12-1

Chapter 12 Thermoelectric Transport Properties for

Contents

Chapter 12 Thermoelectric Transport Properties for Electrons...... 12-1 Contents ...... 12-1 12.1 Boltzmann Transport Equation ...... 12-2 12.2 Simple Model of ...... 12-4 12.2.1 Electric ...... 12-4 12.2.2 Electrical Conductivity ...... 12-5 Example 12.1 Relaxation Time of ...... 12-7 12.2.3 Seebeck Coefficient ...... 12-8 Example 12.2 Seebeck Coefficient of Gold ...... 12-10 12.2.4 Electronic Thermal Conductivity ...... 12-10 Example 12.3 Electronic Thermal Conductivity of Gold ...... 12-12 12.3 Power-Law Model for Metals and ...... 12-12 12.3.1 Equipartition Principle ...... 12-13 12.3.2 Parabolic Single-Band Model ...... 12-15 Example 12.4 Seebeck Coefficient of PbTe ...... 12-17 Example 12.5 Material Parameter ...... 12-24 12.4 Electron Relaxation Time ...... 12-25 12.4.1 Acoustic Phonon Scattering ...... 12-25 12.4.2 Polar Optical Phonon Scattering ...... 12-26 12.4.3 Ionized Impurity Scattering ...... 12-26 Example 12.6 Electron Mobility ...... 12-27 12.5 Multiband Effects...... 12-28 12.6 Nonparabolicity...... 12-30 Problems ...... 12-33 References ...... 12-35

12-2

12.1 Boltzmann Transport Equation The flow of electrons in solids involves two characteristic mechanisms with opposite effects: the driving force of the external fields and the dissipative effect of the scattering of the electrons by phonons and defects. The interplay between the two mechanisms is described by the Boltzmann transport equation. The distribution function f gives the probability of finding an electron at time t, at position r, with momentum p. The Boltzmann transport equation [1] accounts for all possible mechanisms by which f may change. The electron flow in a can be affected by applied fields, temperature gradients, and collisions (scattering). Consider the electron distribution f (t, r, p). We expand the total derivative of f (t, r, p) as

f f f df  dt  dr  dp (12.1) t r p

With collisions, using p  k , we have [2]

df f f f  f    k  r    (12.2) dt t k r  t coll

This is the celebrated Boltzmann transport equation, which is very difficult to solve. In many problems the collision terms may be treated by introduction of the electron relaxation time  as

 f  f  f     o (12.3)  t coll 

This is called the relaxation time approximation (RTA). Here f and fo are the perturbed and unperturbed distribution functions. The latter is the Fermi-Dirac distribution in thermal equilibrium. The electron relaxation time  is the average flight time of an electron between successive collisions (scattering events) with electrons, phonons, or impurities.

We must solve this equation to obtain f – fo in terms of the internal electric field and temperature gradient. It will be assumed that the conductor is isotropic and that the electric field and flows are in the direction of the x axis so that r denotes x. And there is no magnetic field applied. The 12-3

momentum of a free electron is related to the wavevector bymv  k . In an electric fieldΕ with charge e, the force F on an electron is

dv dk F  m    eΕ (12.4) dt dt

which is the Coulomb force (an external electric field Ε causes electrons to move to the opposite direction). Using Equations (11.6) and (12.4), the force related term in Equation (12.2) is expressed as

f f k  eΕ  v (12.5) k E

Since f is a function of E  EF  kBT as shown in Equation (11.27), we introduce

  E  EF  kBT . Then,

 1 EF E  EF    2 (12.6) T kBT T kBT

And

f f T f  T v  v  v (12.7) x T x  T x

Using these equations, we have [3]

f  f f  E E  E T  o  v eΕ  F  F  (12.8)  E  x T x 

We may relate the gradient of the EF x to the electric field Ε if the system is on

open circuit in thermal equilibrium (no temperature gradient and no electron flow, f  fo  0 ). Then, the gradient of the Fermi energy is notably a form of an electric field. But the temperature 12-4 gradient also causes the gradient of the Fermi energy if the system is on open circuit with a temperature gradient which can be seen in Equation (12.8). Therefore, the total electric field Ε is the sum of the external electric field and the gradient of the Fermi energy as (Ashcroft and Mermin (1976) [3]

1 E Ε  Ε  F (12.9) e x

When no external electric filed is applied, the total electric field Ε becomes

1 E Ε  F (12.10) e x

In general it is known that the difference (f – fo) between the perturbed and unperturbed distributions is relatively small compared to fo. Then, the f in the right-hand side of Equation (12.8) may be replaced by fo. It is also known that the term fo in the left-hand side will not make any contribution. Bearing in mind that the electric field and temperature gradients lie along the x-axis, Equation (12.8) reduces to

f  E E  E T  f  v o  F  F  (12.11) E  x T x  which is a different version of the Boltzmann transport equation (BTE) with the relaxation time approximation (RTA).

