TEST BANK FOR Fabrication Engineering at the Micro- and Nanoscale 3rd Edition By Stephen A. (Solution manual)

Exam (elaborations) TEST BANK FOR Fabrication Engineering at the Micro- and Nanoscale 3rd Edition By Stephen A. (Solution manual) Fabrication Engineering at the Micro and Nanoscale S. A. Campbell Solutions Manual Version 1.1b – Third Edition 2.1) The nearest neighbor Ga atoms are at (-a/4, a/4, -a/4), (-a/4, -a/4, a/4), (a/4, -a/4, -a/4), and (a/4, a/4, a/4). The distance 31/2 a/4 = 0.254 nm. The ionic lengths are given as rGa +1 ~ 0.081 nm and rAs -3 ~ 0.22 nm. Then the sum of the ionic distances is slightly larger than the a spacing in the crystal. 2.2) For the Ga atom at (a/4, a/4, a/4), the for nearest neighbors are at: (0, 0, 0), (a/2, a/2, 0), (a/2, 0, a/2) and (0, a/2, a/2). They are the As atoms on the faces of the unit cell. 2.3a) Referring to the phase diagram for GeSi, at 1100 oC, the equilibrium concentration in the melt is given as 15%. b) The entire charge melts at 1190 oC. c) If the material is in equilibrium, about 50% of the solid is silicon. 2.4) According to the phase diagram for GaAs, and excess Ga will tend to precipitate out as a liquid (pure Ga) if the temperature is above 29.8 oC. Since typical growth temperatures are much higher than this, droplets will form on the surface. When the material is then lowered to room temperature, these droplets should be slowly absorbed back into the stoichiometric GaAs where they solidify. 2.5) Solid solubility is an equilibrium value. It is possible, and in fact is often desirable, to incorporate an impurity concentration well above the solid solubility. Such a mixture will tend to precipitate over time, but at room temperature the time scales involved may be so long as to preclude any detectable amount of precipitation. 2.6 According to Equation 2.1, No = 5*1022 cm−3e−2.6eV / kT = 2*1010 cm−3 V Then 28.5 2*10 2.6 ln 5*10 10 22 = = kT eV Solving K C eV K T eV 1058 785o 28.5*8.62*10 / 2.6 5 = = = − One can use this temperature to solve the problem as E E kT i V e v i n N+ 2*1010cm−3 p ( + − ) / = From Fig 3.4, ni=2*1018 cm-3. Since NBoronni, p= ni. Then 1 E E kT eV V i 0.28 2*10 *ln 10 10 9 + − = = − 2.7) The temperature is unchanged since NV o is unchanged. Thus, T=785 oC. Since NAni, p=NA and E E kT eV V i 0.17 2*10 5*10 2*10 *ln 10 19 17 10 9 − = ⎥⎦ ⎤ ⎢⎣ ⎡ + − = 2.8) Using Eq. 2.9, * 1 0.28sec 0.091 / sec (10 ) 2 1.2 / 3 2 = = − − cm e eV kT t cm According to Eq. 2.8, C cm e kT cm ppm ox = 2*1021 −3 −1.032 / = 3.3*1017 −3 = 6.5 2.9) From Eq. 2.11, erface dx dT L V k int max * ⎥⎦ ⎤ ⎢⎣ ⎡ = ρ Note that k here is the thermal conductivity, not Boltzman’s constant and is a function of temperature. The value in Appendix II corresponds to room temperature. It is better therefore to use the value given in Table 2.2. C cm cm cm hr gm cm cal gm J cal V W cm C o o *100 / 0.0071 / sec 25.6 / 2.4 / *340 / * 4.14 / 0.24 / max 3 = = ⎥⎦ ⎤ ⎢⎣ ⎡ − = 2.11) From the chapter ( ) = (1− )k −1 o C x kC x For boron, k=0.8. At x=0, C(x = 0) = 0.8*C (1)−0.2 = 0.8*C = 3*1015cm−3 o o Solving, Co=3.75*1015 cm-3. Then C(x = 0.9) = 3.5*1015cm−3(0.1)−0.2 = 4.75*1015cm−3 2.12) Initially the melt concentration is = 0.01/1000 =10−5 o C For arsenic, k=0.3, so using Eq. 2.13 0.933 6.67 (1 ) 0.3*10 (1 ) 5*10 ( ) 10 0.7 5 0.7 22 3 18 3 = = − = = − − − − − − x x X cm C x cm Or 93.3% of the boule is usable. 