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(E) 0 Ψt(E) Ut Now let us estimate the d value for a 1eV energy barrier. The k2 value is k = ) ( 2m u − E =2 ~ 2× 9.1 − e 31 1.6 × 1e − 68 e −19 ~ 4 e m-1, 9 2.57 Fall 2004 – Lecture 14 96 and the characteristic length fo...
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incident angle on the scanned surface is larger than the critical angle and thus it will be totally reflected in normal cases. However, when a scanning probe approaches the other side of the surface, there will be tunneling between the surface and the probe and the topography is obtained as in a STM. Tunneling θ>θ...
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the image quality. Additionally, thermal imaging normally requires high vacuum to avoid air conduction and convection effects. Large effective contact area 5.4 Bragg reflector In practice, the reflectivity and transmissivity of multilayers can be controlled quite accurately with various thin-film deposition techn...
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E R 1 0.8 0.6 0.4 0.2 0 1 Stop Band 1.5 2 2.5 3 WAVENUMBER=1/WAVELENGTH (µm -1) It is interesting to compare the stop band with electron band gap in semiconductors. In the latter case, no electrons exist in the forbidden energy levels, while in the reflector no photons exist in the stop band. The idea of f...
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12 3m ⎡ q → = ∑ ⎢ p=1 ⎣⎢V1 kx1 =−∞ k y1 =−∞ kz1 =0 ⎤ 1 )⎥ , ⎦⎥ where Te1 represents the temperature of the phonons coming towards the interface and f(Ε1,Te1) is the Bose-Einstein distribution for phonons at Te1, and τ12 is the phonon transmissivity from medium 1 into medium 2. The unit of energy flux q is W/m2...
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However, we also notice that velocity vz1 depends on k and transmissivity τ depend on (k . To deal with this, first we note k , k ) , x1 12 y1 x1 z1 dk dk dk = k sin θ θ ϕ = k dkd Ω = k dkd d x1 y1 z1 2 2 2 dk dE dEdΩ , where the solid angle Ω = d k 2 sin d dθ θ ϕ 2 k = sin d d θ θ ϕ in the spherica...
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2 =Te1 . In nonequilibrium situations, this relationship still holds true because we have the same expression for q → (T ) . This idea is comparable to the Kirchoff's Law in radiation, in which we consider the energy exchange with a blackbody surface to derive ε( , ) λθ = ( ,α λθ ) . →2 ( e2 ) , T 2 1 e2 1 2 2...
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12 ranges from 0 to 1. In the Landauer formalism, E1 ∂f ∂T )D E dE ( has the physical meaning of volumetric specific heat C. For normal materials, the magnitude of C is 106 (mass-based specific heat can differ a lot) and v is on the order of 103. Thus K~vz1C~1e3 × 1e6=1e9 W/m2 K. (2) Based on one-dimensional heat ...
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exponentially on both sides of the surface) exists, the maximum heat flux can be higher than n2 due to tunneling, because the density of states of the surface wave is much higher. 3) Quantum conductance (Universal conductance) In the following figure, electrons flow between two points 1 and 2. Under the assumption ...
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2.57 Nano-to-Macro Transport Processes Fall 2004 - Lecture 15 Guest lecture by Prof. Dresselhaus 1. Outline -Overview -Synthesis -Structures and Symmetry -Electronic Properties -Transport Properties -Phonon Properties -Resonant Raman Effect -Applications 2. Unique Structure and Properties of Single Wall Car...
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ubes are formed along with other products. Image removed for copyright reasons. Y. Saito et al, Phys. Rev. 48 1907 (1993) (2) Laser Ablation The experimental setup is shown as following, where Nd-Yb-Al-garmet Laser is utilized at 1200 ℃. Image removed for copyright reasons. A. Thess et al. Science 273 483 (1996...
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its electronic structure is similar to empty SWNTs especially near the Fermi level. By Heating at higher termperatures, C60 fullerenes inside the nanotubes can merge and form double wall carbon nanotubes (DWNTs). 4. Structures and Symmetry ) Ch = na1 + ma2 ≡ ( m : primitive lattice a , a 2 1 n , vectors I...
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Symmorphic (mirror symmetry) –Armchair Nanotube (n,n), n=m –Zigzag Nanotube (n,0), m=0 • Non-Symmorphic (axial chirality) –Chiral Nanotube (n,m), n≠m In the above figure, the chiral angle θ (0<θ<π/6) is defined as the included angle between Ch and a1. The diameter dt of SWNTs can be calculated from the length of ...
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summarized in the embedded equation of the following figure. Image removed for copyright reasons. metal semiconduc tor R. Saito et al., Appl. Phys. Lett. 60, 2204 (1992) ⎧ n m = − ⎨ 3 ⎩ 3 p p ±1 The width of DOS split depends on the chirality of the SWNT. In the following figures, we find the Zigzag SWNT ...
