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 = 0.3×1017(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 as r= 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 sil icon 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 × 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 ρ = 0.01 Ω–cm is 9×1018 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 . 4π
5. The cross-sectional area of the seed is 0.55 2 π = 0.24 cm 2 The maximum weig that can be suorted 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 corresonding weight of a 200-mm-diameter ingot with length l is 2 20.0
( 2.33g/cm )π
l = 480000 g
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
∴ 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 2 1 − 8 × 10 − 3 ln
8 × 10 3
1 − 5 × 1015 / 5 × 1016
10. We have from Eq.18 Cs = C0[1 − (1 − ke ) exp( ke x / L)] − 4
= 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 Ω-cm to 155 Ω-cm. The co esponding doping concent ation va ies f om 2.5×1013 to 4×1013 cm-3. The efo e the ange of b eakdown voltages of p+ - n junctions is given by VB ≅ ε 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 / 7250 = ±30% 2 For th nutron irradiatd silicon, ρ = 148 1.5 Ω-cm. ncent ation is 3×1013 ( 1%). The ange of b eakdown voltage is 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 ∆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 l Thrfor, th fraction of liquid rmaind f can b obtaind as following 5
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 incr asd. 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 0 173 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 = 7 × 10 24 × −94.7 /(T / 300 ) = 5.27 × 10 −17 at 27oC = 300 K =2.14×1014 at 900oC = 1173 . 16. 37 × 4 = 148 chips In trms of litho stppr 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 (withτ=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. 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 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 oxida ion, he value ofτcan be ob ained as followed: (0.02)2 + 0.166(0.02) = 0.01 (0 +τ) or τ= 0.372 hr. For an oxida ion 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 (ga e 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 on ly for he field oxide). 3. x2 + Ax = B ( + τ ) (x + A 2 A2 ) − = B (t + τ ) 2 4 (x + A2 8
A 2 ) =B
+ ( + τ )
2
4B
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 3 7a Initialize si licon substrate <100> 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
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 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 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 <100> 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 thic ness of 0.2291 µm on the etched side, which mean s an additional 0.44 10
* 0.2291 µm = 0.1008 µm of silicon has been consumed during the process. The total d istance from the Si/SiO2 interface to the original Si surface on this side is th erefore 0.0479 µm + 0.1008 µm = 0.1487 µm. The unetched side is simulated using the SU PREM input file: TITLE COMMENT INITIALIZE COMMENT DIFFUSION Problem 3 7c Initialize silicon substrate <100> Silicon Oxidize the wafer for 60 minutes at 1100 C in dry O2 Time=60 Temperature=1100 DryO2 COMMENT PRINT STOP Oxidize the wafer for 30 minutes at 1000 C in wet O2 Layers End Problem 3 7c DIFFUSION 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 d istance 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 fi nal structure. 11
The step heights on the surface and in the substrate are 0.0818 µm and 0.0212 µm, re spectively. 12
CHAPTER 4 1. With reference to Fig. 2 for class 100 clean room we have a total of 3500 par ticles/m3 with particle sizes ≥ 0.5 µm 21 3500 = 735 particles/m2 with particle size s ≥ 1.0 µm 100 4.5 3500 = 157 particles/m2 with particle sizes ≥ 2.0 µm 100 Therefore, ( a) 3500 735 = 2765 particles/m3 between 0.5 and 1 µm (b) 735 157 = 578 particles/m 3 between 1 and 2 µm (c) 157 particles/m3 above 2 µm. 2. Y = Π e − D1 A n =1 9
A = 50 mm2 = 0.5 cm2 Y = e −4( 0.1 0.5) e −4 ( 0.25 0.5) e −1(1 0.5) = e −1.2 = 30.1% . 3. The available exposure energy in an hour is 0.3 mW2/cm2 3600 s =1080 mJ/cm2 F or 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 projecti on system is given by l m = 1 λ NA = 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 reso ution. We can adopt reso ution enhancemen t techniques (RET) such as optica proximity correction (OPC) and phase-shifting Masks (PSM). We can a so deve op new resists that provide ower k1 and higher k 2 for better 13
