Fabrication Engineering at the Micro- and Nanoscale. Solution manual [3 ed.] 978-0-19-532017-6

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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, N Vo = 5 *10 22 cm −3 e −2.6 eV / kT = 2 *1010 cm −3 Then 2.6eV 5 *10 22 = ln = 28.5 kT 2 *1010 Solving 2.6eV = 1058K = 785o C T= 28.5 * 8.62 *10 −5 eV / K One can use this temperature to solve the problem as p + NV+ = 2 *1010 cm − 3 e( Ev − Ei ) / kT ni 18 -3 From Fig 3.4, ni=2*10 cm . Since NBoronni, p=NA and ⎡ 109 5 *1017 ⎤ + EV − Ei = kT * ln ⎢ = −0.17eV 10 19 ⎥ ⎣ 2 *10 2 *10 ⎦ 2.8) Using Eq. 2.9,

t=

(10−3 cm) 2 1 * −1.2eV / kT = 0.28 sec 2 0.091cm / sec e

According to Eq. 2.8, Cox = 2 *1021 cm −3e −1.032 / kT = 3.3 *1017 cm −3 = 6.5 ppm 2.9) From Eq. 2.11,

⎡ k dT ⎤ Vmax = ⎢ * ⎥ ⎣ ρL dx ⎦ int erface 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. ⎡ ⎤ 0.24W / cm − o C Vmax = ⎢ * 100o C / cm⎥ = 0.0071cm / sec = 25.6cm / hr 3 ⎣ 2.4 gm / cm * 340cal / gm * 4.14 J / cal ⎦

2.11) From the chapter

C ( x) = kCo (1 − x) k −1

For boron, k=0.8. At x=0, C ( x = 0) = 0.8 * Co (1) −0.2 = 0.8 * Co = 3 *1015 cm−3 Solving, Co=3.75*1015 cm-3. Then C ( x = 0.9) = 3.5 *1015 cm −3 (0.1) −0.2 = 4.75 *1015 cm −3 2.12) Initially the melt concentration is Co = 0.01 / 1000 = 10−5 For arsenic, k=0.3, so using Eq. 2.13 10 18 cm −3 C ( x) = = 0.3 *10 −5 (1 − X ) − 0.7 −3 22 5 *10 cm 6.67 = (1 − x) − 0.7 x = 0.933 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 1.2% = 0.8 * 0.5%(1 − X ) −0.2 x = 0.996 20 -3 2.13c) Since CS=6*10 cm , CL= CS /k=6*1020 cm-3/0.8 = 7.5*1020 cm-3. C ( x) = kCo (1 − x) k −1 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) -4 0.1 3.7*10 1.9*1019 0.5 5.5*10-4 2.8*1019 0.9 1.6*10-3 7.8*1019

2.14) From the chapter

2.15a) C ( x = 0) = kCo = 0.35 * 0.01% = 3.5 *10−5% = 1.75 *1018 cm −3 2.15b) C ( x) = 2kCo = kCo (1 − x) k −1

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, (kCo )Ga (1 − 0.5) k −1 = 2(kCo ) P (1 − 0.5) −0.65 (0.5) k −1 = 2(0.5) − 0.65 = 3.14 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 1 dN D( N ) = − * 2t dx

∫ xdN N

0

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 cm 2 − 3.44 eV / kT cm 2 − 4.05eV / kT + 1 *12 D = 0.066 e e sec sec cm 2 cm 2 cm 2 = 1.60 *10−15 + 1.12 *10−15 = 2.72 *10−15 sec sec sec 21 -3 3.3b) For N=10 cm , 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: cm 2 −3.44 eV / kT cm 2 3.1*10 20 − 4.05eV / kT + *12 D = 0.066 e e sec sec 1019 cm 2 cm 2 cm 2 = 1.60 *10 −15 + 3.48 *10 −14 = 3.6 *10 −14 sec sec sec 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

C ( x, t ) =

1 .5 * 10 15 cm −2

D = 7 * 10 − 6

πDt

* e−x

δ-Layer

2

/ 4 Dt

, where

cm 2 cm 2 * e −1.2 eV / 1073 K = 1 .6 * 10 −11 sec sec

GaAs

and

⎡ ⎤ QT 4 Dt * ln ⎢ ⎥ = 4 .2 μm The junction is much thinner than the δ-layer. ⎣ C sub πDt ⎦ 3.5b) The surface concentration is given as 1 .5 * 10 15 cm −3 = 8 .6 * 10 18 cm −3 CS = πDt 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. xJ =

3.6a) This is a drive-in diffusion with a dose of 1018 cm-3 * 2*10-8 cm = 2*1011 cm-2 2 QT * e − x / 4 Dt , where C ( x, t ) = πDt

D = 0 .037

cm 2 cm 2 cm 2 * e −3.46 eV / 1273 k + 0 .41 * e −3.46 eV / 1273 k = 9 .7 * 10 −15 sec sec sec

