Week 10 Study Toolkit · Nanoelectronics

1This week's plan

Read
Datta Part A, Chapters 13–14 (pp. 171–193): thermoelectricity and phonon transport.

Run
Colab notebook week10_thermoelectric_C5.ipynb: Part 1 before class, Part 2 (C5 lab and design task) in the circular session.

Practise
Quiz (aim for 8/10), the practice problems, and prepare your C5 design defence (section 10).

By the end of the week you should be able to

Explain why ΔT drives electrons hot → cold in n-type and cold → hot in p-type, with no positive charges.
Compute G₀, G_S, S = −G_S/G₀, the power factor and ZT.
Write the heat current, Π = TS and G_K, and explain why a delta-function channel has G_K = 0.
Choose the level position that maximises the power factor (≈ 2kT from µ₀) and explain why ZT prefers a different one.
Compute κ_ph from (Mλ/A)_ph, the ballistic T³ law, and ZT ≈ (Mλ/A)/(Mλ/A)_ph.

Optional video: Datta's nanoHUB-U course, Unit 4: L4.2–L4.5 (thermoelectricity, phonons).

2Key ideas in five sentences

Same current formula. I = (1/q)∫G(f₁ − f₂)dE also works when the contacts differ in temperature; f₁ − f₂ is then positive above µ and negative below.
The sign of S comes from dG/dE. n-type (G rising) gives electrons hot → cold and S < 0; p-type gives S > 0. No positive particles needed.
Figures of merit. Power factor S²G₀ (maximum power, Eq. 13.10) and ZT = S²G₀T/G_K (efficiency, Eq. 13.12). Metals have |S| ≪ k/q; the best place for µ₀ is within a few kT of a band edge, S ≈ 2k/q.
Heat currents. Every electron carries heat E − µ; Π = TS (Kelvin); a single sharp channel has G_K = 0, but real materials also conduct heat through phonons.
Phonons follow the same elastic-resistor formula with Bose functions: κ_ph = (π²k²T/3h)(Mλ/A)_ph; ZT ≈ (Mλ/A)/(Mλ/A)_ph: "electron crystal, phonon glass".

3Equation sheet

I = (1/q)∫dE G(E)(f₁ − f₂)Eqs. 13.1–13.3

f₁, f₂ with their own µ and T.

I = G₀(V₁ − V₂) + G_S(T₁ − T₂)Eq. 13.4

Linear response; V = µ/q is the electron voltage.

G₀ = ∫(−∂f₀/∂E)G dE, G_S = ∫(−∂f₀/∂E)((E − µ₀)/qT)G dEEqs. 13.5–13.6

Conductance and thermoelectric coefficient.

S = V_OC/(T₁ − T₂) = −G_S/G₀Eqs. 13.7–13.9

n-type: S < 0.

P_max = S²G₀ΔT²/4Eq. 13.10

Matched load R_L = 1/G₀; S²G₀ = power factor.

ZT = S²G₀T/G_K = S²σ₀T/κEqs. 13.11–13.12

Thermoelectric figure of merit.

I_Q1 = (1/q)∫dE((E − µ₁)/q)G(f₁ − f₂)Eqs. 13.13–13.15

I_Q1 + I_Q2 + (V₁ − V₂)I = 0.

S = −G_S/G₀, Π = −G_P/G₀, G_K = G_Q − G_PG_S/G₀Eqs. 13.16–13.18

G_P = ∫(−∂f₀/∂E)((E − µ₀)/q)G, G_Q = ∫(−∂f₀/∂E)((E − µ₀)²/q²T)G.

Π = TSEq. 13.19

Kelvin relation (G_P = TG_S).

S = −(ε − µ₀)/qT, Π = −(ε − µ₀)/q, G_K = 0Eq. 13.21

Delta-function thermoelectric.

