L2 · Chapter 7

The energy ledger of thermoelectric conversion: why phonon conduction limits efficiency

Sandbox Physics · 2026-10-30

A thermoelectric material with its ends at different temperatures produces a voltage, and with a load attached it generates power with no moving parts. Its efficiency is set by a dimensionless figure of merit whose denominator is the thermal conductivity: some heat leaks straight from the hot end to the cold end without ever being converted. This chapter keeps a complete energy ledger for one thermoelectric leg, derives the familiar maximum-efficiency formula from it, says when that formula overestimates, and shows where the phonon conduction of earlier chapters comes in.

Seebeck voltage and Peltier heat

A conductor with $\Delta T$ across it has an open-circuit voltage $S\Delta T$ , where $S$ is the Seebeck coefficient. A current $I$ crossing a junction at temperature $T$ also carries away or releases the Peltier heat $STI$ . In a generator, the Peltier heat at the hot end is the heat pulled into the circuit to become electrical work.

Two losses run alongside. One is conduction: a leg of thermal conductance $K$ passes $K\Delta T$ straight from hot to cold whether or not current flows. The other is Joule heating, $I^2R$ , generated inside the leg.

Source: H. S. Kim · Wikipedia: Thermoelectric materials

The ledger at the hot end

Kim and colleagues write the heat taken in at the hot end, in the constant-property model, as conduction plus Peltier heat minus the half of the Joule heat that flows back to the hot end. With properties independent of temperature, exactly half the Joule heat goes to each end.

Attach a load $R_L$ and write $m=R_L/R$ . The current is $I=S\Delta T/\bigl(R(1+m)\bigr)$ , the output power $P=I^2R_L$ , and the efficiency is the output power over the heat taken in.

Q_h=K\Delta T+S\,T_h\,I-\tfrac12I^{2}R,\qquad P=I^{2}R_L,\qquad \eta=\frac{P}{Q_h}

Energy conservation makes

Q_h-P

the heat released at the cold end, a check on the ledger.

Source: H. S. Kim

The figure of merit and the maximum-efficiency formula

Maximizing the efficiency over $m$ gives the best load ratio $m=\sqrt{1+Z\bar T}$ , with $Z=S^2/(\rho\kappa)$ and $\bar T$ the mean of the two end temperatures. Putting it back gives the maximum-efficiency formula below. As Wikipedia points out, its first factor is the Carnot efficiency, which a thermoelectric device, being a heat engine, cannot exceed; the second factor depends on the figure of merit and tends to 1 as the figure of merit grows without bound.

The material figure of merit is $zT=\sigma S^2T/\kappa$ . The numerator $\sigma S^2$ is the power factor and depends on electrons; the denominator $\kappa$ includes heat carried by both electrons and phonons.

If you want the most power rather than the best efficiency, the load should equal the internal resistance; the efficiency is then a little below its maximum.

\eta_{\max}=\frac{T_h-T_c}{T_h}\cdot\frac{\sqrt{1+Z\bar T}-1}{\sqrt{1+Z\bar T}+T_c/T_h}

Both Wikipedia and Kim and colleagues stress that it is exact only when the properties do not change with temperature.

Source: H. S. Kim · Wikipedia: Thermoelectric materials

When the formula overestimates

Kim and colleagues compared the usual formula with numerical calculations that use measured, temperature-dependent properties. With the hot side at 700 °C the formula overestimated the efficiency of SnSe more than twofold, because its figure of merit changes strongly with temperature and one value at the mean temperature no longer stands in for the whole curve. They also note that temperature-dependent properties bring in Thomson heat, and the share of Joule heat returning to the hot end is no longer one half.

Wikipedia gives another reality check: commercial thermoelectric refrigerators have coefficients of performance between 0.3 and 0.6, about one sixth of those of conventional vapour-compression refrigerators.

Source: H. S. Kim · Wikipedia: Thermoelectric materials

Where phonons enter the ledger

Phonon conduction sits in the denominator of the figure of merit. Halve it without hurting the power factor and the figure of merit nearly doubles. Chapter 3 showed that a 40 nm gray silicon film carries only 43% of the flux the bulk conductivity predicts; Chapter 6 showed that in a 200 nm stack one interface takes 80% of the temperature drop.

These are ways to cut phonon conduction, and the course models phonons only. In real materials boundaries and interfaces scatter electrons too and may lower the power factor along with the conductivity, so whether a structure really raises the figure of merit depends on how much each carrier is affected, a question that needs a model of the electrons as well.