reso ution and depth of focus. (c) PSM technique changes k1 to improve reso utio n. 5. (a) Using resists with high γ value can result in a more vertical profile bu t throu hput decreases. (b) Conventional resists can not be used in deep UV lith o raphy process because these resists have hi h absorption and require hi h dose to be exposed in deep UV. This raises the concern of dama e to stepper lens, lo wer exposure speed and reduced throu hput. 6. (a) A shaped beam system enables the size and shape of the beam to be varied, thereb y minimizin the number of flashes required for exposin a iven area to be patt erned. Therefore, a shaped beam can save time and increase throu hput compared t o a Gaussian beam. (b) We can make ali nment marks on wafers usin e-beam and etch the exposed marks. W e can then use them to do ali nment with e-beam radiation and obtain the si nal from these marks for wafer ali nment. X-ray litho raphy is a proximity printin litho raphy. Its accuracy requirement is very hi h, therefore ali nment is diffi cult. (c) X-ray litho raphy usin synchrotron radiation has a hi h exposure flux so Xray has better throu hput than e-beam. 7. (a) To avoid the mask dama e problem associated with shadow printin , project ion printin exposure tools have been developed to project an ima e from the mas k. With a 1:1 projection printin system is much more difficult to produce defec t-free masks than it is with a 5:1 reduction step-and-repeat system. (b) It is n ot possible. The main reason is that X-rays cannot be focused by an optical lens . When it is throu h the reticle. So we can not build a step-and-scan X-ray lith o raphy system. 8. All of the above values can be entered from the Parameters menu or by clickin on the appropriate 14
icon on the toolbar. The resultin resist profile is shown in the fi ure below. In comparison to Example 3, we see that no resist feature is printed under the m odified process conditions. The combination of the lon pre-bake time and low ex posure dose prevents the feature from bein defined. 15
CHAPTER 5 1. As shown in the fi ure, the profile for each case is a se ment of a circle with ori in at the initial mask-film ed e. As overetchin proceeds the radius of curv ature 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) plane Wb = 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) plane Wb = W0 − 2l = 1.5 − 2 0.4 = 0.93 µm (c) 60 sec 0.6 1 = 0.6 µm….(100)plane 0.6/16 1 = 0.0375 µm…. (110) plane 0.6/100 1= 0.006 µm…..(111) plane Wb = 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 µm 16
(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 fr om 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%=(18842 1000 2) /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 fr om the back. The width of each square mask centered with resect 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 10 3 = n/V 0.082 273 n/V = 4.42 10 7 mol e/liter = 4.42 10 7 6.02 1023/1000 =2.7 1014 cm 3 mean–free–path λ = 5 × 10 −3 / P cm = 5 10 3 1000/ 7.52 = 0.6649 cm = 6649 µm 150Pa = 1128 m Torr PV = nRT 17
1128/ 760 10 3 = n/V 0.082 273 n/V = 6.63 10 5 mole/liter = 6.63 10 5 6.02 1023/10 = 4 1016 cm 3 mean free path λ = 5 × 10 −3 / P cm = 5 10 3 1000/1128 = 0.0044 cm = 44 µm. 6. Si Etch Rate (nm/min) = 2.86 10 13 n F T 1 2
e − Ea RT = 2.86 10
13 3 10 (298) 15 1 2
e −2.48 103 1.987 298 = 224.7 nm/min. 7. SiO2 Etch Rate (nm/min) = 0.614 10 Etch selectivity of SiO2 over Si =
13 3 10 (298) 15 1 2
e −3.76 103 1.987 298 = 5.6 nm/min 5.6 = 0.025 224.7 0.614 ( −3.76+ 2.48)1.987 298 e = 0.025 . 2.86 Or etch rate (SiO2)/etch rate (Si) = 8. A three–step process is required for polysilicon gate etching. Step 1 is a nons elective 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 polysilico n 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 phot oresist maintains 3. The minimum photoresist thic ness required is 18
(1+ 30%) 1 µm/3 = 0.433 µm = 433.3 nm. 11. ωe = qB me 9
1.6 × 10 −19 × B 2π × 2.45 × 10 = 9.1 × 10 −31 B = 8.75 × 10 2(tesla) = 875 (gauss). 12. Tr al RIE generates lo density lasma (109 cm 3) ith high ion energy. ECR and ICP generate high density lasma (1011 to 1012 cm 3) ith lo ion energy. Advantage s of ECR and ICP are lo etch damage, lo microloading, lo asect ratio deende nt etching effect, and simle chemistry. Ho ever, ECR and ICP systems are more c omlicated than traditional RIE systems. 13. The corrosion reaction requires the resence of moisture to roceed. Therefore, the first line of defense in contro lling corrosion is controlling humidity. Lo humidity is essential,. esecially if coer containing alloys are being etched. Second is to remove as much chlori ne as ossible from the afers before the afers are exosed to air. Finally, ga ses such as CF4 and SF6 can be used for fluorine/chlorine exchange reactions and olymeric encasulation. Thus, Al Cl bonds are relaced by Al F bonds. Whereas Al Cl bonds ill react ith ambient moisture and start the corrosion rocess , A l F bonds are very stable and do not react. Furthermore, fluorine ill not catal yze 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 ×10 4, C(5× 10 6) = 3.6 × 1019 atom s/cm3; x = 0.075 ×10 4 , C(7.5×10 6) = 9.4 ×1018 atoms/cm3; x = 0.1×10 4, C(10 5) = 1.8 × 1018 atoms/cm3; x = 0.15× 10 4, C(1.5×10 5) = 1.8× 1016 atoms/cm3. The x j = 2 Dt (erfc 1 C sub ) = 0.15µm Cs Total amount of doant introduced = Q(t) = 2 C s L = 5.54 × 1014 atoms/cm2. π − 3.46 − Ea −14 2 2. D = D0 ex , C S = C (0, t ) = S
= 0.76 ex
= 4.96 × 10 cm /s −5 kT
8.614 ×