12.2 Simple Model of Metals

12.2.1 Electric Current Density If n electrons with charge e move in the x-direction with a velocity, the electric current density j is

12-5

j  nevx (12.12)

From Equation (11.28) in thermal nonequilibrium, we have

 j   ev g E f E dE  x     (12.13) 0

From Equation (12.11), we have

 f  E E  E T  j   ev2 g E o F  F dE  x     (12.14) 0 E  x T x 

Using the of Equation (11.24) and also assuming a cubic isotropic structure

2 2 2 2 2 ( vx  2E 3m and v  vx  vy  vz ), the electrical current density j is

22m1 2 e  3 f  E E  E T  j   E 2 o F  F dE 2 3    (12.15) 3  0 E  x T x 

12.2.2 Electrical Conductivity The electrical conductivity  is obtained from Equations (11.1) and (12.10) in the absence of a temperature gradient.

j j    Ε 1 EF (12.16) e x

Equation (12.15) to

 f    E o dE (12.17) 0 E where

12-6

22m1 2 3  E  e2 E 2 (12.18) 3 23

We use the classical asymptotic formula (Taylor series) for EF kBT >>1 (metals) in [4]

  2n f 2n d E    E 0 dE  E  2 C k T F     F   2n  B  2n (12.19) 0 E n1 dEF where

 2 7 4  S1 C  C  1 2 12 4 720 (12.20) C2n   2n , so that , S1 S

Then, Equation (12.17) can be expressed as

 2 2 fo  2   E   E dE  EF  kBT  (12.21)  E 6 E 2 0 EEF

Neglecting the second order term (linearized assumption), the electrical conductivity  is expressed as

22m1 2 3    E   e2E 2 (12.22) F 3 23 F

This tells us that the electrical conduction can take place only nearby the Fermi energy, where the electrons in the conduction band can move from one energy state to another (partially filled). Using Equation (11.24), Equation (12.22) can be expressed as

   E   e2 gEv2  F x EEF (12.23)

2 Using Equation (11.33a) and vx  2E 3m, we have

12-7

ne2    ne (12.24) m which is known as the Drude model, and the electron mobility  becomes

e   (12.25) m

Note that, in this simple model of Equations (12.24) and (12.25), all free electrons contribute to the electrical current. The electron mobility is proportional to the relaxation time and inversely proportional to the (effective) mass m. The relaxation time at room temperature is typically 10-14 to 10-15 sec.

Example 12.1 Electron Relaxation Time of Gold Estimate the electron relaxation time  and the electron mean free path  for gold if its electrical conductivity  of 4.55 × 105 (cm)-1 is given.

Solution: From Equation (12.22),

3 23    3 1 2 2 2 22m e EF

We assume that the effective mass in gold is equal to the electron mass and the Fermi energy EF in gold is taken from Example 11.1, which is 8.84 × 10-19 J.

2 34 7 1 3 6.6210 Js 2  14   4.5510 m 1 2 2 3 2  2.7410 s 229.10910-31 kg 1.6021019 C 8.841019 J From Example 11.1, the velocity of electron is assumed to be the Fermi velocity of 1.39 × 106 m/s. From Equation (11.36), the electron mean free path  for gold is obtained as

12-8

6 14 9   vF  1.3910 m/ s2.7410 s 38.2110 m

Comments: This calculation is fairly crude, giving the electron mean free path of 38.2 × 10-9 m for gold, which may be compared with the phonon mean free path in Chapter 13 (about 6.4 × 10-9 m for PbTe). Nothing can be concluded because of the different materials.

12.2.3 Seebeck Coefficient The Seebeck coefficient (or thermopower)  is obtained with j = 0 from Equation (11.1) as

E   (12.26) T x

From Equation (12.14), using Equation (12.22), we have

1  f  E E  E T  j    E o  F  F dE (12.27) e 0 E  x T x 

We approximate the second term of the right-hand side using the asymptotic formula of Equation (12.19).

 2 f  2  E  E  E  E o dE  k T  (12.28)  F E 3 B E 0 EEF

From Equation (12.27),

2  1 E 1 T  2  E  j   F  E  k T   (12.29) e x F eT x 3 B E  EEF 

From Equation (12.26) using Equation (12.10), the Seebeck coefficient (thermopower) is

 2 ln E    k 2T (12.30) 3e B E E EF 12-9

This is the well-known Mott formula. In spite of the simplifications made it gives a good interpretation and has been widely used in literature.

Using Equation (12.24), we can have a different version of the Seebeck coefficient as

 2  gE 1      k 2T   (12.31) 3e B  n  E   EEF which is an interesting expression where the Seebeck coefficient  is only meaningful near EF and proportional to the density of states g(EF) but inversely proportional to the electron concentration n. When we insert Equation (12.23) into Equation (12.30), we have

 2  1 gE 1 v2     k 2T  x  B  2  (12.32) 3e gE E vx E  EEF

As an approximation, we shall assume that the relaxation time can be expressed in the form of

r   0E where  0 and r are constant for a given scattering process. In many , it seems that for scattering by acoustic-lattice vibrations, r is equal to ˗1/2. And also, for scattering by ionized impurities, r equal to 3/2. We use the proportionality from Equations (11.24) and so on as

1 gE E 2 (12.33)

2 vx  E

  E r

Equation (12.32) becomes with Equation (12.33)

 3  r    2k 2 (12.34)    B   3e EF kBT 12-10

Using Equation (11.33) for the Fermi energy with r = ˗1/2, the Seebeck coefficient is expressed as

2 2 2k    3    B mT   (12.35) 3e2  3n  which is another version of the Mott formula for metals (degenerate). But it is also used for heavily doped semiconductors (nondegenerate).

Example 12.2 Seebeck Coefficient of Gold Estimate the Seebeck coefficient of gold at room temperature.

Solution:

Using Equation (12.34) assuming r = -1/2 and the Fermi energy in Example 11.1,

2 2 2 23 2  kBTo  1.3810 J K 300K V     19 19  1.328 3eEF 31.60210 C8.8410 J  K

The Seebeck coefficient of -1.328 V/K for gold is small compared to the typical value of -200 V/K for . Therefore, gold is not a good material for thermoelectrics.

