Problem Set 9 Solutions 1. Mobility in extrinsic semiconductors is affected by phonon scattering and impurity scattering. Thoroughly explain the mobility plots for the following figures from your textbook and the Course Notes provided to you titled MSE 410-ECE340 Ch5 Ext SC Part 2-Phonons & Dark Matter Det.pdf. a. Figure 5.18 Figure 5.18 shows the temperature dependence on mobility, μ. As temperature decreases, μ will decrease with doping concentration because the impurity concentration increases. Ionized impurities will scatter carriers due to Coulomb interaction. At high impurity concentrations, μi goes by a power law dependence of T 3/2 for heavily doped semiconductors. At high temperatures (> 300K), the mobility T 1.5 10 70 100 800 Temperature (K) decreases primarily because of carrier scattering due to electron-phonon interaction. A secondary effect that is observed at higher temperatures is scattering by ionized dopant impurities. The phonon-limited mobility, μp, goes by a power law dependence of T -3/2. Electron Drift Mobility(cm 2 V -1 s -1 ) 50000 10000 1000 100 N d =10 14 N d =10 16 N d =10 17 N d =10 18 N d =10 19 Ge N d =10 13 L T 1.5 Fig. 5.18: Log-log plot of drift mobility vs temperature for n-type Ge and n-type Si samples. Various donor concentrations for Si are shown. N d are in cm -3. The upper right inset is the simple theory for lattice limited mobility whereas the lower left inset is the simple theory for impurity scattering limited mobility. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 2002) http://materials.usask.ca Si b. Figure 5.19 Figure 5.19 demonstrates the mobility 1000 in Si at 300K as a function of doping for both electrons and holes. The mobility Holes decreases with increasing dopant 100 concentration because, as we pointed 50 out earlier, dopant impurities will cause Dopant Concentration, cm -3 scattering of carriers. Note that electrons have higher mobility than http://materials.usask.ca holes. This can be explained by examining the relationship between mobility and mass, m, given by: e m Drift Mobility(cm 2 V -1 s -1 ) 2000 Electrons 10 15 10 16 10 17 10 18 10 19 10 20 Fig. 5.19: The variation of the drift mobility with dopant concentration in Si for electrons and holes From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 2002) 1
where e is the fundamental charge and τ is the time between scattering events. Note that mobility is inversely proportional to mass. The mass of electrons in Si are lighter than that of holes hence the greater mobility of electrons, thus the mobility of electrons will be greater. 2. Phonons: In the Course Notes provided to you titled MSE 410-ECE340 Ch5 Ext SC Part 2-Phonons & Dark Matter Det.pdf, specifically covering phonons, find the dispersion relations of phonons in a 1D periodic chain of atoms with alternating masses (i.e., diatomic). Sketch these dispersion relations of phonons and label the phonon modes (types of phonons). Furthermore, describe why the optical phonon mode is higher energy than the acoustical mode. Hint: the answers are all in the notes on phonons. 1 2 1 1 2c m 1 m 2 2c m 2 2c m 1 a a Whether a vibrational mode is longitudinal or transverse, a given atomic amplitude of motion requires much more energy for a long wave optical than acoustic mode. Why? Because the optical modes minimize changes in 2nd nearest neighbor separation by maximizing separation between nearest neighbors. Nearest neighbor interaction is greater than any other higher order nearest neighbor interaction, therefore, the energy required for nearest neighbor separation is greater than any other separation. The figure and figure caption below helps visualize this. 2
3. Below are phonon dispersion relations for crystalline Si and two other Group IV crystalline semiconductors of unknown identity. 3
a. Calculate the gap (in both linear frequency and energy) at the L-point between the LA and TO modes for each dispersion relation of each material both in linear frequency (in Hz or THz) and energy (in mev). 4
Using the mathematical relationship between energy and frequency, in which Planck s constant is the proportionality constant, we can calculate the energy. E = hv. Calculations were performed using Mathematica: b. Calculate the energy (in mev) at the point where the LO and TO modes meet at the Γ-point for each dispersion relation of each material. c. Using part a and b, identify the two materials of the unidentified dispersion relations. Explain and defend your choices using part a and b and include a plot that includes your numerical answers to the calculations of part a and b and a physical parameter of the Group IV semiconductors (e.g., mass, density, AMU, atomic number, melting point, lattice constant, etc.) and a fitting of the data (your choice 5