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 = 0.1×10 4, C(10 5) = 1.41×1019 atoms/cm3; x = 0.2×10 4, C(2×10 5) = 6.79×1018 atoms/cm3; x = 0.3×10 4, C(3×10 5) = 2.65×1018 20
atoms/cm3; x = 0.4×10 4, C(4×10 5) = 9.37×1017 atoms/cm3; x = 0.5×10 4, C(5×10 5) = 1.87×1017 atoms/cm3; x = 0.6×10 4, C(6×10 5) = 3.51×1016 atoms/cm3; x = 0.7×10 4, C(7×10 5) = 7.03×1015 atoms/cm3; x = 0.8×10 4, C(8×10 5) = 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 constant total doant 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 ram do n s ituation, the furnace temerature T is given by T = T0 rt here T0 is the init ial temerature and r is the linear ram rate. The effective Dt roduct during a ram do n 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
1 T = 1 T0 − rt ≈ 1 T0 (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 0
here 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 ) ex 0 ∞ − rE a t kT0 2 dt = D (T0 ) kT0 rE a 2 Thus the ram do n rocess results in an effective additional time equal to kT02 /rEa at the initial diffusion temerature T0. For hoshorus diffusion in silico n at 1000°C, e have from Fig. 6.4: D(T0) = D (1273 K) = 2× 10 14 cm2/s r= 1273 − 773 20 × 60 = 0.417 K / s Ea = 3.66 eV Therefore, the effective diffusion time for the ram do n rocess is kT 2 0 a rE = 1.38 × 10
−23 (1273) 2 −19 0.417( 3.66 × 1.6 × 10 ) = 91s ≈ 1.5 min . 5. For lo concentration drive in diffusion, the diffusion is given by Gaussian distribution. The surface concentration is then C (0, t ) = S S E = ex a πDt πD0 t 2kT 22
dC S C
E
− t 3 / 2 = ex
= −0.5 ×
a
dt πD 0
2kT
2
t
or dC C = −0.5 × dt t
hich means 1% change in diffusion time ill induce 0.5% change in surface conce ntration. 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 T
hich means 1% change in diffusion temerature ill cause 16.9% change in surfac e concentration. 6. At 1100°C, ni = 6×1018 cm 3. Therefore, the doing rofile for a surface concentr ation of 4 × 1018 cm 3 is given by the “intrinsic” diffusion rocess: x C ( x, t ) = C s erfc 2 Dt
here Cs = 4× 1018 cm 3, t = 3 hr = 10800 s, and D = 5x10 14 cm2/s. The diffusion length is then Dt = 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 follo s
xj
1015 = 4 × 1018 erfc
xj = 1.2× 10 4 cm = 1.2 µm. 23
4.64 × 10 −5
7. At 900°C, ni = 2× 1018 cm 3. For a surface concentration of 4×1018 cm 3, given by t he “extrinsic” diffusion rocess D = 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 lo er than the intrinsic carr ier concentration ni at the diffusion temerature. Extrinsic diffusion is for do ant concentration higher than ni. 9. The SUPREM inut file for this roblem is: TITLE COMMENT INITIALIZE COMMENT DIFFUSIONT COMMENT DIFFUSIONT PRINT PLOT STOP P roblem 6 9 Initialize silicon substrate Thickness = 5 <100> Silicon Phoshor Con centration=1e16 Diffuse Boron ime=15 Temerature=850 Boron Solidsol Perform Driv e In ime=360 Temerature=1175 Active Net Cmin=1e12 End Problem 6 9 Layers Active Concentration Phoshorus Boron Net Note that the thickness of the structure has been increased to 5 µm (in the INITIA LIZE statement) to accommodate an anticiated deeer junction. The resulting lo t is sho n 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 C OMMENT DIFFUSION PRINT PLOT Problem 6 10 Initialize silicon substrate Thickness = 5 <100> Silicon Phoshor Concentration=1e16 Boron Prede ime=15 Temerature=85 0 Boron Solidsol Boron Drive In Time=360 Temerature=1175 Phoshorus Prede Time =30 Temerature=850 Phoshor Solidsol Phoshorus Drive In Time=30 Temerature=10 00 Active Net Cmin=1e13 Layers Active Concentration Phoshorus Boron Net 25
STOP End Problem 6 10 The resulting lot is sho n 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 is It 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 in dicates the σp i 20 nm. Therefore, the ion concentration i S σ p 2π 2. = 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 i S σ p 2π From Eq. 1, = 2 × 1015 62 × 10 −7 2π = 1.29 × 10 20 cm −3 − (x j − R )2 xj = 0.53 µm.
1015 = 1.29 × 10 20 ex
2 2σ p
3. Do e 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 su rfac. Also, it is at (140 25) = 115 nm from th Si SiO2 intrfac. 27
4. Th total implantd dos is intgratd from Eq. 1 QT = ∫ ∞ 0 S σp − (x − Rp )2
R p
S S
S
)
= [2 − erfc( 2.3)] = × 1.9989 exp
dx =
1 +
The total do e 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 =
the ratio of do e 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× 10 18 cm3 total damage layer = 150/0.25 = 600 dislaced atom for one layer = 15/15 = 1 da mage density = 600/VD = 2×1020 cm 3 2×1020/5.02×1022 = 0.4%. 6. The higher the temera ture, 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 oxid smiconductor surfa c by ion implantation. COX is a paralll plat capacitanc pr unit ara givn by C ox = εs d (d is th oxid thicknss, ε s is th prmittivity of th smiconductor) 28
1
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 × 1 012 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 ff ctiv n ss of th photorsist mask is only 16%. Sd 1 1 − 0.6 = rfc = 0.023 S 2 0.2 2 Th ff ctiv n ss of th photorsist mask is 97.7%. 10.