12.2.4 Electronic Thermal Conductivity The thermal conductivity k is the sum of the electronic and lattice thermal conductivities.

k  ke  kl (12.36)

The lattice thermal conductivity will be discussed in a later chapter. The electronic thermal conductivity is given by

12-11

qe ke  (12.37)  T x

The heat current density (heat flux) qe is the product of electron concentration, drift velocity and total energy transported by an electron, when the current is zero.

qe  nvx E  EF  (12.38)

Then, using Equation (11.28),

 q   v E  E g E f E dE e  x  F      (12.39) 0

Similar to the process of the electrical conductivity , we have

1  f  E E  E T  q   E  E  E o F  F dE e 2  F      (12.40) e 0 E  x T x 

From Equation (12.35), using Equation (12.19), we have

2 2    kBT  EF  1  ke  T    (12.41) 3e2  x  T x E  EEF 

The Lorentz number Lo is defined as

2 2 ke   kB  Lo     (12.42) T 3  e 

For metals the Lorentz number is 2.44 × 10-8 -W/K2. This is known as the Wiedemann-Franz law, which states that the ratio of the thermal to the electrical conductivity is the same for all metals at any particular temperature. Using Equations (12.10), (12.30) and (12.42)), we have

12-12

2 2    kBT  EF  1  ke  T    (12.43) 3e2  x  T x E  EEF 

Using Equation (12.26), this reduces to

2 ke  LoT  T (12.44)

In metals or heavily doped semiconductors, the second term is small and usually neglected.

Example 12.3 Electronic Thermal Conductivity of Gold Estimate the electronic thermal conductivity of gold at room temperature if the electrical conductivity  of 4.55 × 105 (cm)-1 is given.

Solution:

From Equation (12.44) and the result of Example 12.2 for   1.328V K

8 2 7 1 6 2 7 1 ke  2.4410 W / K 4.5510 m 300K  1.32810 V K  4.5510 m 300K  W W W  3.334  2.406104  3.335 cmK cmK cmK

Comments: Note that the second term in Equation (12.44) is negligible. The electronic thermal conductivity of gold of 3.335 W/cmK is large compared to the typical value of 0.01 W/cmK in semiconductors.

12.3 Power-Law Model for Metals and Semiconductors The Fermi energy is much greater than zero (conduction band edge) in metals while it is much less than zero in semiconductors. The former are often called degenerate materials while the latter are called the nondegenerate materials. In this section, we look for a generic parabolic single-band 12-13 model covering both the degenerate and nondegenerate materials. The power-law model assumes that the electron relaxation time is a function of energy as

r    constE (12.45)

where r is called the scattering parameter, and const is independent of energy but it may be dependent on effective mass and temperature. There are three fundamental scattering mechanisms: r = -1/2 for the acoustic phonon scattering (most materials), r = 3/2 for ionized impurity scattering, and r = 1/2 for polar optical phonon scattering.

12.3.1 Equipartition Principle When we considering Equation (11.15), we may introduce the conductivity (or inertial) effective

 mass mc . The kinetic energy of an electron depends only on the temperature and is independent of mass.

1 3 E  mv2  k T (12.46) 2 c 2 B

Based on the equipartition principle, we have

1 1 1 m v2  m v2  m v2 (12.47) 2 x x 2 y y 2 z z

Using p = mv = ħk and Equation (11.15), we have

2 2 2 2 2 2 2 2 2 2 2 2 2 2  k 1  2  kx  ky  kz mx vx myvy mz vz E    mcv       (12.48) 2mc 2 2mx 2my 2mz 2mx 2my 2mz which is equal to

1 1 1 E  m v2  m v2  m v2 (12.49) 2 x x 2 y y 2 z z which, from Equation (12.47), reduces to 12-14

 1 2  E  3 mxvx  (12.50)  2 

We know that the average velocity is

2 2 2 2 v  vx  vy  vz (12.51)

Manipulating this,

 1 2 1 2 1 2   mxvx myvy mzvz  v2  2 2  2  2  (12.52) m m m  x y z    and

 1  1 1 1  v2  2 m v2      x x   (12.53)  2  mx my mz 

Using Equation (12.50), we have

1 1 1  1 1 1  E       v2 (12.54) 2 3 m m m    x y z 

 The conductivity effective mass mI is defined as

1 1  1 1 1  1  1 2                (12.55) mc 3 mx my mz  3 ml mt  and

1 E  mv2 (12.56) 2 c 12-15

2 2 Under an assumption of v  3vx , we have

2 2E vx   (12.57) 3mc

 The conductivity effective mass mc has an expression in Equation (12.55) while the density-of-

 states effective mass md has an expression in Equation (11.25). Their mathematics is rather intractable while their quantities can be calculated as shown. Closed expressions for them is discussed in Appendix G.

