Is it possible to power a peltier with 12 V and 1 or 2 A?

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    \$\begingroup\$ Read the datasheet? hebeiltd.com.cn/peltier.datasheet/TEC1-12706.pdf Look at the graphs, refine your question. The possibilities are endless Benjj!. To answer your question - NO .. not with the type you refer to. The resistance is "fixed" .. so on average you will run 5-6 amps at 12V for this type \$\endgroup\$ – Sorenp Aug 6 '19 at 6:02

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Here's the relevant graph from the datasheet. This is a 40mm x 40mm x 3.9mm Thermoelectric Cooler (TEC). I added yellow highlight on the 1.5A curve. assuming that you're driving the TEC from a constant-current 1.5A source, with available compliance voltage of at least 6V. This typical operating characteristic curve shows that under all of these conditions:

  • the hot side has a temperature of 50°C
  • and the current through the junction is 1.5A

the resulting performance is

  • up to a maximum of about 55°C delta C (so cold side might get to about 0°C)
  • burden voltage across the junction is about 5, maybe 6 volts

Note that the line stops between 50°C and 60°C for the 1.5A drive current, this implies that is the condition where the TEC can't achieve any further heat removal. At higher currents the delta-T endpoints increase, but with diminishing returns. The TEC has to move not only the hot-side heat but also its own internal heat.

If the TEC is instead powered with a constant voltage 12V supply that was capable of delivering at least 4.5A, then the drive current would be somewhere around 4.0A to 4.5A, depending on the temperature difference.

If the power supply is rated 12V at 2A maximum, something has to happen when the load tries to draw 4.5A. Either the supply will shut itself down (because it thinks the load is a short circuit), or the output voltage will reduce until the load only demands 2A. Or worse, the supply might oscillate, repeatedly shutting down and trying to restart. I don't think the 12V 2A power supply is a good match for this particular TEC.

Just a couple of general words of advice about TEC and heat removing applications:

  • The hot side needs a heat sink and a fan
  • There needs to be good thermal contact between the TEC hot side and the heatsink
  • There needs to be good thermal contact between the TEC cold side and the cooling target
  • There needs to be good thermal insulation between the cooling target and the heatsink
  • The power supply that drives the TEC will also be generating some waste heat, so that needs a path to ambient as well
  • For an application like this, the design of the thermal contacting, thermal isolating, and airflow parts of the system are just as important as the electrical parts
  • Don't cut the TEC or drill any holes into it. Mount it using a thermal conducting epoxy or adhesive.

After the TEC has been running for awhile and the heatsink is warm, and then the TEC drive current is shut down, heat will flow back from the (now very warm) heatsink through the TEC back into the cold side. The cold side is now cooler than ambient, and the hot side is much warmer than ambient. So unless the cold chamber has a great deal more thermal mass than the heatsink, then as soon as you switch off the power, your homemade refrigerator becomes an oven. Surprise! Eventually the heat sink and cold chamber reach thermal equilibrium at ambient temperature. For a CPU cooler that's probably acceptable, but if it's cooling some food or medicine that would be spoiled by high temperature, that could be an issue to watch out for. Adding some thermal mass (like water bottles) to the cold side would help avoid this issue, but the tradeoff is that it will also take longer for the system to cool down.


It's possible to pass 1 or 2 amps from a constant current source through a 12706, at which point the voltage across it will be somewhat less than 12v. Many easily obtainable buck converters have a constant current output mode that's settable to do this.

Alternatively, you can supply a lower voltage, such that it draws 1 or 2 amps, again using a buck converter to set the lower voltage.

At the low current, the 12706 will run at a much better COP (coefficient of performance, aka efficiency), but will not achieve as high a temperature difference or pump as much heat as it would at the higher voltage.

  • \$\begingroup\$ How much temperature differance would it achieve? \$\endgroup\$ – Benjj Aug 6 '19 at 6:12
  • \$\begingroup\$ Max temperature difference (at zero heat pumped) scales roughly with applied voltage. \$\endgroup\$ – Neil_UK Aug 6 '19 at 6:14
  • \$\begingroup\$ How much would that be at 1 A and 12 V? \$\endgroup\$ – Benjj Aug 6 '19 at 6:17
  • \$\begingroup\$ @Benjj: It wouldn't be 1A and 12V. To get lower current through the device, you have to lower the voltage. \$\endgroup\$ – JRE Aug 6 '19 at 7:43
  • \$\begingroup\$ Read my answer, you only get lower current at lower voltage, or vice versa. To a good approximation (because TECs are so incredibly inefficient), a TEC is more or less a resistive load. To get decreased current, decrease the volts. \$\endgroup\$ – Neil_UK Aug 6 '19 at 7:59

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