of equation to fit the data) to support your argument. The plot and fit should reveal trends that you can then use to defend your argument. From parts a and b, Table 1 was created that lists the frequency and energy values for (1) gap at the L-point between the LA and TO modes for each dispersion relation of each material and (2) the point at which the LO and TO modes meet at the Γ-point for each dispersion relation of each Group IV semiconductor. Listed also are the identities of the materials based on this data. More on that later. Table 1: Frequency and Energies of certain aspects of the dispersion relations for three Group IV semiconductors. IV Semiconductors (1) Freq & Energy at Γ-point for LO & TO modes (THz/meV) (2) Freq. Gap & Energy Gap between LA-TO modes at L-Point (THz /mev) AMU (g/mole) Diamond 6/24.8 40/165.4 12.011 Silicon 1/4.14 15.5/64.1 28.0855 Germanium 0.75/3.1 9.25/38.3 72.59 Tin 0.75/3.1 9.25/38.3 118.69 The fact that the dispersion relations are nearly identical except for their frequency ranges indicates that the crystal structure and the basis of the Group IV semiconductors are the same. From Table 1, note that the magnitude of the frequency and energy values scale the same for both (1) and (2) relative to the material (i.e., semiconductor). The trends in the frequency and energy dependence per dispersion relation suggests a difference in mass in the Group IV semiconductors in the following manner. Because phonons are atom vibrations, we can enlist a simple ball-spring model to envision the oscillatory motion of the atoms. We recall from 1 st semester physics that the frequency of a ball-spring harmonic oscillator is proportional to the inverse of the square root of mass and the proportionality constant is the square root of the spring constant (spring constant, K, from Hooke s law). That is: K v [1] m where m is the mass. Because energy and frequency are directly proportional with Planck s constant as the proportionality constant, the 6
energy of the ball-spring harmonic oscillator will similarly scale and the relation is given by: K E hv. [2] m or 1 E C m [3] where h is Planck s constant and C is just a constant that contains the spring constant and Planck s constant. Hence, the lower the frequency or energy, the greater the mass. The fact that frequency and energy should vary inversely to the square root of mass, the likelihood that the crystal structure and basis are the same because the dispersion relations are so similar, and that the dispersion relations of the unknowns show that one dispersion relation is on average higher in frequency than Si and the other dispersion relation is on average lower than Si suggests the following. The identity of the Group Semiconductor with a dispersion relation that is on average higher than Si is diamond as it has a smaller mass. The identity of the Group Semiconductor with a dispersion relation that is on average higher than Si is either Ge or Sn because their masses are heavier than Si as listed accordingly in Table 1. Note that the mass for each Group IV semiconductor in Table 1 is given by the AMU value, which is the mass normalized to one mole of the semiconductor. To further examine whether the energy should vary inversely to the square root of mass, one can plot the energy of either the (1) energy gap at Γ-point LO & TO or (2) energy of the LA-TO modes at L-point and fit the data with equation [3] with either the mass of Ge or the mass of Sn. If the fit follows the data sufficiently with either Ge or Sn, such an outcome would further support the reported identity of the Group IV semiconductors as shown in Table 1. Figure 1 demonstrates that the fit follows the data quite well. 7
Figure 1: The energy of the TO- LO modes at the Γ- Point for two unknowns and Si with the masses of diamond, Si and Sn Group IV semiconductors as a function of AMU in which the fit shows an inverse square root dependence on the AMU (i.e., mass). Goodness of Fit Statistics for Figure 1: Statistics of Fit R 2 0.946886 Adjusted R 2 0.920329 The statistics of the fit shows that the goodness of fit is better than 0.9 for both R 2 and the adjusted R 2, which indicates that the fit to equation [3] describes the data sufficiently well. 8
Figure 2: The energy of the TO- LO modes at the Γ- Point for two unknowns and Si with the masses of diamond, Si and Sn Group IV semiconductors as a function of AMU in which the fit shows an inverse square root dependence on the AMU (i.e., mass). Goodness of Fit Statistics for Figure 2: We see that the goodness of fit is slightly better when using the mass of Sn rather than the mass of Ge which suggests that the unknown Group IV semiconductor is Sn. Alpha-Sn is indeed a semiconductor. Another way to determine whether or not the unknown Group IV semiconductor is Sn is to examine the dispersion relation for alpha-sn which is shown below. If one examines the frequency at which the TO- LO modes intersect at the Γ- Point, the frequency is 6 THz which is over 3 THz lower than the TO- LO modes Γ- Point frequency in the third unknown dispersion relation. Obviously, the third unknown is not Sn and must be lighter than Sn but heavier than Si which leaves Ge as the unknown Group IV semiconductor. So the unknown Group IV 9