12.3.2 Parabolic Single-Band Model

Density of States   We like to use the reduced energy E  E kBT and the reduced Fermi energy EF  EF kBT since the electron energy is approximately the order of kBT . The electron relaxation time  in Equation (12.45) becomes

r r r   const kBT  E  0E (12.58)

The density of states g(E) in Equation (11.24) with the degeneracy is expressed as

3  2 1 1 Nv  2md   g E    k T 2 E 2 (12.59)   2  2   B  2    where Nv is the degeneracy (multiple valleys or the number of bands). The electron concentration n in Equation (11.30) is expressed as

3 1  N  2mk T  2 E2 v d B  (12.60) n  2  2    dE    E EF  2    0 e 1 12-16

For the sake of simplicity, define the Fermi integrals as

 s E  Fs    dE (12.61)  E EF  0 e 1

Using the Fermi integrals, the electron concentration n is expressed as

3 N  2m k T  2 n  v  d B  F (12.62) 2  2  1 2 2   

Electrical Conductivity

In a similar manner in Section 12.2, the electrical conductivity  using Equations (12.57) and (12.61) is expressed as

3 N e2  2m k T  2  3    v 0  d B  r  F (12.63) 2   2    r1 2 3 mc     2 

For this equation, we used an integral relation below:

 f E  E  E 0 dE   f E dE , where  0  0     0     (12.64) 0 E 0 E

Equation (12.63) is expressed in terms of n, e, and  as

 3   2 N  2m k T  e 2  3  Fr1 2     v  d B  F e 0 r   (12.65)  2  2  1 2     2     mc 3  2  F1 2   

Equivalently,

  ne (12.66) 12-17 where  is here a different form of the electron mobility as

e     (12.67) mc where  is the average relaxation time.

2  3  Fr 1 2   0 r   (12.68) 3  2  F1 2

Equation (12.67) is the electron mobility for the power-law model also shown in Equation (12.25).

Seebeck Coefficient

The Seebeck coefficient  is derived in a similar manner of Section 12.4 as

 5   r  Fr 3 2   kB   2        EF  (12.69)  e  3   r  Fr 1 2  2  

The value of kB e is 86 V/K, which gives an idea of the order of the magnitude of the Seebeck Coefficient in many thermoelectric materials.

Example 12.4 Seebeck Coefficient of PbTe PbTe ( telluride) is a widespread thermoelectric material at mid-range temperatures, which is doped by Na () at a doping concentration of 1.3 × 1019 cm-3, having that the degeneracy of the conduction valleys is 4 and the DOS effective mass is 0.12 me. Assuming that the acoustic phonon scattering (r = -1/2) is a dominant mechanism of the relaxation time, determine the Seebeck coefficient of PbTe at room temperature.

Solution:

34 23 19 Physical constant:   6.62610 Js 2 , kB  1.3810 J K , and e  1.602110 C 12-18

25 3  31 Information given: n  1.310 m , Nv = 4, and md  0.12(9.110 kg)

From Equation (12.69), the Seebeck coefficient is given by

 5   r  Fr 3 2   kB   2        EF   e  3   r  Fr 1 2  2  

In order to calculate the Fermi integrals, we need the Fermi energy for the integrals. Since the 19 -3 high doping of 1.3 × 10 cm is used, the criteria of E  EF  kBT >> 1 in Section 11.6 cannot be applied and Equation (11.41) is not used. Instead, we solve Equation (12.60) for the Fermi energy.

3 1   2 2 Nv  2md kBT  E  n  2  2    dE    E EF  2    0 e 1

We want to use Table C-1 for the Fermi integral. Therefore,

3 1   2  2 2 2  2md kBT  E  n  2   2.761   dE    E EF  Nv    0 e 1

From Table F-1, we find by interpolating the 2.761 between 2.5025 and 2.7694 for s = ½ as

 EF  2.2

 Then, the values of the Fermi integrals of F0 and F1with EF  2.2 can be obtained by interpolation between the values in Table F-1. Then, we have

F0  2.3051 and F1  3.9571

The Seebeck coefficient is finally obtained by

 1 5     3.9571 1.381023 J K  2 2  V       2.2  106.4  1.60211019 C   1 3  K    2.3051   2 2  

Comments: The absolute value of -106.4 V/K for PbTe (semiconductor) is much greater than the value of -1.328 V/K for gold (metal). If we use Equation (11.41) for the Fermi energy (this is not correct since the criteria of >> 1 cannot be applied), we have 12-19

3     n  mk T  2  E  k T ln  d B  F B   2    2Nv  2       3 2  25  31 23   23 1.310  0.129.110 1.3810 300   1.3810 300ln 2  0.029eV  24  2 6.6261034 2        

 EF 0.029eV which leads to EF    1.136 kBT 0.026eV

 Then, the values of the Fermi integrals with EF  1.136 are obtained by interpolation between the values in Table F-1. Then, we have

F0  1.415 and F1  1.992

The Seebeck coefficient is finally obtained by

 1 5     1.992 1.381023 J K  2 2  V      1.136  144.7  1.60211019 C   1 3  K    1.415   2 2  

This value of -144.7 V/K differs from -106 V/K by an error of 44%, which can be seen in Figure 12.2.

Electronic Thermal Conductivity

After a lengthy algebra, it comes up with the Lorentz number of the power-law model as

2  7   5    2 r  Fr 5 2 r  Fr 3 2    kB   2   2   Lo         (12.70) e  3   3     r  F  r  F     r 1 2   r 1 2   2   2   

And the electron thermal conductivity of the power-law model is expressed as

12-20

ke   T  Lo (12.71)

 The Lorentz number Lo is plotted as a function of the reduced Fermi energy EF for three different values of r in Figure 12.1. It is seen that, at the high Fermi energies in metals, they converge on the Wiedemann-Franz law of 2.44 × 10-8 -W/K2. On the other hand, as decreases below zero

-8 2 in semiconductors, Lo drops to about 1.5 × 10 -W/K for r = -1/2 which is the case of acoustic- phonon scattering. Note that Lo for the acoustic-phonon reveals the lowest electronic thermal conductivity among the three as shown in Equation (12.71).

