This thermoelectric module for cooling system is new for me. I conducted an experiment on Peltier Thermoelectric Module. I want to test how it behaves when I provide certain level of voltage and what would be the current looks like and calculate the power used before designing suitable circuit to control it.

Peltier Thermoelectric Module with max rating: 100W, 20V, 8.5A, deltaT 71DegC
http://uk.rs-online.com/web/p/peltier-modules/6935107/

Test Setup: As shown in the picture. Cold side of Peltier faced upwards exposing to ambient. Hot side faced downwards and fan+heat sink was used to dissipate the heat from there.



DC power supply is used to generate DC voltages 12V, 16,V and 18V to Peltier and to measure the current.

Measurement Results:
At 12V, I = 3.54A, P=42.28W
At 16V, I = 4.1A, P=65.6W
At 18V, I=4.5A, P= 81W
At 20V, I= 4.9A, P=98W

Observation:
1. When voltage increases, the current also increases. Thus the calculated power is higher.
2. When first turn on the DC power supply, eg. at 12V, the current reading at first is around 4.1A then gradually drops to 3.54A after 1 minute, and it keeps on decreasing slowly as well. Same observation happened at other voltages.
3. When I touch at the cold side, the current increases. Thus I think that more current is consumed to cool down the heat (from my finger)
4. Eventually one of the thermoelectric modules broken down, found it was open, high impedance. But the last reading was 12V and approx 3.3A. The cold side has turned hot when the moment current reading became 0A.

Questions:
Q1: Why the measured current is higher when the voltage increases? Assume the ambient is about the same, so the power used to cool down the ambient heat should be the same, right?

Q2. I didn't not exceed the max voltage 20V, and the current measured was 3.3A still far below 8.5A. What makes the thermoelectric module broken? Is it because I must always put hot object at cold side to keep this module working?

Q3. Is this how the thermoelectric module behaves? My application is to use this module to cool down water tank. So when the water tank is hot, then only I turn on the PWM to drive the thermoeletric module?

Hope to get some answer to help me understand more on its behavior. Thanks.

 

Q1: Why the measured current is higher when the voltage increases? Assume the ambient is about the same, so the power used to cool down the ambient heat should be the same, right?

The module is roughly a resistor with nearly (but not exactly) constant resistance. A value of 1-2Ω is typical. So the current will roughly obey Ohms law, V = I•R or I =V/R

Q2. I didn't not exceed the max voltage 20V, and the current measured was 3.3A still far below 8.5A. What makes the thermoelectric module broken? Is it because I must always put hot object at cold side to keep this module working?

The behavior of a TEC varies with the ∆T across it's two sides, and with the absolute T of the hot side. I suspect that you drove the TEC outside of of its specifications and allowed the hot side to get far too hot. Without a good heat sink and plenty of air flow, that is the predictable result.

Q3. Is this how the thermoelectric module behaves? My application is to use this module to cool down water tank. So when the water tank is hot, then only I turn on the PWM to drive the thermoeletric module?

I don't understand your question, so I'll just make some general comments. You need to find the properties of your TEC, to model how it will perform at every condition of T and ∆T that you expect to encounter. One thing you will learn is that efficiency is greatest when the TEC is operated far below its specified maximum current an ∆T. So power is used efficiently at low ∆T and low current. Of course it will operate far below the rated capacity under these conditions.

It may be obvious but I'll say it anyway: All the power you supply, plus the heat you move from the cold side to the hot, appears on the hot side of the TEC. Its crucial you remove and dissipate all that heat. At maximum current, the power you supply the TEC may be 10 times greater than the amount moved from hot to cold. So if you need 10W of cooling delivered to your water vessel, you might have to remove 110W of heat from the TEC.

Failure to cool the hot side can cook your TEC and destroy it.

[update] I looked again at your setup and it looks adequate. Did you use heatsink compound between the TEC and the heat sink? I still suspect you overheated.

 

Good points from wayneh.

I have used TECs in a few projects. They are a bit tricky and the usual "linear" design has to be tweaked.

A TEC is like a thermocouple (Seebeck effect) operated in reverse. Applying a current changes the temperatures around the junction.

One last point: If you are going to use a PWM signal to drive the TEC, make sure it's much faster than the time constant of the TEC. If the frequency is too slow you can damage the construction of the junction.