Effet Nernst et la figure de mérite thermomagnétique dans les semi-métaux

Transcription

1 Effet Nernst et la figure de mérite thermomagnétique dans les semi-métaux Kamran Behnia Laboratoire Photons et Matière Ecole Supérieure de Physique et de Chimie Industrielles - Paris Alexandre Pourret, Romain Bel & Marie-Aude Méasson (Paris) Collaborateurs: Jacques Flouquet & Pascal Lejay(Grenoble) Hideyuki Sato (Tokyo) Yakov Kopelevich (Sao Paulo)

2 Thermoelectric coefficients In presence of a thermal gradient, electrons produce an electric field. B r E r x Seebeck and Nernst effect refer to the longitudinal and the transverse components of this field. hot E r y J Q cold r T S= Ex xt N Ey xt = [ Ey ν = B T ] z x

3 Set-up for monitoring thermal(κ xx, κ xy ), thermo-electric (S, N) and electric (σ xx, σ xy ) conductivity tensors Heater SC wires Thermometers dv (nv) T(s) DC voltages of the order of 1 nv resolved! 20 mm

4 Thermoelectricity in extreme conditions (a network) LPEM-ESPCI, Paris (down to 0.15 K, up to 12 T) SPSMPS-CEA, Grenoble (J. Flouquet) (down to 0.1 K, up to 16 T) LNCMP, Toulouse (C. Proust) (down to 1.5 K up to 62 T[pulsed field]) LCMI, Grenoble (user facility) (down to 0.15 K up to 28 T) NHMFL, Tallahassee (user facility) (down to 0.35 K up to 33 T[45T planned])

5 Nernst effect in a single-band metal Absence of charge current leads to a counterflow of hot and cold electrons: Almost undetectable in gold!(<0.5 nv/kt) B r e - J Q 0 ; J e = 0 ; E y = 0 e - J Q E y r T In an ideally simple metal, the Nernst effect vanishes! («Sondheimer cancellation», 1948)

6 Ambipolar Nernst effect The Nernst contribution os of hole-like and electron-like carriers add up! B r e- J q h + R N / H (µv/kt) H (10-10 m 3 / C) T 5T NbSe 2 J Q 0 ; J e = T(K)

7 What does «Sondheimer cancellation» mean? T E T J T E J Q e = = r r r r r r κ α α σ J e =0 If the Hall angle, Θ H, does not depend on the position of the Fermi level, then the Nernst signal vanishes!

8 The Nernst coefficient can be large in some metals 1000 ν (absolute value) (µv K -1 T -1 ) PrFe 4 P 12 URu 2 Si 2 CeCoIn 5 CeRu 2 Si 2 NbSe T(K) 10 50

9 The enigmatic order of URu 2 Si 2! Wiebe et al., 04 Palstra et al., 85 A lot of entropy is lost, but only a tiny magnetic moment appears!

10 Exceeds by an order of magnitude the signal in high-t c superconductors!

11 Pr-filled Skutterudites: a dazzling variety of ground states! Sato 03 What happens at T Q in PrFe 4 P 12?

12 How can quasi-particles produce a Nernst coefficient of this size? A crude estimation : N= 285 µv/k X Θ H X k B T/ ε F A low Fermi energy and a large Hall mobility produce a giant Nernst effect!

13 Back to 1886! ν (absolute value) (µv K -1 T -1 ) Bi PrFe 4 P 12 URu 2 Si 2 CeCoIn 5 CeRu 2 Si 2 NbSe T(K) The Nernst effect in semi-metallic Bismuth is still more than one order of magnitude larger!

14 Roughly, the Nernst coefficient tracks ω c τ/ε F N ~ π 2 /3 k B /e ω c τ / Ε F Bismuth URu 2 Si 2 PrFe 4 P 12 k F (nm -1 ) m* (m e ) n( per f.u.) ω c τ (1T) E F (K) ω c τ /E F

15 Thermomagnetic figure of merit: ZT ε = 2 N σt κ Τ T = max ZT ε 2

16 Infinite stage Ettingshausen cooler T. C. Herman et al., Applied Physics Lett. 4, 77 (1964) ( ) = z( L ) (0)[ z(0) x z x z ] x L x 100 degrees of cooling! But, with B =

17 Ettingshausen cooling with a permanent magnet K. Scholz et al. J. Appl. Phys. 75, 5406 (1994) Bi 0.97 Sb 0.03 B=0.75 T T= 42 K at T=160K

18 What sets the magnitude of ZT ε in a metal at low temperatures? Let us forget phonons! κ L = σ T = V K [The Wiedemann-Franz law] 2 2 ZT ε = 2 N σt κ N L [ = 156µ V / K] ZT 0 ε 1

19 In PrFe 4 P 12, this threshold is approached! An interesting thermomagnetic material at cryogenic temperatures!

20 Peltier cooling with CeB 6 at cryogenic temperatures Harutyanyan et al., 2003

21 Why Bismuth does NOT qualify [for low T Ettingshausen cooling]? In presence of a magnetic field, Bismuth behaves like an insulator! Magnetoresistance is very large! Heat is almost entirely carried by phonons! Theultimatefault: The lightness of electrons leading to a large cyclotron frequency! ω c = eb m* ρ ( ω τ ) c 2

22 A comparison of the thermomagnetic figures of merit at 1.2 K 0.20 ZT ε 0.15 N(µV/K) T=1.2 K Bi PrFe 4 P B(T) (N 2 σ T )/ κ E-3 PrFe 4 P 12 Maximum ZT ε ~ 0.19! 1E-4 Bi B(T) 10

23 Heavy-electron semi-metals The combination of small k F and large m* is not very common CeNiSn PrFe 4 P 12 URu 2 Si 2 Carrier density /f.u. Effective mass : m e emerge as promising candidates for Ettingshausen cooling!

24 Ettingshausen vs. Peltier cooling Possibility of using an infinite stage cooler ZT ε depends on the electronic mean-free-path But needs magnetic field! 1T (the field provided by permanent magnets) is an important threshold!

25 Summary Low Fermi energy and high electronic mobility are the ingredients to produce a giant Nernst effect! Semi-metals with heavy electrons may prove to be useful for thermomagnetic cooling at low temperatures The thermomagnetic figure of merit of PrFe 4 P 12 appears large enough to attain subkelvin temperatures without He 3