S²G₀ = G(ε)(Δε/kT)(k/q)²F(x), F = x²eˣ/(eˣ + 1)²Eqs. 13.22–13.23

Maximum at x = 2.40 (Datta ≈ 2): S ≈ 2k/q.

n(ω) = 1/(e^ħω/kT − 1)Eq. 14.5

Bose function, no µ.

G_K = (k²T/h)∫dx(Mλ/(L + λ))_phx²eˣ/(eˣ − 1)²Eqs. 14.6–14.8

Phonon heat conductance.

G_K/A = π²k⁴T³/(10ħ³c_s²)Eqs. 14.9–14.10

Ballistic phonons: T³ law, I_Q ∝ T₁⁴ − T₂⁴.

κ_ph = (π²k²T/3h)(Mλ/A)_ph, ZT ≈ (Mλ/A)/(Mλ/A)_phEqs. 14.2–14.4, 14.11

π²k²T/3h = 284 pW/K at 300 K.

4Physics refresher: heat, temperature and conductivity

Heat vs charge current

Charge current: C/s. Heat current: J/s, counted from µ, so an electron at energy E carries heat E − µ. Fourier's law I_Q/A = −κ dT/dz is the thermal Ohm's law; G_K = κA/L.

Electrons and phonons

κ = κ_e + κ_ph. Metals: κ_e dominates and κ_e/(σT) = (π²/3)(k/q)² = 2.44 × 10⁻⁸ WΩ/K² (Wiedemann–Franz). Semiconductors and insulators: phonons usually dominate.

Where does (E − µ₀)/qT come from?

∂f₀/∂T = ((E − µ₀)/T)(−∂f₀/∂E): raising T empties states below µ₀ and fills states above, exactly like raising µ but with opposite signs on the two sides. That turns the T-derivative into Eq. 13.6.

Fermi vs Bose

f = 1/(e^(E−µ)/kT + 1) lies between 0 and 1 (exclusion). n = 1/(e^ħω/kT − 1) can be ≫ 1 at low ω (many phonons in one mode) and has no µ because phonons are not conserved.

5Glossary

Term	Meaning
Seebeck coefficient S	Open-circuit voltage per kelvin; negative for n-type.
Peltier coefficient Π	Heat carried per unit charge current; Π = TS.
Power factor	S²G₀ (or S²σ): sets the maximum power from a given ΔT.
ZT	Dimensionless figure of merit S²σT/κ; sets the efficiency.
Wiedemann–Franz law	κ_e/(σT) = (π²/3)(k/q)² in metals.
Phonon	Quantum of lattice vibration; obeys Bose statistics.
Debye temperature θ_D	Above it all phonon modes are excited.
Electron crystal, phonon glass	Material design: long λ for electrons, short λ for phonons.

6Calculators

Delta-function thermoelectric (Eqs. 13.21–13.23)

(ε − µ₀)/kTT (K)G(ε)Δε/kT (mS)ΔT (K)

S = −(ε − µ₀)/qT

Π = TS

S²G₀ = G(ε)(Δε/kT)(k/q)²F(x)

P_max = S²G₀ΔT²/4

Material figure of merit

|S| (µV/K)σ (S/m)κ_e (W/m·K)κ_ph (W/m·K)T (K)

Power factor S²σ

ZT = S²σT/(κ_e + κ_ph)

Wiedemann–Franz κ_e = LσT

Π = TS

Phonons (Ch. 14)

(M/A)_ph (m⁻²)λ_ph (nm)T (K)Sound velocity c_s (km/s)electron M/A (m⁻²)electron λ (nm)

κ_ph = (π²k²T/3h)(Mλ/A)_ph

ballistic G_K/A (T³ law)

σ = (q²/h)(Mλ/A)

ZT ≈ 4(k/q)²Tσ/κ_ph

7Self-check quiz

Click an answer: you get instant feedback. Score: 0 / 0

8Practice problems

k/q = 86.17 µV/K, (k/q)² = 7.43 × 10⁻⁹ V²/K², π²k²T/3h = 284 pW/K at 300 K, q²/h = 38.74 µS.

