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]I have 6 EC1-12706 in series connected to a variable DC Power Supply (rated with maximum 30V and 10A). I set the power supply to 30V and 1.50A.

My problem is that the DC Power Supply when the output is enabled keeps the voltage to 30V but the ampere goes down to 1.29A.

Why does the power supply cannot provide the 1.5A desired?

My calculus is that in series the 30V should be divided by 6 (5V each TEC) and the amp distributed among the 6 equally.

Schematic:

enter image description here

Photo:

enter image description here

EC1-12706 DataSheet: https://datasheetspdf.com/pdf-file/634368/HB/TEC1-12706/1

DC Power Supply: https://www.amazon.com/gp/product/B07Y5XVTQL/

Testing data: I tested with a single EC1-12706 unit. I set the variable DC Power Supply to 12V and 5A, the power supply says (and the multimeter) says: 12V, 2.4A (cannot reach 5A, hence it seems that the 1.98ohms of the spec sheet is wrong?

Inside: enter image description here

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  • \$\begingroup\$ I set the power supply to 30V and 1.50A. ... what does that mean? ... what parameter are you actually setting? \$\endgroup\$ – jsotola Sep 21 '20 at 1:41
  • \$\begingroup\$ The Ampere setting is a Limiter function... it does not make sense to use it here. How many Watt the supply has? at 30V at any case with 6 in a row the resistance is so low that more than 2A "want" to flow \$\endgroup\$ – schnedan Sep 21 '20 at 10:18
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How much voltage drop is there in the wiring?

What deltaT do you have? The required voltage changes with the temperature difference.

Don't forget to take account the resistance of the meter (and it's leads). Even the ammeter will have resistance that will reduce the current, the current when you connect the TEC directly will be somewhat more.

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  • \$\begingroup\$ Thank you Kevin for guiding me. I am very beginner to I will do my best. After 5 minutes, with the heatsink in place on the hot side, I am probing 23C and 31C (74F and 84F). Ambient temperature is 26.5C \$\endgroup\$ – Patrick Desjardins Sep 20 '20 at 22:03
  • \$\begingroup\$ After 10 minutes, the temperature is 14C and 28C. The module resistance from the spec sheet is 1.98 Omhs \$\endgroup\$ – Patrick Desjardins Sep 20 '20 at 22:13
  • \$\begingroup\$ @PatrickDesjardins - don't forget that the hotside temperature of the TEC will be a bit higher than that of the heat sink. The hot side only being 3C above ambient seems low since there are ~40W+ being dissipated. \$\endgroup\$ – Kevin White Sep 20 '20 at 22:32
  • \$\begingroup\$ "My calculus is that in series the 30V should be divided by 6 (5V each TEC)" - can you measure the actual TEC voltages? Please show us a photo of your setup. "Why does the power supply cannot provide the 1.5A desired?" - if the power supply continues to put out 30V under load then it can deliver more current, but the load's resistance is limiting the current. \$\endgroup\$ – Bruce Abbott Sep 20 '20 at 23:32
  • \$\begingroup\$ @BruceAbbott I added in the question a schema and a picture. When I connect the power supply to a single TEC I can set to 12V and only 2.4A even if I set the power supply to 12V and 6A which is what the spec sheet says could handle. \$\endgroup\$ – Patrick Desjardins Sep 21 '20 at 0:25
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My problem is that the DC Power Supply when the output is enabled keeps the voltage to 30V but the ampere goes down to 1.29A.

Why does the power supply cannot provide the 1.5A desired?

Your power supply is working properly. It can supply up to 1.5A when the current limit is set to this value. If the load draws less than 1.5A the power supply will not force it to draw more.

Your problem is that your TEC modules have higher than expected resistance, and so are drawing less current than expected. This suggests that they are 'fakes' ie. lower spec devices with fake markings.

The site What TEC do I have? shows how to measure the size of one semiconductor pellet in the TEC to determine its likely current rating.

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Solution is simple:

When its cold, you adjust the current to be 1,5A. (Ri is 6 × 1,98 Ω.)

When its hot resistance changes to 6 × 2,3 Ω.

1,5 / 2,3 × 1,98 = 1,291 A.

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  • \$\begingroup\$ So you mean that I could raise the AMP by adding resistance in the circuit? What do you mean "when it's hot" or "when it's cold" ? \$\endgroup\$ – Patrick Desjardins Sep 21 '20 at 16:20
  • \$\begingroup\$ @PatrickDesjardins You know how a ThermoelectricCooler works? You have read the datasheet you linked to? And no, you can not raise AMP by adding resistance... \$\endgroup\$ – schnedan Sep 21 '20 at 16:31
  • \$\begingroup\$ I know the basics of thermoelectric cooler works (as mentioned I am new in the field). I am learning by doing this exercise in which I have this question. I have read the datasheet but it is not very clear. I thought I could by modifying the voltage and amperage the temperature. Nothing talks about "when cold" or "when hot" in any tutorial or datasheet I have read. Sorry. \$\endgroup\$ – Patrick Desjardins Sep 21 '20 at 16:44
  • \$\begingroup\$ I think I understand what you mean now. When the device gets cold, the resistance is 1.98 but when it gets warm (delta T in the datasheet diagram) it makes the resistance increase. The more it gets warm, the more there is resistance. What is do not understand is still why if I use 1 unit at 12v and 5 amp that I still get 1.3 amp and not the total 5. The device is cold and 1 unit. The resistance should be 1.98, So 12/1.98 = about 6 amp, right? \$\endgroup\$ – Patrick Desjardins Sep 21 '20 at 16:47
  • \$\begingroup\$ Well you have 30V and 6*2,3 Ohm, because you choose series connection. this is 13,8 Ohm and 30 V / 13,8 Ohm is 2,173 A. It will never consume 5. If you connect 2 times 3 modules in series connection parallel, you can half the resistance and double the current. So with 2 times 3 in a row you get ~4,2A \$\endgroup\$ – schnedan Sep 21 '20 at 17:09

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