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Thermoelectric Cooler Selection Procedure

Thermoelectric Cooler Selection Procedure

Outlined below is a simplified selection procedure devised to allow the user to obtain initial designs and estimates of performance for single- and two-stage thermoelectric coolers. Because of the non-linear behavior of TECs and the number of variables involved in analyzing them, they can be designed and modeled more accurately by our experienced engineers using II-VI Marlows' internally developed computer software.

For selection of coolers with more than two stages, or if more precision is required, please consult one of our application engineers. Once the decision to use a thermoelectric cooler has been made, the actual selection of suitable modules is relatively simple.

The following pages outline a systematic procedure that will take you through determining your heat load, required DT and the number of stages required to meet the DT.

Once you have completed the analysis, you will have narrowed the field of suitable TECs to two or three. You may then proceed to Steps 5 through 10 to determine the performance of the selected TECs within your application requirements. 

A. Estimating High Loads


Before the cooler or heat sink can be selected, the cooling requirements must be defined. This includes determining the amount of heat to be pumped. Minimizing the heat load allows the cooler to achieve colder temperatures or reduces the power required to reach the defined cooling level. The following describes the techniques used to estimate active and passive heat loads and applies only to steady state heat loads. If the heat load is of a transient nature, or involves more complex factors such as air or fluid flow, we suggest that you call one of our applications engineers for assistance.

1. Heat Load

The heat load may consist of two types: active or passive, or a combination of the two. An active load is the heat dissipated by the device being cooled. It generally equals the input power to the device. Passive heat loads are parasitic in nature and may consist of radiation, convection or conduction.

2. Active Heat Load

The general equation for active heat load dissipation is: 
Qactive = V2/R = VI = I2R 

Qactive = active heat load (W) 
V = voltage applied to the device being cooled (V) 
R = device resistance 
(W) I = current through the device (A) 

For example, a typical lead selenide (PbSe) infrared detector is operated at a bias voltage of 50 volts and a resistance of 0.5 megohms. Therefore, the active load is 0.005 watts.

3. Radiation

When two objects at different temperatures come within proximity of each other, heat is exchanged. This occurs through electromagnetic radiation emitted from one object and absorbed by the other.

The hot object will experience a net heat loss and the cold object a net heat gain because of the temperature difference. This is called thermal radiation.

Radiation heat loads are usually considered insignificant when the system is operated in a gaseous environment, since the other passive heat loads are typically much greater in magnitude. Radiation loading is usually significant in systems with small active loads and large temperature differences, especially when operating in a vacuum environment.

The fundamental equation for radiation loading is: Qrad = F e s A (Tamb4 - Tc4) 

Qrad = radiation heat load (W) 
F = shape factor (worst case value = 1) 
e = emissivity (worst case value = 1) 
s = Stefan-Boltzmann constant (5.667 X 10-8W/m2K4) 
A = area of cooled surface (m2) 
Tamb = Ambient temperature (K) 
Tc = TEC cold ceramic temperature (K) 

Example calculation:

A Charge Coupled Device is being cooled from an ambient temperature of 27°C (300 K) to -50°C (223 K). 

The detector surface area (includes 4 edges + top surface) is 
8.54 X 10 -4 m2 and has an emissivity of 1. Assume the shape factor = 1 

From the equation above: 
Qrad = (1)(1) (5.66X10-8 W/m2K4) (8.54 X 10-4 m2) [(300 K)4 - (223 K)4]
= 0.272 W

4. Convection

When the temperature of a fluid (in this case, a gas) flowing over an object differs from that of the object, heat transfer occurs. The amount of heat transfer varies, depending on the fluid flow rate.

Convective heat loads on TECs are generally a result of natural (or free) convection. This is the case when gas flow is not artificially induced as with a fan or pump, but rather occurs naturally from the varying density in the gas caused by the temperature difference between the object being cooled and the gas. 

The convective loading on a system is a function of the exposed area and the difference in temperature between this area and the surrounding gas. Convective loading is usually most significant in systems operating in a gaseous environment with small active loads or large temperature differences. 