Source: H. S. Kim · Wikipedia: Thermoelectric materials

Worked example: the ledger of one leg

Problem: a leg 1 mm² in cross-section and 1 mm long has a Seebeck coefficient of 200 µV/K, a resistivity of $10^{-5}\,\Omega\,\mathrm m$ and a thermal conductivity of 1.5 W/(m K), with its hot end at 400 K and its cold end at 300 K. The properties are assumed for this exercise and do not change with temperature. Find the maximum efficiency and write out the ledger at the best load.

$Z=(2\times10^{-4})^2/(10^{-5}\times1.5)=2.667\times10^{-3}\,\mathrm{K^{-1}}$ , $Z\bar T=0.933$ , and the best load ratio is $m=\sqrt{1.933}=1.390$ . The Carnot efficiency is 0.25 and the maximum efficiency $\eta_{\max}=0.25\times0.390/(1.390+0.75)=0.0456$ .

The internal resistance is $R=0.01\,\Omega$ , the conductance $K=1.5\times10^{-3}\,\mathrm{W/K}$ , the open-circuit voltage 0.02 V, and the current $I=0.02/(0.01\times2.390)=0.8367\,\mathrm A$ . The hot-end ledger: conduction 0.150 W, Peltier heat $200\times10^{-6}\times400\times0.8367=0.0669\,\mathrm W$ , minus half the Joule heat, 0.00350 W, for a total of $Q_h=0.2134\,\mathrm W$ .

The output power is $P=I^2mR=0.00973\,\mathrm W$ and $P/Q_h=0.0456$ , matching the formula. The ledger shows that about seven tenths of the heat taken in is conduction that leaks straight to the cold end, which is why lowering the conductivity raises the efficiency.

Source: H. S. Kim · Wikipedia: Thermoelectric materials

Try it in the Lab

Let interfaces block phonons

Choose “Thin stack: the interface dominates” and read the interface jump as a share of the total drop.
Increase the stack thickness step by step and watch that share fall; think about how closely interfaces would have to be packed in a thermoelectric material to block phonon heat flow.

How far a film cuts phonon conduction

Choose “Thickness near the mean free path” and read the settled film conductivity over the bulk value.
Assume electrons are not affected at all and use this chapter’s formulas: multiply only the phonon conductivity by that ratio and estimate the new figure of merit of the worked example. Then say why the assumption is too optimistic.

Exercises with explained answers

If the leg in the worked example drives a load equal to its internal resistance, what are the output power and efficiency?
Show the answer
The current is $I=0.02/0.02=1.00\,\mathrm A$ , the power $I^2R=0.0100\,\mathrm W$ , the hot-end heat $0.150+0.0800-0.00500=0.225\,\mathrm W$ , and the efficiency 0.0444. The power is a little higher than at the best-efficiency load, the efficiency a little lower.
If the thermal conductivity is halved and nothing else changes, what is the maximum efficiency of the worked example?
Show the answer
$Z\bar T$ doubles to 1.867, $\sqrt{2.867}=1.693$ , and $\eta_{\max}=0.25\times0.693/(1.693+0.75)=0.0709$ , about 1.56 times the original.
What percentage of the Carnot efficiency does the worked example reach?
Show the answer
$0.0456/0.25=18.2\%$ . Even with these idealized assumed properties, most of the available heat is lost to conduction and Joule heating.
Why does exactly half of the Joule heat return to the hot end in the constant-property model?
Show the answer
With constant properties every length of the leg produces the same Joule heat, which adds a symmetric parabola to the linear temperature profile, so heat leaves the middle equally towards both ends. Kim and colleagues note that the share departs from one half when properties change with temperature.

Sources and what we recomputed

H. S. Kim, W. Liu, G. Chen, C.-W. Chu and Z. Ren, Relationship between thermoelectric figure of merit and energy conversion efficiency, PNAS 112(27), 8205 (2015), doi:10.1073/pnas.1510231112
Main text read on PubMed Central; the SI Appendix, where the derivations sit, was not opened. Used here: the conventional maximum-efficiency formula from the constant-property model, written as the Carnot efficiency times a factor that depends on the figure of merit at the average temperature; the heat input as conduction plus Peltier heat minus half the Joule heat; the load ratio and its optimum; and the finding that the conventional formula overestimates the efficiency of SnSe more than twofold at a hot side of 700 °C because its properties change strongly with temperature. Equation numbers are not quoted.
Wikipedia, Thermoelectric materials, revision 1375991214 (21 September 2026)
Partly read: the lead and the sections on the figure of merit, device efficiency and power factor. Used here: zT as the electrical conductivity times the Seebeck coefficient squared times temperature over the thermal conductivity; the maximum-efficiency formula with the Carnot efficiency as its first factor, exact only for temperature-independent properties; and coefficients of performance of 0.3 to 0.6 for commercial thermoelectric refrigerators. The materials sections were not read. Wikipedia is a secondary source.