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gate voltage is used to change the electronic structure of a SWNT. Image removed for copyright reasons. S.J. Tans et al. Nature, 393, 49 (1998) Resonant tunneling transport is demonstrated in the left bottom figure. The energy levels in a SWNT can be compared with a particle in a box. By changing the applied gate ...
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This inelastic scattering can occur with a change in vibrational, rotational or electronic energy of a molecule. Since Resonant Raman Effect is diameter selective (also chirality dependent), it is used to determine the size of SWNTs. Image removed for copyright reasons. dR = 2(m N = + nm) 2 + 2n dR if n− m = 3d ...
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2004 – Lecture 15 112 diameter and chirality. For applications, it is significant to develop method for large-scale, cheap synthesis, and improve nanotube characterization and manipulation. Separation processes reported till now include: (1) Precipitation of SWNTs non-covalently functionalized with ODA (2) Ion-exc...
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→ t x Consider a plane traveling along x-direction Ae− i(ω t − kx) . The velocity v =ω / k is NOT the speed of signal or energy propagation. We see that the plane wave represented by above equation extends from minus infinite to plus infinite in both time and space. There is no start or finish and it does not rep...
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one at frequency ωο−∆ω/2 and another at frequency ωο+∆ω/2 propagating along positive x-direction [figure (a)]. The superposition of these two waves gives the electric fields as y ( , E x t {−i ⎡(ω + ∆ω)t − (k − ∆ k ) = a exp ⎤} ) ⎦ x ⎣ o o = 2 cos (∆ω• t− ∆k x ) exp ⎡−i(ωt a o ⎣ {−i ⎡(ω − ∆ω)t − ( ko − ∆ k ...
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the time-averaged Poynting vector Re ( E H ) to get the real energy flux. k x o − o a JG JJG * ⎣ × • 1 2 This yields another wave propagating at the speed 2.57 Fall 2004 – Lecture 16 115 v = g x, ∆ω dω , = ∆ k dk which means that the energy is propagating at the speed of vg,x rather than the p...
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In most cases, the energy velocity is just the group velocity. However, for photons the group velocity can be larger than the speed of light in special cases. We can appreciate this from the calculation of Poynting vector, where we assumed that ∆ω is much smaller 2.57 Fall 2004 – Lecture 16 116 than ωo. In the case...
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but have the same energy before and after the scattering, or inelastic in which both direction and energy of electrons are changed. The scattering of electrons by impurities and at the boundaries is elastic. The elastic scattering itself does not destroy the phase but the random locations of the impurities and the s...
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2167.8 µm.K, very close to the Wien’s displacement law. For electrons, lc = 100-1000 Å, while it is 1-10 Å for phonons. 2) Spatial & temporal coherence Reflected 3 2 1 Incident Transmitted (a) (b) (c) The coherence length as a measure of the wave packet size gives an indication whether the phase information ...
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such as by phonons. In this case, we can ignore the phase information and the particle picture is valid. Crystal Inelastic MFP When size effects appear, the characteristic length is normally less than the inelastic mean free path. This can be the case for nanowires. However, the particle treatment may be still va...
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um State Ω ri, Pi ri, Pi Now each particle can be described by the generalized coordinate r and momentum p. For example, the generalized coordinates of a diatomic molecule, r1, include the position (x1,y1,z1), the vibrational coordinate (the separation between the two atoms, ∆x1), the rotational coordinates (po...
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∆r(n ∆r(n) =∆r(1)∆r(2)…∆r(n) =∆r1∆r2…∆rN )∆p(n) ) , where and ∆p(n =∆p1∆p2…∆pN =∆p(1)∆p(2)…∆p(n) . ) ),p(n) The time evolution of f(N (t,r(n ) in the phase space is governed by the Liouville equation, which can be derived based on the fact that the traces of systems in the ensemble do not intersect. Consider a tub...
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volum e r(i ) r(i) r(i)+∆r(i) Note: The Boltzmann equation reduces variables from the Liouville equation, while the linear response theory focuses on small perturbation. They are both simplified form of the Louiville equation. 2.57 Fall 2004 – Lecture 16 121 2.57 Nano-to-Macro Transport Processes Fall 2004 ...
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involves a huge number of variables, which makes its impractical in terms of the boundary and initial conditions, as well as analytical and numerical solutions. One way to simplify the Liouville equation is to consider one particle in a system. This is a representative particle having coordinate r1 and momentum p1, ...
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variables of the ⎛ ∂f ⎞ ⎟ ⎝ ∂t ⎠c • ∇ f = ⎜ p • ∇ f + r dp dt dr dt + gradient. The scattering term ⎛ ∂f ⎞ ⎟ ⎜ ⎝ ∂t ⎠ c will be discussed in details later in this lecture. The above equation is the Boltzmann equation or Boltzmann transport equation. Note: The derivative dr dt has the meaning of velocity...