The fundamental equation that describes convective loading is: 
Qconv = h A (Tair - Tc) 

Qconv = convective heat load (W) 
h = convective heat transfer coefficient (W/m2 °C) (typical value for a flat, horizontal plate in air at 1 atm) = 21.7 W/m2 °C 
A = exposed surface area (m2) 
Tair = temperature of surrounding air (°C) 
Tc = temperature of cold surface (°C) 

Example calculation:

A square plate is being cooled from 25°C to 5°C. The top and four sides are exposed surfaces. The plate is 0.006 meters thick and each side is 0.1 meters long. 

From the convection equation: 
Qconv = (21.7 W/m2°C) (0.0124 m2)(25°C - 5°C)
= 5.4 W

It is very important to avoid allowing condensation to form when cooling below the dew point. This problem may be avoided by enclosing the cooling system in a dry gas or a vacuum environment.

5. Conduction

Conductive heat transfer occurs when energy exchange takes place by direct impact of molecules moving from a high temperature region to a low temperature region.

Conductive heat loading on a system may occur through lead wires, mounting screws, etc., which form a thermal path from the device being cooled to the heat sink or ambient environment. 

The fundamental equation that describes conductive loading is: 
Qcond =k A DT

Qcond = conductive heat load (W) 
k = thermal conductivity of the material (W/m °C) 
A = cross-sectional area of the material (m2) 
L = length of the heat path (m) 
DT = temperature difference across the heat path (°C) (usually ambient or heat sink temperature minus cold side temperature)

Table I 
Thermal Conductivities of Various Wire Materials 

Material Thermal Conductivity (W / m°C)
Aluminum 205
Copper 386
Gold 315
Platinum (90%) Iridium (10%) 31.1
Platinum 70.9
Manganin 22.2

Example calculation:

A TEC is used as a black body reference. A temperature sensor is attached to the cold surface of the TEC. It has two platinum leads, which have a diameter of 25mm and are 12mm long. These leads are attached to pins on the heat sink. The cold plate is at -20°C while the heat sink is at 30°C. 

The known parameters are: 
k = 70.9 W/m°C, from Table I 
DT = [30°C - (-20°C)] = 50°C 
A(1 wire) = pd2 / 4 = p (25 mm)2 / 4 
= 4.91 X 10 -10 m2 
A(2 wires) = (2)(4.91 X 10 -10m2) = 9.82 X 10 -10 m2 
L = 12 mm = 0.012 m 

From the equation above: 
Qcond = [(70.9 W/m°C) (9.82 X 10-10 m2)] (50°C) / (0.012 m) 
= 0.0003 W 

Since the conductive load is inversely proportional to the length of the wire, it can be reduced by using longer wires.

6. Combined Convection & Conduction

The following equation can be used for estimating heat losses due to convection and conduction of an enclosure. 

Q passive = (A x DT)/(x/k + 1/h) 

Qpassive = heat load (W) 
A = total external surface area of enclosure (m2) 
x = thickness of insulation (m) 
k = thermal conductivity of insulation (W/m °C) 
h = convective heat transfer coefficient (W/m2 °C) 
DT = Temperature change (°C) 

Table II 
Typical Values of Convection Heat Transfer Coefficient 

Process h (W / m2 °C)
Free Convection – Air 2 – 25
Forced Convection – Air 25 – 250

Table III 
Typical Values of Thermal Conductivity for Insulation 

Product Thermal Conductivity (W / m°C)
Styrofoam .031
Polystyrene .037
Polyurethane .039


7. Transient Temperature

Some designs require a set amount of time to reach the desired temperature. The following equation may be used to estimate the time required: 

t = [(r) (V) (Cp) (T1 - T2)]/Q 

t = time (seconds) 
r = density (g/cm3) 
V = volume (cm3) 
Cp = specific heat (J/g °C) 
T1-T2 = temperature change (°C) 
Q = (Qto + Qtt) / 2 (J/s; Note: 1 J/s = 1 W) 

Q(to) is the initial heat pumping capacity when the temperature difference across the cooler is zero. Q(tt) is the heat pumping capacity when the desired temperature difference is reached and heat pumping capacity is decreased. Q(to) and Q(tt) are used to obtain an average value. 