5

At the surface of the wafer, x=0

C ( 0, t ) =

3.6b)

xJ =

QT = 2 .7 * 10 16 cm −3 πDt

⎡ QT 4 Dt * ln ⎢ ⎢⎣ C sub πDt

⎤ ⎥ = 0 .15 μm ⎥⎦

3.7) This is also a drive-in diffusion. Ignoring heavy doping effects, 2 QT * e − x / 4 Dt , where C ( x, t ) = πDt

cm 2 cm 2 cm 2 − 3 .46 eV / 1273 K + 0 .41 *e * e − 3.46 eV / 1273 K = 9 .0 * 10 −15 D = 0 .037 sec sec sec

3.7a)

⎡ ⎤ QT 4 Dt * ln ⎢ ⎥ = 0 .019 μm π C Dt ⎣ sub ⎦ At the surface of the wafer, x=0 QT C ( 0, t ) = = 1 .9 * 10 21 cm − 3 πDt xJ =

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. QT = 10 18 cm −3 * 10 −7 cm = 10 11 cm −2 D = 0.019

cm 2 − 2.6 eV / kT cm 2 e = 3.7 *10 −13 sec sec

*3600 sec − z 2 / 4*3.7*10 −13 2 2 1011 cm −2 sec e C ( z) = = 1.5 *1015 cm −3 e − z / 0.53 μm 2 cm * 3600 sec π * 3.7 *10 −13 sec To find the depth, set C(z)=1014, then 2 1014 cm −3 = C ( z ) = 1.5 *1015 cm −3 e − z / 0.53 μm cm 2

z 2 = 0.53μm 2 * ln[0.067] z = 1.20μm

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 2 QT = C (0, t ) Dt

π

For intrinsic diffusion, n=ni, so

6

D = 3 .9

cm 2 cm 2 cm 2 cm 2 * e −3.66 eV / kT + 4 .4 * e − 4.0 eV / kT + 44 * e − 4.37 eV / kT = 1 .39 * 10 −14 sec sec sec sec

Then

QT = 4 .6 * 10 15 cm − 2 The drive-in produces a junction depth of xJ = Dt =

⎡ QT 4 Dt * ln ⎢ ⎣ C sub πDt x J2

=

⎤ ⎥ , so ⎦

4 * 10 −8 cm 2 ⎡ 0 .026 cm ⎤ ln ⎢ ⎥ Dt ⎦ ⎣

⎤ ⎡ QT 4 ln ⎢ ⎥ ⎣ C sub πDt ⎦ 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, cm 2 n cm 2 cm 2 cm 2 n D = 0 .066 * e −3.44 eV / kT + 12 * e − 4.05 eV / kT = 1 .6 * 10 −15 + * 1 .1 * 10 −15 ni sec sec ni sec sec

For : x = 0

D = 2 .4 * 10 −14

cm 2 sec 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, Cmax = 1.9 *10 22 cm −3e −0.453 / kT = 3.06 *10 20 cm −3 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.

∫

∫

∫

x x ⎤ ⎡ ⎡ ⎤ Q = C ( x)dx = 10 erfc ⎢ dx = QT − 10 21 erfc ⎢ dx −7 ⎥ −7 ⎥ 4.3 *10 ⎦ 4.3 *10 ⎦ ⎣ ⎣ t t 0 ∞

oc

∞

oc

t oc

21

∫

Using the erfc expansion given in the appendix of this book, 2 ⎡ 10 − 6 ⎤ ⎤ ⎤ t oc ⎡ ⎡ −6 −6 −⎢ ⎥ ⎡ 10 ⎤ 1 ⎢ x 10 ⎡ ⎤ 4.3*10 − 7 ⎥⎦ ⎥ ⎥ Q = QT − 10 21 erfc ⎢ dx = QT − 10 21 ⎢⎢ erfc ⎢ 1 − e ⎢⎣ + −7 ⎥ −7 −7 ⎥ ⎥⎥ 4.3 *10 π ⎢ ⎣ 4.3 *10 ⎦ ⎣ 4.3 *10 ⎦ 0 ⎢⎣ ⎥⎦ ⎥⎦ ⎢⎣ 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 2 t ox + At ox = B(t + τ ) t=

t ox + At ox −τ B 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 2 t ox + At ox (0.1μm) 2 + 0.1μm * .226μm = = 0.114hr = 6.8 min t= B .287 μm 2 / hr This solution is closer to linear than parabolic, but is still in the transition regime. 4.3) For the first oxidation, (0.05μm) 2 + 0.165μm * .05μm t= − 0.37 hr = 0.55hr = 33 min .0117 μm 2 / hr For the second oxidation, (0.05μm) 2 + 0.165μm * .05μm − (0.55hr + 0.37 hr ) = 1.34hr t= .0117 μm 2 / hr 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) t (hr) B (μm2/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, (0.003μm) 2 + 0.5μm * .003μm t= = 7.43 *10 −3 hr = 26.8 sec 2 .203μm / hr 4.5b) Since B is proportional to Pg and t is inversely proportional to B, 640torr t = 26.8 sec* = 225 sec 76torr 4.6a) Since there is no rapid growth regime, from Eq. 4.10, Hk s Pg Hk s Pg dt ox = ≈ dt N1 [1 + k s / h + k s t ox / D ] N1 [1 + k s t ox / D ] 9