u T= = 10 −5 u 2 π 1 ∴u = 3.02 d = R + 4.27 σ p = 0.53 + 4.27 × 0.093 = 0.927 µm 2 29
11. The SUPREM input file for thi problem i : TITLE Problem 7 11 COMMENT Initialize ilicon ub trate INITIALIZE <100> Silicon Pho phor Concentration=1e14 COMMENT Implant Boron IMPLANT Boron Energy=30 Do e= 1e13 PRINT Layer Active Concentration Pho phoru Boron Net PLOT Active Net Cmin =1e11 STOP End Problem 7 11 The re ulting doping profile i hown in the figure below. Examining thi figure and the SUPREM output file give : (a) The peak of the impl anted boron occur at a depth of 0.11 µm. (b) The boron concentration at the peak i 8.59e17 cm 3. (c) The junction depth i 0.4492 µm. 30
12. The SUPREM input parameter that mu t be determined are the do e and implant energy. The do e can be determined from Eq. 11 in Chapter 6 a Q(t ) ≅ 1.13C Dt where C can be read directly from the SUPREM output file for Example 3 in Chapt er 6 a 4.6e19 cm 3, D ≈ 2.3e − 16 cm2/ for boron at 850 oC ( ee Figure 6.4), and t = 900 . Sub tituting the e number into the above expre ion give a do e of 2 .36e13 cm 2. The implant energy required can be approximated by matching the dif fu ed and implanted concentration profile at the urface (x = 0) and at the jun ction and u ing Eq. 1 to olve for Rp and σp imultaneou ly. Note from the SUPREM output file corre ponding to Example 3 in Chapter 6 that the junction occur at xj = 0.0555 µm, at which point the doping concentration i 1016 cm 3. A tated be fore, the urface concentration i 4.6e19 cm 3. The e equation cannot be olved analytically, but after everal iteration , the approximate value of Rp and σp r equired are found to be 0.011 µm and 0.008 µm, re pectively. The e value corre pond to an implant energy of 5 keV (extrapolating from Figure 7.6a). The require SUP REM input file i therefore: TITLE COMMENT INITIALIZE COMMENT IMPLANT PRINT PLOT STOP Problem 7 12 Initialize ilicon ub trate <100> Silicon Pho phor Concentration=1e16 Implant Boron Boron Energy=5 Do e=2.36e13 Layer Active Concentration Pho phoru Boron Net Active N et Cmin=1e11 End Problem 7 12 The re ulting doping profile appear in the figure below. 31
32
CHAPTER 8 1. ν av = ∫ ∫ ∞ 0 ∞ 0 vf v dv f v dv = 8kT πM 3/ 2 4
M
ν 2 ex Mν 2
Where fν =
π
2kT
− 2kT
M: Molecular mass k: Boltzma co sta t = 1.38×10 23 J/k T: The absolute temeratu re ν: Seed of molecular So that ν 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 close acki g arra ge, there are 3 ie shaed sectio s i the equilateral tr ia gle. Each sectio correso ds to 1/6 of a atom. Therefore 1 3× umber of atoms co tai ed i the tria gle 6 = Ns = area of the tria gle 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 2.9×10 1 Pa for Ga a d 13 Pa for As2. The arri val rate is give by the roduct of the imri geme t rate a d A/πL2 : P A Arrival r ate = 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 determi ed by the Ga arrival rate a d is give by (2.9×1015)×2.8/(6×1014) = 13.5 Å/s = 810 Å/mi . (b) The ressure at 700ºC or ti is 2.66×10 6 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 molecula If Sn atoms are fully incorporated and active in the Ga sublattice of GaAs, we h ave 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 InAs GaAs system is f = (5.65 − 6.05) 6.05 = −0.066 And for Ge Si system is f = (5.43 − 5.65) 5.65 = −0.39 . 7. (a) For SiNxHy Si 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 inc eases, the ases exponentially.
esistivity dec e
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 A 3t 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 thn ε 1ε 0 A1 3t = 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 thn ε 1ε 0 A 3t d= = ε 2ε 0 A d = 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 silan oxygn raction. Thrfor for T1 = 698 K 1 r (T2 ) 1
= 2 = xp
0.6
kT − kT
r (T1 ) 2 1 ln 2 = 0.6 4 T2 =1030 K= 757 ℃. 12. We can useenergy-enhanced CVD me hods such as using a fo cused energy source -glass whi or UVlamp. Ano her me hod is o use boron doped ch will reflow a empera ures less han 900 . 13. Modera ely low empera ures a re usually used for polysilicon deposi ion, and silane decomposi ion occurs a l ower empera ures han ha for chloride reac ions. In addi ion, silane is used for be ercoverage over ma erials amorphous such SiO2. 14. There are wo reason s. One is o minimize he hermal budge of he wafer, reducing dopan diffusion and ma erial degrada ion. In addi ion, fewer gas phase reac ions occur a lower i empera ures, resul ing in smoo her and be er adhering films. Ano her reason s ha he polysilicon will have small grains. The finer grains are easier o ma sk and e ch o givesmoo hand uniform edges. However, uresless ha for empera n 575 C he deposi ion ra e is oo low. 15. The fla -band vol age shif is Qo C0 ∆VFB = 0.5 V ~ C0 = ε ox d = 3.9 × 8.85 × 10 −14 = 6.9 × 10 −8 F/cm − 2 . 500 × 10 −8 ∴ Numbr of fixd oxid charg is 37
