On this post, the question presents a problem of a Peltier cooler heating up after operating for a while. The explanation was that the Peltier cooler is only able to maintain a fixed temperature difference between the cold side and the hot side. Because of this, as heat is taken from the cold side and deposited onto the hot side, the temperature of the hot side begins to rise. Because the cooler is only able to sustain a fixed difference between the sides, if the temperature of the hot side begins to rise so does the temperature of the cold side. So in order to cool the cool side, the hot side's heat needs to be dissipated.

That's my convoluted reinterpretation, but I think I got the idea. Ultimately, I need a way to expel the heat generated by the Peltier cooler. I've seen many people invest in large heatsinks with multiple fans, but I'm kinda low on money:P I was wondering if ice would work well. The heat generated would be, just as in a heatsink, sucked out from the hot side of the cooler. So after some preparation, I took my device and slapped the hot side onto a slab of ice. Unfortunately, the same issue rose: the temperature dropped to the point where little flecks of snow began to precipitate on the cool plate; however, it soon started to heat up, melting the snow and, along with it, my hope. I don't understand, wouldn't the heatsinks used in this application be at higher temperatures than the ice? How come people using heatsinks are reaching lower temperatures instead of when the device is cooled by ice? Furthermore, if the heatsinks are at higher temperatures than the ice, then this doesn't align with the temperature differential spec of the Peltier cooler, which I guess means the unit could be defective. But I have a feeling my understanding of how these units work is wrong. Any help would be appreciated:)

  • \$\begingroup\$ If there is no warm water created on the ice and the gap is closed, that would help. But Used CPU coolers might be free if you had access to old AMD’s \$\endgroup\$ Sep 24, 2019 at 3:07
  • \$\begingroup\$ I don't know where you live, but if there is a waste collection site around, you certainly can get decent heatsinks and as many coolers as you need, for free. In addition, Greta will be proud of you (well, this also depends on how long you plan to let your contraption running). \$\endgroup\$
    – dim
    Sep 24, 2019 at 10:59

3 Answers 3


You still have to ensure thermal coupling to the ice. Once your ice melts, it forms a thin layer of water, which is warmer towards your TEC element. Ice by itself is a decent thermal insulator (look at igloos), so the heat from water doesn't have anywhere to go (at least not fast enough). And the layers of water create a gradient with warmest part being on your TEC surface, which eats into your thermal budget.

Forced air cooling works better despite poorer thermal conductivity of air (as opposed to the water) and higher temperatures than the ice purely because they can ensure that warm air gets blown away.

I think it could be more effective to try using just a water bath as a coolant instead of putting TEC directly on ice. That way heat mixes into the whole bath gradually heating it up. Warmer layers of water next to TEC should "push" their heat away to cooler layers (via convection currents) and this process wouldn't be blocked by the insulating properties of ice. Even better, if you could throw in a ventilator to circulate the water around. If you have access to large amounts of distilled water, you don't have to worry about short circuiting either, not unless water becomes dirty enough to start conducting.


I took my device and slapped the hot side onto a slab of ice.

Right. And how did you ensure that the hot side continued to contact the ice? Unless you mounted the TEC so that the hot side was forced into contact (spring-loaded or gravity loaded), the hot side got hot, melted the top of the ice, and the water flowed away. Now the hot side is not making contact, so it gets very hot, and the cold side follows.


When feeding a peltier a voltage (or current), it heats up because of resistive losses. The BiTe material (usually on most peltiers) is resistive and it functions like a resistor (seen as Rm below in the peltier circuit model)

enter image description here
Source: https://www.researchgate.net/publication/50406971_TESTING_AND_SIMULATION_OF_SOLID_STATE_HEATING_AND_COOLING/figures?lo=1

The problem is the peltier generates heat, but the peltier also moves heat from the cold side to the hot side. This means that whatever heat the peltier is generating ends up on the hot side along with whatever heat it pulled from the cold side.

The heat needs to go somewhere however (a peltier is part of a larger thermal circuit). This means you need a large heatsink (air is not a good conductor of heat).

This also means that the deltaT generated by the peltier is controlled by the hot side. For example:

If a heatsink (and hot side of the peltier) is at 30C and the deltaT (calculated from the datasheet) is 20C then I could run the cold side at 10C

If a heatsink (and hot side of the peltier) is at 10C and the deltaT (calculated from the datasheet) is 20C then I could run the cold side at -10C.

If a heatsink (and hot side of the peltier) is at 30C and the deltaT (calculated from the datasheet) is 35C then I could run the cold side at -5C

This is assuming the heatsink can keep a steady temperature and not heat up when the peltier dumps heat into it.

If you do use a heatsink, it needs to have adequate thermal grease (thin as possible) between the peltier and the heatsink

  • \$\begingroup\$ I see, thanks for the reply! I noticed at the very end you said the thermal grease should be as thin as possible. Are you saying thicker layers of thermal grease reduces proper heat transfer? Thanks again!! \$\endgroup\$
    – Espresso
    Sep 26, 2019 at 4:08

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