4.6b) For short oxidation times, tox~(B/A)(t+τ), so dt ox B = = Ke − E A / kT dt A where K is some constant. From the problem, and taking a ratio to eliminate K, 1.0nm / sec e − E A / 1273k = =2 0.5nm / sec e − E A / 1173k Solving, EA=0.89 eV, and K=3320 nm/sec, so dt ox = (3320nm / sec)e − 0.89eV / 1073k = 0.22nm / sec dt

4.7a) This is clearly in the linear regime so B 0.287 1 t ox = t = = 0.042μm = 42nm A 0.226 30 4.7b) If τ is not zero, t 0.06 (t + τ ) = ox = = 0.047 hr = 2.83 min B / A 0.287 / 0.226 Then τ is 0.83 min. 4.8) Ignoring rapid growth 2 t ox + At ox 0.04 2 + 0.165 * 0.04 =t = = 0.7hr 0.0117 B Including rapid growth 2 t ox + At ox 0.04 2 + 0.165 * 0.04 −τ = t = − 0.37 = 0.33hr 0.0117 B 4.9) In this case we are diluting the oxygen and so changing B (see Eq. 4.12). Then 2 t ox + At ox .002 2 + 0.165 * 0.002 =B= = 2.0 *10 −3 μm 2 / hr t 10 / 60 The pressure is proportional to the ratio of the B parameters 0.002 = 0.17atm P = 1atm 0.0117 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 2 t ox + At ox = B(t + τ ) t ox =

− A + A 2 + 4 B (t + τ ) − 0.09 + 0.09 2 + 4 * 0.027(1.076) = = 0.131μm = 131nm 2 2

10

4.11) From Eq. 4.12,

AO − ≈ 2

2 DHPg 2 D AO2 = ≈ 2 BO ; BO − = 2 2 RS N1 5

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, (0.1μm) 2 + 0.1μm * .033μm t= − τ = 0.57 hr − τ .0234 μm 2 / hr 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 (0.01μm) 2 + 0.01μm * .165μm t= = 1.5hr = 90 min .00117 μm 2 / hr 4.14) From Equation 4.16, when C2~0, dt ox B = + C 2 e −tox / L2 dt A Then tox t dt ∫−τ dt = ∫0 B + Coxe −tox / L2 2 A ⎡ B + C2 e −tox / L2 ⎤ A A ⎥ t + τ = tox + L2 ln ⎢ A B B ⎢ B + C2 ⎥ A ⎦ ⎣ − t L / 2 ⎤ ox ⎤ ⎡ B A ⎢ ⎡ A + C2 e −tox / L2 ⎥ ⎥ ⎢ τ ln ln t + = L2 − e ⎥ B ⎢ ⎢ B + C2 ⎥ A ⎦ ⎦ ⎣ ⎣ ⎡ B etox / L2 + C2 ⎤ ⎡ B + C2 e −tox / L2 ⎤ A A A ⎥ ⎥ ⎢ = L2 ln ⎢ A t + τ = L2 ln B ⎢ B + C2 ⎥ ⎢ B + C2 e −tox / L2 ⎥ B A ⎦ ⎣ ⎦ ⎣ A 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

e

B t +τ A L2

⎡ B e tox / L2 + C 2 ⎤ ⎥ =⎢ A ⎢ B + C2 ⎥ A ⎣ ⎦

[B A + C ]e

[

]

B t +τ A L2

− C 2 = B e tox / L2 A

2

B t +τ

⎡1 + AC 2 + C ⎤ e A L2 − AC 2 = e tox / L2 2⎥ B ⎢⎣ ⎦ B B ( t +τ ) ⎡ AC ⎡ − ⎤⎤ B AL2 2 ⎢ t ox = (t + τ ) + L2 ln 1 + ⎢1 − e ⎥⎥ A B ⎣⎢ ⎢⎣ ⎦⎥ ⎥⎦ The equation for τ can be readily found from the second equation by setting tox=t0 at t=0 ⎡ B + C2 e −t0 / L2 ⎤ A ⎡ B et0 / L2 + C2 ⎤ A A A ⎢ ⎥ ⎥ = L2 ln ⎢ A τ = t0 + L2 ln B B ⎢ B + C2 ⎥ B ⎢ B + C2 ⎥ A A ⎣ ⎣ ⎦ ⎦

