lecture notes on thermoelectricity

111Equation Chapter 1 Section 1K Thursday, august 22, 2013

KAM office hours = T at 2 PM, or T 3-4.

KAM became interested in thermoelectricity in Saclay, France. Europe is very interested in thermoelectricity. You use a temperature-difference to make a voltage. If you know something about thermoelectricity that KAM does not know, bring it to class!

Old textbooks on thermoelectricity = not useful; they do not take the nano approach.

Current application: space travel. Desired application: as common as being in your car.

C. Goupil, Ch. 13; DOI 105772/12988 - SSP - thermodynamics of thermoelectricity.2003 - Caltech - Snyder and Ursell - Thermoelectric Efficiency and Compatibility.

Thermodynamics of thermoelectricity

Theremoelectric engines are currently inefficient. Associated with TE engine is figure of merit.

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Consider an ideal gas in a box of volume V. pressure of gas can be computed. However, if gas-particles are charged, you will have chemical potential. Chemical potential then becomes related to the bias across the gas.


Problem: Qualitatively describe how a temperature-gradient (maintained by a hot and cold reservoir separated by a thermoelectric junction) elicits a voltage-gradient. Solution: You have hot vs. cold reservoir, and the thermoelectric junction, appearing as,


The temperature-gradient elicits a density gradient. You have a compressible Fermi-gas of electrons that is hot, and a similar gas of electrons that is cold. The resulting density-gradient, in turn, elicits a number-current (diffusion).

Problem: find an order-of-magnitude estimate for the current across the junction 14. solution: Consider

transmission of a particle of energy to be an event with probability . Consider left-center hopping to

have a parameter , and center-right hopping to have a parameter . Let the density of states

for left-right transmission be a Lorentzian function, , in which , the

license to switch the variables being a consequence of time-reversal symmetry. We realize the simplest

satisfying this symmetry is . Then, the left-right transmission function (which has no dimensions), using the pi-theorem, is approximately,


The non-interacting transmission function is subsequently computed using the series-conductor-addition-law,

, and this appears as,


Finally, the thermodynamic probability is given by , which implies that we treat the electron gas as a grand canonical ensemble. By Kirchoff’s rule, we subtract left-center and center-right currents to constitute the net current,


That the number-current is a function of the indicated thermodynamic variables allows us to establish derivative-relations, if we wanted to.

Problem: Show that for , you have zero number current. Solution: Let the chemical potential gradient

be entirely due to a temperature gradient. Then, , and 17 immediately yields .

Note, also, that causes the following integral1 to vanish,

1 This is not because the Lorentzian is an even function; replacing with a cosine-function yields nonzero current.


Equilibrium: when left and right chemical potentials balance out. That is when the thermoelectric effect ceases.

So, you are basically making a out of a . These are encapsulated by .

Problem: Let and , and similarly and . Series-expand the

about and , truncating at 1st order. Solution: we shall need the derivatives, in which ,


We shall also need the fact that the 1st term of any series expansion that is between the difference of two functions always vanishes,


Thus, the series-expansion appears as,


Plotting the prefactor/dimensionless-scaling function of , we have,

10 5 5 10x k B T

0.2

0.1

0.1

0.2

f sc a le x 0.25 xSech x22


Problem: write down the units of number-flux, and energy-flux. Solution: the units of flux of [stuff] is [stuff] per unit time per unit (normal) area, so we have,


Problem: consider a quasi-static system. Write an expression for the entropy flux. Solution: write,


Problem: introduce so-called thermodynamic “forces”: and , which elicit the indicated currents. Let there be coupling between the forces. Write a linear “constitutive relation”; e.g., Fick’s Law of diffusion, for each current. Solution: since there is coupling between the forces, we will have terms making each current which is due to the other elicitor of the force. Specifically,


Together, we have the linear relations,


Problem: consider , and consider the new constitutive relations,


Find the in terms of the . Solution: Writing , we

compute of 116 and then 117 as,


From 118, considering the variations and as independent, we read off the relations,

19119\* MERGEFORMAT(.)

Carrying out the same procedure for (favouring occurrences of ), we have,


From 120, we read off,


Problem: Write the inverse-transform of the expressions 119 and 121. solution: The is already solved for, so we just need to invert a 3x3 system. However, be careful to make sure all elements of the column-vector have matching-dimensions, or else the coefficient-matrix won’t make sense,


Problem: Ohm’s Law is . Compute the isothermal ( ) electrical conductivity in the

system described by 117 (where we have the symmetry ), realizing . Solution: computing

from 118,


Problem: Fourier’s law is . Compute the iso-electric thermal conductivity2 . Solution: using 120

in which ,


Problem: Fourier’s law is . Compute the thermal conductivity , this time using . Solution:

the condition , using 118, allows us to eliminate in favour of . Repeating 124,


Problem: the Seebeck coefficient3 is defined as ; re-express this in terms of the .

