Neophytos Neophytou School of Engineering, University of Warwick, Coventry, U.K. Example class ES2D6 – Revision Exercise 1: DOS – Si VB 2 Find the density of states effective mass mDOS* of the valence band of Si. The valence band of Si consists of a doubly degenerate heavy hole band and a single degenerate light hole band. Although the heavy hole band is warped, as indicated in the figure, we an effective mass of m*hh= 0.54 m0 is assumed. For the light-hole, the effective mass is assumed to be: m*lh= 0.16 m0. (We ignore the split-off band with mass mS0=0.29 m0 that is located 0.044 eV further below the valence band edge. WHY ?). Exercise 1: DOS m* – Si VB 3 Find the density of states effective mass mDOS* of the valence band of Si. The valence band of Si consists of a doubly degenerate heavy hole band and a single degenerate light hole band. Although the heavy hole band is warped, as indicated in the figure, we an effective mass of m*hh= 0.54 m0 is assumed. For the light-hole, the effective mass is assumed to be: m*lh= 0.16 m0. (We ignore the split-off band with mass mS0=0.29 m0 that is located 0.044 eV further below the valence band edge. WHY ?). 2/3 * 3/2 3/2 p * p 0 2 0.54 0.16 0.9m m m Solution: * 3/2 1/2 DOS 3 2 3 2 ( )D m E g E To find the total DOS, we need to add the DOS of each valley VB VB-HH VB-LH * 3/2 1/2 * 3/2 1/2 hh lh 2 3 2 3 * 3/2 * 3/2 hh lh 2 3 *3/2 * 3/2 * 3/2 p hh lh 2/3 * * 3/2 * 3/2 p hh lh ( ) ( ) ( ) 2 2 2 2 2 2 2 g E g E g E m E m E E m m m m m m m m Exercise 2: Conductivity m* - ellipsoid 4 A specific semiconductor material has a conduction band with lowest energy valleys along the X-direction centered at the edge of the Brillouin zone. The longitudinal effective mass along the [100] equivalent direction is ml-[100]=1.2 m0. The transverse ellipsoidal mass perpendicular to the [100] equivalent directions is mt-[001]= 0.6 m0. Find the density of states effective mass and the conductivity effective mass for the conduction band of this specific semiconductor. Exercise 2: Conductivity m* - ellipsoid 5 A specific semiconductor material has a conduction band with lowest energy valleys along the X-direction centered at the edge of the Brillouin zone. The longitudinal effective mass along the [100] equivalent direction is ml-[100]=1.2 m0. The transverse ellipsoidal mass perpendicular to the [100] equivalent directions is mt-[001]= 0.6 m0. Find the density of states effective mass and the conductivity effective mass for the conduction band of this specific semiconductor. How will these values change is the valleys in the X-direction are centered in the middle of the Brillouin zone? Solution (DOS mass): * 3/2 1/2 DOS 3 2 3 2 ( )D m E g E To match the DOS, we need to form a spherical band that has the same k-space volume as the complex multi-degeneracy, ellipsoidal band. Then, the volume of a sphere and the volume of the ellipsoids are equated: For ellipsoidal bands, we take the geometric mean for the overall DOS mass – it gives the same volume in k-space 2 2mE k We use: 3 3 . 1 2 3 . 1 2 3 3 1/23/2 DOS-sph 1 2 3 DOS-sph 1 2 3 1/32/3 DOS-sph 1 2 3 4 4 3 3 eff sph eff sphk N k k k k N k k k m N m m m m N m m m m N m m m Volume of sphere = volume of ellipsoid 2/3 1/3 DOS-sph 0 6 1.2 0.6 0.6 1.57 m 2 m Exercise 2: Conductivity m* - ellipsoid 6 Solution (conductivity mass): The velocities (which define the conductivity) in a specific direction, follow still the mass of the individual states, not the DOS mass. To get a conductivity effective mass, we have to average the conductivity masses using the harmonic mean as: * cond * cond 0 0 0 0 1 1 1 1 3 3 3 0.72 1 1 1 1 1 1 0.6 0.6 1.2 t t l t t l m m m m m m m m m m m m Solution (if valleys are in the middle of the Brillouin zone along from the Γ-X): 1/32/3 DOS-sph 1 2 3m N m m m 2/3 1/3 DOS-sph 06 1.2 0.4 0.4 2.5 mm * cond 00.72 m m remains the same Exercise 3: Thermoelectrics 7 A certain thermoelectric material the mean-free-path (mfp) for electrons is λe = 10 nm and the mfp for phonons is λp = 50 nm. The material is nanostructured by introducing a special type of superlattice layers, such that phonons undergo an additional scattering every distance d = 20 nm, but electrons are not affected. Find an estimate for the change in the ΖT figure of merit of the material due to nanostructuring. You can ignore any changes in the Seebeck coefficient, and you can consider that the phonon part of the thermal conductivity is much larger compared to the electronic part of the thermal conductivity. Compare the ZT improvement if both electrons and phonons undergo this additional scattering. Exercise 3: Thermoelectrics 8 A certain thermoelectric material the mean-free-path (mfp) for electrons is λe = 10 nm and the mfp for phonons is λp = 50 nm. The material is nanostructured by introducing a special type of superlattice layers, such that phonons undergo an additional scattering every distance d = 20 nm, but electrons are not affected. Find an estimate for the change in the ΖT figure of merit of the material due to nanostructuring. You can ignore any changes in the Seebeck coefficient, and you can consider that the phonon part of the thermal conductivity is much larger compared to the electronic part of the thermal conductivity. Compare the ZT improvement if both electrons and phonons undergo this additional scattering. 2 2 l e l S S ZT Solution – 1st case: (1) We start from the definition of the ZT figure of merit (ignore ke): (2) The ratio of the ZT under nanostructuring and before is, thus: 2 2 NS NS NS l NS l NS NS l l NS ll S ZT SZT (3) The conductivities can be estimated to be proportional to the total mfp, which can be computed using Matthiessens rule: for electrons, in the first case we still have : 1 1 1 1 1 14 nm 50 20 ph NS ph NS ph d for phonons: (4) The change in ZT is then: 10x50 3.57 10x14 e NS phNS NS l l NS e l NS ZT ZT 10 nme Exercise 3: Thermoelectrics 9 A certain thermoelectric material the mean-free-path (mfp) for electrons is λe = 10 nm and the mfp for phonons is λp = 50 nm. The material is nanostructured by introducing a special type of superlattice layers, such that phonons undergo an additional scattering every distance d = 20 nm, but electrons are not affected. Find an estimate for the change in the ΖT figure of merit of the material due to nanostructuring. You can ignore any changes in the Seebeck coefficient, and you can consider that the phonon part of the thermal conductivity is much larger compared to the electronic part of the thermal conductivity. Compare the ZT improvement if both electrons and phonons undergo this additional scattering. 2 2 l e l S S ZT Solution – 2nd case: (1) We start from the definition of the ZT figure of merit (ignore ke): (2) The ratio of the ZT under nanostructuring and before is, thus: 2 2 NS NS NS l NS l NS NS l l NS ll S ZT SZT (3) The conductivities can be estimated to be proportional to the total mfp, which can be computed using Matthiessens rule: 1 1 1 1 1 6.67 nm 10 20 e NS e NS e d for electrons: 1 1 1 1 1 14 nm 50 20 ph NS ph NS ph d for phonons: (4) The change in ZT is then: 6.67x50 2.38 10x14 e NS phNS NS l l NS e l NS ZT ZT Exercise 4: Subthreshold swing: 60 mv/dec 10 a) Show that the charge in a MOS-capacitor that is biased in the depletion to inversion regime increases by an order of magnitude (10x) every 60 mV that is applied on the gate electrode. b) Based on your result, what can you infer about the increase in the drain current of a MOSFET in the OFF- state as the gate bias increases by 60 mV? c) Under high performance operation, a microprocessor heats up to even up to 50C. Find the voltage on the gate needed at those temperatures to increase the drain current by 10x. Comment on the operating voltage and power consumption of the circuitry as a consequence of your answer. Exercise 4: Subthreshold swing: 60 mv/dec 11 So, the charge increases exponentially