Figure 12.1 Lorentz number versus the reduced Fermi energy for r = -1/2, 1/2, and 3/2.

Since the power-law model covers the entire range from metals to semiconductors, it is worthwhile to examine all the thermoelectric transport properties against the Fermi energy. In order to calculate those properties, for example for PbTe, using the material inputs (r = -1/2, NV  4 and

 -14 md  0.22me ) and the constant relaxation time  0 of 6.2 × 10 s. Then the Seebeck coefficient  can be computed using Equation (12.69), the electrical conductivity using Equation (12.63), and the electronic thermal conductivity ke using Equation (12.71). The results are shown in Figure 12.2. One faces the fundamental challenges in having a high dimensionless figure of merit ZT because, as decreases,  increases while  decreases. The net improvement in the ZT can be examined by introducing the power factor  2 . 12-21

Figure 12.2 The Seebeck coefficient, the electrical conductivity, and the electronic thermal conductivity as a function of the Fermi energy at room temperature for PbTe assuming a constant -14 relaxation time  0 of 6.2 × 10 s (realistic value) and r = -1/2 (see Equation (12.42)).

Dimensionless Figure of Merit and Material Parameter

The dimensionless figure of merit is given as

 2T ZT  (12.72) ke  kl

Using Equation (12.71),

 2T  2T ZT   (12.73) TLo  kl TLo  kl 

Inserting Equations (12.63), (12.69), and (12.71) into (12.72) yields

12-22

2  5   r  Fr 3 2   3   2   r    EF  Fr 1 2  2  3   r  Fr 1 2  2   ZT  2 (12.74)  5  2 r   Fr 3 2 1  7   2   r  Fr 5 2    2   3  r  Fr 1 2  2 

where  is called the material parameter as

2 푘퐵 ( ) 휎0푇 (12.75) 훽 = 푒 푘푙

Equivalently,

3 N  2mk T  2 k 2T   v  d B  B 0 (12.76) 2  2   3    mc kl

Note that 0 is defined without the Fermi integral Fr1 2 , so that 0 does not depend on the reduced

Fermi energy. The dimensionless material parameter  was first introduced by Chasmar and Stratton [5]. In Figure 12.3, we show how the dimensionless figure of merit ZT varies with the reduced

 Fermi energy EF for different values of the material parameter . We have supposed that the scattering parameter r has a value of -1/2, as for acoustic phonon scattering. We see that, as  becomes larger, the optimum value for becomes more negative. Thus, if  were large enough, we could use classical statistics in our calculations. However, the best materials that are used in today’s thermoelectric modules,  is about 0.4 and we hardly expect it ever to approach the highest valve in Figure 12.3. It can be seen that the optimum Fermi energy has a range of little more than kBT for a wide range of values for the material parameter 

12-23

Figure 12.3 The dimensionless figure of merit plotted against the reduced Fermi energy for different values of the material parameter . The scattering parameter is r = -1/2.

It is well known that the optimal electronic performance of a thermoelectric semiconductor depends primarily on the weighted mobility [1, 5, 7]. The weighted mobility is,

3  m  2  d  (12.77)   Nv     me 

 This weighted mobility implies that high degeneracy Nv, high density-of-states effective mass md , and high mobility  seem to improve the performance of ZT. This has been widely understood in the literature and industry. Nevertheless, in many semiconductors acoustic-phonon scattering dominates and that has the proportionality

1   0 3 (12.78)  2 Nv md me 

Considering this, the material parameter is expressed as

12-24

e       (12.79) mc

For materials dominated by acoustic phonon scattering, the material parameter depends only on the electron mobility as shown in Equation (12.79). Therefore, for anisotropic materials, the

 direction of lightest mc is preferred for thermoelectric transport (in cubic crystals, is close to

 md [8]), which is actually oppose to Equation (12.77) and the following statement. It is not sufficient to select a particular semiconductor or compound if a high ZT is required. It is necessary to specify the carrier concentration which can, of course, be adjusted by changing the number of donors or acceptors. In this section, we discuss the problem of achieving the optimum concentrations of charge carriers. It is instructional to proceed first with a calculation in which it is assumed that the semiconductor obeys classical statistics. We also suppose that there is only one type of carrier in a parabolic band, and we ignore the possibility of a bipolar effect. Thus, the thermoelectric parameter can be expressed by Equations (12.63) – (12.71).

Example 12.5 Material Parameter PbTe (lead telluride) is a thermoelectric material at mid-range temperatures with a band degeneracy of 4, DOS effective mass of 0.12 me and acoustic-phonon scattering for electrons with constant relaxation time of 1.55 × 10-13 s. Assuming the material has lattice thermal conductivity of 1.2 W/mK, determine the material parameter  at room temperature. Solution:

Since the acoustic-phonon scattering for electrons is assumed, the scattering parameter should be r = -1/2.

Information given:

W N  4 , m  0.12m ,  1.551013 s , and k  1.2 v d e 0 l mK

  From Equation (12.76), removing the Fermi integral (cancelled) and assuming that mc  md , 12-25

3 N  2m k T  2 k 2T   v  d B  B 0  0.246 2  2   3    mc kl

Comment: Most thermoelectric materials show about 0.1 ~ 0.5 for .

12.4 Electron Relaxation Time The relaxation time is the average flight time of electrons between successive collisions or scattering events with the lattice or impurities. The relaxation time plays the most important role in determining the transport properties such as electron mobility, electrical conductivity, thermal conductivity, and Seebeck coefficient. In this section, three fundamental scattering mechanisms are introduced to account for electron relaxation time.