P1. A single levelEASY

A molecule conducts through one level 2kT above µ₀ at 300 K. Find S and Π.

Solution

S = −2k/q = −172 µV/K; Π = TS = −51.7 mV.

P2. Open-circuit voltageEASY

An n-type leg has S = −200 µV/K with ΔT = 50 K. How large is V_OC, and which end is positive (conventional)?

Solution

|V_OC| = 200 µV/K × 50 K = 10 mV; the hot end is positive (electrons pile up at the cold end).

P3. Maximum powerEASY

The same leg has resistance 1 Ω. Find the maximum power into a matched load.

Solution

P_max = S²G₀ΔT²/4 = (2 × 10⁻⁴)² × 1 × 2500/4 = 25 µW.

P4. ZTMEDIUM

S = 200 µV/K, σ = 10⁵ S/m, κ = 1.5 W/m·K at 300 K. Find ZT.

Solution

ZT = 4 × 10⁻⁸ × 10⁵ × 300/1.5 = 0.80.

P5. Wiedemann–FranzMEDIUM

Copper has σ = 5.96 × 10⁷ S/m at 300 K. Estimate κ_e.

Solution

κ_e = 2.443 × 10⁻⁸ × 5.96 × 10⁷ × 300 = 437 W/m·K (measured ≈ 400).

P6. Phonon conductivityMEDIUM

(M/A)_ph = 4 × 10¹⁷ m⁻², λ_ph = 15 nm at 300 K. Find κ_ph. What λ_ph would halve it?

Solution

κ_ph = 284 pW/K × 4 × 10¹⁷ × 15 × 10⁻⁹ = 1.70 W/m·K; λ_ph = 7.5 nm.

P7. Ballistic phononsMEDIUM

Find the ballistic phonon heat conductance per area at 10 K for c_s = 5 km/s (Eq. 14.9). By what factor does it grow at 20 K?

Solution

π²k⁴T³/(10ħ³c_s²) = 1.22 × 10⁶ W/m²K; T³ ⇒ 8× at 20 K (well below θ_D).

P8. Peltier coolerMEDIUM

An n-type leg with S = −200 µV/K carries 2 A at 300 K. How much heat does it pump from the cold side (ignoring Joule heat and back-conduction)?

Solution

|Π|I = 300 × 200 × 10⁻⁶ × 2 = 0.12 W.

P9. Delta-function power factorHARDER

A level with G(ε) = 1 mS and Δε = 0.1kT sits 2.4kT above µ₀ at 300 K. Find S²G₀.

Hint

Use Eq. 13.22 with F(2.4) = 0.439.

Solution

S²G₀ = 10⁻³ × 0.1 × 7.43 × 10⁻⁹ × 0.439 = 3.26 × 10⁻¹³ W/K² (S = −207 µV/K).

P10. Kelvin relationHARDER

Using Eqs. 13.6 and 13.17, show that G_P = TG_S and hence Π = TS.

Solution

G_P = ∫(−∂f₀/∂E)((E − µ₀)/q)G dE = T∫(−∂f₀/∂E)((E − µ₀)/qT)G dE = TG_S. So Π = −G_P/G₀ = −TG_S/G₀ = TS.

9Common mistakes

MISTAKE"p-type samples show the opposite thermopower because holes are positive."

CORRECTOnly electrons move; G(E) decreasing with E makes the part below µ win (Fig. 13.4).

MISTAKE"The Hall effect and the Seebeck effect change sign for the same reason."

CORRECTHall: m = p/v (needs E(p)). Seebeck: the slope of G(E) around µ (no E(p) needed).

MISTAKE"Good conductors make good thermoelectrics."

CORRECTMetals have |S| ≪ k/q because states above and below µ₀ cancel.

MISTAKE"Maximum power factor and maximum ZT occur at the same doping."

