University of Kurdistan logoNanoelectronics: A New Perspective on TransportWeek 10 · Thermoelectricity, phonons · C5
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WEEK 10 · THERMOELECTRICITY AND PHONONS · C5

Nanoelectronics:
A New Perspective on Transport

Seebeck and Peltier effects, figures of merit, phonon heat transport; Weeks 9–10 lab
Prof. Mohammad RazaghiUniversity of Kurdistan · Graduate course in Electronics Engineering
Based on S. Datta, Lessons from Nanoelectronics, Part A: Basic Concepts, Chapters 13–14
Part 1 is the fixed lecture (Ch. 13–14). Part 2 is circular session C5: a Colab lab on Weeks 9–10 and the design task "choose the best thermoelectric level position and defend it".

RecapWhere we are

Week 9 gave us

  • Smart contacts: separate µ for groups of states
  • Hot scatterers drive current with no voltage: qVOC = ħω(1 − T/T₀)
  • Spin valves and spin potentials

Today's question

Make the contacts different in temperature instead of in µ. Which way do electrons flow, why does the answer differ for n- and p-type, and how good can a heat-to-electricity converter be?

Week 10Today's goals

  1. Explain why ΔT drives electrons hot → cold in n-type and cold → hot in p-type.
  2. Compute S = −GS/G₀, the power factor S²G₀ and ZT.
  3. Write the heat current, the Peltier coefficient Π = TS and GK.
  4. Optimise a delta-function thermoelectric: level ≈ 2kT from µ₀.
  5. Treat phonons with the same formula: Bose window, T³ law, κph, ZT ≈ (Mλ/A)/(Mλ/A)ph.

Physics refresherRefresher: heat current vs charge current

  • Charge current: coulombs per second. Heat current: joules per second, counted from µ: an electron at E carries heat E − µ.
  • Temperature gradient drives heat: IQ/A = −κ dT/dz; conductance GK = κA/L.
  • Electrons and phonons both carry heat: κ = κe + κph.
Cu: κ = 400 W/m·K; Si: 150; Bi₂Te₃: ≈ 1.5–2; glass: ≈ 1.
Wiedemann–Franz: in metals κ_e/(σT) = L = 2.44×10⁻⁸ WΩ/K²; Datta's formulas reproduce it deep inside a band (notebook).

Datta Sec. 13.1Hot and cold contacts

I = (1/q)∫dE G(E)(f₁ − f₂)Eq. 13.1 · f₁, f₂ at T₁, T₂ (Eqs. 13.2–13.3)
  • Same µ, T₁ > T₂: f₁ − f₂ > 0 above µ, < 0 below µ (Fig. 13.3).
  • The sign of the current depends on where G(E) is larger.
  • No new physics: still "current flows because f₁ ≠ f₂".

Datta Fig. 13.4n-type and p-type: no positive charges needed

  • n-type: D(E), G(E) increase with E ⇒ the part above µ wins ⇒ electrons go hot → cold.
  • p-type: G(E) decreases with E ⇒ the part below µ wins ⇒ electrons go cold → hot.
  • Hall sign comes from m = p/v (Ch. 11); thermoelectric sign from dG/dE. Molecules with no E(p) show it too.

Datta Sec. 13.2The Seebeck coefficient

I = G₀(V₁ − V₂) + GS(T₁ − T₂)Eq. 13.4
G₀ = ∫(−∂f₀/∂E)G dE,  GS = ∫(−∂f₀/∂E)((E − µ₀)/qT)G dEEqs. 13.5–13.6
S = VOC/ΔT = −GS/G₀Eqs. 13.7–13.9 · n-type: S < 0

V = µ/q is the electron voltage; the hot side of an n-type sample is positive in conventional terms.

Energies above and below µ0 contribute with opposite signs to G_S: for a large S put all the states on one side of µ0.

Datta Sec. 13.3Figures of merit

Pmax = S²G₀(T₁ − T₂)²/4Eq. 13.10 · matched load RL = 1/G₀; S²G₀ = power factor
ZT = S²G₀T/GK = S²σ₀T/κEqs. 13.11–13.12

Two different goals

Power factor: most power when heat is free (waste heat from an engine). ZT: best efficiency, the ratio of electrical output to heat drawn; essential for coolers.

Metals: µ₀ deep in a band, |S| ≈ few µV/K ≪ k/q = 86 µV/K. Semiconductors win.

InteractiveExplore: the thermoelectric window

Thermoelectric window: Seebeck coefficient, power factor and ZT vs μ₀  (Datta Ch. 13, Figs. 13.3–13.9)

Two contacts at the same μ₀ but different temperatures: f₁ − f₂ is positive above μ₀ and negative below (Fig. 13.3). Whether electrons flow hot → cold or cold → hot depends on where the channel conducts. Move the band (or level) relative to μ₀ and watch S, the power factor S²G₀ and ZT (Eqs. 13.5–13.18). Energies in units of kT; conductance in units of Gc = G(E) at 1 kT inside the band.

