SOLUIONS: Homework: Week hermoelectricity from Atoms to Systems Mark Lundstrom, nanohub- U all 13 here are seven HW questions below. his homework will be graded in multiple choice format no partial credit. Doing the homework before the solutions and tutorial are posted is a great way to test your understanding of the material covered. Some problems may be challenging, but give them a try, and review the tutorial and solutions later. 1) We have discussed the coupled current equations for 3D electrons: E x ρj x + S d dx V/m J Qx π J x ( +κ L ) d dx W/m Consider a nanowire with 1D electrons. Instead of the current density, J x, in A/m, we use the current, I, in A, and instead of the heat current density, J Qx, in W/m, we use the heat current, I Q, in W. Deduce the units of each of the four 1D transport coefficients. e he units of ρ 1D, S 1D, π 1D, and κ 1D, are: Solution: a) Ω-m,V/K,V,W/ ( m-k) b) Ω,V/K,V,W/K c) Ω/m,V/K,V,W-m/K d) Ω/m,V-K,V-K,W/( m-k) e) Ω-m,V/K,V K,W-K/m he coupled current equations in 1D are: E x ρi x + S d dx V/m I Qx π I x ( +κ L ) d dx W (i) (ii) In terms of units, we can write these equations as: nanohub- U all 13 1
Homework: Week hermoelectric Materials and Devices (continued) V V/m A-m A + V K K m W ( V)A + W-m K K m (iii) (iv) So, ρ Ω m S V K π V κ W-m K Answer is c) Ω/m,V/K,V,W-m/K ) he expression for the short circuit (electronic) thermal conductivity is: E E κ σ ( E)dE q where σ ( E), the differential conductivity, is given by σ ( E) q h λ E f M ( E) A E. Evaluate this expression assuming that the ermi level is located above the middle of the gap, so that only the conduction band need be considered. You may assume that the mean- free- path for backscattering is independent of energy, λ E λ, and parabolic energy bands so that in 3D: M ( E) A m* ( E E π C ) H ( E E C ), where H ( E E C ) is the Heaviside step function. Your answer should be expressed in terms of ermi- Dirac integrals. or a tutorial on ermi- Dirac integrals see: Notes on ermi- Dirac Integrals, 3 rd Ed. http://nanohub.org/resources/5475/ nanohub- U all 13
Homework: Week hermoelectric Materials and Devices (continued) he expression for the short- circuit thermal conductivity of 3D electrons in a semiconductor with parabolic energy bands in terms of the normalized ermi energy, ( E E C ) k B is: a) κ k B q { } b) κ k B q 1 { ( ) } c) κ k B 1 q ( ) η d) κ k B q q h λ m * k B π 6 e) κ k B q q h λ m * k B π 6 { ( ) + η ( )} { ( ) 4 1 ( ) + η ( )} Solution: κ E E q σ ( E)dE Substituting in for the differential conductivity, we find: E E κ q q h λ M E ( A) f E de, and then for the number of channels: E E κ q q h λ m * E E ( π C ) f E de. Pull the constants out front: 1 q κ q h λ m * π ( E E ) ( E E C ) f E de. (i) nanohub- U all 13 3
Homework: Week hermoelectric Materials and Devices (continued) Work on the integral first: 1 q κ q h λ m * π I (ii) I ( E E ) ( E E C ) f E de. Add and subtract, E C : I ( E E C + E C E ) ( E E C ) f E de. Now change variables: ( η E E C ) k B ( E E C ) k B de k dη B 4 ( η ) η f I k B I k B 4 η η + I k B E dη η f E dη 4 η 3 f E dη η η f E dη + η η f E dη 4 I k B 3 I k B η 3 f E dη E η 3 f η dη η η f dη + E η f dη + η f dη η f dη nanohub- U all 13 4
Homework: Week hermoelectric Materials and Devices (continued) 3 I k B Γ( 4) η 3 Γ( 3) η + Γ( ) η 1 3 6 ( ) 4 1 ( ) + η ( ) I k B. Now insert this result in (ii) above to find: 1 q κ q h λ m * π k B 3 6 ( ) 4 1 ( ) + η ( ) κ k B q he answer is e). q h λ m * k B π 6 { ( ) 4 1 ( ) + η ( )} Please see the Appendix of Near- Equilibrium ransport: undamentals and Applications, by Lundstrom and Jeong, for a list transport coefficients worked out for 1D, D, and 3D conductors. his is eqn. (A34). Additional exercise for those who are interested: Assume that the mean- free- path is energy- dependent according to λ ( E E C ) k B λ E r. Work out the analytical expression and explain physically why r > increases the magnitude of the Seebeck coefficient. nanohub- U all 13 5
