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I'm planning an addition to my liquid cooling system for my computer. I would like to use two TEC-12715 peltiers, which pull up to 15a @ 16v, and two CPU fans, probably pulling around 5W total power. However, for pricing and other reasons I was considering this power supply, which supplies 30a @ 12v.

I know that the peltiers will work at 12v, though I suspect at a lower efficiency and higher amperage(?). What my novice electronics understanding is struggling with, though, is understanding what will happen given that my devices can pull more amps than the power supply can provide.

I suspect all four devices (2x peltiers, 2x fans) will just get less power than they can handle, and the peltiers will dissappate less heat than their capacity and the fans will run at a slightly lower RPM. Is this correct? If so, will the power loss be a constant percentage of total power across all devices (everything runs at 90% power), or will the fans still get all the power they need (since their load is so small) and the peltiers will just run at lower power? Or is this all a really bad idea that will cause the PSU and devices to fail in bad ways?

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  • \$\begingroup\$ what is a peltair? Do you mean peltier module? \$\endgroup\$
    – Pierre
    Mar 6, 2017 at 14:53
  • \$\begingroup\$ if it is, here is a short answer, you will use a lot of power, and this power will make a lot of heat, on the peltier module, and on the power supply, which is not what you want. \$\endgroup\$
    – Pierre
    Mar 6, 2017 at 15:00
  • \$\begingroup\$ @Pierre the title says something about TEC's - which is synonymous with a peltier element/cell/module. \$\endgroup\$
    – Morten Jensen
    Mar 6, 2017 at 15:01
  • \$\begingroup\$ @Pierre Yes, that was an embarrassing mistake. The spelling in the question has been corrected. \$\endgroup\$
    – Nicholas
    Mar 6, 2017 at 15:04
  • \$\begingroup\$ @MortenJensen yes, i did notice that after my comments. But it doesn't change anything about the word "peltair" which google and I have never seen. But it's not a problem any more because the author did change it in the question. \$\endgroup\$
    – Pierre
    Mar 6, 2017 at 15:06

1 Answer 1

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TL;DR:

It's ALWAYS a bad idea to operate outside the current rating of any PSU.

Don't say I didn't warn you.

However, as you will be operating it at 12V, and after looking at the datasheet, your highest current consumption will probably be 12.5A per cell. 12.5A x 2 = 25A, so you'll have 4A headroom after 1A for the fans. It may be safe.

Long-form answer:

If you PSU reaches it current limit, it will start to reduce the voltage supplied. It can easily drop to almost zero in an overcurrent condition.

A good PSU should be protected against this happening and be able to recover itself. OTOH, the modules/devices you connect to it (especially its interfacing circuitry) may not be well protected against an undervoltage condition.

Even if nothing gets permanently damaged, just upset, you may briefly lose functionality while the PSU, peltier cells and fans recover from the overcurrent event. This could be tolerable or not in your application. Maybe your cooling system has enough thermal inertia to keep temperatures low until everything recovers. Or maybe not and you end up with performance glitches as the processor backs off due to temperature rising. That's for you to discover.

However, as you will be operating it at 12V, and after looking at the datasheet, your highest current consumption will probably be 12.5A per cell. 12.5A x 2 = 25A, so you'll have 4A headroom after 1A for the fans. It may be safe.

Anyway, try to be sure that the PSU can supply that much power safely (you may need some degree of forced convection). Otherwise, it's not a good idea.

N.B. - calculation of \$I_{max}\$ at 12V:

Qc vs. supply voltage C.O.P. vs. supply voltage

Worst case (maximum) current consumption at 12V for which we have data happens when \$T_h=50ºC\$ and \$DT=0ºC\$. Looking up the corresponding values from the curves above we obtain:

$$ \begin{align} Q_c \approx &\ 150\ W\\ C.O.P.\approx &\ 1 \end{align} $$

On the other hand, \$C.O.P.=\dfrac{Q_c}{V\cdot I}\$ thus:

$$ I=\frac{Q_c}{V\cdot C.O.P.} \approx \frac{150\ W}{12\ V \cdot 1} = 12.5\ A $$

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  • \$\begingroup\$ Thank you for the explanation. I've read warnings against "overpowering" peltier modules, but your answer indicates that the peltier will self-regulate it's amp usage. Is that the case? If so, how would one overpower a device anyway? This was part of my fear against going for a higher amp PSU. \$\endgroup\$
    – Nicholas
    Mar 6, 2017 at 17:21
  • \$\begingroup\$ That's a very good but entirely different question. You may want to open a new question focused on peltier cells (if there aren't any already) to find out. \$\endgroup\$
    – Enric Blanco
    Mar 6, 2017 at 17:50
  • \$\begingroup\$ I will do that; thank you. If you feel it appropriate, would you be able to amend the answer to explain how you determined 12.5A usage at 12v based on the datasheet? I spent 20 minutes staring at it trying to logic it out but don't see what information would have allowed you to calculate that. \$\endgroup\$
    – Nicholas
    Mar 6, 2017 at 22:27
  • \$\begingroup\$ I've updated the answer as per your request. \$\endgroup\$
    – Enric Blanco
    Mar 6, 2017 at 23:18
  • \$\begingroup\$ Fantastic; thank you for taking the time to expand upon your original answer. In this form doesn't just answer the question, but helps anyone who stumbles across it to learn the concepts and gain a more complete understanding. \$\endgroup\$
    – Nicholas
    Mar 6, 2017 at 23:47

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