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in the Schrödinger equation H 0Ψ = EΨ . By treating H’ as a small perturbation to unperturbed Hamiltonian Ho, the solution of the Schrödinger equation for the new H can be obtained through perturbation method and expressed in terms of the wave functions Ψ of the unperturbed two-particle system with a Hamiltonian Ho...
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scattering from other quantum states into the quantum state under consideration. The other is the decrease of the number of particles due to scattering from the current quantum states to other quantum states. We again take two-particle scattering process as an example. The initial wave vector of the particle is k an...
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4 – Lecture 17 124 ' d 3k1 ' , a similar argument, the above equation can be simplified by noticing in the equilibrium status W (k k1 → k ' k1 ') W ( , This leads to ⎛ ∂f ⎞ ⎜ ⎝ ∂t ⎠c ( r k1 ' t)⎤ k d , ⎦ k k ' k k ) , 1 , → k ,k1 ') ( , ,t ) ⎡ f r k ⎣ f r k1 ) ( , ,t − ...
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v f = − F m o . r Note: In fluid mechanics, we have a counterpart of the Boltzmann equation, which is called the Krook equation. E k k k’ Now let us consider the general rules of the phonon-phonon scattering. First we have energy balance as ) ( ) E = E ⇒ E k + E k = E k ' + E k' . Secondly we have momentu...
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and can be proved to be negligible for optical lights ( λ~ 0.5 mµ ). Therefore, the arrow points upward almost vertically. Absorptivity Direct semiconductor Indirect semiconductor Frequency 2.57 Fall 2004 – Lecture 17 126 For silicon (indi...
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1 τ = ∑ 1 τ j , which is used in the expression of f − fo τ . For phonon scattering, the two-particle collision is drawn as following. k2,ν2 ν3 =ν1+ν2 k3 k1,ν1 (a) 2.57 Fall 2004 – Lecture 17 ν1,k1 ν2,k2 ν3,k3 (b) 127 For the merging of two phonons in...
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⎤ 1 d U ⎥ 2 2 ⎢ dx' ⎥ ⎦ x'= x o (x'−x ) + Ο[(x'−x ) ] 3 2 o o can be considered as the perturbation from the harmonic potential. In the umklapp process, the randomly distributed vector G plays a similar role as O[(x’-xo)3]. Energy Repulsion Harmonic Force Approximation Interatomic Distance Attraction Equili...
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other particle. This is the so called the molecular chaos assumption. Additionally, the particles have no memory of its history and its final state is unaffected by its very beginning status. (3) The Boltzmann equation cannot include explicitly the wave effects such as interference and tunneling. We use the particl...
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⎪e k T −1 ⎪⎪ 1 f = ⎨ E −µ ⎪ k TBe ⎪ ⎪ ⎛ ⎪n ⎜ ⎪ ⎝ ⎩ +1 0 3/ 2 ⎞ m ⎟ e k T 2π B ⎠ − 2 m -v u 2k TB Displaced Maxwell velocity distribution (molecules) Note: The relaxation time is due to combined factors and can be evaluated numerically by adding all possible influence together. This idea can be used in ca...
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such as electron-electron scattering (energy conserved). It is not accurate for electron-phonon scattering. In this situation, we ( e − p ) . (3) In semiconductor devices, Te can be thousands of ,k) as τ(ω) . (2) The time τ ω( o + g T T have = − f ∂f ⎞ ⎛ ⎟ ⎜ ⎝ ∂t ⎠ s f − τ degrees, while Tp is only hundreds d...
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of fo, and similarly, g is much smaller than fo. Under these assumptions, the above equation becomes ⎛ ⎝ g = −τ⎜ v • ∇r fo + ⎞ • ∇ v fo ⎟ ⎠ F m or 2.57 Fall 2004 – Lecture 18 131 ⎛ f = f −τ⎜ v • ∇ f + ⎝ r o o F m • ∇ ⎞ f ⎟ . ⎠ v o From the distribution funct...
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/ V x v =ωfdk dk dk π/ L 3 ) z ( / 2 y x ∞ ∞ ∞ )∑ ∫ ∫ ∫ x p −∞ −∞ −∞ ω = max ∫ 0 π ⎡ 2π⎧ ∫ 0 ∫ 0 ⎨ v cosθ ωf dω⎢ ⎢⎣ ⎩ = ky φ θ kz ) D ( ⎤ ⎫ ω sin θ θ⎬ d ⎥ d ϕ 4π ⎥⎦ ⎭ kx Note: Here θ varies from 0 to π because both positive and negative vx are considered. 2.57 Fall 2004 – Lecture 18 132 ...