300 3
0.5C 0 0.5 × 6.9 × 10 −8 = = 2.1 × 1011 cm 2 −19 q 1.6 × 10 To rmov ths chargs, a 450℃ hea rea men forabou 30 minu es is in hydrogen 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 laye s TiSi2 on the gate elect ode is mo e esistant to high-field-induced hot-elect on deg ada tion. Disadvantage: b idging effect occu s. La ge Si consumption du ing fo mati on of TiSi2 Less the mal stability Fo CoSi2: Advantage: low esistivity High te mpe atu e stability No b idging effect A selective chemical etch exits Low shea fo ces Disadvantage: not a good candidate fo polycides 38
19. (a) R = ρ εA d L 1 = 2.67 × 10 − 6 × = 3.2 × 10 3 Ω −4 −4 A 0.28 × 10 × 0.3 × 10 = C= εTL S = 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 × 1 (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= εA d = εTL S = 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
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 thickn ss (1µm). Th total rsistanc is 0.16 0.04 + −4 R = ρ 1 + 2 = 3 × 10 − 6 − 4 −4 −4 10 × (0.5 × 10 ) 39
10 ×
= 72 Ω. The limiting cu ent I is given by the maximum allowed cu ent density tim es c oss-sectional a ea 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 d op ac oss the whole conducto is then V = RI = 72Ω × 2.5 × 10 −3 A = 0.18V. 22. 0.5 µm 40 m 0.5 µm Cu = Al 60 m
h: height , W : width , t : thick ess, assume that the resistivities of the clad di g layer a d 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 U shae sectio (refer to the figure) has a area of 2500 µm × 8 µm = 2 × 104 µm2. Th erefore, there are (2500)2/2 ×104 = 312.5 U shaed sectio . Each sectio co tai s 2 lo g li es with 1248 squares each, 4 cor er squares, 1 bottom square, a d 2 ha lf squares at the to. Therefore the resista ce 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 MΩ 2. The a ea 41
equi ed on the chip is
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 l vls (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 cont act window and form th MOS capacitor). Bcaus of th rgistration rrors, an a dditional 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, fo r on turn). Assum thr ar n turns, from Eq.6, L ≈ µ0n2r ≈ 1.2 × 10 6n2r, 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 cross sction of a clampd transistor ar show n in (a) and (b), rspctivly. 6. (a) Th undopd polysilicon is usd for isolation. (b) Th polysilicon 1 is usd as a solid phas diffusion sourc to form th xtrinsic bas rgion and th bas lctrod. (c) Th polysilicon 2 is usd as a solid phas 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 for 45
boron ar th sam in Si and SiO2 th pak concntration is givn by S 2π ∆R p = 8 × 1 011 2π (34 × 10 ) −7 = 9.4 × 1016 cm −3 The amou t of boro io s 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
= 750
2 − rfc
2 × 340
= 7.88 × 1011 cm − 2 Assum that th implantd boron ions form a ngativ sht charg nar th Si Si O2 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) orie ted silico has lower (~ o e te th) i terface traed cha rge a d a lower fixed oxide charge. (b) If the field oxide is too thi , it may ot rovide a large e ough threshold voltage for adequate isolatio betwee eigh bori g MOSFETs. (c) The tyical sheet resista ce of heavily doed olysilico ga te is 20 to 30 Ω /□, which is adequate fo MOSFETs with gate lengths la ge than 3 µm. Fo sho te gates, the sheet esistance of polysilicon is too high and will cau se la ge RC 47
delays. We can use ef acto y metals (e.g., Mo) o silicides as the gate mate ia l to educe the sheet esistance to about 1 Ω /□. (d) A self-aligned gate can be obt ained by fi st defining the MOS gate st uctu e, then using the gate elect ode as a mask fo the sou ce/d ain implantation. The self-aligned gate can minimize pa asitic capacitance caused by the sou ce/d ain egions extending unde neath the gate elect ode (due to diffusion o misalignment). (e) P-glass can be used fo i nsulation between conducting laye s, fo diffusion and ion implantation masks, a nd fo passivation to p otect devices f om impu ities, moistu e, and sc atches. 9. The lowe insulato has a dielect ic 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 n m. Upon application of a positiv voltag VG to th xtrnal gat, lctric fil d E1 and E2 ar stablishd in th d1 and d2 rspctivly. W hav, from Gauss’ la w, that ε1E1 = ε2E2 +Q and VG = E1d1 + E2d2 whr Q is th stord charg on th floa ting gat. From ths abov two quations, w obtain E= 1 VG Q + d1 + d 2 (ε 1 / ε 2 ) ε 1 + ε 2 (d1 / d 2 ) 48