Heat loading may occur through one or more of four modes: active, radiation, convection or conduction. By utilizing these equations, you can estimate your heat loads. The resulting information can then be used to select a suitable TEC for your application.

B. Calculating Heat Loads


Type of Load Example Your Case:

Active 8.0 W W

Radiation 0.2*W W

Convective 0.8*W W

Conductive 0.0*W W

Total Heat Load(Q) 9.0 W W

* Refer to "Estimating Heat Loads" for information on determining these loads.

C. Define Temperatures


Component Temp.  Example Your Case:

TEC hot side (Th) 27°C _______ °C 
TEC cold side (Tc) -8°C _______ °C 
DT(Th-Tc) 35°C _______ °C

D. Determine Number of Stages Required


Select the minimum number of stages from Table IV necessary to meet the required DT. 

Table IV 
Typical maximum obtainable DT Values for Single- & Multi-stage TECs

 Of Stages DTmax in Dry N2 @ 1 atm
(°C) DTmax in a Vacuum
1 64 67
2 84 91
3 95 109
4 -- 115
5 -- 121
6 -- 127

In this example, a single-stage TEC will suffice since 64°C is greater than the desired 35°C DT. If the number of stages required exceeds two, the following selection process is not applicable. These calculations are only accurate for one- or two-stage TECs.

For three-stage and above, call one of our applications engineers for assistance.

E. Select an Appropriate TEC


The performance graphs used in this brochure have been normalized to provide a universal curve for use with any single- or two-stage TEC for which the "maximum" values are known. By using ratios of actual to "maximum" performance values, performance may be estimated over a wide range of operating conditions.

  • a. Determine the ratio of DT/DTmax.

Example Your Case:

Record DT from step 2 35°C _______ °C 
Record DTmax from step 3 64°C _______ °C
Calculate DT/Tmax 35°C/64°C = 0.55 _______

  • b. On the performance graph (Figure 2) draw a horizontal line on the graph corresponding to the calculated value of DT/DTmax (0.55 in this example).
  • c. Obtain the optimum value of Q/Qmax at the intersection of the horizontal line just drawn and the diagonal Optimum Q/Qmax line. Interpolation between curves may be necessary.

ExampleYour Case:

Optimum Q/Qmax = 0.25

  • d. Obtain the maximum value of Q/Qmax at the intersection of the horizontal line (drawn in Step 4b) and the right vertical axis.


Maximum Q/Qmax = 0.45

  • e. Divide the total heat load (from Step 1) by the Q/Qmax ratios above to calculate the optimum and maximum Qmax.


Optimum Q/max = 9.0W / 0.25 = 36W ¸ = W
Maximum Q/Qmax = 9.0W / 0.45 = 20W ¸ = W

  • f. Select a TEC from Marlow's Standard Thermoelectric Coolers product list with a Qmax greater than the maximum Qmax (20 watts in this example), but less than the optimum Qmax (36 watts in this example).


Keep in mind that within this range, a TEC with a Qmax close to the optimum Qmax will provide maximum efficiency, and a Qmax close to the maximum Qmax will yield smaller and possibly less expensive TECs.

Reading down the Qmax column results in the selection of the following TECs:

Example TEC Values Your Case TEC Values:

Range Location Model# Qmax Imax Vmax Model# Qmax Imax Vmax
Near Optimum RC6-6 30 5.6 8.2 ______ _____ ____ _____
Nearer Maximum RC6-4 20 3.7 8.2 ______ _____ ____ _____

For this example, let us assume maximum efficiency is desired. Thus, the 5.6 amp, 8.2 volt cooler is selected, because between these two potential TECs, its Qmax (30 watts) is closest to the optimum Qmax (36 watts).

Thermoelectric Cooler

F. Lead Wire & External Metalization Options


For volume production, various lead wire configurations are available to meet customer-specific applications. For initial samples, Marlow offers standard lead wire lengths of 4 inches for the RC product line. 

