PELTIER APPLICATION NOTE - CUI Inc

1 PELTIER . APPLICATION NOTE. PELTIER APPLICATION Note Early 19th century scientists, Thomas Seebeck and Jean PELTIER , first discovered the phenomena that are the basis for today's thermoelectric industry. Seebeck found that if you placed a temperature gradient across the junctions of two dissimilar conductors, electrical current would flow. PELTIER , on the other hand, learned that passing current through two dissimilar electrical conductors, caused heat to be either emitted or absorbed at the junction of the materials. It was only after mid-20th Century advancements in semiconductor technology, however, that practical applications for thermoelectric devices became feasible.

2 With modern techniques, we can now produce thermoelectric modules that deliver efficient solid state heat-pumping for both cooling and heating; many of these units can also be used to generate DC power at reduced efficiency. New and often elegant uses for thermoelectrics continue to be developed each day. PELTIER STRUCTURE PELTIER THEORY. A typical thermoelectric module consists of an array When DC voltage is applied to the module, the of Bismuth Telluride semiconductor pellets that have positive and negative charge carriers in the pellet been doped so that one type of charge carrier array absorb heat energy from one substrate surface either positive or negative carries the majority of and release it to the substrate at the opposite side.

3 Current. The pairs of P/N pellets are configured so The surface where heat energy is absorbed becomes that they are connected electrically in series, but cold; the opposite surface where heat energy is thermally in parallel. Metalized ceramic substrates released, becomes hot. Reversing the polarity will provide the platform for the pellets and the small result in reversed hot and cold sides. conductive tabs that connect them. Ceramic Released Released Substrate Heat Heat N-Type Bismuth P-Type Bismuth Telluride Telluride P-Type Absorbed Absorbed Semiconductor Heat Heat Conductor Pellets - +.

4 Tabs N-Type DC Power Source Semiconductor Negative (-) Positive (+) Pellets FIGURE 2. THEORY. FIGURE 1. CONSTRUCTION. page 2. PELTIER APPLICATION Note HOW TO READ THE TECHNICAL DATA. REGARDING PELTIER MODULE SPECIFICATIONS. The maximum electric current (Imax) and the maximum voltage (Vmax) values are not the absolute maximum rated values. Instead, considering performance coefficients and heat radiation design, it is recommended that products are used to around 70% of the maximum electric current and voltage values. If products are used with voltages and currents which exceed the maximum values, heat absorption will decrease and Joule heating will increase.

5 As a result, not only will efficiency be reduced, but the increase in temperature will have an adverse effect on the soldering connecting the semiconductor and could lead to a break down and reverse diffusion. REGARDING PELTIER MODULE FUNCTION DIAGRAMS. The maximum temperature difference ( Tmax) is the temperature difference between the sides of the semiconductor when the heat absorption is 0(W). Also, the maximum heat absorption, Qmax, is attained when the temperature difference between the sides of the semiconductor is 0. Even though these are both not actual values but theoretical figures, please use these as a guide for choosing modules.

6 For the relationship between electric current, voltage, temperature difference and heat absorption, please consult the function diagrams in the design examples. page 3. PELTIER APPLICATION Note DESIGN EXAMPLES. EXAMPLE 1 (CP60440). What is the heat absorption (Qc) and supplied current (I) when Th=50 C, Tc=10 C, Voltage=12 Vdc? 1) Find T. T = Th-Tc = 50-10=40 C. 2) Find the supplied voltage from the function diagram From the diagram, at Th=50 C I= A. 3) Find the heat absorption (Qc) from the function diagram The current found in [2] is A, so from the diagram at Th=50 C Qc= W.

7 (Th=50 C) Tc (C ) [2] I = 16 12V. 1 Supplied Voltage (V) 8 2. 0. 60. 4 Qc Heat Absorption 40. Qc (W) 3. 20 [3] Qc = 0. 0 70 60 50 40 30 20 10 0. Temperature Di erence (C ) 5 T = Th - Tc [1]. Semiconductor FIGURE 3. FUNCTION DIAGRAM. ceramic Tc electrode T element electrode 1) Potential difference across the PELTIER Module (V). Th ceramic 2) The connection between T ( C) for each supplied current and voltage (V). 3) The connection between T ( C) for each supplied current and heat absorption (W). 4) Heat absorption of the PELTIER Module (W). 5) The temperature difference ( T) shows the difference between the hot side and cool side of the PELTIER module and the electrodes.

8 It is not the difference between the cool side and the background temperature. page 4. PELTIER APPLICATION Note EXAMPLE 2. How to choose a cooler unit When choosing a cooler unit that suits the functions you require, you need to know the heat absorption rate and temperature difference. These two things can be calculated easily. Internal dimensions of box 500 200 100 (mm) 10 liters Example of heat calculation Conditions: External dimensions 560 260 160 (mm). when cooling in an enclosed Thermal insulation 30 (mm) urethane foam environment Internal temperature 5 C. Ambient temperature 30 C.

9 1) Thermal conductivity . = (W/m C): Thermal conductivity of urethane foam t = (m): Thickness of thermal insulation 2) Surface area of box (at center of thermal insulation) S. S = ( ) 2 + ( ) 2+( ) 2 = (m2). 3) Overall heat transfer rate K. Thermal conductivity of external surface: h1 = 20 (W/m2 C). Thermal conductivity of internal surface: h2 = 10 (W/m2 C). K = 1/{(1/h1) + (1/h2) + t/ }|. = 1/{(1/20) + (1/10) + }|. = (W/m2 C). 4) Amount of heat entering the insulated box from external sources Q1. Q1 = S K T. = (30-5). = W. 5) Necessary heat absorption Q. Internal thermal loading (in the case of loading from an internal heat source): Let Q2 = 5 (W).

10 Q = Q1 + Q2 = (W). 6) Choice of cooler unit Adding a safety margin of 25% to the necessary heat absorption give (W). In other words, a cooler unit that provides heat absorption of over 16 (W) and a temperature difference of at least 25 is necessary. Ex. CP30238 can easily give a 30 difference at 6 W of heat absorption, so W / 6 W means roughly three units are necessary. page 5. PELTIER APPLICATION Note EXAMPLE 3. Example of heat calculation Conditions: Internal dimensions of box 250 200 100 (mm) 5 liters when cooling in an enclosed External dimensions 310 260 160 (mm). Thermal insulation 30 (mm) urethane foam environment Cooling time 1 hour Initial water temperature 20 C.