Solution: the condition allows use of 118 to eliminate in favour of . Repeating 125, it is trivial to write,


Problem: the Peltier coefficient is defined as ; re-express this in terms of the . Solution: using 118 and 120, we effect direct computation and obtain,

27127\*MERGEFORMAT (.)

Interlude: By 123, 124, 125, 126, and 127, we can write,

2 Actually, there are two different thermal conductivities, since you are holding different thermodynamic variables constant.3 The symbol for the Seebeck coefficient in


Problem: Compute the entropy flux density, re-expressing in terms of the . Solution: we write,


Problem: Assume a continuity equation for the entropy flux density 129, and compute ; this is computed as,

. Solution: let the continuity equation appear as ; we directly compute, and use the continuity for

particle-number and energy, , and we get,


The thermoelectric engine: consider engine: a device between a hot and cold reservoir. A load is put between the hot and cold reservoirs. It is equivalent to a battery V0, a resistor Rin, and a resistor RL, all in series, assuming the linear/quasi-static4 regime. These resistances are the internal and load resistances, respectively,


Limits: you have , the broken-circuit limit, where the maximum voltage V0 is across the

load. The opposite limit is , the short-circuit limit, where the maximum current flows through the entire loop.

4 Completely-unrealistic; a battery is manifestly a non-equilibrium object.

IV (current-voltage) characteristics: We are interested in power: . Maximum power occurs

when , by maximum-power-transfer. We have,


Problem: The thermoelectric figure of merit is defined as . Write this in terms of the ,

and also show that . Solution: directly computing, we have,


Having established 133, we can begin from the LHS of , and use 128 throughout, as,


We also write,


Making 133 as large as possible, even so we’re in the linear/equilibrium-regime, is equivalent to maximizing the non-equilibrium efficiency 12.

Result: In 133, we immediately see that we want a high electrical conductivity and a low thermal conductivity.

Wiedemann Franz law is a problem: The problem is that we have the correlation ,

in which we have the Lorenz number . This requires 3 possible approaches (1) use non-Fermi liquids (2) use the Kondo-regime, (3) use nano-systems. All three of these approaches are the topic of intense research.

Thermodynamic potential: G. Snyder and T. Ursell, Phys Rev. Lett 91, 148301 (2003),

Problem: compute for a non-radiating closed system. Solution: that the solution is closed means we have the following statements of mass and energy conservation,


Poynting’s theorem (another statement of conservation) for a non-radiating electromagnetic system says

, in which, for a chemical voltage and , the conservation laws 136 imply,


Problem: compute the relative efficiency for the thermoelectric device 14. solution: Using

(from 137), , , and considering a quasi-static/reversible/linear thermoelectric device so

that , the relative efficiency5 is computed as,


Problem: introduce a new thermodynamic potential, , and use this to define

another thermodynamic potential, . Show that . Solution: using Fourier’s law 125

for , we have,


Problem: re-express the entropy current in terms of the potentials and , and subsequently compute

. Solution: using 139,


Problem: compute the “relative efficiency”, , in which (the electric power), and in

which . Solution: Using 139 in 138, and using , and , and

, we have,


5 This is the efficiency relative to the Carnot efficiency: , in which .

Problem: Re-write the relative efficiency 141 in terms of the thermodynamic potential of 139. solution:

recalling that , we solve the definition of u, 139, for by inserting (see 136) and

(see 128), yielding,


Problem: Show that corresponds to , the Carnot efficiency. Solution: it is trivial to write,


Not merely power and efficiency, but also “compatibility” for a good heat engine; we write,



462Equation Section (Next)Tuesday, august 27, 2013

Problem: draw a schematic of a thermodynamic engine, operating between a hot and cold reservoir. Solution: just draw,


Problem: plot the Fermi functions for low and high temperature, but with matching chemical potential.

Solution: let the matching chemical potential be . Then,

2 1 0 1 2 W

0.2

0.4

0.6

0.8

1.0

f FD f FD ,T

F 0.5

Hot k B TW 8

Cold k B TW 0.1


Problem: recall the order-of-magnitude estimate for current from the previous lesson, 17. Compute the current

resulting from . Then, write down a thermodynamic condition for . Solution: by direct computation,


The condition for the current to vanish is for the Fermi-function difference to vanish at ,


Problem: consider 247 as a single-loop electric circuit. Write a schematic illustrating the action of an ammeter

measuring current, and another schematic illustrating a voltmeter measurement across the load . Solution: the current-measurement and voltage measurement respectively appear as,

vs. 51251\* MERGEFORMAT (.)

The measurement is indicative of the voltmeter’s infinite internal resistance.

