Lecture 11: Coupled Current Equations: and thermoelectric devices

Transcription

1 ECE-656: Fall 011 Lecture 11: Coupled Current Euations: and thermoelectric devices Professor Mark Lundstrom Electrical and Computer Engineering Purdue University, West Lafayette, IN USA 9/15/11 1 basic euations of thermoelectricity E x = ρ n J x + S n d L dx J x = π n J x κ n d L dx Four transport coefficients: 1) resistivity (Ω-cm) = 1/conductivity (S/cm) ) Seebeck coefficient (V/K) 3) Peltier coefficient (W/A) 4) Electronic heat conductivity (W/m-K) Note: hese euations describe electric and heat currents due to electrons in the diffusive limit and in 3D.

2 physics of Peltier cooling electrons absorb thermal energy, E - E F1 energy channel electrons dissipate energy, E - E F electrons enter contact 1 at the Fermi energy, E F1 contact 1 Net power dissipated: P D = IV contact electrons leave contact at the Fermi energy, E F 3 a closer look at contact 1 energy channel λ E evaporation of the electron liuid contact 1 f 0 4

3 Monte Carlo simulation It is interesting that the thermoelectric cooling and heating regions are contained in the highly doped contact layers. 5 Mona Zebarjadi, Ali Shakouri, and Keivan Esfarjani, hermoelectric transport perpendicular to thin-film heterostructures calculated using the Monte Carlo techniue, Phys. Rev. B, 74, (006). outline 1) Introduction ) Coupled flow euations 3) hermoelectric devices 4) Discussion 5) Summary his work is licensed under a Creative Commons Attribution- NonCommercial-ShareAlike 3.0 United States License. 6

4 for bulk 3D semiconductors J x = σe x σ S d L J x = L σ SE x κ 0 d L E x = ρj x + S d L dx J x = π J x κ e d dx dx (diffusive transport) dx σ E ( ) = κ 0 = L ( ) σ = σ E de π = L S h λ E ( ) M ( E ) A f 0 E S = k E E B F σ ( E) de L σ ( E) de E E F σ ( E) de L 7 transport parameters (3D, diffusive) 1) Assume parabolic energy bands ) Assume power law scattering ( ) = λ 0 ( E E C ) L λ E r Ionized impurity scattering: r = Acoustic phonon scattering: r = 0 σ E ( ) = κ 0 = L ( ) σ = σ E de S = π = L S h λ E ( ) M ( E ) A f 0 E E E F σ ( E) de L σ ( E) de E E F σ ( E) de L 8

5 transport parameters σ = h S = ( ) g V m * L π F 0 η F ( r + )F r +1 η F F r ( η F ) ( ) ( ) Γ λ Γ r + 0 ( ) η F ( ) ( η F ) F r η F F 0 κ 0 = L h g m * L λ V π 0 Γ( r + 4)F r + η F ( ) η F Γ r + 3 ( )F r +1 η F ( ) + η F Γ r + ( )F r η F ( ) 9 transport parameters in a different form σ = h M λ σ = h M λ S = E E F L σ ( E) de ( ) σ E de S = E E F L ave π = L S κ 0 = L E E F σ ( E) de L κ 0 = σ L E E F L ave 10

6 Wiedemann-Franz Law κ 0 σ = L E E F = L L L ave Wiedeman Franz Law Wiedeman Franz Law 11 electronic heat conductivity κ n = σ n L L L is the Lorenz number he Lorenz number depends on details of bandstructure, scattering, dimensionality, and degree of degeneracy, but for a constant mfp and parabolic energy bands, it is useful to remember: L k B non-degenerate, 3D semiconductors L π 3 fully degenerate e.g. 3D metals 1 a rule of thumb not a law of nature (G.D. Mahan and M. Bartkowiak, Appl. Phys. Lett., 74, 953, 1999)

7 Peltier and Seebeck coefficient Kelvin relation Onsager relations for coupled flows 13 Onsaeger relations df J x = L n 11 dx + L d 1 1 dx L df J x = L n 1 dx + L d 1 dx L 14 J 1 = L 11 J = L 1 B ( ) F 1 + L 1 ( B ) F J, J 1 B ( ) F 1 + L ( B ) F generalized fluxes generalized forces Onsager relation

8 Onsaeger relations: example 1) temperature differences produce heat currents ) pressure differences produce matter currents 3) heat flow per pressure difference = matter flow per temperature difference Lars Onsaeger, Nobel Prize in Chemistry, http.en.wikipedia.org/wiki/onsaeger_reciprocal_relations 15 outline 1) Introduction ) Coupled flow euations 3) hermoelectric devices 4) Discussion 5) Summary his work is licensed under a Creative Commons Attribution- NonCommercial-ShareAlike 3.0 United States License. 16