Seebeck S = −GS/G₀ (Eq. 13.9)–
G₀ (Eq. 13.5) · power factor S²G₀–
ZT: electrons only · with phonons–
Peltier Π = TS (Eq. 13.19) · VOC for ΔT = 10 K–

Try this

  1. n-type, edge at +2kT: is S negative? Which colour wins in the channel?
  2. Switch to p-type: same electrons, opposite sign of S.
  3. Press Best power factor (3D band): read S. Close to 2k/q?
  4. Press Best ZT: why is it at a different place?

Scroll inside the panel for all controls. Swipe or use the arrow buttons outside it to change slide.

Energies in units of kT, conductance in units of G_c = G at 1 kT inside the band.

Datta Sec. 13.4Heat current and the Peltier effect

IQ1 = (1/q)∫dE ((E − µ₁)/q) G (f₁ − f₂)Eq. 13.13 · IQ1 + IQ2 + (V₁ − V₂)I = 0
S = −GS/G₀,  Π = −GP/G₀,  GK = GQ − GPGS/G₀Eqs. 13.16–13.18
Π = TSEq. 13.19 · Kelvin relation

A current through an n-type channel cools the source and heats the drain: thermoelectric (Peltier) coolers.

Datta Sec. 13.5The delta-function thermoelectric

S = −(ε − µ₀)/qT,  Π = −(ε − µ₀)/q,  GK = 0Eq. 13.21 · f₁(ε) = f₂(ε) at open circuit
S²G₀ = G(ε)(Δε/kT)(k/q)²F(x), F = x²eˣ/(eˣ + 1)²Eq. 13.22
  • F(x) peaks at x = 2.40 (Datta: "≈ 2"): S ≈ 2k/q ≈ 170 µV/K (Eq. 13.23).
  • Best materials: µ₀ just below a band edge, so G·(−∂f₀/∂E) looks delta-like.

Datta Sec. 13.5Why GK is usually not zero

  • Open circuit: total charge current zero.
  • With two energy channels, equal and opposite currents can flow in each.
  • The upper channel carries more energy: a net heat current remains.
  • Only a single sharp channel gives GK = 0, and then phonons still conduct heat.

Datta Sec. 14.2Phonons: same formula, new particle

n(ω) = 1/(eħω/kT − 1)Eq. 14.5 · Bose, no µ
GK = (k²T/h)∫dx (Mλ/(L + λ))ph x²eˣ/(eˣ − 1)²Eq. 14.8 · x = ħω/kT
ballistic: GK/A = π²k⁴T³/(10ħ³cs²)Eq. 14.9 · phonon Stefan–Boltzmann, IQ ∝ T₁⁴ − T₂⁴
M_ph = 3ω²A/(4πc_s²): three polarizations. Photons have two: the Stefan–Boltzmann prefactor is 2/3 of the phonon one at the same velocity.

Datta Sec. 14.3Thermal conductivity and ZT

κph = (π²k²T/3h)(Mλ/A)ph,  σ = (q²/h)(Mλ/A)Eqs. 14.1–14.2 · 284 pW/K vs 38.7 µS
ZT ≈ (Mλ/A)/(Mλ/A)phEq. 14.4 · with S ≈ 2k/q

Bi₂Te₃: (M/A)ph ≈ 4 × 10¹⁷ m⁻², λph ≈ 15 nm ⇒ κ ≈ 1.7 W/m·K.

InteractiveExplore: phonon transport

Phonons carry heat too: windows, ballistic T³ law and ZT ≈ (Mλ/A)/(Mλ/A)ph  (Datta Ch. 14)

Phonons obey the same elastic-transport formula as electrons with q → ℏω and Fermi → Bose functions (Eq. 14.6). Left: the two window functions (Fig. 14.1). Right: the ballistic phonon heat conductance per area (Eq. 14.9), with a Debye cut-off. Bottom: thermal conductivity κph = (π²k²T/3h)(Mλ/A)ph and electrical σ = (q²/h)(Mλ/A) set ZT (Eqs. 14.1–14.4).

Ballistic GK/A at T (Debye cut-off) · T³ law–
κph = 284 pW/K × (T/300 K) × (Mλ/A)ph–
σ = (q²/h)(Mλ/A)–
ZT ≈ 4(k/q)²Tσ/κph (Eq. 14.3) · ratio (Eq. 14.4)–

Try this

  1. Compare the two windows: which particles conduct at x → 0?
  2. Lower T below θD/10: is the ballistic curve T³?
  3. Press Bulk silicon, then Nanostructured silicon: how much does κph drop?
  4. Press Electron crystal, phonon glass: what ZT do you get?

Scroll inside the panel for all controls. Swipe or use the arrow buttons outside it to change slide.

ZT here uses S = 2k/q (Eq. 14.3); the ratio form drops 12/π².