12.4.1 Acoustic Phonon Scattering The wavelength of a free electron of thermal energy is large compared with the lattice constant. Such electrons interact only with acoustical vibration modes of comparable long wavelength. The local deformations produced by the lattice waves are similar to those in homogeneously deformed crystals. Longitudinal acoustic phonons may deform the electric band structure leading to electron scattering due to the deformation potential. The deformation potential is determined by a shift in the energy bands with dilations of the crystal produced by thermal vibration. The theory of acoustic-phonon scattering was originally provided by Bardeen and Shockley (1950)[9]. The relaxation time for the acoustic-phonon scattering is

4 2 1 1 2 v d     s E 2  E 2 a 2  3 2   0   (12.80) a 2md kBT 

  where E is the reduced energy ( E  E kBT ), d the mass density and a the acoustic deformation potential.

12-26

12.4.2 Polar Optical Phonon Scattering Polar optical phonon scattering is of considerable importance at low electron concentrations, although its effect is expected to diminish at high electron concentrations because free-electron screening will reduce the electron-phonon interaction. Polar materials are partly ionic compounds. When two atoms in a unit cell are not alike, the longitudinal optical phonons produce a crystal polarization that scatters free electrons. The interaction between electrons and optical phonons cannot generally be represented in terms of a relaxation time. However, Ehrenreich (1961)[10] developed an expression for the relaxation time assuming that at high temperatures (T D ) the energy change after collision is small compared to the electron energy. This allows the use of the relaxation time. With a formula of Callen (1949)[11] for the effective ionic charge, the relaxation time by the optical polar phonons is expressed as

82 1   E 2 po 2  1 2 1 1   (12.81) e 2md kBT   o 

where  o and   are the static and high frequency permittivities. Equation (12.81) has been widely used, giving good estimates for even lower than  D . The 8 is added to the magnitude of the Ehrenreich’s formula from the work of Nag (1980)[12] and Lundstrom (2000)[13].

12.4.3 Ionized Impurity Scattering Ionized impurity scattering becomes most important at low temperatures, where phonon effects are small. An ionized impurity produces a long-range (larger than the phonon wavelength) Coulomb field, which forms a screening and scatters electrons. Conwell and Weisskopf (1950)[14], Brooks (1951)[15], Blatt (1957)[16], and Amith (1965)[17] studied the ionized impurity scattering suggesting the Brooks-Herring formula to take account of the screening effect as

(2푚∗ )1⁄2휀2(푘 푇)3⁄2 휏 = 푑 0 퐵 퐼 푏 (12.82) 휋푁 (푍푒2)2 [푙푛(1 + 푏) − ] 퐼 1 + 푏 12-27

6 m k T 2 where b  o d B , N is the concentration of ionized impurities, n is the electron ne22 I concentration and Z is the vacancy charge. The impurity concentration and electron concentration are assumed to be equal.

Total Electron Relaxation Time The electron scattering rate is the reciprocal of the relaxation time. The total relaxation time can be calculated from individual relaxation times according to Matthiessen’s rule as

1 1   (12.83)  i  i

Matthiessen’s rule assumes that the scattering mechanisms are independent of each other.

Example 12.6 Electron Mobility PbTe (lead telluride) is a widespread thermoelectric material at mid-range temperatures, which is doped by Na (sodium) at doping concentration of 1.3 × 1019 cm-3, having the following features: the velocity of sound is 1.45 × 105 cm/s, the mass density is 8.65 g/cm3, the deformation potential is 11.4 eV, the degeneracy of the conduction valleys is 4 and the DOS effective mass is 0.12 me. Assuming that the acoustic phonon scattering is a dominant mechanism for the relaxation time, show that the electron mobility for the PbTe at room temperature is 1270 cm2/Vs.

Solution:

34 23 19 Physical constant:   6.62610 Js , kB  1.3810 J K , and e  1.602110 C

25 3 3 3 3 Information given: n  1.310 m , vs 1.4510 m s , dPbTe  8.6510 kg m , a  11.4eV ,  31 Nv = 4, and md  0.12(9.110 kg)

The relaxation time with r = -1/2 for the acoustic phonon scattering is assumed. From Equation (12.80),

24v2d   s 1.551013 s 0 2  3 2 a 2md kBT  12-28

From Equations (12.67) and (12.68), the mobility is expressed as

e 2  3  Fr 1 2     0 r   mc 3  2  F1 2

In order to calculate the Fermi integral, the reduced Fermi energy should be provided. From Example 12.4, the Fermi energy is

 EF  2.2

 Then, the values of the Fermi integrals of F0 and F1/2 with EF  2.2 can be obtained by interpolation between the values in Table C-1. Then, we have

F0  2.3051 and F1/ 2  2.7694

  Now, assuming that mc  md , the electron mobility is calculated as

19 2 e 2  1 3  F0 1.602110 C 13 2  2.3051 m     0     31 1.5510 s    0.127 mc 3  2 2  F1 2 0.12 (9.110 kg) 3  2.7694 Vs

Comments: The reduced Fermi energy of 2.2 for this type of semiconductor (PbTe) shows a positive value because of the high doping concentration of 1.3 × 1019 cm-3. The mobility of 1270 cm2/Vs is comparable to the experimental value of 1100 cm2/Vs at room temperature.