semiconductors are diamond (Eg is a bit too large to be a semiconductor, but we shall ignore that in this case) and Ge. To see how well equation [3] describes the data when Sn is included from its dispersion relation, we can then plot the energy of the TO- LO modes at the Γ- Point for Sn along with those of Si and diamond and their respective masses (Figure 4). We see that the data is represented by the fit quite well. When examining the goodness of fit statistics, R 2 and Adj. R 2 are above 0.9, but not as good as the previous goodness of fit statistics shown in the fits of Figure 1 and 2. The question arises as to why the model fits the incorrect data slightly better (figure 2) than the correct data (figures 1 and 3) when using the Goodness of Fit parameters to assess the fits (summarized in Table 2). There may be two reasons. The first reason is that the model is too simple. The second is that we are only fitting data that is taken from one point of the dispersion relation the intersection of the TO- LO modes at the Γ- Point. Although further investigation is warranted, it is intriguing how well this simple model fits the data. Figure 3: Dispersion relation for alpha-sn which shows considerable similarities to the other dispersion relations. 10
Figure 4: The energy of the TO- LO modes at the Γ- Point for diamond, Si and Sn and their respective masses. The fit shows an inverse square root dependence on the AMU (i.e., mass). Goodness of Fit Statistics for Figure 4: Table 2: Group IV S/Cs and Values of Masses and Dispersion Relations and respective Goodness of Fit Parameters Group IV S/C Masses Used Group IV S/C Dispersion Relations Used R 2 Adjusted R 2 C, Si, Ge C, Si, Ge 0.9469 0.9203 C, Si, Sn C, Si, Ge 0.9569 0.9353 C, Si, Sn C, Si, Sn 0.9454 0.9180 References: [1] C. de Tomas, A. Cantarero, A. F. Lopeandia, F. X. Alvarez, Tomas Thermal conductivity of group-iv semiconductors from a kinetic- 11
collective model, Proceedings of the Royal Society A, 470 (2014) 20140371. [2] http://wolf.ifj.edu.pl/phonondb/sn.htm 4. Consider Cu and Ni with their density of states as schematically sketched in Figure 4.61 in your textbook by Kasap. Both have overlapping 3d and 4s bands, but the 3d band is very narrow compared to the 4s band. In the case of Cu the band is full, whereas in Ni, it is only partially filled. Based on this description, figure 4.61, and your observations, answer the following questions. a. In Cu, do the electrons in the 3d band contribute to electrical conduction? Explain. In Cu the 3d band is full, so the electrons in this band do not contribute to conduction. b. In Ni, do electrons in both bands contribute to conduction? Explain. In Ni both the 3d and 4s bands are partially filled so electrons in both bands can gain energy from the field and move to higher energy levels. Thus both contribute to electrical conductivity. c. Do electrons have the same effective mass in the two bands? Explain. No, because it is highly likely that the band structure (i.e., E versus k diagram) of the 3d and 4s states are different. Recall that the effective mass is inversely proportional to the second derivative of energy with respect to k. Furthermore, the energy distributions in the two bands are different. In the 4s band, the concentration of states is increasing with energy whereas in the 3d band, it is decreasing with energy. One would therefore expect different inertial resistances to acceleration and thus different effective mass. d. Can an electron in the 4s band with energy around EF become scattered into the 3d band as a result of a scattering process? Consider both metals. Not in copper because the 3d band is full and cannot take electrons. In Ni the electrons can indeed be scattered from one band to the other, e.g. an electron in the 4s band can be scattered into the 3d band. Its mobility will then change. Electrons in the 3d band are very sluggish (low drift mobility) and contribute less to the conductivity. e. Scattering of electrons from the 4s band to the 3d band and vice versa can be viewed as an additional scattering process. How would you expect the resistivity of Ni to 12
compare with that of Cu, even though Ni has two valence electrons and nearly the same density as Cu? In which case would you expect a stronger temperature dependence for the resistivity? Ni should be more resistive because of the additional scattering mechanism from the 4s to the 3d band (Matthiessen's rule). This scattering is called s-d scattering. One may at first think that this s-d scattering de-emphasizes the importance of scattering from lattice vibrations and hence, overall, the resistivity should be less temperature dependent. In reality, electrons in Ni also get scattered by magnetic interactions with Ni ion magnetic moments (Nickel is ferromagnetic; Ch. 8 in the textbook) which has a stronger temperature dependence than T. Figure 4.61: Density of states for Cu and Ni. Kasap. 13
5. Draw by hand the following metal-semiconductor junctions in the following conditions. Label the band diagrams thoroughly. a. Schottky (i.e., rectifying or blocking) contacts/junctions on n-type Si. i. Equilibrium condition ii. Flatband condition 14
b. Schottky (i.e., rectifying or blocking) contacts/junctions on p-type Si. i. Equilibrium condition ii. Flatband condition 15
c. Ohmic contacts/junctions on n-type silicon. i. Equilibrium condition ii. Flatband condition 16
d. Ohmic contacts/junctions p-type silicon. i. Equilibrium condition ii. Flatband condition 17