Problem: Let the load’s resistance be constituted by a chemical potential gradient. Then, the condition of

“dead thermoelectric battery” ( ) is characterized by being a maximum:

. Note that also happens when the condition 250 is satisfied for the special case of a delta-function potential, and when an analogous condition for other transmission functions is satisfied. Plot

vs. for a fixed temperature-difference. Solution: we use the following commands,


This then yields,

0.0 0.5 1.0 1.5 2.00 W

0.05

0.10

0.15

0.20

0.25

I N 0

Fresh battery

I N 0

Fresh batteryI N 0 IN = 0 --> dead battery


One can also see,


Without loss of generality, we can stick to case 1 of 254, which is less confusing.

Problem: Write down chemical potential and load-resistances for three separate conditions: (1) fresh battery, (2) discharging/ operation, and (3) dead battery. Solution: write,


Problem: efficiency is defined as . Use the 2nd law of thermodynamics to derive the Carnot efficiency. Solution: the 2nd law says,


Directly computing the efficiency, we get,


Interlude: For a non-interacting system out of equilibrium6, you have number and thermal currents. The following equations can be heuristically derived by using the reasoning discussed in 15 through 17, although they are the correct expressions,


Problem: compute the power output and efficiency using 258. solution: write,


Problem: compute and for and , and try integrating for a “toy” transmission function

. Solution: looking at 111, and putting this into 258,



Problem: write the coupled linear response formulae 117 in terms . Solution: write,


Interlude: In contrast to the dimensions that may be indicated in 262, the transport coefficients scattered throughout 128, in terms of these new integral-functions, are,


6 Our system 247 is manifestly out of equilibrium, since a disparity in temperature and chemical potential is what drives the effect.





Interlude: target ZT for commercial viability,


Problem: compute the chemical potential difference in the case . Solution: the zero

current condition, from 262 (e.g., from 117) says ; we then write,


Problem: consider the case , so that there is an arbitrarily-small (but nonzero) number current.

In this case, compute the quasi-static efficiency. Solution: we can use to simplify

the definition of efficiency 259; using the abbreviation , we write,


Problem: maximize the efficiency 270 with respect to x. State the large- and small- of the extreme-value of

you find. solution: extremize 270 by taking a derivative of the relative7 efficiency;

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Then, the efficiency in all cases is,

7 Efficiency scaled to the Carnot efficiency; this is simpler.


733Equation Section (Next)Thursday, august 29, 2013

“thermodynamic efficiency at maximum power”, C van der Broeck, PRL 95, 190602 (2005)“thermodynamic bounds on efficiency for systems with broken TRS”, C. Benenti et al, PRL 106, 230602 (2011)“Strong bounds on onsager coefficients and efficiency for 3 terminal thermoelectric transport in magnetic field” K brandner et al., PRL 110 070603 (2013).Dubi and DiVentura, RMP, 83, (2011)Rego and kirezenow, PRL 81, 232, (1998).Schwab et al, Nature, 404, 974, (2000).1996 - Tennesee - Mahan and Sofo - The best thermoelectric

Claim: increase efficiency if time-reversal symmetry broken?

3-terminal transport is also way to increase TE efficiency.

Preliminaries: Efficiency is work done (electricity) divided by heat flow associated with this work,


Thermodynamic efficiency at maximum power: you have,


Problem: rewrite all thermodynamic forces and fluxes in such a manner that one can write power as

proportional to . Solution: write the linear relations 262 (or 117) as,


Then, the power appears as,


Problem: consider the condition . Compute the condition on resulting from this. Solution: you

have ; this is rearranged as,


Problem: relate to the which maximizes the power (found in 1st homework). solution: write,


You can’t reach the Carnot efficiency, even with the maximum figure of merit.

Interlude - Curzon-Ahlborn limit: Not possible to reach Carnot efficiency, . In fact, if you maximize power, you do not8 maximize efficiency, and vice versa.

Non-linear efficiency: the following is not a very rigorous definition, but check to see if you agree. Let our

engine operate between two temperatures , as,


Now: let temperature be a function of the spatial coordinate y, , which runs between the limits indicated in 380. Rewrite the schematic 380 as,


Let an infinitesimal of heat energy produce an infinitesimal of work energy, for all time. Then, ,

which in turn implies . Series-expanding the function for ,

8 Recall: you can have . We are in nano-systems, where Wiedemann-Franz law is no longer valid

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Then write,


Eventually, you will show in your homework the following functional form for the mazimized efficiency,


Breaking time-reversal symmetry using magnets

Problem: On turning on a magnetic field, you can increase the efficiency by breaking time reversal symmetry. Draw a schematic of a thermoelectric junction with such an applied magnetic field. Solution: Both electric and heat currents flow along the horizontal axis. The system is in contact with left and right reservoirs at temperatures TL and TR and chemical potentials L and R.