Datta Sec. 14.1Electron crystal, phonon glass

  • Good thermoelectrics need µ₀ near a band edge ⇒ small (M/A) for electrons.
  • Phonons: (M/A)ph ~ 1 nm⁻² at room temperature ⇒ the ratio of M/A's ≲ 0.1.
  • Win back with λ: keep electrons long, make phonons short (alloys, nanostructures, inclusions).
ZT ≈ 1 for decades, recently ≈ 3; ZT ≈ 4–10 would transform applications.

CheckConcept check

One minute with a neighbour, then click to reveal.

1. Only electrons move. Why opposite thermoelectric signs for n and p? (click)
f₁ − f₂ changes sign at µ; n-type G(E) weights the part above µ, p-type the part below.
2. A level 3kT above µ₀ at 300 K: S and Π? (click)
S = −3k/q = −259 µV/K; Π = TS = −77.6 mV.
3. Why are metals poor thermoelectrics? (click)
States above and below µ₀ nearly cancel in GS: |S| ≪ k/q.
4. λph in Si drops from 500 to 10 nm. κph? (click)
50 times smaller: ≈ 142 → 2.8 W/m·K with (M/A)ph = 10¹⁸ m⁻².

Wrap-upFixed session summary

Take-home

  • Same I = (1/q)∫G(f₁ − f₂), now driven by T₁ ≠ T₂
  • S = −GS/G₀; Π = TS; ZT = S²G₀T/GK
  • Best: level ≈ 2kT from µ₀, S ≈ 2k/q
  • Phonons: κph = 284 pW/K·(Mλ/A)ph; ZT ≈ ratio of Mλ/A

Next: circular session C5

A Colab lab on Weeks 9–10 and a design task: choose the best thermoelectric level position and defend it. Open week10_thermoelectric_C5.ipynb (Part 2) and the Week 10 study toolkit.

PART 2

Circular session C5

Weeks 9–10 lab · thermoelectric design task

A · 10 min
Recap: two reservoirs that differ
B · 40 min
Lab: p-n, spin valve, S and PF vs µ₀
C · 30 min
Design: the best level position
D · 10 min
Presentations
Groups of 3–4. Every group submits one plot and a three-sentence defence of its design.

Circular session C5Block A · Two reservoirs that differ

DeviceThe reservoirs differ inThe channel discriminates byKey result
p-n junction / solar cellµ (and T₀ of the scatterers)which band each contact feedsI ∝ X − 1, qVOC = ħω(1 − T/T₀)
Spin valvewhich spin each magnet favoursr vs R interfacesMR = P²/(1 − P²)
ThermoelectricTG(E) above vs below µ₀S = −GS/G₀, ZT
Phonon conductorT(Mλ/A)phκph = 284 pW/K·(Mλ/A)ph

In pairs: for each row, what would make the device stop working?

Circular session C5Block B · Lab: Weeks 9 and 10 in one notebook

B1. p-n I–V dark and lit: slope 59.5 mV/dec; VOC = 1.045 V (Datta) and 0.767 V (diluted sun).

B2. Spin valve MR vs Rch and spin flips: why do the interfaces have to win?

B3. A Bi₂Te₃-flavoured 3D band (m = 0.3m₀, 6 valleys, λ = 15 nm, κph = 1.2 W/m·K): S, σ, S²σ and ZT vs µ₀.

Best power factor ≈ 3.4 mW/m·K² at µ₀ ≈ 0.67kT inside the band; best ZT ≈ 0.63 at µ₀ ≈ 0.24kT below the edge.

Circular session C5Block C · Design task

Choose the best level position and defend it

  1. Where do you put µ₀ relative to Ec (the doping)?
  2. Is nanostructuring worth it: κph × ½ but electron λ × 0.7?
  3. Would a narrow-band, delta-like material be better? What limits it?

Your defence (3 sentences)

Name the goal (maximum power or maximum efficiency), quote S, σ, κe, κph, ZT, and explain the S–σ trade-off with the window −∂f₀/∂E. Use evaluate(eta, ...) in the notebook.

Instructor notes are in the notebook (collapsed) and in the Week 10 toolkit.

Circular session C5Block D · What a good answer looks like

Maximum power (waste heat is free) (click)
µ₀ ≈ 0.67kT inside the band: S ≈ −167 µV/K, σ ≈ 1.2 × 10⁵ S/m, S²σ ≈ 3.4 mW/m·K², ZT ≈ 0.56.
Maximum efficiency (a cooler) (click)
µ₀ ≈ 0.24kT below the edge: S ≈ −220 µV/K, κe halves to 0.31 W/m·K, ZT ≈ 0.63.
Nanostructuring (click)
ZT ≈ 0.63 → 0.81 although S²σ falls ≈ 30%: κph dominated the denominator.

Before Week 11

  • Finish both notebook parts
  • Week 10 toolkit: quiz and problems
  • Preview: Datta Ch. 15–16, the second law and the fuel value of information
Prof. Mohammad Razaghi · University of Kurdistan · after S. Datta, Lessons from Nanoelectronics