10 × 10 7 −7 Q
5
+ J = σE = 10
= 0.2 − 2.26 × 10 Q
1 10
10 + 100 4 −14
10 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 write Q = ∫ σE dt ' ≈ 0.2∆t = 0.2 × (0.25 × 10 − 6 ) = 5 × 10 −8 C 1 t 0 ∆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 ≅ 8.84×10 7 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 by C ox = ε SiO d 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 0.5µm 51
and th maximum allowabl wir rsistanc is 0.1 V/5 mA, or 20Ω. Then, the length of the wi e must be L≤ R × A ea ρ = 20Ω × 10 −8 cm 2 = 0.074 cm 2.7 × 10 −8 Ω − cm or 740 µm. This is a lo g dista ce comared to most device saci g. Whe drivi g s ig als betwee widely saced logic blocks however, mi imum feature sized li es w ould ot be aroriate. 12. 52
13. To solve the short cha el effect of devices. 14. The device erforma ce wil l be degraded from the boro e etratio . There are 53
methods to reduce this effect: (1) usi g raid thermal a eali g to reduce the t ime at high temeratures, co seque tly reduces the diffusio of boro , (2) usi g itrided oxide to suress the boro e etratio , si ce boro ca easily combi e with itroge a d becomes less mobile, (3) maki g a multi layer of olysilico to tra the boro atoms at the i terface of each layer. 15. Total caacita ce o f the stacked gate structure is : C= ε1 d1 × ε2 d2 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 V T shift, (2) bird’s bak, (3) not a planar surfac, (4) xhibits oxid thinning f fct. Advantags of shallow trnch isolation: (1) planar surfac, (2) no high t mpratur 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 la cks of high quality insulating film. 19. Answrs will vary. If w ignor th contributions of isolation rgion procs sing, th structur can b simulatd using four SUPREM input dcks. Th first d ck simulats 54
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 compl ts all procssing in th activ rgions. This allows th doping profil throug h th mittr to b plottd (for part b). Th third dck is similar to th scon d, xcpt it liminats th mittr implant and facilitats plotting of th dopi ng 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 plotti ng th doping profil through th collctor rgion (for part c). Th complt pr ocss squnc is as follows: 1) Bgin with a high rsistivity, <100>, p typ si licon 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 cm 2. Driv in th burid layr for 5 hours at 1150 oC. 5) Etch th silicon dioxid f rom th surfac. 6) Grow a 1.6 µm arsnic dopd 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 pi layr. 55
11) Implant boron in th fild rgions with a dos of 113 cm 2 at an nrgy of 50 kV. 12) Oxidiz th fild rgions to a thicknss approximatly on half that of th pi layr. 13) Implant th bas rgion with boron at a dos of 114 cm 2 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 cm 2 at an n rgy of 100 kV. 16) Driv in th arsnic and activat th bas diffusion. Th SU PREM input dcks ar as follows: TITLE COMMENT COMMENT INITIALIZE COMMENT DIFFUSION COMMENT ETCH COMMENT IMPLANT DIFFUSION DIFFUSION COMMENT ETCH COMMENT EPITAXY BJT – Deck 1 I itial Active Regio Processi g I itialize silico substrate <100> Silico Boro Co ce tratio =5e14 Grow maski g oxide for o active regio s Time=100 Temerature=1150 WetO2 Etch oxide over buried layer regio s Oxide Imla t a d drive i a timo y buried layer A timo y Dose=5e14 E ergy=120 Time=15 Temerature=1150 DryO2 Time=300 Temeratu re=1150 Etch the oxide Oxide Grow 1.6 µm of arse ic doed ei Temerature=1050 Tim e=4 Growth.Rate=0.4 Arse ic Gas.Co c=5e15 56
COMMENT DIFFUSION COMMENT DEPOSITION SAVEFILE STOP Grow 400 A ad oxide Time=20 Temerature=1060 DryO2 Deosit itride to mask the field oxidatio Nitride Thick ess=0.08 Structur File ame=bjtactivei it.str E d B JT 1 TITLE COMMENT COMMENT INITIALIZE COMMENT DIFFUSION DIFFUSION DIFFUSION DIFFUSION DIFFUSION COMMENT ETCH ETCH ETCH COMMENT IMPLANT COMMENT ETCH COMMENT IMPLANT C OMMENT DIFFUSION PRINT PLOT STOP BJT – Deck 2 Fi al Active Regio Processi g for Emitter Profile Start with reviou s results Structur=bjtactivei it.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 a d itride layers Oxide Nitride Oxide Imla t boro base Boro Dose=1e14 E ergy=50 Remove oxide from emitter regio Oxide Iml a t arse ic emitter a d collector co tacts Arse ic Dose=5e15 E ergy=100 Drive i emitter a d collector co tact regio s Time=20 Temerature=1000 Layers Chemical Boro Arse ic Phoshor Net E d BJT 2 DryO2 Time=10 Temerature=1100 t.rate= 30 TITLE COMMENT COMMENT BJT – Deck 3 Active Regio Processi g for Base Profile Start with revious results 57