9 thermoelectric power generation hot N-type P-type cool (ambient) cool (ambient) 1) How much heat can be converted into electricity? (what determines the efficiency?) 17 thermoelectric cooling cool N-type P-type hot hot 1) What determines the maximum temperature difference? ) How much heat can be pumped? 3) What is the coefficient of performance? 18

10 inside view of heat absorbtion/emission cool N P hot 19 simplified E device (one leg) hot LH N-type LC cool 0

11 simplified E cooling device (one leg) 1) heat extracted from the cold side hot LH N-type = ) heat pumped by Peltier effect cool LC - 3) heat diffusing down the thermal gradient - 4) heat generated by Joule heating 1 simplified E cooling device (one leg) hot LH N-type LC cool Δ max = 1 Z LC Z = S σ n n κ E figure of merit (FOM)

12 cooling efficiency (COP) hot LH N-type LC cool COP at maximum cooling power: η = Q C max P in = f ( LC, LH, Z ) 3 E power generation hot Similarly, an analysis of the power conversion efficiency, N P cool (ambient) shows that it is also determined by the E figure of merit, Z. 4

13 FOM dimensionless figure of merit he higher the Z figure of merit, the more efficient a E device. 1) What material properties are needed for a high Z? ) Given a material, how can we optimize Z? 5 FOM Numerator: Mostly determined by position of band edge and E F. Similar for most materials. Need a large no. of channels (large M, E F ). Need large m.f.p. (mobility) Denominator: 6 Mostly determined by lattice thermal conductivity. (Lecture 9 in notes)

14 FOM: PF vs. Fermi level E E E F E F k k 7 FOM: PF vs. Fermi level he peak PF occurs when E F is near the band edge. 8

15 outline 1) Introduction ) Coupled flow euations 3) hermoelectric devices 4) Discussion 5) Summary his work is licensed under a Creative Commons Attribution- NonCommercial-ShareAlike 3.0 United States License. 9 basic euations of thermoelectricity E x = ρ n J x + S n d L dx J x = π n J nx κ n d L dx We can write these euations in vector notation as: E = ρ n J + Sn L J = π n J κn L or in indicial notation as: 30 E i = ρ n J i + S n i L J i = π n J i κ n i L i = x, y, z

16 transport tensors E = ρj + S L J = π J κ L n E = ρ J + S J = π J κ n E x E y E z ρ 11 ρ 1 ρ 13 = ρ 1 ρ ρ 3 ρ 31 ρ 3 ρ 33 J x J y J z E i = 3 ρ ij J j j=1 E i ρ ij J j summation convention 31 coupled current euations again E = ρ J + S J = π J κ n J Q i = π ij J j κ n ij j For isotropic materials, the tensors are diagonal. Kronecker delta J i = π 0 J j K 0 e j 3

17 form of the transport tensors E i = ρ ij J j + S ij j J i = π ij J j κ ij e j For isotropic materials, such as common, cubic semiconductors, the tensors are diagonal (under low-fields). For a given crystal structure, the form of the tensors (i.e. which elements are zero and which are non-zero) can be deduced from symmetry arguments. (See Smith, Janak, and Adler, Chapter 4.) he transport tensors can be readily computed by solving the Boltzmann ransport Euation (BE). 33 measuring S thermocouples to measure temperature V meas = ΔV s ΔV l sample ΔV s = S s Δ ΔV l = S l Δ contact 1 contact heater V meas = ( S s S l )Δ 34

18 what about the valence band? E i = ρ ij J j + S ij j J i = π ij J j κ ij e j 35 what about the valence band? 36

19 treating both bands: conductivity 37 treating both bands: S E E E F σ ( E)dE E L S tot = 1 S tot = E 1 σ tot E V E E F L E E E σ ( E)dE F σ ( E)dE L σ tot E C S tot = S n σ n + S p σ p σ tot 38

20 treating both bands: κ κ n = σ tot L L tot L tot >> L n + L p bipolar thermodiffusion H.J. Goldsmid, hermoelectric Refrigeration, outline 1) Introduction ) Coupled flow euations 3) hermoelectric devices 4) Discussion 5) Summary his work is licensed under a Creative Commons Attribution- NonCommercial-ShareAlike 3.0 United States License. 40

21 summary E i = ρ ij J j + S ij j J i = π ij J j κ ij e j his work is licensed under a Creative Commons Attribution- NonCommercial-ShareAlike 3.0 United States License uestions 1) Introduction ) Coupled flow euations 3) hermoelectric devices 4) Discussion 5) Summary 4