Homework: Week hermoelectric Materials and Devices (continued) 3) In Lecture we worked out the approximate values of the four thermoelectric transport coefficients for lightly doped n- type Ge. or practical E devices, the material would be doped so that E E C. Work out the four thermoelectric transport coefficients for n- type Ge doped at N D 1 19 cm -3. You may assume that 3 K, that the dopants are fully ionized, and that the mean- free- path for backscattering, λ, is independent of energy. Use the material parameters presented in Lecture, but use a mobility of µ n 33 cm /V-s (from http://www.ioffe.ru/sva/nsm/semicond/ge/igs/3.gif. You may assume non- degenerate carrier statistics (but realize that this assumption may not well- justified for E E C, which is the case here, so we will only obtain estimates). he approximate values of λ, ρ, S, π, and, are: a) 11 nm.19 Ohm- cm, - 175 µv/k, -.5 V,.4 W/(m- K) b) 11 nm.19 Ohm- cm, - 175 µv/k, -.5 V,.4 W/(m- K) c) 11 nm.19 Ohm- cm, - 175 µv/k, -.5 V,.4 W/(m- K) d) 33 nm.19 Ohm- cm, - 175 µv/k, -.5 V,.4 W/(m- K) e) 33 nm.19 Ohm- cm, - 175 µv/k, -.5 V,.4 W/(m- K) Solution: rom Lecture : υ k B π m * 1.55 1 7 cm/s D n k B q µ n 8.6cm s D n υ λ cm /s λ D n υ 11.1 1 7 cm λ 11.1 nm ρ 1 ( n qµ n ) 1 ( 1 19 1.6 1 19 33).19 Ω-cm ρ.19 Ω-cm S k B E c E q k B + δ n ( E c E ) k B ln( N C n ) N C 1.4 1 19 cm -3 nanohub- U all 13 6
Homework: Week hermoelectric Materials and Devices (continued) k B ln 1.4 1 19 E c E S k B E c E q k B S 175µV/K ( 1 ) 19 3.9 1 δ n + δ n 86 µv/k { 3.9 1 + } 175µV/K π S.5 V σ L L ρ L ( k B q) (We are using the factor of because we assume nondegenerate carrier statistics.) ( q ) k B ρ.4 W/m-K.4 W/m-K Answer is c). 4) Perhaps we should use ermi- Dirac statistics for thermoelectric calculations when E E c. Repeat problem 3), but this time use ermi- Dirac statistics to determine the approximate values of λ, ρ, S, π, and. You might find it useful to know that σ 3D q h λ g V m * k B π and S k B 1 q 1 ( ) a) 11 nm.19 Ohm- cm, - 175 µv/k, -.5 V,.4 W/(m- K) b) 11 nm.19 Ohm- cm, - 175 µv/k, -.5 V,.4 W/(m- K) c) 11 nm.19 Ohm- cm, - 175 µv/k, -.5 V,.4 W/(m- K) d) 13 nm.19 Ohm- cm, - 186 µv/k, -.6 V,.4 W/(m- K) e) 13 nm.19 Ohm- cm, - 186 µv/k, -.6 V,.4 W/(m- K) nanohub- U all 13 7
Homework: Week hermoelectric Materials and Devices (continued) Solution: he conductivity does not change from prob. 3): σ 3D 1 ρ 1.19 56 S/cm rom: σ 3D q h λ m * k B π conductivity:, we can solve for the MP in terms of the λ q h σ 3D g V m * k B π o proceed, we must find using: n 1 19 N C 1/ ( ) 1.4 1 19 1/ 1 1 19 1/ 1.4 1 19 1.96 1/.97 (computed with the iphone app or with the nanohub tool: http://nanohub.org/resources/11396) λ q h σ 3D g V m * k B π 5.6 1 4 S/m 7.71 1 ( 5 ).97 6.37 1 16.13 1 7 m where we used the distribution of modes effective mass, m * 1.18m. or a discussion of distribution of modes (DOM) effective mass, see: Changwook Jeong, Raseong Kim, Mathieu Luisier, Supriyo Datta, and Mark Lundstrom, On Landauer vs. Boltzmann and ull Band vs. Effective Mass Evaluation of hermoelectric ransport Coefficients, J. Appl. Phys., Vol. 17, 377, 1. λ 13 nm a bit longer than for MB statistics nanohub- U all 13 8
Homework: Week hermoelectric Materials and Devices (continued) ρ.19 Ω-cm same as before S k B 1 q S 186 µv/k ( ) 86 1.97 1 6.97.97 186µV/K π S.6 V σ L L ρ L π 3 k B q (We are using the fully degenerate Lorenz number, for simplicity.) π 3 k q B ρ.4 W-m/K.4 W-m/K Answer is d). 5) We have discussed two different electronic thermal conductivities one measured under short circuit conditions, κ, and one measured under open circuit conditions,. he two are related according to: κ σ S Using the estimated E transport coefficients for Ge doped such that E E C (from problem 4) deduce the ratio, κ. or this example, the ratio, κ, is closest to: a). b).9 c) 1.3 d).4 e) 5.5 nanohub- U all 13 9
