I have an unknown (unmarked) peltier module. I'd like to determine the characteristics of the peltier itself. The only things I know about it are its size and the fact that it is drawing 6 amps at 12 volts. I'd like to determine it's output wattage if possible. Anyone have ideas as to how I could do this?
"Eric Braunhauffer" wrote in news:mNFJi.919$Wo4.916@trnddc03:
A TEC made for 12V is probably set to be efficient at that voltage. Such TEC's usually have a Vmax of about 14.5V, so assuming 2 ohms as a basis for assuming 7.25 amps as Imax (you can test that briefly to see if it's true) then consumed power will be 105 watts. Qmax will be about half of that.
Pelteriers get hot on one side and cld one the other. They cannot go below 0C. You need to heatsink them well and maybe cool the hot side )with fan and heatsink)., Maybe having a fan runnning over it, and the other side of the Peltier gets cold.They change polarity (heat and cold) depending on the way you wire then up. I used 4mm perspex tp make a decent heat exchanger. It is worth the effort.
firstname.lastname@example.org (Hal Murray) wrote in news:aZednZuMbtzUmGfbnZ2dnUVZ email@example.com:
Yes, they do.
TEC's can have delta T of 67°C with a hot side clamped at 25°C, some as much as 71°C, so around -45°C is usually possible with a tiny load, like a sensor. The main reason to avoid going below 0°C is ice, if water vapour gets in there, the expansion on freezing will easily break a TEC, but water will damage the thermocouples anyway, so it should always be excluded.
wrote in message news: firstname.lastname@example.org...
Peltier devices, are effectively electrical heat pumps. They only get cold on one side, if the heat being generated on the other is removed. This heat comprises the heat being pumped, and the heat generated by the Peltier itself. They can go way below 0C. There is a balance between how much heat they can move, and the temperature difference between the two sides of the module. Typically modules can achieve up to about 70C delta, but at this, they are able to move almost no heat. Conversely, depending on the size of the module, there will be a point where the temperature difference reaches zero, for a given amount of heat to move (Qmax). Typically a Peltier will draw about double the power, that they can move. So, in the example, the element is drawing about 72W. This would be typical of a module able to move perhaps 40W of heat at a 0C delta. Add a heatsink, with a temperature rise of perhape 0.1C/W, fan cooled to the 'hot' side, and this will need to dissipate about 112W of heat in this 'worst case'. It'll then run up to about 11C above ambient. If you have something that only generates 15W of heat, and attach this to the 'cold' side, expect the delta C to be perhaps 45C, and the component to cool to about 34C below the ambient. The actual delta will depend on the module, but this would be a typical value for a module this size. The 'lowside' temperature can be well below zero. There is nothing 'magic' for a Peltier module about 0C. For even higher delta's, cascade two modules. Remember though, that the 'hot' module, has to pump not only the original heat, but the heat being added by the first module... I have one sitting here on the bench, which is currently holding a couple of semiconductor junctions at -40C. They are only generating about 1W of heat, but to maintain this temperature (currently 62C below ambient), with the two stage cooler attached, over 100W of power is being fed into the cooler module, and the heatsink size reflects this... The best way to 'characterise' the module, would be to measure the temperatures on both sides of the junction, with known amount of heat being applied to the cold side (resistor fed from a controlled voltage source). You can then measure the delta T, for different power inputs, and build your own graph for the behaviour of the module.