INITIALIZE COMMENT DIFFUSION DIFFUSION DIFFUSION DIFFUSION DIFFUSION COMMENT ETC H ETCH ETCH COMMENT IMPLANT COMMENT ETCH COMMENT DIFFUSION PRINT PLOT STOP Structur=bjtactivei it.str Field oxide growth Time=30 Temerature=800 t.rate=10 Time=15 Temerature=1000 DryO2 Time=210 Temerature=1100 Wet02 Time=15 Temeratu re=1100 Etch the oxide a d itride layers Oxide Nitride Oxide Imla t boro base Boro Dose=1e14 E ergy=50 Remove oxide from emitter regio Oxide Drive i emitt er a d collector co tact regio Time=20 Temerature=1000 Layers Chemical Boro A rse ic Phoshor Net E d BJT 3 DryO2 Time=10 Temerature=1100 t.rate= 30 TITLE COMMENT COMMENT INITIALIZE COMMENT DIFFUSION DIFFUSION DIFFUSION DIFFUSION DIFFUSION COMMENT ETCH ETCH ETCH COMMENT ETCH BJT – Deck 4 Active Regio Processi g for Collector Profile Start with revious re sults Structur=bjtactivei it.str Field oxide growth Time=30 Temerature=800 t.ra te=10 Time=15 Temerature=1000 DryO2 Time=210 Temerature=1100 Wet02 Time=15 Tem erature=1100 Etch the oxide a d itride layers Oxide Nitride Oxide Remove oxide from emitter regio Oxide DryO2 Time=10 Temerature=1100 t.rate= 30 58
COMMENT IMPLANT COMMENT DIFFUSION PRINT PLOT STOP Imla t arse ic emitter a d collector co tacts Arse ic Dose=5e15 E ergy=100 Driv e i emitter a d collector co tact regio s Time=20 Temerature=1000 Layers Chemi cal Boro Arse ic Phoshor Net E d BJT 4 The resulti g doi g rofiles though the base, emitter, a d collector (arts a, b, a d c), resectively, are show i the followi g three figures: (a) 59
(b) 60
(c) 20. A swers will vary. For the sake of simlicity, we will ig ore isolatio rela ted rocessi g. The structure ca be simulated usi g four SUPREM i ut decks (o e for each requested rofile). The comlete rocess seque ce 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. Imla t boro well at a dose of 5e14 cm 2 at 50 keV. Drive i the well for 6 hours at 1150 oC. Remove oxide i PMOS source/drai r egio s. Imla t boro for PMOS source/drai at a dose of 1e14 cm 2 at 20 keV. Dr ive i the PMOS source/drai regio s for 2.5 hours at 1100 oC. Etch oxide i NMO S source/drai regio s. 10) Imla t hoshorus for NMOS source/drai at a dose of 1e14 cm 2 at 20 keV. 1 1) Drive i the NMOS source/drai regio s for 2.5 hours at 1100 oC. 12) Etch oxi de i gate areas. 13) Grow 500 Å gate oxide. 14) Deosit a d atter olysilico g ates. 15) Grow assivatio oxide. 16) Deosit a d atter metallizatio . The SUP REM i ut decks are a d correso di g oututs as follows: TITLE COMMENT COMMENT INITIALIZE CMOS – Deck 1 PMOS source/drai I itialize silico substrate <100> Silico Phoshorus Co ce tratio =5e15 Thick ess=5 62
COMMENT DIFFUSION COMMENT ETCH COMMENT DIFFUSION COMMENT ETCH COMMENT IMPLANT CO MMENT DIFFUSION COMMENT DIFFUSION COMMENT ETCH COMMENT DEPOSITION PRINT PLOT STO P Grow field oxide Time=120 Temerature=1100 WetO2 Etch oxide after well imla t Oxide P well drive i Time=900 Temerature=1150 DryO2 Etch the oxide rior to P MOS source/drai imla t Oxide PMOS source/drai imla t Boro Dose=1e14 E ergy= 20 PMOS source/drai drive i Time=150 Temerature=1100 DryO2 NMOS source/drai drive i a d gate oxidatio Time=150 Temerature=1100 DryO2 Etch oxide Oxide De osit metal Alumi um Thick ess=1.0 Layers Chemical Boro Phoshor Net E d CMOS 1 63
(a) TITLE COMMENT COMMENT INITIALIZE COMMENT DIFFUSION COMMENT ETCH COMMENT DIFFUSIO N CMOS – Deck 2 PMOS Gate I itialize silico substrate <100> Silico Phoshorus Co ce tratio =5e15 Thick ess=5 Grow field oxide Time=120 Temerature=1100 WetO2 Et ch oxide after well imla t Oxide P well drive i Time=900 Temerature=1150 Dr yO2 64
COMMENT ETCH COMMENT DIFFUSION COMMENT DIFFUSION COMMENT DEPOSITION COMMENT DIFF USION PRINT PLOT STOP Etch the oxide rior to PMOS source/drai imla t Oxide PMOS source/drai drive i Time=150 Temerature=1100 DryO2 NMOS source/drai drive i a d gate oxidatio Time=150 Temerature=1100 DryO2 Deosit olysilico Polysilico Thick ess=0.5 G row assivatio oxide Time=30 Temerature=1000 DryO2 Layers Chemical Boro Phos hor Net E d CMOS 2 65