CORRECTZT also contains κ_e, which grows as µ₀ moves into the band; the ZT optimum sits closer to the band edge.

MISTAKE"No charge current means no heat current."

CORRECTTrue only for a single sharp channel; with several channels opposite currents carry net heat.

MISTAKE"Phonons need a µ like electrons."

CORRECTPhonons are not conserved: n(ω) has no µ, and no exclusion principle.

10Prepare for circular session C5

Block B lab (notebook Part 2)

B1 p-n I–V dark and lit (Week 9); B2 spin-valve MR vs R_ch and spin flips; B3 a Bi₂Te₃-flavoured 3D band (m = 0.3m₀, 6 valleys, λ = 15 nm, κ_ph = 1.2 W/m·K): S, σ, S²σ, ZT vs µ₀.

Block C design task

Choose the best thermoelectric level position and defend it. Decide (a) µ₀ relative to E_c, (b) whether nanostructuring (κ_ph × ½, electron λ × 0.7) pays, (c) whether a delta-like band is better.

A defence template (three sentences)

Goal: "We optimise for maximum power / maximum efficiency because …"
Numbers: "At (E_c − µ₀)/kT = …, S = … µV/K, σ = … S/m, κ_e = …, κ_ph = … W/m·K, so ZT = …"
Trade-off: "Moving µ₀ up raises σ but lowers |S| and raises κ_e, because the window −∂f₀/∂E then sees states on both sides of µ₀."

Reference numbers (open after your group has decided)

Best power factor ≈ 3.4 mW/m·K² at E_c − µ₀ ≈ −0.67kT (S ≈ −167 µV/K, ZT ≈ 0.56). Best ZT ≈ 0.63 at E_c − µ₀ ≈ +0.24kT (S ≈ −220 µV/K, κ_e ≈ 0.31 W/m·K). Nanostructured: best ZT ≈ 0.81 at ≈ +0.40kT, although the power factor drops ≈ 30%.

11Explore

Files are in the course folder (course_materials).

Week 10 slides Widget: thermoelectric window Widget: phonon transport Colab notebook Week 9 toolkit

Mini-investigations (15 min each)

In the thermoelectric widget, find the best power factor for α = 0, 0.5 and 1. How does the best position of µ₀ move, and why?
Set the phonon conductance very small: where does the best ZT go? Very large?
In the phonon widget, change θ_D from 165 K to 645 K (silicon). At what temperature does the ballistic conductance leave the T³ law?

12Can I…?

Tick what you can do without notes. Your ticks are saved in this browser.

explain the sign of S for n- and p-type without holeswrite G₀, G_S and S = −G_S/G₀compute the power factor, P_max and ZTwrite the heat current and prove Π = TSexplain why a delta-function channel has G_K = 0explain why the best ZT and best power factor differcompute κ_ph from (Mλ/A)_ph and the ballistic T³ lawdefend a thermoelectric design with numbers

13Looking ahead: Week 11

The second law and the fuel value of information (Datta Ch. 15–16): where the e^{−ħω/kT₀} of Week 9 and the Bose and Fermi functions come from, entropy, and why erasing a bit costs kT ln2.

Why is absorbing energy from a bath harder than giving it up?
A solar cell and a thermoelectric are both heat engines. What limits their efficiency?
Can information be used as fuel?

Thermoelectricity, phonon transport and circular session C5

1This week's plan

By the end of the week you should be able to

2Key ideas in five sentences

3Equation sheet

4Physics refresher: heat, temperature and conductivity

5Glossary

6Calculators

Delta-function thermoelectric (Eqs. 13.21–13.23)

Material figure of merit

Phonons (Ch. 14)

7Self-check quiz

8Practice problems

9Common mistakes

10Prepare for circular session C5

A defence template (three sentences)

11Explore

Mini-investigations (15 min each)

12Can I…?

13Looking ahead: Week 11