4.15) From Eq. 4.17, and using 5 nm/min = 0.3 μm/hr, ⎡ 0.0117 ⎡ B e tox / L2 + C 2 ⎤ 0.165 e100 / 70 + 0.007 ⎤ A 0 . 165 A ⎥ ⎥= * 0.007 ln ⎢ t + τ = L2 ln ⎢ 0.0117 B ⎢ ⎥ ⎢ B + C 2 ⎥ 0.0117 + 0.3 A ⎣ ⎦ 0.165 ⎣ ⎦ 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, ⎡ 0.00117 e100 / 70 + 0.007 ⎤ 0.165 0 . 165 ⎢ ⎥ = 0.07 hr = 4.2 min t +τ = * 0.007 ln 0.00117 0.00117 ⎢ ⎥ + 0.3 0.165 ⎣ ⎦

[

]

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

Q t 1 QMI t ox ΔVT 1 = − = MI ox ε SiO2 t ox 2 2ε SiO2 2

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 Q t QMI 2.5 *10 −6 cm ΔVT 2 = − MI ox = −0.015V = − − 3.9 * 8.84 *10 −14 F / cm ε SiO2 QMI = 2.3 *10 −9 Coul / cm 2 , so N = 1.4 *1010 Ions / cm 2

For the first case, one gets the same threshold shift so the density of ions must be twice as much.

13

5.1a) From Fig 5.9, RP=100 nm and ΔRP=38 nm. The dose is 1012 cm-2. 5.1b) At the peak concentration the depth, x=RP so N ( x = RP ) =

φ = 1.05 *1017 cm −3 2π ΔRP

5.1c) At x=300 nm, 2 2 N ( x = 300nm) = 1.05 *1017 cm −3 e − ( 300 nm−100 nm ) / 2*(38 nm ) = 1.0 *1011 cm −3 This is essentially zero for every practical device. 5.1d) A channeling tail exists, increasing the concentration for the deep portion of the profile. 5.2) From Fig 5.9, if RP=300 nm, the energy is about 90 keV. Then ΔRP=68 nm. The dose can be found from the peak concentration as N ( x = RP ) = 1017 cm −3 =

φ , then 2π ΔRP

φ = 2π * 6.8 *10 −6 cm *1017 cm −3 = 1.7 *1012 cm −2

If the bulk is 1015cm-3,

⎡1015 ⎤ − ( x − RP ) 2 ln ⎢ 17 ⎥ = 2 2ΔRP ⎣10 ⎦ Solving, x = RP +/- 206 nm = 94 nm or 506 nm.

5.3a) From Fig 5.9, RP=180 nm and ΔRP=50 nm. At the peak concentration the depth, x=RP so N ( x = RP ) =

φ = 8 *1019 cm −3 2π ΔRP

5.3b) To find the position, 2 2 N ( x) = 1016 cm −3 = 8 *1019 cm −3 e − ( x −180 nm ) / 2*(50 nm )

⎡1016 cm −3 ⎤ = −( x − 180nm) 2 / 2 * (50nm) 2 ln ⎢ 19 ⎥ ⎣ 8 *10 ⎦ Solving x=392 nm = 0.39 μm.

5.4a) From Fig 5.9, RP=340 nm and ΔRP=72 nm for boron and RP=60 nm and ΔRP=12 nm for arsenic. If we set the two concentrations equal, and recognizing that the doses are equal, ⎡ ⎡ 2 ⎤ 2 ⎤ 2 2 φ φ =⎢ e − ( x − RP ) / 2 ΔRP ⎥ e − ( x − RP ) / 2 ΔRP ⎥ ⎢ ⎣⎢ 2π ΔR P ⎦⎥ Boron ⎣⎢ 2π ΔR P ⎦⎥ Arsenic

14

ΔRPB

ΔRPAs

=

e − ( x − RPB )

e −( x − RPAs )

( x − RPB ) 2 2ΔRPB

2

2

/ 2 ΔRPB 2

2

/ 2 ΔRPAs 2

⎡ ΔRPB ⎤ ( x − RAs ) 2 − = ln ⎢ ⎥ 2 2ΔRPAs ⎣⎢ ΔRPAs ⎦⎥

(ΔRPAs − ΔRPB ) * x 2 − 2 x( RPB ΔRPAs − RPAs ΔRPB ) + ΔRPAs * RPB − ΔRPB * RPAs 2

2

2

2

2

2

2

2

⎡ ΔRPB ⎤ 2 2 = 2ΔRPB ΔRPAs ln ⎢ ⎥ ⎣⎢ ΔRPAs ⎦⎥

5040nm 2 * x 2 − 524000nm3 * x + 4691000nm 4 = 0 Solving

524000 + / − 524000 2 − 4 * 5040 * 4691000 = 9.9nm,94nm x= 10080 The other root is negative and so is not physical. 5.5a) To get an implant peak at 0.2 μm requires an energy of about 60keV 5.5b) At 60 keV the maximum concentration is

φ 2π ΔRP

N ( x = RP ) =

φ = 2π * N ( x = RP ) * ΔRP = 2.5 *1012 cm −2 5.6) From Fig 5.9, RP=100 nm and ΔRP=30 nm. Since most of the implant will be in silicon, we will take this to be the implant parameters for the entire profile. Then

