How to Choose: TEC / Peltiers

Three factors influence the temperature you will be running at. These parameters are the hot surface temperature (T_h), the cold surface temperature (T_c), and the heat load to be absorbed at the cold surface (Q_c).

As a basic rule of thumb, you can estimate the temperature of the hot side by using one of the following three statements :

Natural Convection : 20°C to 40°C [Natural Convection refers to a heatsink not coupled to a fan, this will probably not apply to your system.]
Forced Convection : 10°C to 15°C [Forced Convection refers to aircooling your peltier with a heatsink and fan, like the Swiftech 360 for example.]
Liquid Cooling : 2°C to 5°C [Liquid Cooling is of course, in reference to water-cooling your system with a radiator and waterblock.]

Take the temperature from the list that applies to you, and ADD it to the current room temperature, and presto, you have instant hot-side temperature. For water-cooled rigs, the 2-5°C rise is in comparison to water temperature, and not the room temp.

The cold-side temperature is easily obtainable, through the following operation. T = T_h - T_c
This operation is then transformed to put the "known" variables together, and the T_c by itself. T_h - T = T_c
[This formula is only applied when Vmax and Imax are present, with a room temperature of 25°C]

Figure 1 : The Basic functionning of a TEC

Now you would like to know if your peltier will be able to pump out all your processor's heat don't you. Here is where it gets a little more complicated.

Estimating Q_c, the heat load in watts absorbed from the cold side is difficult, because all thermal loads in the design must be considered. Among these thermal loads are :

The ambient air. Some of the watts of heat dissipated by the cold side will be used at dissipating the surrounding's heat.
· Conductive materials. There will be some sort of waste occurring from actually cooling the wires, connectors and screws.
· The amount of heat produced by the TEC getting current flowing through it.
· Lastly, if the TEC is not completely isolated from outside air, the dissipation characteristics will often drop.

Here is an example application of the formulaes to calculate if a TEC is right for you.

In this example we will be using the Melcor #CP 1.4-127-06L Module to cool a CPU that produces 30 Watts of heat, down to say 5°C.We will water-cool that TEC, meaning the hotside will only gain about 5°C from room temp (30°C). Here are this module's characteristics. :

I_max (Amps) 6.0
Q_max² (Watts @ 25°C) 51.4
V_max (Volts) 15.4
T_max (°C) 67

The parameter T follows directly from T_h and T_c. Since the cold side of the thermoelectric is in direct contact with the object being cooled, T_c is estimated to be 5°C. Assuming a 5°C rise above ambient for the water-cooled system, T_h is estimated to be 35°C. Without knowing the power into the thermoelectric, an exact value of T_h cannot be found. The following equation gives the temperature difference across the thermoelectric:

T = Th - Tc = 35°C - 5°C = 30°C

Figures 2 and 3 show performance curves for the CP1.4-127-06L at a hot side temperature of 35°C. Referring to Figure 3, the intersection of Q_c and T show that this thermoelectric can pump 30 watts of heat at a T of 30°C with an input current of 6 amps (I_max).

T = Th - Tc = 35°C - 5°C = 30°C

Figures 2 and 3 show performance curves for the CP1.4-127-06L at a hot side temperature of 35°C. Referring to Figure 3, the intersection of Q_c and T show that this thermoelectric can pump 30 watts of heat at a T of 30°C with an input current of 6 amps (I_max).

Figure 2: T vs. Voltage


Figure 3 : T vs. Q_c

These values are based on the estimate T_h = 35°C. Once the power into the thermoelectric is determined, 2 equations can be used to solve for T_h and to determine whether the original estimate of T_h was appropriate.

The input power to the thermoelectric, P_in, is the product of the current and the voltage. Using the 6 amp line in Figure 2 for the current, the input voltage corresponding to T = 30°C is approximately 15.2 volts.

Using these equations, T_h can now be calculated.

Q_h = Q_c + P_in
Q_c = 30Watts (Cpu heat)
P_in = 6Amps * 15.2 Volts = 91,2 Watts (electrical input for the TEC)
Q_h = 30 Watts + 91,2 Watts = 121,2 Watts.

T_h = T_amb + (O) (Q_h)
T_amp = 30°C (Ambient temperature)
O = 0.10°C/watt (thermal resistance of heat exchanger, IE Coldplate and waterblock...value estimated here)
Q_h = 121,2 Watts
T_h = 30°C + (0.10°C/watt)*(121,2 Watts)
T_h = 30°C + 12,12°C
T_h = 42,12°C

Now, since the Calculated T_h is 7,12°C higher than our anticipated T_h of 35°C, it means that the TEC used here will NOT be able to cool the CPU down to 5°C. It will probably be more realistic to think that the T_c would be around 10°C.

Conclusion
I hope that this article has helped you understand even more about peltiers, and about knowing WHICH one is right. I have learned alot while writing this article, and it is my goal to make you understand how to calculate a TEC's acceptable load, and how to estimate T_h and T_c temperatures before having to even use the peltier.

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