Homework: Week hermoelectric Materials and Devices (continued) Solution: he relation between the two electronic thermal conductivities is: κ σ S κ + σ S Use numbers from problem 4).4 W-m/K σ 1 ρ 1.19 56 S/cm 5.6 1 4 S/m S 186 µv/k -1.86 1 4 V/K κ + σ S.4 + 3 5.6 1 4 ( 1.86 1 ) 4.4 +.55 κ.95 W-m/K κ.38 he answer is d). 6) Using the results of prob. 4), estimate the thermoelectric OM, Z for n- type Ge at 3 K. You may assume that κ L 58 W/m-K. a) Z.1 b) Z.1 c) Z.1 d) Z 1. e) Z 1.8 nanohub- U all 13 1
Homework: Week hermoelectric Materials and Devices (continued) Solution:.4 W-m/K σ 1 ρ 1.19 56 S/cm 5.6 1 4 S/m S 186 µv/k -1.86 1 4 V/K Z ( 1.86 1 4 ) 5.4 1 4 3.1 Z.1.4 + 58 he answer is b). 7) he expressions for conductance and Seebeck coefficient are: G G E de (1) G ( E) q h λ( E) L M E f E () S ( E E ) G ( E)dE q G ( E)dE (3) Assuming 1D electrons, parabolic energy bands, and an energy- independent mean- free- path, λ E λ, eqns. (1) (3) can be evaluated to find (Lundstrom and Jeong, Near- Equilibrium ransport, Appendix, eqn. (A.5), World Scientific, 13) G 1D q h S 1D k B q where λ ( ) L 1 1 ( E E C ) k B. nanohub- U all 13 11
Homework: Week hermoelectric Materials and Devices (continued) or energy independent scattering, the optimum location of the ermi level for maximum power factor is ˆ. Assuming 3D electrons, parabolic energy bands, and an energy- independent mean- free- path, λ E λ, eqns. (1) (3) can be evaluated to find (Lundstrom and Jeong, Near- Equilibrium ransport, Appendix, eqn. (A.34), World Scientific, 13) G 3D A q h S 3D k B q λ L m * k B π 1 or energy independent scattering, the optimum location of the ermi level for a 3D material is ˆ +.668. Assuming 3 K, m * m, λ 1 nm, and L 1 cm, compare the power factors in 1D and 3D for a 1 cm cross- sectional area. Determine how many 1D nanowires would need to be placed in this 1 cm area to equal the power factor of the 3D material. he center- to center spacing of the nanowires is: Solution: a).1 nm b) 1.5 nm c).4 nm d) 5.7 nm e) 1.3 nm he optimum power factor of the nanowire is: 1D q λ P max h L 1 k B +1.14 q 1 he optimum power factor of the 3D material is: 3D q λ P max A h L m * k B π.668 k B 1.668 q.668.668 nanohub- U all 13 1
Homework: Week hermoelectric Materials and Devices (continued) he number of nanowires needed to give the same power factor as the 3D structure is 3D N P max 1D P max A q h λ L q h m * k B π.668 λ L 1 k B k B.668 +1.14 1.668 q.668 q 1 3D N P max 1D P max A m* k B.668 π 1.668 ( ) +1.14 1.668.668 1 A 1 cm 1 4 m m * k B π 5.39 116 m - 1.8.775 1 1 1.49.43 1 1.668 1.668 1 1.49 N 1 4 5.39 1 16 1.8.4 1.8.668.78.4 +1.14 If these are placed in an N x N array, the on a side we have 1.7 1 13 nanowires N 1.7 1 13 4.15 1 6 nanowires Since these are placed in a 1 cm by 1 cm square, the nanowires must be placed a distance d 1 N 1.4 nm 6 4.15 1 nanohub- U all 13 13
Homework: Week hermoelectric Materials and Devices (continued) he answer is c). Even though a nanowire is inherently more efficient than a bulk thermoelectric (on a per channel basis), they must be packed very closely just to equal the power factor of the bulk device. or more discussion of this topic, see: Raseong Kim, Supriyo Datta, and Mark S. Lundstrom, Influence of dimensionality on thermoelectric device performance, J. Appl. Phys., Vol. 15, 3456, 9. End of homework assignment. his assignment contains 7 questions. nanohub- U all 13 14