(b) TITLE COMMENT COMMENT INITIALIZE COMMENT DIFFUSION COMMENT ETCH COMMENT CMOS – Dec k 3 NMOS source/drai I itialize silico substrate <100> Silico Phoshorus Co c e tratio =5e15 Thick ess=5 Grow field oxide Time=120 Temerature=1100 WetO2 Etch oxide i well regio Oxide P well imla t 66
IMPLANT COMMENT DIFFUSION COMMENT DIFFUSION COMMENT ETCH COMMENT IMPLANT COMMENT DIFFUSION COMMENT ETCH COMMENT DEPOSITION PRINT PLOT STOP Boro Dose=5e14 E ergy=50 P well drive i Time=900 Temerature=1150 DryO2 PMOS s ource/drai drive i Time=150 Temerature=1100 DryO2 Etch the oxide rior to NMO S source/drai imla t Oxide NMOS source/drai imla t Phoshorus Dose=1e14 E er gy=20 NMOS source/drai drive i a d gate oxidatio Time=150 Temerature=1100 Dr yO2 Etch oxide Oxide Deosit metal Alumi um Thick ess=1.0 Layers Chemical Boro Phoshor Net E d CMOS 3 67
(c) TITLE COMMENT COMMENT INITIALIZE COMMENT DIFFUSION COMMENT ETCH CMOS – Deck 4 NMOS gate I itialize silico substrate <100> Silico Phoshorus Co ce tratio =5e15 T hick ess=5 Grow field oxide Time=120 Temerature=1100 WetO2 Etch oxide i well regio Oxide 68
COMMENT IMPLANT COMMENT DIFFUSION COMMENT DIFFUSION COMMENT ETCH COMMENT DIFFUSI ON COMMENT DEPOSITION COMMENT DIFFUSION PRINT PLOT STOP P well imla t Boro Dose=5e14 E ergy=50 P well drive i Time=900 Temerature=11 50 DryO2 PMOS source/drai drive i Time=150 Temerature=1100 DryO2 Etch the oxi de rior to NMOS source/drai imla t Oxide NMOS source/drai drive i a d gate oxidatio Time=150 Temerature=1100 DryO2 Deosit olysilico Polysilico Thick ess=0.5 Grow assivatio oxide Time=30 Temerature=1000 DryO2 Layers Chemical Bo ro Phoshor Net E d CMOS 4 69
(d) 70
CHAPTER 10 1. x chart: Ce ter = µ = 0.75 V UCL = µ + 3σ 3(0.1) = 0.75 + = 0.845 V n 10 LCL = 3 3σ = 0.655 V n
chart: Center = = c4σ = 0.9727(0.1) = 0.0973 V UCL = + 3 σ 1 − c = 0.973 + 3(0.1 )(1 − 0.9727) = 0.167 V 2 4 1 2 2 LCL = 3 σ 1 − c4 = 0.028 V 2. x chart: Center = x = 0.734 V UCL = x + 3 c4 n = 0.846 V LCL = x − 3 c4 n = 0.622 V
chart: 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 = expo ure do e, 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 14 92 20 6 6 40 0 22 4 4 4 4 4 4 4 23 5 1.5 1.5 10 0 5.5 ED DT ED x DT BT ED x BT DT x BT ED x DT x BT 4. Let: day = block (i.e., n = 3) proce e = 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 52 4 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) Re idual (SR) Sum o f 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 – 1 i =1 k n 2 S T = n∑ ( y t − y ) , DF = k – 1 t =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 = 90.48 72
2 2 T / R = 16.55 A. Significance level for the null hypothe i that the block are the ame i ve ry low, ince P( F 2,8 > 90.48) ~0 B. The ame i true for the hypothe i that the treatment are the ame, ince: P ( F4,8 > 16.55) ~ 0 I. II. The proce e are ignificantly different. The proce ing date have ignificant difference . 5. From Eq. 33: Y = exp( AcD0)N = exp( NAcD0) where: Y = 0.95 N = 100,000 Ac = W L = (10e 4 cm)(1e 4 cm) = 1e 7 cm2 => D = − ln Y = 5.13 cm 2 NAc ∞ 6. Murphy’s Yield I tegral (Eq. 34): Y = ∫ e − Ac D f ( D)dD 0 U iform defect distributio : f ( D) = 1 / 2 D0 for 0 ≤ D ≤ 2 D0 Y= 2 D0 ∫ 0 e − Ac D 1 e − Ac D dD = 2 D0 2 D0 − Ac 2 D0 0 => Yu iform = 1 − e −2 D0 Ac 2 D0 Ac Tria gular defect distributio : f ( D) = D / D02 for 0 ≤ D ≤ D0 =− D 2 + for D0 ≤ D ≤ 2D0 2 D0 D0 73
Y = ∫e 0 D0 − Ac D D dD + D02 2 D0 D0 ∫e − Ac D D 2 2 D0
− 2 + dD
D D0
0
− 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 2 D0 − Ac D0 => Ytria gular 1 − e − D0 Ac =
D A 0 c
2 Exo e tial Defect Distributio : f ( D) = ∞
− D
1 ex
D
∞ − D(1 + Ac D0 ) 0 D0 0 D0 D0 0
D0
0
0 − D(1 + Ac D0 )
1 − Ac D − D / D0 1 1 ex
Y =∫ e e dD =
∞ => Yex o e tial = 1 1 + D0 Ac 7. Use Murhy’s Yield I tegral (Eq. 34): Y = ∫ e − AD f ( D)dD , where: A = 100cm2, f( D) = 100D+10 0.1 Y= 0.05 ∫e −100 D
dD =
(− 100 D + 10)dD 0.1 = 100 0.05 −100 D −100 D ∫ De dD + 10 ∫ e dD 0.05 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 runnr L A RC = Rsquar ε ox
W
d
1
= 30
− 4
69.03 × 10 −14 = 2.07 × 10 −
( ) Thrfor th polysilicon runnr’s RC tim constant is 150 tims largr than th a luminum runnr. 2. Whn w combin th logic circuits and mmory on th chip, w nd multipl s upply voltags. For rliability issu, diffrnt oxid thicknsss ar ndd fo r diffrnt supply voltags. 3. (a) 1 1 + 1 C total = C Ta 2 O5 C nitrid + 10 = 17.3 Å 7 hnc EOT 3.9 = 75 25 (b) EOT = 16.7 Å. 75