∫

φ e −( x − R 2π ΔRP

15

QSiO2 QT QSiO2 QT QSiO2 QT QSiO2 QT

= =

0

1

π

∫

P)

2

/ 2 ΔRP 2

φ

(15− RP ) / 2ΔRP

e −u du

− RP / 2ΔRP

∫

2

∫

− RP / 2ΔRP ⎡(15− RP ) / 2ΔR2P ⎤ 2 1 ⎢ −u = e du − e −u du ⎥ ⎥ π ⎢ 0 0 ⎣ ⎦ 1 ⎡ ⎡15 − 100 ⎤ ⎡ − 100 ⎤ ⎤ = ⎢erc ⎢ ⎥ ⎥ ≈ 0.19% ⎥ − erc ⎢ π ⎣ ⎣ 2 * 30 ⎦ ⎣ 2 * 30 ⎦ ⎦

15

5.7)

kg 2 * 28amu *1.67 *10 −27 * 20000V 2mVext m amu = = 0.36T B= V = φ qr 1.6 *10 −19 coul * (0.3m) 2 where T is the MKS unit Tesla. Since N2 has a mass that is twice that of atomic nitrogen (14 amu), it would be hard to distinguish it from silicon which is also 28 amu. Of course, the vapor pressure of silicon is very low so one does not expect to see high concentrations of mass 28, even if one uses silane (SiH4) as a gaseous source. 5.8a) From Fig 5.9, RP=75 nm and ΔRP=40 nm. The concentration is a maximum at x=75 nm. 5.8b) At that position, N ( x = RP ) =

φ = 1018 cm −3 2π ΔRP

5.8c) Since the peak position is in the GaAs, most of the implant will be in the GaAs. One can picture the profile as a conventional Gaussian, where the peak is at 75 nm, while the AlGaAs only extends 50 nm 5.9a) From the problem, RP=200 nm and ΔRP=50 nm. From Eq. 3.1, J = -D dC/dx = -D dN/dx At x=RP, dC/dx=0, so there is no net flux. There is an equal flux to the left and the right. 5.9b) At some other position, −4 −4 x − RP ⎡ 1015 cm −2 dC − ( 0.3− 0.2 ) 2 / 2*0.52 ) ⎤ 0.2 *10 cm − 0.3 *10 cm * = = − N ( x) * e ⎥ ⎢ −6 dx (0.05 *10 −4 cm) 2 ΔRP2 ⎦ ⎣ 2π * 5 *10 cm

= −4.3 *10 24 cm −4 The flux moves in the direction of decreasing concentration (i.e. away from the peak). In this case it is away from the surface of the wafer. The concentration (bracket term) is 1.1*1019 cm-3, compared to an intrinsic carrier concentration, ni=1019 cm-3. Following the calculations given in Example 3.1, n=1.6*1019 cm-3. Using Table 3.2,

cm 2 D = 44 sec

⎡n⎤ cm 2 − 4 .37 eV / kT + 4 .4 ⎢ ⎥ *e sec ⎣ ni ⎦ 2

⎡ n ⎤ − 4.0 eV / kT cm 2 + 3 .9 * e − 3.66 eV / kT ⎢ ⎥*e sec ⎣ ni ⎦

cm 2 ⎡1 .6 ⎤ cm 2 ⎡1 .6 ⎤ − 4.0 eV / kT cm 2 − 4 .37 eV / kT + 4 .4 + 3 .9 D = 44 *e *e * e − 3.66 eV / kT ⎢ ⎥ ⎢ ⎥ sec ⎣1 .0 ⎦ sec ⎣1 .0 ⎦ sec cm 2 cm 2 cm 2 cm 2 + 1 .04 * 10 −15 + 1 .28 * 10 −14 = 1 .44 * 10 −14 D = 5 .7 * 10 −16 sec sec sec sec 10 -2 -1 Then J=6.18*10 cm sec 2

5.10a) From the problem, RP=100 nm and ΔRP=30 nm.

16

2 2 ⎤ ⎡ 4 *1014 cm −2 e −( 0.13−0.10 ) / 2*0.03 ) ⎥ = 3.2 *1019 cm −3 N =⎢ −6 ⎦ ⎣ 2π * 3 *10 cm 18 5.10b) The intrinsic carrier concentration, ni is 7*10 cm-3, so n~N. Using Table 3.2, cm 2 cm 2 n cm 2 cm 2 + 2 .2 * 10 −16 D = 0 .066 * e −3.44 eV / kT + 12 * e − 4.05 eV / kT = 1 .1 * 10 −16 sec sec ni sec sec

D = 3 .3 * 10 −16

cm 2 sec 5.10c) From Eq. 3.1, J = -D dC/dx = -D dN/dx . 2 x − RP ⎡ 0.13 *10 −4 cm − 0.10 *10 −4 cm 19 −16 cm ⎤ −3 3 . 3 * 10 * 3 . 2 * 10 * = cm J = D * N ( x) * ⎢ sec ⎥⎦ (0.03 *10 −4 cm) 2 ΔRP2 ⎣