2.13a) If the boule is quenched, one might exceed the solid solubility. From Fig. 2.4, at 1400 oC, the solid solubility is approximately 6*1020 cm-3. 2 2.13b) 6*1020 cm-3 corresponds to approximately 1.2 atomic percent (6*1020/5*1022 ) impurity. Then 0.996 1.2% 0.8*0.5%(1 ) 0.2 = = − − x X 2.13c) Since CS=6*1020 cm-3, CL= CS /k=6*1020 cm-3/0.8 = 7.5*1020 cm-3. 2.14) From the chapter ( ) = (1− )k −1 o C x kC x For phosphorus, k=0.35. For this problem Co is 10-3. Then C(x) = 3.5*10−4 (1− x)−0.65 Inserting different values of x, x C N (cm-3) 0.1 3.7*10-4 1.9*1019 0.5 5.5*10-4 2.8*1019 0.9 1.6*10-3 7.8*1019 2.15a) C(x = 0) = kC = 0.35*0.01% = 3.5*10−5% =1.75*1018cm−3 o 2.15b) ( ) = 2 = (1− )k −1 o o C x kC kC x 2 = (1− x)k −1 = (1− x)−0.65 2−1/ 0.65 =1− x Therefore x=0.66. Since the boule is 1 m long, the doping concentration is double 0.66 m from the top. 2.15c) At x=0, (kCo)Ga=(kCo)P. At x=0.5, (0.5) 2(0.5) 3.14 ( ) (1 0.5) 2( ) (1 0.5) 1 0.65 1 0.65 = = − = − − − − − k o P k o Ga kC kC Solving this would require a k of -0.65 which is not physical. 2.16) In Bridgeman growth, the boule is in contact with the crucible for an extended period of time and goes through several melt/solidification cycles. 3 3.1) One would need to measure the diffusivity as a function of temperature, then plot the data as an Arrhenius function (log D vs 1/T). Ideally this should be done with various doping concentrations to extract charge effects. If D is concentration dependent, the Boltzman-Matano method can be used (see J. Appl. Phys. 8 p. 109, 1933). Then = − ∫ N xdN dx dN t D N 0 * 2 ( ) 1 3.2) 4 3.3a) In this case, N=1015 cm-3, and from Fig. 3.4 ni=1019 cm-3. Then n~ni, and sec 2.72*10 sec 1.12 *10 sec 1.60*10 sec 1*12 sec 0.066 2 15 2 15 2 15 4.05 / 2 3.44 / 2 cm cm cm D cm e eV kT cm e eV kT − − − − − = + = = + 3.3b) For N=1021 cm-3, n is limited by the maximum electrically active concentration. From Eq. 3.23, n = Cmax = 1.9e+22 * exp(-0.45/kT) = 3.1e+20 cm-3. Then: sec 3.6*10 sec 3.48*10 sec 1.60*10 10 3.1*10 sec *12 sec 0.066 2 14 2 14 2 15 4.05 / 19 2 20 3.44 / 2 cm cm cm D cm e eV kT cm e eV kT − − − − − = + = = + 3.4) The much larger As atom strains the silicon lattice where it is incorporated at high concentration. This strain increases the point defect concentration. The increase in vacancies can increase the diffusivity due to vacancy exchange. 