12.5 Multiband Effects So far it has been assumed that only one type of charge carrier (electron or hole) is present in the conductor. We now consider a conductor in which there are both electrons and holes. In an intrinsic semiconductor there are equal numbers of negative electrons and positive holes. Similarly, if an impurity semiconductor is taken up to a high enough temperature, a certain number of electron- hole pairs are excited across the forbidden gap. From the electrodynamic relations (Equation (11.1)), we have

j Ε  T (12.84)  or 12-29

T j  Ε -  (12.85) x

Considering two bands, the problem will be discussed for both electron and hole carriers (represented by the subscripts 1 and 2, respectively)  T   T  (12.86) j1  1Ε - 1 and j 2   2 Ε - 2   x   x 

The total current must be

j  j1  j2 (12.87)

Then,

T j     Ε -       (12.88) 1 2 1 1 2 2 x

Therefore, comparing this with Equation (12.85), the total electrical conductivity is expressed as

  1   2 (12.89) and also the total Seebeck coefficient is expressed as

      1 1 2 2 (12.90) 1  2

From the electrodynamic relations (Equation (11.2)), we have

T q  Tj  k (12.91) e e x

For two bands, we have

12-30

T T q   Tj  k and q   Tj  k (12.92) e1 1 1 e1 x e2 2 2 e2 x

The total electronic thermal conductivity is expressed as

 T   T  qe  qe1  qe2  1Tj1  ke1   2Tj2  ke2  (12.93)  x   x 

Inserting Equation (12.86) into (12.93) and using Equation (12.34) finally gives

1 2 2 ke  ke1  ke2  2 1  T (12.94) 1  2

The remarkable feature of Equation (12.94) is that the total electronic thermal conductivity is not merely the sum of the thermal conductivities of the separate carriers. There is an additional term associated with the bipolar flow that is the third term in the right-hand side of the equation.

12.6 Nonparabolicity Nonparabolic Density of States

For the simple parabolic model the energy dispersion is given as

2 2 2 2 ℏ 푘 푘푦 푘 퐸 = ( 푥 + + 푧 ) (12.95) 2 푚푥 푚푦 푚푧 where mx, my, and mz are the principal effective masses in the x-, y-, and z-directions and here k is the magnitude of the wavevector.

2 2 2 2 푘 = 푘푥 + 푘푦 + 푘푧 (12.96)

For high applied fields, electrons may be far above the conduction band edge, and the higher order terms in the Taylor series expansion cannot be ignored. For the nonparabolic model the energy dispersion is given by

12-31

2 2 2 2 퐸 ℏ 푘 푘푦 푘 (12.97) 퐸 (1 + ) = ( 푥 + + 푧 ) 퐸퐺 2 푚푥 푚푦 푚푧 which is known as Kane model. Let 훾(퐸) = 퐸(1 + 퐸⁄퐸퐺) and introduce a new wavevector 푘′ and an effective mass 푚′ as

2 2 2 (12.98) 퐸 ℏ 푘′ ℏ 2 2 2 훾(퐸) = 퐸 (1 + ) = = (푘′푥 + 푘′푦 + 푘′푧) 퐸퐺 2푚′ 2푚′

Equating Equations (12.97) and (12.98), we have a relationship between the original wavevector and the new wavevector as

푚 (12.99) 푘 = 푘′ √ 푥 푥 푥 푚′

푚푦 푘 = 푘′ √ 푦 푦 푚′

푚 푘 = 푘′ √ 푧 푧 푧 푚′

In k-space of Figure 12.4, we have

푚푥푚푦푚푧 푚푥푚푦푚푧 (12.100) 푑푘 = 푑푘 푑푘 푑푘 = √ 푑푘′ 푑푘′ 푑푘′ = √ 4휋푘′2푑푘′ 푥 푦 푧 푚′3 푥 푦 푧 푚′3

12-32

(a) (b) Figure 12.4 A constant energy surface in k-space: (a) three-dimensional view, (b) lattice points for a spherical band in two-dimensional view.

The volume of the smallest wavevector in a crystal of volume L3 is (2/L)3 since L is the largest wavelength. The number of states between k and k + dk in three-dimensional k-space is then obtained (see Figure 12.4)

2 2 ∙ 4휋푘′ 푚푥푚푦푚푧 (12.101) 푁(푘)푑푘 = √ 푑푘′ (2휋⁄퐿)3 푚′3 where the factor of 2 accounts for the electron spin (Pauli Exclusion Principle). Now the density of states g(k) is obtained by dividing the number of states N by the volume of the crystal L3.

2 푘′ 푚푥푚푦푚푧 (12.102) 푔(푘)푑푘 = √ 푑푘′ 휋2 푚′3

From Equation (12.98), we have

1 (12.103) 2 1 푑푘′ (2푚′) − = 훾 2 푑훾 2ℏ

1 1 1 − (12.104) 2 1 2 2 2 푑푘′ 푑푘′ 푑훾 (2푚′) − (2푚′) 퐸 2퐸 = = 훾 2훾′ = (퐸 + ) (1 + ) 푑퐸 푑훾 푑퐸 2ℏ 2ℏ 퐸퐺 퐸퐺

∗ 1⁄3 Using 푚푑 = (푚푥푚푦푚푧) and including the degeneracy of valleys 푁푣 , 푚′ is cancelled out. The nonparabolic density of states is finally obtained as 12-33

3 1 (12.105) ∗ 2 2 2 푁푣 2푚푑 퐸 2퐸 푔(퐸) = 2 ( 2 ) (퐸 + ) (1 + ) 2휋 ℏ 퐸퐺 퐸퐺

Electron Group Velocity

From Equation (11.10), the group velocity of electrons is given by

1 휕퐸 1 휕훾⁄휕푘′ (12.106) 휐(퐸) = = ℏ 휕푘′ ℏ 휕훾⁄휕퐸

Using Equations (12.48), (12.51) and (12.55), the group velocity of electrons is obtained from

퐸2 (12.107) 2 (퐸 + ) 2 퐸퐺 푣푥 = 2 ∗ 2퐸 3푚푐 (1 + ) 퐸퐺 where

−1 (12.108) ∗ 1 1 1 1 푚푐 = [ ( + + )] 3 푚푥 푚푦 푚푧

The nonparabolic two-band model of thermoelectric transport properties and scattering rates for electrons and phonons are further discussed in Chapters 15 and 16.