[

]

= 3.5 *10 9 cm −2 sec −1 The flux moves in the direction of decreasing concentration (i.e. away from the peak). In this case it is away from the surface of the wafer. 5.11) Since

φ * Area =

I ion * time, q

φ * Area * q

1018 cm −2 * π * (7.5cm) 2 *1.6 *10 −19 Coul = = 1.4 *10 4 sec time = −3 I ion 2 *10 Coul / sec This corresponds to just under four wafers per hour. 5.12) ke increases sharply as zi increases. Se is therefore dominant for implanting light species like Boron. 5.13) Referring back to Fig. 5.9, both As and Sb can be implanted with shallow (1.0 mm. Also assume that εlamp and εwafer are not functions of wavelength. Then the fraction of light absorbed is

εk

F=

∫ λ (e

1μm

∫ λ (e 5

0 ∞

k

5

C1

C2 / λT

C1

C 2 / λT

0

− 1)

− 1)

dλ

dλ

=

ε

∫ λ (e

∫ λ (e

1μm

5

0

C

1 C2 / λT

σT 4

where k is the geometry (view) factor. For T=2000K, F=

εC1

1μm

W 9.07 *10 2 m 5

F = ε * 4.09 *10

− 22

5

0

1

C 2 / λT

∫ λ (e

1μm

m

4

5

0

− 1) 1

C 2 / λT

− 1)

dλ

dλ

− 1)

dλ

Over the range of interest, the factor of one in the denominator can be neglected. Then the integral can be evaluated as F = ε * 4.09 *10

let.x = 1 / λ

F = ε * 409 μm

− 22

∫

1μm

m

4

0

∫x e ∞

4

3 −7.2 x

e −C 2 / λT

λ5

dλ

dx

⎡ ⎤⎤ ⎡ x3 x2 x F = 409 μm * 3!*⎢e −7.2 x ⎢ − − − ...⎥ ⎥ = 5.7% 3 2 ⎦⎦ ⎣ 3!*(−7.2) 2!*(−7.2) 1!*(−7.2) ⎣ Therefore 94.3% is has wavelengths above 1 micron. 1

The short wavelength radiation can either be reflected or absorbed. Assuming an absorptivity of 0.7, the amount absorbed is 3.9%. The amount reflected is 1.7%. The long wavelength light can be reflected or transmitted. The amount transmitted is 0.7*94.3%=66%. The amount reflected is 28%. So, the total reflected is 30%, transmitted is 66%, and absorbed is ~4%. 6.2) If the absorptivity were constant and the heat capacity temperature independent, the temperature would increase linearly with time. If the wafer is nearly transparent at low temperature, however, the heating rate would be slow initially. As the temperature increased, the absorptivity would increase and so the heating rate would rise until the wafer is nearly opaque to the radiation.

19

6.3) Slip occurs because the thermoplastic stress exceeds the yield strength. (Actually one has to resolve the stress into its components along the crystal planes since the wafer yields along these directions.) Slip lines nucleate at the edges of the wafer because stress is largest there and because it is easiest to nucleate slip at the edge of the wafer. If temperature ramping causes slip, the radiation pattern may have been fixed in such a way as to make the temperature uniform when the wafer temperature is fixed. Since ramping requires a different heating pattern, the wafer will be stressed during the ramp. 6.4) Many software packages are not set up to model transient annealing effects, or they handle it in a very empirical manner. Due to the short anneal times in RTP, transient effects are likely to be the dominant factor in the anneal redistribution. 6.5a) We know that P = εσT 4 A A = 2π (10cm) 2 = 200πcm 2 , so

P = 0.7 * 200πcm 2 * 5.67 *10 −12

W * (1223K ) 4 = 5.6kW 4 cm − K 2

where the two in the second equation is used to take into account both surfaces of the wafer. 6.5b) To ramp at 100 C/sec, o dT gm J C P = Vol * Cp * ρ * = 100πcm 2 * 0.07cm * 2.33 3 * 0.7 * 100 o dt sec cm gm − C P = 3.58kW This is in addition to the power being radiated. 6.6) Little work was done on wet RTO because RTO is intended to grow gate oxides and silicon oxidizes more rapidly in a wet ambient. Furthermore, it has been generally assumed that wet oxides are lower quality since they incorporate hydroxyls (OH groups) leading to a lower bridging density. It has been shown that wet RTO can actually improve oxide integrity in some cases. This is due to the ability of water vapor to volatize any nitride residue left on the edges of the field regions during a LOCOS process (see chapter 15). If the LOCOS is done properly, however, this is not a problem and thin wet oxides are not generally desired. 6.7) Multizone heating is used to provide a more uniform temperature across the wafer. It allows one to compensate for the fact that different process conditions require different radiant profiles. It is particularly essential for large diameter wafers where temperature differences can be large enough to create slip.

