns. (b) For a polysilicon runner RC R 3 7 ns square 4 L ε ox W A d 4 ( 69.3 ) 3.9 8.85.7 7 s 4 4 ( ).5 Therefore the polysilicon runner s RC time constant is 5 times larger than the aluminum runner. 4. When we combine the logic circuits and memory on the chip, we need multiple supply voltages. For reliability issue, different oxide thicknesses are needed for different supply voltages. 3. (a) C total C Ta O + C 5 nitride hence EOT 75 + 7. 3 3.9 5 7 Å (b) EOT 6.7 Å. 75