3.5a) The situation is as shown at right. The gate prevents all out diffusion so this is a drive-in if the δ-layer is sufficiently thin. Let’s assume this for now. QT=1.5*1015 cm-2. Then, Gate δ-Layer sec * 1.6 *10 sec 7 *10 ( , ) 1.5*10 * , 2 2 1.2 /1073 11 2 6 / 4 15 2 D cm e cm e where Dt C x t cm eV K x Dt − − − − − = = = π and GaAs m C Dt x Dt Q sub T J μ π 2 . 4 ln * 4 = ⎥⎦ ⎤ ⎢⎣ ⎡ = The junction is much thinner than the δ-layer. 3.5b) The surface concentration is given as 18 3 15 3 1.5 *10 8.6 *10 − − = = cm Dt C cm S π 3.5c) The effect of enhanced diffusion is to flatten the profile, making it deeper than the standard and with a lower surface concentration. Enhanced diffusion occurs due to heavy doping effects such as: 1) Internal fields, 2) Strain, 3) The increased concentration of charged vacancies. 3.6a) This is a drive-in diffusion with a dose of 1018 cm-3 * 2*10-8 cm = 2*1011 cm-2 sec * 9.7 *10 sec * 0.41 sec 0.037 ( , ) * , 2 2 3.46 /1273 15 2 3.46 /1273 2 / 4 D cm e cm e cm e where Dt C x t Q eV k eV k T x Dt − − − − = + = = π 5 At the surface of the wafer, x=0 (0, ) = = 2.7 *1016 cm−3 Dt C t QT π 3.6b) m C Dt Q x Dt sub T J μ π 4 * ln = 0.15 ⎥ ⎥⎦ ⎤ ⎢ ⎢⎣ ⎡ = 3.7) This is also a drive-in diffusion. Ignoring heavy doping effects, sec * 9.0 *10 sec * 0.41 sec 0.037 ( , ) * , 2 2 3.46 /1273 15 2 3.46 /1273 2 / 4 D cm e cm e cm e where Dt C x t Q eV K eV K T x Dt − − − − = + = = π 3.7a) m C Dt x Dt Q sub T J μ π 019 . 0 ln * 4 = ⎥⎦ ⎤ ⎢⎣ ⎡ = At the surface of the wafer, x=0 (0, ) = = 1.9 *1021 cm−3 Dt C t QT π 3.8) This is a drive in diffusion since the nitride prevents out-diffusion. We can approximate the 1 nm thick initial region as a δ-layer. Q = 1018 cm−3 *10 −7 cm = 1011 cm−2 T 2 2 2 2 13 *3600sec 15 3 / 0.53 sec / 4*3.7*10 2 13 11 2 2 2.6 / 13 2 1.5*10 *3600sec sec *3.7*10 ( ) 10 sec 3.7*10 sec 0.019 z m z cm eV kT e cm e cm C z cm D cm e cm μ π − − − − − − − = = = = − To find the depth, set C(z)=1014, then [ ] z m z m cm C z cm e z m μ μ μ 1.20 0.53 *ln 0.067 10 ( ) 1.5*10 2 2 14 3 15 3 2 / 0.53 = = − = = − − 3.9) According to Fig. 2.4, the solid solubility limit for P in Si at 1000 oC is 1021 cm-3. The predeposition diffusion provides a dose as Q C t Dt T 2 (0, ) π = For intrinsic diffusion, n=ni, so 6 15 2 4.37 / 14 2 4.0 / 2 3.66 / 2 4.6 *10 sec * 1.39 *10 sec * 44 sec * 4.4 sec 3.9 2 − − − − − = = + + = Q cm Then D cm e cm e cm e cm T eV kT eV kT eV kT The drive-in produces a junction depth of ⎥⎦ ⎤ ⎢⎣ ⎡ = ⎥⎦ ⎤ ⎢⎣ ⎡ = ⎥⎦ ⎤ ⎢⎣ ⎡ = − Dt cm cm C Dt Q Dt x so C Dt x Dt Q sub T J sub T J ln 0.026 4 *10 4 ln 4 * ln , 2 8 2 π π This can be solved iteratively by guessing a value of Dt and inserting it into the right hand side to calculate a revised value of Dt. For example, guess (Dt)1/2=10-4 cm. Solving gives (Dt)1/2=8.5x10-5 cm. Inserting this again gives (Dt)1/2=8.4x10-5 cm, which we take to be converged. At 1100 oC, D=1.56*10-13 cm2/sec, so t=4.6*104 sec = 12.9 hours. Finally, C(0,t)=QT/(πDt)1/2=3.1*1019 cm-3. 