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ωmax ∫ 3 dx 0 max dω ⎧ π ⎨∫ ⎩ ⎧ ⎪ π ∫ 0 dω ⎨ ⎪⎩ 0 {∫ ω dω max π 0 1 dT 3 dx ∫ 0 = − and 2 dfo ⎫ ) 2 τv s inθ cos × = D ω dθ⎬ dT ⎭ θ ω ( 2 τv d (=ωD(ω) f ) dT o ⎫ ⎪ dθ⎬ ⎪⎭ ( τv2 C ω dθ ) } k = ∫ τv Cdω in the Fourier law. 2 1 3 Note: The equilibrium fo term drops out in the integral, which ...
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m P(v , v , v ) = ⎜ ⎜ ⎝ 2πκ T x z y 2 ⎡ 3 / 2 − m⎢(v x − u) + v y + v z ⎥ / 2κ B T ⎞ ⎟ . ⎟ B ⎠ 2 ⎤ ⎦ e ⎣ 2 where u is the average velocity along the x-direction (spatially changed). U o u y vy vx x Assuming that the number density of particles is n, the number density of particles having velocity v i...
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∂ ∂ . Compared with the law τxy = µ ∂ u ∂ y have µ= n k Tτ B . (x is the force direction, y is the normal direction), we Following a similar procedure, we can also calculate the energy flux due to molecular heat conduction and obtain the thermal conductivity for a gas as k = ⎜ B ⎟ nτ kB T = ⎜ B ⎟µ. 5 ⎛ k ⎞ ...
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� fo(vx,vy,vz)=nP(vz,vy,vz) = n⎜ ⎜ ⎝ 3 / 2 − m (v − u)2 ⎞ ⎟ ⎟ B ⎠ m 2πκ T e ⎡ ⎢ ⎣ x ⎤ + v 2 + v 2 ⎥ z y ⎦ / 2κ T B From f = f −τ⎜ v • ∇ f + r o o ⎛ ⎝ F m ⎞ ⎠ • ∇ v o f ⎟ (F is zero here because this is no external fields here and gravity is neglected), the distribution function is ⎛ ∂ f ⎝ ∂x ( x z ...
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xdv ydvz ∞ ∞ ∞ = − ∂ u ∞ ∂ y ∞ ∫ −∞ ∫ −∞ ∫ −∞ τv y mv x ∞ 2 ∂ fo ∂u dv xdv ydvz = µ ∂ u ∂ y The dynamic viscosity can be τv (mv µ= − dv 2 x y x ∫ ∫ ∫ ∂ fo ) ∂u dv y dv z 2.57 Fall 2004 – Lecture 19 136 2 = m nτ m 2πκ T 3/ 2 ∞ ⎞ − mv /(2κ...
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) f a = ) +∞ ∫−∞ 2 −ax e dx = π a leads to f '( ) = − a +∞ ∫−∞ 2 2 −ax x e ' ⎛ π ⎞ dx = ⎟⎟ ⎜⎜ a ⎠ ⎝ = − π −3/ 2 a 2 . Ohm’s Law x εx Now fo obeys the Fermi-Dirac distribution o ( f E E, ) ,T = f 1 ⎛ E − µ ⎞ κ T B ⎟ +1 ⎠ exp ⎜ ⎝ , and f = f −τ⎜ v • ∇ f + r o o ⎛ ⎝ F m • ∇ f ⎟ be...
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thus f = f o − ⎛ τ v ⎜ x ⎝ ∂fo ∂x m ∂E ∂v qεx ∂fo ∂E ⎞ ⎟ x ⎠ + = f − τv o ⎛ ∂fo x ⎜ ⎝ ∂x q + ε x ∂fo ⎞ ⎟ ∂E ⎠ . The current density (A/m2) is then ∞ ∞ 1 Jq = ∑ ∑ ∑ q x V kz =−∞ ky =−∞ kz =−∞ ∞ v f = 1 ∞ ∞ ∞ ∫ ∫ ∫ π) −∞ −∞ −∞ 3 (2 qv fdk dk dk z x x y Changing to spherical coordinate syste...
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� B where Ek denotes the kinetic energy. E − µ ∂f 0 ∂Ek , ∂f 0 ∂µ 0 ∂f ∂Ek = − , we have Noticing ∂f 0 ∂T = − k T ∂f ∂µ ∂f ∂T + ∂µ ∂x ∂T x∂ 0 0 ∂f 0 ∂x = = − ∂f ∂µ E − µ ∂f ∂T k 0 0 ∂Ek ∂x T ∂Ek ∂x − Note: Defining y = k E − µ κBT , f E E T = , f , ) ( o 1 exp y( , we can observe ) +...
