I had recently ordered a TEC1-12706 Peltier module from amazon. I tested the module as soon as I recieved with a 1.2V Ni-Cd to make sure its working and also to make sure of the hot and cold sides. I purchased it for a small mini cooler project.So, attached the Peltier module to a CPU heat sink with fan and the cold side with a small aluminium block and added the necessary insulation etc. and I was able to cool down a small beaker of boiled hot water to ice cold, though it didn't freeze.I ran the unit for like 45 minutes. The unit was powered with a 12.61V-2400mAh (3 Cell Li-ion ICR18650-24E) pack. The initial current draw was around 3.8A but quickly dropped to 2.6-2-7A and remained stable.Tried bumping up the voltage to 14V and current draw increased to 4.3A but reduced it back since I didn't was to ruin the module. So, is it normal for the device to run at just 2.7A current draw ? because most people on the internet who brought 12706 say the current draw is 4A and above at 12V.

Update: Guys finally I managed to get my issue fixed. the current draw of 6A is infact for Vmax. At 12V the initial current draw of 3.8-4A is okay. I finally managed to get ice. Got a 30ml aluminium cylindrical vessel containing hot water at 60C in my mini cooler to ice in 15minutes!:)..The main issue was the cheap thermal paste I had used initially that I brought for just $0.95.So, purchased another brand ZP Heatsink Compound $3 thermal paste(Thermal conductivity:1.22W/m.k& Thermal Resistance:0.201C-in2/w),removed my old paste and applied this to both hot and cold side.Everything is working awesome.Never thought a bad thermal paste could cause this much bad performance."Gotta change the thermal paste on my i7 too or its gonna fry too!!:rolleyes:"

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    \$\begingroup\$ Most likely the current depends on the junction temperatures. \$\endgroup\$ – skvery Feb 16 '17 at 19:00
  • \$\begingroup\$ Do you have a heatsink / fan on the hot side? \$\endgroup\$ – CHendrix Feb 16 '17 at 19:00
  • \$\begingroup\$ Yes, I have added a heat sink with a fan attached to it I had salvaged from an old PC.The peltier is really cold to the touch with water condensation on top, though I didn't see any ice forming. \$\endgroup\$ – RSB Feb 16 '17 at 19:02
  • \$\begingroup\$ tried first with the following heat sink with fan attached:ebay.com/p/HP-Dc7900-SFF-Processor-Heatsink-480368-001/… \$\endgroup\$ – RSB Feb 16 '17 at 19:09
  • \$\begingroup\$ currently using amazon.com/INTEL-A65061-002-SOCKET-478-A65061-002/dp/B00CD44IU4 \$\endgroup\$ – RSB Feb 16 '17 at 19:10

There are a bunch of these devices available at very cheap prices.... not unexpectedly though many have either intermittent or broken connections to some of the elements. When you see the prices vary from close $2 a piece through to $15 a piece it's hardly a surprise that defects and dropouts make it into the market. You will see devices quoted from QC of 50W through to close to 100W.

Sounds like you have some intermittent connections in your device.

There are a bunch of datasheets too ...here's one.

  • \$\begingroup\$ Thanks. I guess I'll have to buy a new one,probably the $10-15 one itself. But the current one is still usable.So, I'll use it for cooling till it fails meanwhile going to order a good one. \$\endgroup\$ – RSB Feb 16 '17 at 19:41
  • \$\begingroup\$ I use these for an Ice clamp for my CNC ...I bought the cheap Ebay ones and used 4 where I could have used one. It still worked out cheaper than buying prime parts. \$\endgroup\$ – Jack Creasey Feb 16 '17 at 21:46
  • \$\begingroup\$ Mate the peltier does work but I wanted to know if there is something wrong with my module or my setup. Like the cold side does get really cold but does not produce ice like in thiose Youtube videos. \$\endgroup\$ – RSB Feb 17 '17 at 10:28

The current in a Peltier cooler is not determined by voltage alone (it is not a resistor). It depends on applied voltage, AND on the temperature difference between the plates. If you put a voltmeter on the device while the plates are at disparate temperatures, you will find that it generates an internal voltage; it operates in reverse, turning a temerature difference into electrical energy. This is the Seebeck effect, usually known as the thermocouple effect.

The applied terminal voltage, minus the internal thermocouple voltage, determines the current flow. As your cold side gets colder, the thermocouple voltage opposes the applied voltage input, and current drops.

  • \$\begingroup\$ Yes, I know that the current drops as the thermocouple gets colder. But it was the initial current draw of only 3A from the rated 5A that got me thinking if my module was defective.Mmm I'm going to try lacing the heat sink in an ice bath and see the I vs T(cold side) and also if there is ice formation. \$\endgroup\$ – RSB Feb 17 '17 at 10:01

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