20

7.1) According to Eq. 6.5,

0.2898cm − K T 2.898 *10 −6 nm − K = 14490 K T= 200nm The internal energy of the gas is 3/2*kT*N, where N is the number of electrons (which is the same as the number of ions) in the discharge. atoms 0.1L N = 6.02 *1023 = 2.64 *1021 electrons L mole 22.8 mole Then the energy is 791 J. 7.1 (revised) 2.898 *10−6 nm − K = 9660 K T= 300nm The internal energy of the gas is 3/2*kT*N, where N is the number of electrons (which is the same as the number of ions) in the discharge. . atoms 0.1L N = 6.02 *1023 *10−4 = 2.64 *1017 atoms mole 22.8 L mole Then the energy of the electrons is 53 mJ. Most of the energy in the tube is in the heated compressed gas, even though the energy of the individual atoms and ions is small.

λmax =

⎡ ⎡ 2πxW ⎤ ⎤ ⎡ ⎡ 2πyL ⎤ ⎤ 2 ⎢ sin ⎢ ⎥ ⎥ ⎢ sin ⎢ ⎥⎥ ⎡ 2W * 2 L ⎤ ⎢ ⎣ λ * g ⎦ ⎥ ⎢ ⎣ λ * g ⎦ ⎥ 7.2) I ( x, y ) = ε o (0) ⎢ ⎥ ⎣ λ * g ⎦ ⎢ ⎡ 2πxW ⎤ ⎥ ⎢ ⎡ 2πyL ⎤ ⎥ ⎢ ⎢ ⎥ ⎥ ⎢ ⎢ ⎥ ⎥ ⎣ ⎣ λ * g ⎦ ⎦ ⎣ ⎣λ *g ⎦ ⎦ When x λ g 2 + r 2 In this case, the lines and spaces are 1.0 μm wide so for the central line, r=0.5 μm. Since rλg, or W>>( λg)1/2=3.3 μm. Thus the Fresnel criterion is not met. The minimum feature size depends on the value of k that is assumed. For k~1, the criteria are the same. 7.7) For an i-line source (365 nm), W>>3.02 μm. For an ArF laser, the criteria changes to 2.2 μm. Thus the criteria is not even met for the laser source. 7.8) For an i-line source,

νo =

NA = 1.8μm −1 0.61* λ According to Fig. 7.22, for S=0.5 and MTF=0.3, the normalized spatial frequency is 0.59νo = 1.06 μm-1. Then Γ=1/ν=0.94 μm. The minimum feature size is Γ /2=0.47 μm.

23

8.1) From Eq. 8.6,

CMTF =

νo (μm-1) Wmin(μm) 0.7 0.93 2.64 1.11 248 3.4 0.33 2.10 0.43 313 3.6 0.31 1.80 0.47 365 3.6 0.31 1.50 0.56 436 For NA=0.4, one calculates νo as shown above. Then for S=0.5, the normalized spatial frequencies corresponding to the CMTF’s are found as given in the table. Finally the minimum feature sizes are determined. Note that, as expected, as the wavelength is reduced from g-line, the minimum feature size shrinks until 248 nm, when the loss in resist contrast leads to poor resolution. Obviously, one needs a better resist at this wavelength.

Then

γ

101 / γ − 1 101 / γ + 1

8.2a)

8.2b)

8.2c) Since

CMTF

γ=

1 1 = = 3.05 ⎡ D100 ⎤ ⎡ 85 ⎤ log ⎢ ⎥ log ⎢⎣ 40 ⎥⎦ ⎣ D0 ⎦

101 / γ − 1 = 0.36 CMTF = 1 / γ 10 + 1

γ=

1 β + αTR One can solve for the parameters. With a bit of math, α=0.28 μm-1 and β=0.16. Then the highest possible contrast corresponds to the resist thickness going to zero. In that case the contrast is just the inverse of beta or 6.3. 8.3)The contrast decreases as short wavelength because the photon energy is sufficient to drive other reactions than the PAC decomposition. In particular, for DQN, a great deal of the energy is absorbed by the matrix material at short wavelength. 8.4)

From the second equation,

D100 − D0 , and D100 + D0 1 γ= ⎡D ⎤ log ⎢ 100 ⎥ ⎣ D0 ⎦

CMTF =

D100 = D0 *101 / γ

24

Inserting this into the first equation,

CMTF =

8.5a)

Then

γ=

D0 *101 / γ − D0 101 / γ − 1 = D0 *101 / γ + D0 101 / γ + 1

D 1 , so 100 = 101 / γ D0 ⎡D ⎤ log ⎢ 100 ⎥ ⎣ D0 ⎦

D100 = D0 101 / γ = 10

mJ 1 / 7 mJ 10 = 13.9 2 2 cm cm 8.5b) Lower dose exposures tend to give more rounded features – See Fig. 8.8 for an example.

