What sets the cooling capacity of a Peltier device?

Current, the temperature difference between cold and hot sides, and device properties (Seebeck coefficient, electrical resistance, thermal conductance). More current increases Peltier heat pumping, but Joule heating rises with current squared, so capacity has a maximum. Larger temperature differences let more heat flow back from the hot side, reducing capacity.

What COP does a Peltier device achieve?

It varies widely with temperature difference and current. Pumping 20 W with the example module gave COP 2.3 at 10 K difference, 1.3 at 20 K and 0.71 at 30 K. Below COP 1, more electrical power is used than heat pumped, and the hot side must reject over twice the cooling load.

What matters in heat-sink design for Peltier coolers?

As the hot-side temperature rises, the temperature difference grows and capacity drops sharply. In the example with a 15 °C cold side and 25 °C ambient, a 0.1 K/W heat sink allowed up to 41 W of cooling, but 0.5 K/W only 22 W. Size the heat sink to reject cooling load plus input power, not cooling load alone.

Thermoelectric (Peltier) Cooling Analysis

Category: Electronics Cooling | Revised 2026-10-01

CAE visualization for thermoelectric cooling theory - technical simulation diagram

Thermoelectric Cooling (Peltier Element)

Theory: Peltier effect versus Joule heating

Overview

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I want to cool a sensor slightly below room temperature and am considering a Peltier device. Does more current always mean more cooling?

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It doesn't work that way. Current through a Peltier device pumps heat at the junctions (the Peltier effect), but it also generates Joule heat inside the device. Pumped heat is proportional to current while Joule heat rises with its square, so cooling peaks at a certain current and falls beyond it. Heat also conducts back from the hot side. The basics are to view cooling capacity, input power and hot-side rejection as functions of current and temperature difference.

Basic equations

$$ Q_c = \alpha I T_c - \tfrac{1}{2} I^2 R - K\,\Delta T, \qquad P = \alpha I\,\Delta T + I^2 R, \qquad Q_h = Q_c + P, \qquad COP = \frac{Q_c}{P} $$

$\alpha$ is the module Seebeck coefficient, $R$ electrical resistance, $K$ thermal conductance, $T_c$ and $T_h$ cold- and hot-side temperatures and $\Delta T = T_h - T_c$. Half the Joule heat is taken to return to the cold side.

$$ Z = \frac{\alpha^2}{R K}, \qquad \Delta T_{max} = T_h - T_c\big|_{Q_c = 0,\ I = \alpha T_c/R}, \qquad Q_{max} = \frac{(\alpha T_h)^2}{2R} $$

$Z$ is the figure of merit; materials with larger $ZT$ reach larger temperature differences and higher COP. Bismuth telluride near room temperature has $ZT$ around 1.

Coffee Break Trivia

A 150-year wait for practical use

The Peltier effect was discovered in 1834 by the French watchmaker Peltier, who found that current through a junction of dissimilar metals heats or cools it depending on direction. In metals the effect was small, overwhelmed by Joule heating and conduction, and impractical. That changed in the 1950s, when semiconductors such as bismuth telluride were found to give far larger effects. Today, with no moving parts, silent operation and precise temperature control, Peltier devices serve in laser temperature control for optical communications, small refrigerators, wine cellars and PCR instruments.

Worked example: a 127-couple module

Assume a 127-couple bismuth telluride module with $\alpha$ = 0.0508 V/K, $R$ = 2.0 Ω and $K$ = 0.5 W/K ($ZT$ = 0.77 at 300 K). With the hot side held at 300 K, the maximum temperature difference is 68.9 K and the maximum cooling at zero difference is 58 W (7.6 A).

Example 1: pumping 20 W

ΔT	Current needed	Input power	COP	Hot-side heat	Cooling at COP-optimal current
10 K	1.96 A	8.7 W	2.31	28.7 W	5.8 W (COP 3.65)
20 K	2.58 A	15.9 W	1.26	35.9 W	9.9 W (COP 1.51)
30 K	3.39 A	28.1 W	0.71	48.1 W	12.2 W (COP 0.80)

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At 30 K difference, pumping 20 W takes 28 W of electricity.

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Yes — and the hot side must reject the 20 W pumped plus the 28 W consumed, 48 W in all. Peltier efficiency falls sharply as the temperature difference grows. The right-hand column shows that running at the COP-optimal current is efficient but pumps little heat per module. So if efficiency matters, use several modules at lower current each; if cost or size matters, use fewer modules at higher current.

Example 2: heat-sink influence

The cold side was held at 15 °C with 25 °C ambient, varying heat-sink resistance. The hot-side temperature rises with the heat rejected:

Heat-sink resistance	Maximum cooling	Hot side then	Cooling at 3 A	Hot side at 3 A	COP at 3 A
0.1 K/W	41.2 W	38.6 °C	27.5 W	29.8 °C	1.36
0.2 K/W	35.1 W	48.2 °C	25.3 W	34.2 °C	1.21
0.5 K/W	22.3 W	61.3 °C	19.4 W	46.1 °C	0.85

A poor heat sink raises the hot side, widening the temperature difference and cutting both capacity and COP. Peltier performance is often set more by the heat sink than by the device itself.

Design workflow

Set the heat load, target temperature and ambient temperature.
Assume a heat-sink resistance and solve together with the hot-side temperature.
Choose module count and current by trading efficiency against size.
Check cold-side insulation and condensation protection (when cooling below the dew point).
Check contact resistance (thermal interface materials) and solder-joint life under thermal cycling.

Coffee Break Trivia

“More current, higher temperature”

A Peltier-cooled device wasn't reaching its target, so the current was raised — and the temperature went up instead. Investigation showed a small heat sink with the hot side already above 60 °C. More current meant more input power, a hotter hot side, a wider temperature difference and less cooling. With a larger heat sink, the original current reached the target. When cooling falls short, measure the hot-side temperature first.

Common mistakes

Mistakes and fixes

Mistake	Effect	Fix
Sizing the heat sink for cooling load	Hot side overheats, cooling fails	Size for load plus input power
Assuming more current cools more	Temperature rises instead	Check current for peak cooling
Designing to catalogue maxima	Real capacity drops with ΔT	Calculate at operating ΔT
Ignoring condensation	Electrical failure	Insulate and seal
Ignoring contact resistance	Extra temperature difference	Evaluate interface materials

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I'd like to try related calculations.

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Try the Peltier calculator, ZT figure of merit calculator and heat sink designer for device and heat-sink calculations. Related topics include thermoelectrics, heat sink design, thermal interface materials and thermal resistance.

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