3.10a) From the figure, CS=C(x=0)=2*1020 cm-3. From Fig. 3.4, ni=1019 cm-3. From Table 3.2, sec 2.4 *10 : 0 sec *1.1*10 sec * 1.6 *10 sec * 12 sec 0.066 2 14 4.05 / 15 15 2 3.44 / 2 2 2 D cm For x cm n e cm n n D cm e cm n i eV kT i eV kT − − − − − = = = + = + 3.10b) For intrinsic diffusion, n=ni, so D=2.7*10-15 cm2/sec. 3.10c) This is a predeposition since the surface concentration is fixed. 3.12a) At 1100 oC the solid solubility of As in Si is about 2*1021 cm-3. According to Eq. 3.23, 22 3 0.453/ 20 3 max C =1.9*10 cm− e− kT = 3.06*10 cm− at 1000 oC. 3.12b) According to the phase diagram, the solubility of As in Si does not change dramatically between 900 and 1100 oC, going from about 2% to about 3.5%. The maximum carrier concentration is much less than the solubility. 3.12c) This difference means that for high concentrations some As may be dissolved in the Si but is electrically inactive. It may reside, for example, in interstitial sites. 3.13) If the wafer is uniformly doped in the depletion region (i.e. reverse bias, but less than breakdown), the capacitance varies as (V+Vbi)-1/2. 3.14) Due to the large difference of boron diffusivity in Si and SiO2, an exact solution must be done numerically. To obtain a rough estimate, one can assume that the 7 diffusivity is the same in both materials. Then, if we ignore heavy doping effects, D=3.2*10-18 cm2/sec and (Dt)1/2=2.2*10-7 cm. ∫ ∫ ∫ ⎥⎦ ⎤ ⎢⎣ = − ⎡ ⎥⎦ ⎤ ⎢⎣ = = ⎡ − ∞ − ∞ oc oc oc t T t t Q C x dx erfc x dx Q erfc x dx 0 7 21 7 21 4.3*10 10 4.3*10 ( ) 10 Using the erfc expansion given in the appendix of this book, ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − + ⎥⎦ ⎤ ⎢⎣ ⎡ − = ⎥⎦ ⎤ ⎢⎣ ⎡ − = ⎥ ⎥⎦ ⎤ ⎢ ⎢⎣ ⎡ − − − − − − − − ∫ 2 7 6 4.3*10 10 7 6 7 6 21 0 7 21 1 1 4.3*10 10 4.3*10 10 10 4.3*10 Q Q 10 erfc x dx Q erfc e T t T oc π Solving, Q~6.2*1010 cm-2. The actual dose would be larger since the diffusivity in Si is larger and so the concentration gradient across the oxide would be larger. 8 4.1) At 1000 oC in dry O2, A=0.165 μm, B=0.0117 μm2/hr, and τ=0.37 μm. Then τ τ − + = + = + B t t At t At B t ox ox ox ox 2 2 ( ) In this case, t=2.26-0.37=1.89 hours. Since A~tox, this process in not in either the linear or the parabolic regime. 4.2) For a wet ambient, A=0.226 μm and B=0.287 μm2/hr. Then 0.114 6.8min .287 / (0.1 ) 0.1 *.226 2 2 2 = = + = + = hr m hr m m m B t tox Atox μ μ μ μ This solution is closer to linear than parabolic, but is still in the transition regime. 4.3) For the first oxidation, 0.37 0.55 33min .0117 / (0.05 ) 0.165 *.05 2 2 − = = + = hr hr m hr t m m m μ μ μ μ For the second oxidation, hr hr hr m hr t m m m (0.55 0.37 ) 1.34 .0117 / (0.05 ) 0.165 *.05 2 2 − + = + = μ μ μ μ Note that, as one would expect, the sum of the answers to the two parts (0.55+1.34) is the same as doing the full oxidation in a single step as in problem 4.1. 