8.6) Bleaching is highly desirable because it allows a more complete and uniform exposure of the resist. If there is no bleaching, the exposure is much higher in the top of the resist than the bottom since the intensity falls off as e-αz where α is the inverse absorption length. 8.7) If the lower level material is spun on like photoresist, and if the film thickness is at least as large as the step height, the lower level will significantly planarize the features, at least locally. This leads to a more uniform imaging (upper) resist layer and therefore better pattern transfer. If the surface of the wafer is reflective, the planarizing layer could also be made of an absorbing material to prevent surface reflections or a thin additional antireflective layer can be added to the stack. 8.8) For γ=3.5,

101 / γ − 1 101 / 3.5 − 1 = = 0.32 101 / γ + 1 101 / 3.5 + 1 Referring to Fig 7.22, and interpolating between the S=1 and S=0.5 lines, ν/νo = 0.58. Then NA pairs 0.6 νo = = = 3.97 μm 0.61λ 0.61 * 0.248μm Then ν=2.30 line pairs per micron, and Wmin=0.22 μm. 8.9) In this case, For γ=3.0,

CMTF =

MTF =

I max − I min 55 − 20 = 0.47 = I max + I min 55 + 20

101 / γ − 1 101 / 3 − 1 = = 0.37 CMTF = 1 / γ 10 + 1 101 / 3 + 1 Since MTF>CMTF, the image can be resolved. The smallest γ will occur when CMTF=MTF. Then 25

101 / γ − 1 CMTF = 1 / γ 10 + 1 1/ γ 0.47 * (10 + 1) = 101 / γ − 1 1.47 = 101 / γ 0.53 2.26 = γ 8.10) If Wmin=0.8 μm, the spatial frequency, given by Eq. 7.21, is ν=0.63 line pairs per micron. For this machine, NA pairs 0.4 νo = = = 1.50 μm 0.61λ 0.61* 0.436μm The ν / νo = 0.42. Referring to Fig 7.22, for S=0.5, CMTF=0.6. Solving 101 / γ − 1 0.6 = 1 / γ 10 + 1 0.6 * (101 / γ + 1) = 101 / γ − 1 1.6 = 101 / γ 0.4 1.7 = γ

8.11)

νo =

pairs NA 0.5 = = 2.2 μm 0.61λ 0.61 * 0.365μm

101 / γ − 1 101 / 4.1 − 1 = = 0.27 101 / γ + 1 10 4.1 + 1 Referring to Fig 7.22, ν/νo = 0.59, so ν=1.3 μm−1 and Wmin=0.39 μm. Then 0.365 λ σ= = = 1.46μm 2 NA 0.5 2

CMTF =

8.12a) From the figure,

MTF =

I max − I min 120 − 20 = = 0.71 I max + I min 120 + 20

8.12b) Intensity (mW/cm2) Dose (mJ/cm2) Position (μm) 0.0 20 20 0.1 55 55 0.2 120 120 8.12c) This will produce a pattern that looks like:

TR/To 0.55 0.2 0.0

26

x=-1/2

x=0

x=+1/2

8.13a) The MTF is (100-20)/100+20)=0.67. 8.13b) The line ends at the point where the dose is 80 mJ/cm2 or D100. For a one second exposure, this corresponds to the point where I is 80 mW/cm2 or +/- 1 μm. Thus the line width is 2 μm. For a 2 second exposure, the required intensity is 40 mW/cm2 or +/- 0.52 μm. (LW=1.04 μm). For a 3 second exposure, the required intensity is 27 mW/cm2 or +/- 0.46 μm. (LW=0.92 μm). For a 4 second exposure, the required intensity is 25 mW/cm2 or +/- 0.45 μm. (LW=0.90 μm). 8.14a) See 8.10, CMTF=0.6. 8.14b) For 0.5 μm lines, ν=1 line pairs per micron, so ν/νο=0.59. From Fig 7.22, CMTF~0.36. Then 101/γ=1.36/0.64=2.1, so γ=3.1. 8.15)

⎡ ⎡ 2πx ⎤ ⎤ 2 ⎢ sin ⎢ ⎥⎥ ⎤ ⎢ ⎣ 4.36 μm ⎦ ⎥ mW ⎡ 2 I ( x, y ) = 100 2 ⎢ cm ⎣ 0.436μm * 10μm ⎥⎦ ⎢ ⎡ 2πx ⎤ ⎥ ⎢ ⎢ ⎥ ⎥ ⎣ ⎣ 4.36μm ⎦ ⎦ ⎡ ⎡ 2πx ⎤ ⎤ ⎢ sin ⎢ ⎥⎥ mW ⎢ ⎣ 4.36μm ⎦ ⎥ I ( x, y ) = 21 2 cm ⎢ ⎡ 2πx ⎤ ⎥ ⎢ ⎢ ⎥ ⎥ ⎣ ⎣ 4.36μm ⎦ ⎦

2

2

This intensity can be plotted using Excel or any commercially available graphing program. The resist remaining is given by ⎡ Dose ⎤ TR = 1.0 μm − γ * log ⎢ ⎥ ⎣ Do ⎦ for values of dose such that 0