The NL products offer various external metallization options depending on the customer’s mounting requirements. Copper metallization with nickel plating and gold plating can be added to either the bottom ceramic or both the top and bottom ceramics. Marlow will also provide pretinning of these external surfaces with indium-based solders to further assist customers in their TE cooler mounting requirements.

G. Estimate TEC Performance


For volume production, various lead wire configurations are available to meet customer-specific applications. For initial samples, Marlow offers standard lead wire lengths of 4 inches for the RC product line. 

The NL products offer various external metallization options depending on the customer’s mounting requirements. Copper metallization with nickel plating and gold plating can be added to either the bottom ceramic or both the top and bottom ceramics. Marlow will also provide pretinning of these external surfaces with indium-based solders to further assist customers in their TE cooler mounting requirements.

1. TEC Performance Parameters


Now that the TEC has been selected, the next step is to determine its performance.

  • a. Refer again to the Table of Standard Thermoelectric Coolers and record the maximum values for the TEC whose performance is being evaluated. Note that if a hot-side temperature other than 27°C is used, the performance parameters must be adjusted to obtain more accurate results. Please call one of our applications engineers for assistance.


Example Your Case:
TEC Minimum Values
DTmax 65° C °C
Qmax 30 W W
Imax 5.6 A A
Vmax 8.2 V V

  • b. Determine the ratio of DT/DTmax:


DT(from step 2) 35° C °C
DTmax (from Step 1a) 65° C °C
DT/Tmax (calculate) 35/65= .54 °C

  • c. On the performance graph (See Figure 3) draw a horizontal line on the upper graph corresponding to the DT/DTmax (0.54 in this example).
  • d. Divide the total heat load (from Step 1 of the TEC Selection Procedure) by the value of Qmax for the TEC selected (from Step 1a):


Q/Qmax = 9.0 W / 30 W = 0.3
= W ¸ W = ____

  • e. At the intersection of the horizontal line (drawn in Step 1c) and the value of Q/Qmax (just calculated in Step 1d), plot a vertical line on the performance graph.
  • f. Record the value of I/Imax at the intersection of the vertical line just plotted and the bottom I/Imax axis.

I/Imax = 0.6 I/Imax = _______


2. Calculate TEC Current

Using the value of Imax (obtained in Step 1a) and the value of I/Imax (just obtained in Step 1f ), calculate the TEC current (I).

TEC current = (Imax) X (I/Imax) = 5.6 A X 0.6 = 3.4 A = A X = A

3. Calculate TEC Voltage

  • a. Draw a horizontal line through each of the intersections of the vertical line (drawn in Step 1e) and the two curves in the lower half of the performance curve in Figure 3.
  • b. Record the values of V/Vmax at the intersection of the two horizontal lines drawn and the left vertical axis. These values determine the range of V/Vmax for the TEC selected.


For large DTs (small values of Q/Qmax), the voltage will correspond to the high end of the range.

For small DTs (large values of Q/Qmax), the voltage will correspond to the low end of the range.

Example Your Case: 
High V/Vmax = 0.68 __________ Low V/Vmax = 0.54 __________


  • c. Multiply Vmax (obtained in Step 1a) by each value of V/Vmax just obtained to get the range of TEC voltage.


High TEC Voltage = Vmax X (V/Vmax) = 8.2 V X 0.68 = 5.6 V V X = V
Low TEC Voltage = Vmax X (V/Vmax) = 8.2 V X 0.54 = 4.4 V ______ V X = V


4. Calculate Maximum TEC Power

Multiply TEC current (from Step 2) by High TEC voltage (from Step 3c).

TEC Power = 3.4 A X 5.6 V = 19.0 W = A X V = W


5. Calculate Power Dissipated into the Heat Sink

The power dissipated into the heat sink is the sum of the total heat load and the input power to the TEC. 

Qh = 9.0 W + 19.0 W = 28.0 W 
= W + W = W 

At this point, you should have several candidate TECs with their approximate performances, and you are now ready to order. 

Heat Sink