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∂f 0 ∂E k ( k ) D E dE ⎫ ⎪⎛ ∂µ k ⎬⎜ ⎪⎝ ∂x ⎭ − qε + ⎧ ∞ ⎞ ⎪1 x ⎟ ⎨ ∫ ⎠ ⎪3 ⎩ 0 2 qτv ⎛ E − µ⎞ ∂f 0 k ⎜ ⎟ ⎠ ∂E ⎝ T k ( k ) D E dE ⎫ ⎪ ∂T k ⎬ ⎪ ∂x ⎭ Discussion To simplify, let L11 = 1 ∞ ∫ 3 0 qτv 2 ∂f ∂Ek ( k ) 0 D E dEk . 1) ∂T / ∂ = (isothermal), we have J = L x 11 ⎜ ⎞ − qε = L q x ...
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m2s-1], n is carrier concentration. In the following figures, we actually measure both terms in the above equation. E µ is continuous x V Note: Recall we use n = ∫ fD E dE to determine µ in Lecture 11. Normally n is related ( ) to µ. In semiconductors, when we increase doping, n will increase while µ will drop...
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the conductor is a uniform material such that S is a constant, the voltage difference does not dependent on the temperature profile. This is the principle behind the thermocouple for temperature measurements. A thermocouple (the following figure) employs two conductors for the easy of measuring the voltage differen...
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circuit condition. We can express this as ∆V S = − T2 − 1 T (V/K), where S > 0 for p-type semiconductors, and S < 0 for n-type materials. In the following figure, we demonstrate the idea of Peltier Effects. Heat Q Heat Q π = Q I I 2.57 Fall 2004 – Lecture 20 141 ...
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cooling process. Therefore, we should focus on the edge of semiconductors. 2.57 Fall 2004 – Lecture 20 142 To increase ZT, we want S ↑,σ↑, k ↓ . There is confliction in satisfying all these requirements. In the figure, we ca...
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Bi2Te3 and alloys with Sb, Se 1950 1960 1970 1980 Year 1990 2000 2010 Image by MIT OpenCourseWare. 2.57 Fall 2004 – Lecture 20 143 The above figure summarizes the recent advances in nanostru...
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1) Figure removed for copyright reasons. Harman et al. J. Electron. mater. Lett. 29, L1 (2000) In the above work, enhancement of ZT at 300 K in a quantum dot superlattice is more than a factor of two relative to best available bulk PbTe because -Favorable carrier scattering mechanism due to PbSeTe quantum dots -L...
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4). Nano Lett., 2 (2002): 83. Nano Lett. 2: 2 (2002): 87-89. 500 nm 2.57 Fall 2004 – Lecture 20 146 The theoretical modeling of this structure includes the following key parameters –Wire diameter: dW –Segment lengths: LA and LB (assum...
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1) New methods for synthesis and assembly of nanowires 2.57 Fall 2004 – Lecture 20 148 The following figure demonstrate the idea of making standing nanowires. Si (100) predeposited layers (adhesive / conductive / patterned) th...
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303, 816 (2004) Carrier profiles and electronic band energies across p-n junction 9. Conclusions 1) Model systems show that: ZT for 0D nanowire superlattice > ZT for 1D quantum wires > ZT for 2D quantum wells > ZT for bulk for same material 2) New research directions now being pursued: -Self assembled bulk com...
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� 12 ⎜ = L T 12 ⎜ ⎛ 1 dΤ ⎞ ⎟ ⎝ T dx ⎠ is similar to S ∆ = dQ T and may be compared with entropy flux. The heat transferred is J q = ∑∑∑ vx ( E − µ) f = L21 ⎜ − 2 V kx k y kz ⎛ dΦ ⎞ ⎝ dx ⎠ ⎟ + L22 dT dx For open circuits, Je=0. We obtain − = S dΦ / dx = dT / dx T − T c V h L 12 , = L 11 where S...
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to the average temperature from 1 to 3. This is different from the normal thermocouples. 1 2 3 1 When dT/dx=0, we have J = L − 21 ⎜ q ⎛ dΦ ⎞ L21 ⎟ ⎝ dx ⎠ L11 = J = Π e J , e where the Peltier coefficient Π = TS , L21=TL12. Note one thermoelectric coefficient (S here) can be used to express all other coeffic...
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dQ/dx The overall energy equation is dT dΦ . q = e dx dx 2 =σJ + k e dJ q dx + J + Thomson term . And thermal conductivity is k = L L / L − L . e 22 21 11 12 The Wiedemann-Franz law states ke σT where the constant L is Lorenz number. 2 = const = L = 2.45E − 8 W Ω / K , Now consider the Boltzmann equ...
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∂fo ∂t on the LHS compared with g τ on the RHS. This indicates ∂fo g ∂t τ  ⇒ t  τ . The theoretical solution of Now consider the above transient process in which an infinite wall is heated suddenly. is T = exp(−x / αt ) , which indicate that T is nonzero at any long distance from the wall. Since the therm...
