2-stages TEC1-12706, heats instead of cool [closed]

Question

I have the following setup: 2 TEC1-12706 stacked with hot side up and cold side down. I cool the hot side with a big heatsink and the cold side with water cooling. All system is clamped with 6 screws and thermal grease has been applied. I have a thermometer at the cold water reservoir, and all system is inside a isolation box (except from the heatsink that is outside through a hole at the top of the box). I have made a lot of experiments and always it turns to heat up the water in the reservoir. The latest experiment was: a speed driver (0-60 V DC converted by 220 AC) to control current and voltage at the cold TEC, and directly to power supply at the hot TEC. I measured at the hot TEC approx 11.9 V with 2.3 A, and I tune the cold TEC to 5.5 V with 1.1 A. The system still heats up the water. I noticed that if I disconnect the cold TEC, then the system cools. The initial temperature conditions for the arrangement was: 45^oC at the heatsink and 30^oC at the water reservoir. Any Idea?

Answer 1

I would suggest the following steps:

1. First i suggest verifying both TEC separatly, so put little load on the "cold" side, and a bit bigger heatsink on the "hot" side. Run SMALL current through. Use SMALL currents, like 500mA. Verify that the "cold" side actually cools down and "hot" side actually heats up. ALso reverse the current and observe the opposite (what you define as "cold" side has to now heat up and "hot" side has to cool down). Don't overdo it with the current. If not sure, use smaller currents. The point of this step is to confirm that TECs are actually turned the right way. Mark the sides in regards to specific polarization of TEC.

2. TECs are current driven. when setting the power to the TEC use current limit and just let the voltage be what ever it will be. OFC, voltage setting should not exceed the max voltage of the tec.

3. Using TEC - based on "generic" data for TEC

Really important thing with TEC are:

a) temperature difference between hot and cold side shouldn't be too much. IF you check datasheet you will see how, too much difference means that you can't use the thermal power that the TEC is rated for. In other words, if you are trying to use 60% of TEC thermal power or heatload, the max temperature difference between the sides of the TEC is 30 K - represented by green curve. This means that the TEC will work the best when hot and cold temperatures will be equal. But you are trying to cool one down, which will heat the other side up, create the difference between the sides and reduce the efficiency of the TEC. This needs to be taken into account and usually when designing systems with TEC around 50% of the thermal heatload (thermal power) is used, which in the case of TEC given means that max temperature difference is around 45C. after this you shouldn't attempt to cool the "cold" side anymore because TEC won't be able to do it, efficiency will drop, which will make the TEC cool less, and difference in the temp between sides will become smaller, which will increase the temperature. If this is not managed, you will get some oscillations. In order to think about this assume for the start that your "hot" side has stable temperature, if you have a heatsink with 1 K/W, and you are using like 10 W TEC, you can think about your heatsink on the hot side as being 10 K over room temperature and being stable, just for a start.

b) The current and selfheating. When you start pushing TEC, you need to provide more current. This means you will burn a lot of power on the TEC, but not all of it will go into cooling the "cold" side. Actaully, TEC isn't very efficient device. So what ever is not used for cooling goes into self heating, bringing temperature up of both hot and cold side. IF you again look at the example given, Qmax (max thermal power) is 2.9 W. but the electrical current used can be 1.3 A where voltage will be around 3.6 V, so electrical power will be around 4.7 W. So around 4.7-2.9=1.8 W goes into heating. if you would go up to 1.6 A it would be even worse, since above 1.3 A there is no efficiency increase, but voltage will increase so you eill get like 1.6 A*4 V=6.4 W of electrical power to get around 3 W of thermal power. So don't over do it on the current. In above calculations i used 100% Qmax, which is impossible because you can see that already at 80% Qmax you ge max 15 K difference between the sides. for 100% this difference will be max 0 K. So again realistically you use TEC at say 50, 60% Qmax. So if I take 60% I get 30 K difference between hot and cold max, but the power required to achieve this is limeted to 60% of Qmax, so 1.74 W. For this I need to run 1.3 A through the TEC which results in about 3.5 V. So thermal power is in reality 1.74 W/(1.3 A*3.5 V)= 38% of electrical power. So 62% of 1.3 A*3.5 W which is 2.8 W goes into heating the TEC.

4. To your example: For simple approach I will name TEC that is cooling down the water TEC1 and the TEC that is cooling down the TEC1 will be named TEC2. What I would suspect is that you are overdoing it with the current of TEC2 or heatsink of the TEC2 is not sufficient or you are using one of them at the limit. TEC2 needs to cool down TEC1 which produces 6 W of power if i assume 30% efficiency of TEC2, the electrical power of TEC2 should be 20 W. This is somehown inline with your numbers, but if TEC1 is toward the end of its power budget it will heat the secondary side a lot. You really need to go step by step. First start with TEC1 and don't put it too high, to some lower current. Then slowly increase the current of TEC2 until you get stable conditions. Then you can increase one an the other. but i would say if you run both of them at half the current of what you are doing now and it works, it's clearly overload issue. IT would help if you would share: PN of both TEC and datasheet, some photos/details about HS and it's K/W capability, water reservoir details (is the water inside heating or is it just passive load system).