with VG Solution: OFF state: the charge is negligible, DOS(EF) is negligible, so CS is negligible, thus: Vchannel follows linearly the change in VG. So the band shifts down linearly with VG. The charge, under non-degenerate conditions, is: F C G C B F C G C B F C G C B B G C0 B ( ) exp ) exp exp exp exp E E V n N k T E E qV n N k T E E qV n N k T k T qV n N k T NEED 60 mV to change the charge n by 10x This is called the subthreshold swing Notice that the band edge EC moves with VG Now, the charge density 10x the initial one (VG1) due to a different VG2, is G2 G1 B B G2 G2 G1 G B B B G1 B 2 C0 C0 G B G G 10 10 ln 10 2.3026 0.026 60 mV/dec qV qV k T k T qV q V V q Vk T k T k T qV k T n N e N e e e e e q V k T V V Exercise 4: Subthreshold swing 12 Solution: b) As the current in the OFF-state of the MOSFET is proportional to the charge, the drain current is also increased by 10x as the VG increases by 60 mV. Remember: G B G G G ln 10 2.3026 0.026 323 / 300 60 mV 323 / 300 ~ 65 mV q V k T V V V D D OX GS T drift D OX GS T eff field 0 0i x i xI WQ x x WQ I WC V V I WC V V E c) At 50C, the absolute temperature is 273K+50K=323 K. Thus, Which means that a larger voltage on the gate is required to achieve the same current. A larger operating voltage, means larger power dissipation. Exercise 5: Shape of TDF 13 Find a simple expression for the energy dependence of the BTE transport distribution function (TD) in the case of acoustic phonon scattering. 2s xE E E g E 2 A B 2 ADP 1 s D k T g E E v 2 2s x E f q g E dE E Exercise 5: Shape of TDF 14 Solution: Since the scattering rates are mostly dependent on g(E), it means that τ(E)~1/g(E) Plugging this into the TD function conductivity formula, and the TD function: 2 2 2 s x x x 1 E E E g E E g E E g E Key findings for the TD function (case of acoustic phonon scattering): - depends only on the velocity, - it is linear with energy, and - inversely proportional to the effective mass of the material Now, the energy dependence of the velocity is: 2 2 * 2 C x * * C C *2 2 2 C x * 2 * 2 * C C C * C / (2 )1 1 2 (which we already know) 2 2 2 , and thus: 2 (linear in energy and inversely proportional to the mass) d k mdE k k E dk dk m m m Ek E E m m m E E m Exercise 6: Shape of mobility versus density 15 Mobility: shows the easiness in transport the electrons/holes experience in a material. The higher it is, the better the transport properties usually It is the freedom in carrier movement. As you can expect, it will be related to the velocity of electrons (or the conductivity mass), and to their scattering. 2 0 * C , in units of cm /V-sq m τ is the scattering time, which is related to the mean-free-path λ as: λ= velocity x τ i.e. it is the distance carriers travel before they scatter Doping dependence Temperature dependence: Jacoboni77 T= 300 K Jacoboni et al., Solid State Electron. 20, 2(1977) 77-89. Dorkel81 Dorkel et al. Solid State Electron. 24, 9 (1981) Using the conductivity as derived from the Boltzmann transport equation, provide an intuitive explanations as to why the mobility is almost constant at low densities. Exercise 6: Shape of mobility versus density 16 Solution: The mobility, is given formally by the following definition – and using the Boltzmann transport equation we can insert the corresponding quantities: 2 2 s x 0 E E f q g dE E qn q g f E dE At low densities, we can approximate the Fermi distribution with the Boltzmann distribution: F F B B BF B 1 1 exp exp 1 exp E E E Ef k T E k T k TE E k T C F Bsince E E k T 2 2F F s x s x B B B B 0 F F B B B 2 s x s * B C 0 B 1 exp exp exp exp exp exp exp e exp E E E E E E E E E E q g E dE q g E dE k T k T k T k T qn E E E E g E dE g E dE k T k T k T E E q g E dE q g E k T m E g E dE k T B B xp exp E E E dE k T E g E dE k T Take EF quantities out EF independent, means density independent How does the density enter the mobility equation? Through EF
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