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raised much higher than that of the phonons and after the relaxation time electrons exchange energy with phonons (the following figure). 2.57 Fall 2004 – Lecture 21 154 Te Tp time An experiment that is often cited as the proof for the validity of the hyperbolic equati...
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with the boundaries. The previous requirement Λ / L  1 is not satisfied in this case. Most reflection on the wall is elastic. Two possibilities occur here: specular reflection, diffuse reflection. First we have the Boltzmann equation 2.57 Fall 2004 – Lecture 21 155 ...
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1 ∑∑∑ fv x =ω. V k k k y z x y vy θ v vx d x vy θ v ϕ vx Temperature Gradient Or Electrical Field (a) vz (b) The Boltzmann equation is G τv ⋅∇ r G g + g = −τ ⎜ v ⋅∇ r G f0 + ⎛ G ⎝ JG F m ⋅∇ v G f0 ⎟ = S0 , ⎞ ⎠ where ∇G f = r 0 g +τv y JG F m v df dT 0 dT dx ∂g ∂y , the x directi...
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� 2 ⎠ , f = f0, g = 0 for θ ∈⎜ d= . At y = 0,θ 0, ∈ ⎜ ⎛ π⎞ , C = -S , g y,θ) = S 1− exp ⎟ ⎝ 2 ⎠ ⎛ 0 ⎜ ⎝ ( 0 ⎞ ⎞ y ⎛ ⎜ − , ⎟ ⎟ ⎝ vτ cos θ ⎠ ⎠ 2.57 Fall 2004 – Lecture 22 157 At y = d, θ∈ , C = -S exp 0 (d vτ / ) cos θ) , g y,θ = S 1− exp ( ⎛π ⎜ ⎝ 2 ⎞ ,...
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= ωmax ∫ 0 2π ∫ 0 π ∫ = ( 0 dω dϕ[ 2 ω v cos ϕsin θ −τv cos ϕsinθ ⎛ )⎜ ⎝ df0 dT ⎛ ⎜ dT dx ⎝ ⎛ exp − ⎜ ⎝ y ⎞ ⎟ vτcos θ ⎠ −1 ⎞ ⎞ D(ω) ⎟ ⎟ 4π ⎠ ⎠ sin d θ θ + ∫π =ω(v cos ϕsin θ) −τv cos ϕsin θ π 2 ⎛ ⎜ ⎝ df0 dT ⎛ dT dx ⎝ ⎛ d − y ⎞ ⎜ exp ⎜ vτcosθ⎠ ⎝ ⎟ −1⎟ ⎟ ⎞ ⎞ D (ω) ⎠ ⎠ 4π sin d ] θ θ Note: A...
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� , Λ =τ . Noting v 2π ∫0 2 cos ϕdϕ π = , the total heat = − π dT ωmax 4 dx ∫0 = ω df0 dT 2 τv D ( ) d ω ω ∫0 [ 1 2 ⎛ 1( − µ dµ Λµ exp ⎜⎜ ⎝ ⎛ ⎜ ⎝ ) ⎛ ξ⎞ ⎜ − ⎟ − µ ⎝ ⎠ 1 ⎞ ⎞ − d ⎟⎟ ⎟ ⎠ ⎠ + 1 ∫0 2 ⎛ (1− µ )dµ⎜ −Λµ⎜1− exp ⎜ ⎝ ⎛ ⎝ ⎛ ξ⎞ ⎞ − ⎜ µ ⎝ ⎞ ⎟ ⎟ − d ⎟] ⎟ ⎠ ⎠ ⎠ = −kd dT dx , whic...
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1 For specular case (above figure), the x-direction momentum is conserved. Following all these procedures, we can finally prove k=kbulk. Note: To determine the mean free path, we should use k = Λ 3 ∫ ω ω ω instead of the C v d simplified k = CvΛ 3 , which gives an underestimation of Λ . This is because the De...
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⎛ y ' ⎞ exp ⎜ ⎟ ⎟ ⎜ τv ⎝ y ⎠ τvy d + ( 0 y C y ' ) , S 0 ( )' y y 0 C = ∫ y y ⎞ ⎟ ⎟ y ⎠ τv . In general, use C y to replace ( ) g y( ) = ( C y 0 y ⎛ ⎞ ) exp ⎜ − ⎟ + ∫y ⎝ τv cosθ⎠ y 0 S 0 ( )' y '− ⎞ y ⎟ ⎟ y ⎠ ⎛ y exp ⎜ ⎜ τv ⎝ τvy dy ' . y = ( ) 0, f = , = 0, C 0 = 0 for θ∈ 0, f g ...
