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

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=