4.4) From Fig. 4.2, B/A=0.226 μm/hr and B=0.4 μm2/hr, so A=0.17 μm. A is independent of pressure while B is proportional to pressure. Then B(5 atm)=2.0 μm2/hr, and B(20 atm) = 8 μm2/hr. Solving the Deal-Grove equations with these parameters, P (atm) B (μm2/hr) t (hr) 1 0.4 2.93 5 2.0 0.59 20 8.0 0.15 4.5a) From Table 4.1, A=0.5 μm, B=0.203 μm2/hr. Then using the Deal-Grove equation with τ=0, 7.43*10 26.8sec .203 / (0.003 ) 0.5 *.003 3 2 2 = = + = − hr m hr t m m m μ μ μ μ 4.5b) Since B is proportional to Pg and t is inversely proportional to B, 225sec 76 = 26.8sec* 640 = torr t torr 4.6a) Since there is no rapid growth regime, from Eq. 4.10, [ ] N [ k t D] Hk P N k h k t D Hk P dt dt s ox s g s s ox ox s g 1 / / 1 / 1 1 + ≈ + + = 9 4.6b) For short oxidation times, tox~(B/A)(t+τ), so ox Ke EA kT A B dt dt = = − / where K is some constant. From the problem, and taking a ratio to eliminate K, 2 0.5 / sec 1.0 / sec /1173 /1273 = = − − E k E k A A e e nm nm Solving, EA=0.89 eV, and K=3320 nm/sec, so (3320nm / sec)e 0.89 / 1073 0.22nm / sec dt dt ox = − eV k = 4.7a) This is clearly in the linear regime so t m nm A t B ox 0.042 42 30 1 0.226 = = 0.287 = μ = 4.7b) If τ is not zero, 0.047 2.83min 0.287 / 0.226 0.06 / ( + ) = = = hr = B A t t τ ox Then τ is 0.83 min. 4.8) Ignoring rapid growth t hr B t Atox ox 0.7 0.0117 2 0.042 0.165*0.04 = + = = + Including rapid growth t hr B t Atox ox 0.37 0.33 0.0117 2 0.042 0.165*0.04 − = + − = = + τ 4.9) In this case we are diluting the oxygen and so changing B (see Eq. 4.12). Then B m hr t t Atox ox 2.0*10 / 10/ 60 .002 0.165*0.002 3 2 2 2 = − μ + = = + The pressure is proportional to the ratio of the B parameters P atm 0.17atm 0.0117 =1 0.002 = 4.9b) At one atm of oxygen and 1000 oC, dtox/dt is about 5 nm/min from Eq. 4.6. To get 2 nm would require 2/5 min or 24 sec. 4.10) At 1100 oC in dry O2, A=0.090 μm, B=0.027 μm2/hr, and τ=0.076 μm. Then m nm A A B t t t At B t ox ox ox 0.131 131 2 0.09 0.09 4*0.027(1.076) 2 4 ( ) ( ) 2 2 2 = = − + + = − + + + = + = + μ τ τ 10 4.11) From Eq. 4.12, 2 2 2 2 2 2 ; 5 2 1 O g O O S O B N DHP B A R A ≈ D = = ≈ − − From the table in the chapter, A(O2 -)=0.033 μm, and B(O2 -)=0.0234 μm2/hr. Then for an oxide of 100 nm at 1000 oC, τ τ μ μ μ μ − = − + = hr m hr t m m m 0.57 .0234 / (0.1 ) 0.1 *.033 2 2 The value of τ is unknown. If one guesses that it is unchanged, t=0.2 hr. 4.12) It is preferred to grow the oxide at higher temperature to minimize the fixed oxide charge, the interface state density, and the interface roughness. This produces the highest inversion layer mobility for the MOSFET. It is likely that these oxides are more robust (i.e. more resistant to damage) than lower temperature thermal oxides. 4.13) Since B is proportional to Pg, at 1000 oC and ignoring rapid growth effects, A=0.165 μm and B=0.0117 μm2/hr. Then 1.5 90min .00117 / (0.01 ) 0.01 *.165 2 2 = = + = hr m hr t m m m μ μ μ μ 4.14) From Equation 4.16, when C2~0, / 2 2 ox C e tox L A B dt dt = + − Then [ ] [ ] [ ] ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + + = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + + = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + + = + + = − − − − − − − ∫ ∫ 2 2 / / 2 2 / 2 2 / 2 / 2 2 2 / 2 2 0 / 2 2 2 2 2 2 2 2 ln ln ln ln ln A C B Ae C B L B A A C e B A C e B L B t A e A C B A C e B L B t A A C B A C e B L B A B t t A A C e B dt dt t L t L t L t L t L t L ox t t L ox t ox ox ox ox ox ox ox ox τ τ τ τ Note that as L2 goes to infinity, C2 goes to zero as expected and the equation reverts to Deal-Grove. Inverting to solve for tox 11 [ ] [ ] ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥⎦ ⎤ ⎢ ⎢⎣ ⎡ = + + + − = − ⎥⎦ ⎤ ⎢⎣ ⎡ + + + − = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + = + − + + + 2 2 2 2 2 2 2 ( ) 2 2 2 / 2 2 / 2 2 2 2 / ( ) ln 1 1 1 AL B t ox L t L t A B L t L t A B t L L t A B e B t L AC A t B e B C e AC B AC Ae C e C B A B A C B Ae C B e ox ox ox τ τ τ τ τ The equation for τ can be readily found from the second equation by setting tox=t0 at t=0 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + = + − 2 2 / 2 2 / 2 0 2 0 2 0 2 ln ln A C B Ae C B L B A A C B A C e B L B A B t A t L t L τ 4.15) From Eq. 4.17, and using 5 nm/min = 0.3 μm/hr, [ ] [ ] ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + + = 0.165 0.3 0.0117 0.165 0.007 0.0117 *0.007ln 0.0117 ln 0.165 100 / 70 2 2 / 2 2 e A C B Ae C B L B t A tox L τ Solving, t-τ=0.046 hours. Presumably, τ=0 since the added terms eliminate the need for the rapid growth term, τ, and the wafer is bare. 4.16) Now B=0.00117. Inserting into Eq. 4.17, [ ] 0.07 4.2min 0.165 0.3 0.00117 0.165 0.007 0.00117 *0.007ln 0.00117 0.165 100 / 70 = = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + + = hr e t τ 4.17) a) The film is not stoichiometric SiO2. This would affect both the dielectric constant and the refractive index. This is very unusual for thermal oxide which is almost always very close to perfectly stoichiometric. b) If the oxide is very thin, one must take into account the finite accumulation layer thickness in the silicon, and in the gate (unless one uses a metal gate electrode). The substrate depletion layer typically adds about 0.4 nm to the measured oxide thickness. c) If the oxide is thin, it may be leaky. Leaky oxides often make for erroneous capacitance measurements. d) There may be an error in the area of the capacitor. e) There may be a poor substrate contact. 4.18) The question does not indicate if this shift is from as-received to +bias or from – bias to +bias. For now assume that latter. In the as-received state, one can assume a random distribution of charge. Then the shift in threshold is 12 2 2 2 2 1 2 1 SiO ox MI ox ox MI SiO T t Q t t V Q ε ε Δ = − = For a positive bias the charge is a delta function at the silicon/oxide interface. This produces a threshold shift that is exactly twice that of the random charge distribution. For a negative bias the charge is a delta function at the gate/oxide interface. This produces no threshold shift. Then the shift between +bias and –bias corresponds to 10 2 9 2 14 6 2 1.4*10 / 2.3*10 / , 3.9