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η ξ )η + ∫ 0 ( θ η' E1 η η − ) ' ) dη' ( where we used dimensionless parameters θ y = ( ) 1 T y( ) − T T T1 − 2 2 ,η= y Λ ,ξ= d Λ . 2.57 Fall 2004 – Lecture 22 160 The temperature profiles for two extreme cases are drawn in the following figure. For ξ→ 0 (note T1 ≠ T2...
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JG F m where the bulk force G G f0 = 0 for phonons, g=f-f0. Noticing v ⋅∇ v ⋅∇G g ≈ vy r ∂g ∂y (d<<x) and ∇G f0 = r df0 dT dT dx , the x direction component gives g +τvy g ∂ y ∂ = −τv df0 dT x dT dx = S0 ( ) . x We can solve g first and then substitute the expression f = g+f0 into any flux equation. U...
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the kernel. The Fredholm integral equation does not have an analytical solution, although approximate solution methods have been developed. . This is the linear, nonhomogeneous, ( ) − u u y u u2 −1 T1 = 2 4 2 4 To solve the equation numerically, first we discretize the equation with 2d T Ti+1 − 2Ti + T dx2 2...
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was discussed ) = in the last lecture. When ξ= d /λ→ 0 , T y ( 1 T T2 + 2 . θ 1 ξ→ ∞ 0ξ→ y/d 1 T1 T1 Teq Te q T2 T2 More generally, θ should be the internal energy of the carriers. For photons, θ( y ) = ) − 2 ( u y u T y T 4 u T2 1 ) − 2 = u2 − 4 ( 4 − T 1 4 . The interpretation of the temperature...
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temperature inside region 1 should equal to that temperature in the film at the boundary. The temperature jump is artificial, and arises from the fact T1 and T2 are only emitted phonon temperature entering the thin film, not the local equilibrium temperature as we use in the Fourier law or solved directly from the ...
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as in the case of photon radiation discussed above. One example is in the experiment determining the quantized conductance of a nanowire. The electrodes are large and the measured voltage drop should represent the differences of electrons entering the channel. Je Chemical potential Superlattice Large electrodes ...
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above left figure, at the interface we have 2.57 Fall 2004 – Lecture 23 166 q = ∑ τ ωvx1 f1 + ∑ τ21 =ωvx2 f2 . = 12 vx1 >0,vy1 ,v z1 vx 2 >0, vy 2 ,v z 2 Note in the right figure, only rightward arrows indicate transport to the second region (V > 0 ). Define τ ' , τ ' as the average transmissivity in each r...
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τ21 ') 1− 2 + τ21 ' 1 (τ23 '+τ22 ') 2 τ23 ' in thin films, while in bulk materials d3 / 4 Λ is dominant. We have similar relationship as previous cases. k kbulk T1 1 T2 2 3 ξ= d Λ 1 Chapter 9 Liquids For gases, after two particles collide, they do not have any memory of previous history. However, th...
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π µ In the left figure above, pressure difference exists in the fluid. For one particle, the osmotic pressure is determined by P = Nk T = nk T . B B 1 V Also, we can use the solution of Stoke’s flow around a sphere to estimate the drag force. The value is F = 3πDµu . For area Ac, the forced balance over the contro...
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. In chapter 6, there is a similar Einstein relation between electron diffusivity and mobility. But Einstein really worked on the Brownian motion. Note: The osmotic pressure can be observed by putting a semi-permeable membrane such that only the base molecules in the solution penetrate the membrane, building up a ...
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( ) ( R t s+ ) • R s = 2 R π δ t o ( ) . ( ) 9.2 Force and potentials For liquids, potential interaction contributes significant to transport. In previous lectures, we have discussed force F = −∇Φ . For the interaction of two charged particles as the JG following left figure, the Coulomb potential is Φ = Q1 ...
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2 ⎞ ⎠ 2 2 1/ 2 , . Under the approximation r>>d, we obtain φ(r,θ) = − qβ cosθ 4πεor 2 , where β=Qd is the dipole moment. Note: (1) The superimposed two fields yield Φ ~ r two dipoles, the potential becomes Φ = − Cr −6 . −2 instead of Φ ~ r −1 . (2) If we have The van der Waals potential between one atom and ...
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from the solution onto a previously uncharged surface. The charges accumulated at the surface are balanced by an equal but oppositely charged region of counterions. Some of these counterions are also bounded to the surface, forming a so-called Stern or Helmholtz layer, which is usually very thin (a few Angstroms). ...
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� ⎠ B Finally we obtain the Debye length, = ∑ 1 δ 2 2 Z e n oi i ε ε κ T i o r B Note: εr ↑ ↑,δ . . Now we consider the force balance inside the liquid. Because the liquid is stationary, the electrostatic force on liquid must balance the pressure force. The pressure inside the electric double layer is highe...