Problems

df f f f  f  12.1 Derive Equation (12.2) of   k  r    . dt t k r  t coll

f  E E  E T  12.2 Derive Equation (12.10) of f  v o  F  F  . E  x T x 

12.3 Estimate the relaxation time  for that has an fcc lattice with lattice constant of 3.61 Å if its electrical conductivity  of 5.88 × 105 (cm)-1 is given.

12-34

12.4 Derive Equation (12.23),    E   e2 gEv2  . F x EEF

ne2 12.5 Derive Equation (12.24),    ne . m  2 ln E 12.6 Derive in detail Equation (12.30),    k 2T . 3e B E E EF

 2  gE 1 l  12.7 Derive Equation (12.31),    k 2T   . 3e B  n  E   EEF

2 2 2k    3 12.8 Derive Equation (12.35),   B mT   . 3e2  3n 

12.9 Estimate the Seebeck coefficient of copper that has an fcc lattice with lattice constant of 3.61 Å at room temperature.

2 12.10 Derive in detail Equation (12.42), ke  LoT  T .

12.11 Provide thermoelectric properties (, , and k) versus electron concentration curves from 17 21 -3 10 to 10 cm for Sn (tin) with the effective mass of 1.3 me at room temperature using the constant relaxation time of 10-14 sec (Figure P12.11). Hint: you may use Equations (11.33), (12.23), (12.34)with r = 0, and (12.44) with neglecting the second term for a metal. For the

plot, you may use vector notation, where n+[+i gives ni of vector notation rather than just subscript in Mathcad to determine the logarithmic interval scale for the plots such as i = 17+0.14×i -1 1,2… 30, ni = 10 . Use V/K for , (cm) for andW/cmK for k.

12-35

a k

s

|a| &s k

n Figure P12.11. TE properties vs. electron concentration.

12.12 Provide the thermoelectric transport property curves against the Fermi energy for PbTe as shown in Figure 12.2 The Seebeck coefficient, the electrical conductivity, and the electronic thermal conductivity as a function of the Fermi energy at room temperature for PbTe

-14 assuming a constant relaxation time  0 of 6.2 × 10 s (realistic value) and r = -1/2 (see Equation (12.42))..

12.13 Derive Equation (12.74) with (12.75).

12.14 Derive the nonparabolic density of states, Equations (12.105), and the group velocity, Equation (107).

12.15 Plot Figure 12.3 with a brief explanation.

References

1. Goldsmid, H.J., Thermoelectric Refrigeration. 1964, New York: Plenum Press. 240. 2. Kittel, C., Introduction-to-Solid-State-Physics 8th Edition. 2005: Wiley. 3. Ashcroft, N.W. and N.D. Mermin, Solid state physics. 1976, New York: Holt, Rinehart and Winston. 4. Wilson, A.H., The theory of metals, 2nd ed. 1953, Cambridge: Cambridge University Press. 12-36

5. Chasmar, R.P. and R. Stratton, The thermoelectric figure of merit and its relation to thermoelectric generators. J. Electron. Control., 1959. 7: p. 52-72. 6. Goldsmid, H.J., Introduction to thermoelectricity. 2010, Heidelberg: Springer. 7. Rowe, D.M., CRC handbook of thermoelectrics. 1995, Boca Raton London New York: CRC Press. 8. Pei, Y., et al., Low effective mass leading to high thermoelectric performance. Energy & Environmental Science, 2012. 5(7): p. 7963. 9. Bardeen, J. and W. Shockley, Deformation Potentials and Mobilities in Non-Polar Crystals. Physical Review, 1950. 80(1): p. 72-80. 10. Ehrenreich, H., Band Structure and Transport Properties of Some 3–5 Compounds. Journal of Applied Physics, 1961. 32(10): p. 2155. 11. Callen, H., Electric Breakdown in Ionic Crystals. Physical Review, 1949. 76(9): p. 1394- 1402. 12. Nag, B.R., Electron Transport in Compound Semiconductors. Springer Series in Solid Sciences. 1980, New York: Springer. 13. Lundstrom, M., Fundamentals of carrier transport, 2nd ed. 2000, Cambridge: Cambridge University Press. 14. Conwell, E. and V. Weisskopf, Theory of Impurity Scattering in Semiconductors. Physical Review, 1950. 77(3): p. 388-390. 15. Brooks, H., Scattering by ionized impurities in semiconductors. Phys. Rev., 1951. 83: p. 879. 16. Blatt, F.J., Theory of mobility of electrons in solids, in Solid State Physics, F. Seitz and D. Turnbull, Editors. 1957, Academic Press: New York. 17. Amith, A., I. Kudman, and E. Steigmeier, Electron and Phonon Scattering in GaAs at High Temperatures. Physical Review, 1965. 138(4A): p. A1270-A1276.