Problem: Even though fluxes are one dimensional, the motion inside the system can be two or three dimensional. Write the linear response equations for the particle and heat fluxes. Solution: write,


In 387, and Jq are the particle and heat currents, B an applied magnetic field or any parameter breaking time

reversibility (such as the Coriolis force, etc.), and,


Under time-reversal, the Onsager reciprocity becomes broken9,


Problem: here, the 2nd law of thermodynamics is,


Write the conditions 390 places on the Onsager coefficients. Solution: the fluctuations and are independent, so we easily derive, for these respective cases,


You also have,


Problem: prove that . Solution: the proof is given in SMT 06, and the final result is in SMT 06 – 367 – [120.11].

Problem: recall ohm’s law, , the isoelectric ( ) Fourier law , the iso-

number current ( ) Fourier law , and the relation ; these constitute four (4)

equations which can be used to find the four in terms of the constitutive coefficients . Do

this. Solution: The Onsager coefficients are related to the familiar transport coefficients as follows,


Note that the Onsager-Casimir relations,


Problem: compute the power from the linear response equations 387, rather than 262 (in other words, use Benenti’s notation). Subsequently, extremize with respet to x1. solution: write,

9 Proof of this, and of the claim in 389, appears in SMT 06 – 367 – [120.11].


Problem: show that the efficiency at maximum power depends on two parameters, x and y,

; 96396\* MERGEFORMAT (.)

You have,

= … 97397\* MERGEFORMAT (.)

You write,


Problem: show that 392 implies a bound on y of 396. solution: the result is,

Now, define


Writing efficiency,



Plotting h(x),

10 5 5 10x

1

1

2

3

hx

1

hx


4 2 2 4x

0.2

0.4

0.6

0.8

1.0

xC

M xx


Extremize, and you find the following expressions for the power and efficiency,


This is the maximum efficiency without TRS. Something provocative about 3104: it seems you want x > 1?

Figure of merit with magnetic field: you have,


Three terminal devices

Problem: Some people are excited about having a 3rd terminal in the thermoelectric junction. Draw a schematic of such a device. Solution: Sketch of a thermoelectric device in the presence of a magnetic field B. The entire dashed box represents the conductor C, which is connected to two reservoirs. For the special case of the three terminal model, the conductor consists essentially of a scattering region and an additional probe terminal.


Problem: the center-site retarded Green function for a 2-terminal junction10 with a t-stub and an interaction

appears as,


Write the Green function for the 3-terminal device shown in 3106. solution: treating 3107 as an isolated Green function, while treating the coupling to the lead P as an interaction, we just need the Green function for the lead P. Indeed, this is gotten from the work of SPH11,

10 See 2013 - nartowt - t-stub thermoelectric junction with e ph interaction.11 See 1992 - OH State - Hershfield, Davies, Wilkins - Resonant tunneling through Anderson impurity – current.


Then, using , we have,


Phonons

Effect of phonons: Looking at , we see phonons are bad. However, in 3106, the 3-terminal device could allow phonons to carry a heat current. They could be “tuned” to actually be a good thing.

Seebeck coefficient: linearized inegrals,


Problem: d

Now, recall Dubi and DiVentura, RMP, 83, 2011. Write Lorentzian transmission function,


However, we have the unphysical limit of yielding a finite value of . Problem: if you go to a scale

where , linear series-expansions break down.

How to get a high ZT (i.e., good thermoelectrics): (1) large variations of T(E) near Fermi energy. (2) violation

of Wiedemann-Franz law; i.e., (in appropriate SI-units), (3) phonon thermal conductivity,


Above = Rego and kirezenow, PRL 81, 232, (1998).

Schwab et al, Nature, 404, 974, (2000).

Problem: when integral 3112 is done, you sum over all phonon modes. You have big . You must take into

account . That must be properly tuned for a good thermoelectric.

The best thermoelectric: 1996 - Tennesee - Mahan and Sofo - The best thermoelectric. Write the figure of merit

with phonon-contribution, using and , as,


It’s clear to see that you want to extremize 3113. The thing that does this is a delta-function T(E),


Howver, this would produce a very low power thermoelectric.

Advantages of nano-systems: (1) the can be nano-engineered, (2) no Wiedemann-Franz law, (3) phonon-conribution to thermal conductivity can be made small, (4) you can have inherently sharp gradients at such a tiny length scale, which mercilessly demands one discard the quasi-static approach we just finished discussing. (5) in bulk, large efficeicny means large power, but these might become independent on the nano-scale. (6) You have nano-currents and nano-powers; you can scale these up as ~ N = 6.02 x 1023.

lecture notes on thermoelectricity

Documents

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number current

density gradient

chemical potential gradient

current application

lorentzian function

electron gas

net current