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in potential so that particles are separated at a distance. The repulse forces between surfaces also created another interesting phenomenon, the disjoining pressure. The following figure shows an example where the disjoining pressure plays an important role. At the base of liquid surface intersection the wall, the ...
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. In the following left figure, charged DNA molecules can be separated by the electric field. In the second case, the solid walls are stationary. The ions will move under an external electric field (called electrokinetic flow). The moving ions will also cause the fluid to move, and form a plug flow (flat profile). T...
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and outside a particle due to surface tension. The other is for the interaction of three surfaces. For the following right case, the Young equation gives γ = γ cosθ +γ . 12 The forces are shown in the figure. 23 13 P’ P” 3 γ 23 2 1 γ 12 13γ Size can affect the phase change processes. One can easily apprec...
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18.413: Error­Correcting Codes Lab February 10, 2004 Lecturer: Daniel A. Spielman Lecture 3 3.1 Analysis of repetition code meta­channel When we specialize our interpretation of the output of a channel to the meta channel formed by encoding using the repitition code and transmitting over another channel, we solve...
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as an experiment, and I now want to determine the probability that w was 1 given the results of both experiments. By the theorem from last class, we have P [w = 1|y1 = b1 and y2 = b2] = P [y1 = b1 and y2 = b2|w = 1] P [y1 = b1 and y2 = b2|w = 1] + P [y1 = b1 and y2 = b2 w = 0] | (3.1) To evaluate this probability...
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1 y1 = b1], we can also comput | | P [y1 = b1 and y2 = b2|w = 0] (1 − p1)P [y1 = b1] (1 − p2)P [y2 = b2] P [w = 0]2 . Combining these equations, and P [w = 0] = P [w = 1] = 1/2, we obtain (3.1) = p1p2 p1p2 + (1 − p1)(1 − p2) . In particular, the terms we don’t know cancel! 3.2 Capacity of meta­channel Conside...
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2 p2 + (1 − p)2 p(1 − p) p(1 − p) + (1 − p)p 2p p2 + (1 − p)2) 2p p2 + (1 − p)2) p(1 − p) p(1 − p) + (1 − p)p (1 − p)2 p2 + (1 − p)2 . = 1/2 = 1/2 Lecture 3: February 10, 2004 3­3 To compute the capacity, we must assume that P [w = 1] = P [w = 0] = 1/2, so we have i (w = 1; (y1, y2) = (1, 1)) = log2 = log...
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p 2 � , � , � 22p (1 − p)2 + p � . 2 We now compute I(w; y1, y2) by summing over all events: I(w; y1, y2) = � a,b1,b2 P [w = a, y1 = b1, y2 = b2] i (w = a; y1 = b1, y2 = b2) = � (1 − p)2 + p 2 � � 1 − H � 2 p )2 + p (1 − p �� . 2 3.3 Prior, Extrinsic, Posterior and Intrinsic Probabilities It is uns...
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w = 1 given the outcome of the experiment. For example, if y is the output of a channel on input w, and b is the value received, then def Pext [w = 1|y = b] = P [y = b|w = 1] P [y = b w = 1] + P [y = b|w = 0] | is the extrinsic probability of w = 1 given the event y = b. Up to now, we have really been computing ext...
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y = b assuming that w is not uniformly distributed, and to observe that one obtains the above formula. We will occasionally also see the term intrinsic probability. This will usually be treated in the same way as the prior, but will be distinguished from the prior in that it will often be determined from previous e...
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6.825 Techniques in Artificial Intelligence First-Order Logic Lecture 5 • 1 At the end of the last lecture, I talked about doing deduction and propositional logic in the natural deduction, high-school geometry style, and then I promised you that we would look at resolution, which is a propositional-logic proof sys...
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about all or some of them without having to name them explicitly. 3 FOL motivation • Statements that cannot be made in propositional logic but can be made in FOL. Lecture 5 • 4 The book has a nice argument for why propositional logic is inadequate in the Hunt- the-Wumpus domain. Here are some examples of the kin...
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. Each one of those guys is dead. All the bacteria are dead now. So you'd like to have a way not only to talk about things in the world, but to quantify over them, to talk about all of them, or some of them, without naming them explicitly. 6 FOL motivation • Statements that cannot be made in propositional logic ...
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first-order logic is that we can talk about things, and so there's a new kind of syntactic element. There's a new kind of syntactic element called a term. And the term, as we'll see when we do the semantics, is a name for a thing. It's an expression that somehow names a thing in the world. There are three kinds of ...
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, y, a • Function symbol applied to one or more terms: F(x), F(F(x)), Mother-of(John) • Sentence • A predicate symbol applied to zero or more terms: on(a,b), sister(Jane, Joan), sister(Mother-of(John), Jane), its-raining() Lecture 5 • 13 In propositional logic we had sentences. Now, in first-order logic it's a ...
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