Are the thermal resistance values of heat sinks/etc reasonably reversible? If I wanted to dissipate an internal ambient temperature rise, could I use an internal heat sink on the enclosure wall coupled to a heat sink on the outside to pull heat out of the box?

Can I use the same thermal resistances of the heat sinks in reverse to calculate how efficient this will be at extracting heat from the air? (Assuming additional resistances of sink #1 -> wall and wall -> sink #2)

I'm developing a product that must be in a sealed enclosure. I'm also trying to have the highest operational temperature range I can (targeting -40 to 85 °C) I'm sinking some heat sources (power bricks, power op amps) directly to the metal enclosure... however I am not currently able to sink the heat from the CPU itself (~1W) directly to the case. I -could- add a wide temperature range heat pipe, but I'd like to avoid it.

We're not talking about that much heat, but I just don't have much room near the top of my targeted ambient range (105 °C Tj and 85 °C Ta)... The CPU would require a heat sink even if it wasn't enclosed.


Okay, to clarify I guess I'm asking if a heat sink's thermal resistance is simply a thermal conductance value that I can use both as a number of how much of a temperature rise a certain amount of power put into it will cause AND the exact opposite in how much power it will conduct given a specific temperature differential across it. (So, a 10 °C/watt heat sink would for 1 watt be 10 degrees above ambient, and that if ambient was 10 °C ABOVE the sink it would pull 1 watt of heat out of ambient)


Say my CPU is putting out 1 watt of power. With a heat sink, the Tjs (die to sink) is 0.5 °C/watt, and the heat sink's Tsa (sink to amibent) is 9.5 °C/watt, for 10 °C/watt total. My CPU is then 10 °C above the immediately nearby internal ambient temperature.

My internal ambient is getting about 1 watt of power sunk into it, and that power needs to go somewhere.

There is about 3" between my CPU heat sink and the enclosure wall. With a thermal conductivity of air at 0.025 W*(m^-1)/°C, there would be a rise of about 3 °C across those 3" to transfer 1 watt. (right? This seems too low to me. Also, it's ignoring any kind of internal convection effects) Call this 3 °C/watt = Tax (internal ambient transfer).

The path is then normally Tac (resistance internal ambient to case) plus TcA (case to external ambient).

If I put heat sinks on the inside and outside, I then have Tah (internal ambient to internal heat sink), Thc (internal heat sink to case), TcH (case to external heat sink), THA (external heat sink to external ambient). (although really I also have Tac and TcA still, for the case surface area not covered by heat sinks))

Say Tah is 3 °C/watt, Thc is 0.25 °C/watt, TcH is 0.1 °C/watt, and THA is 1 °C/watt.

I have 1 watt of power for Tjs, Tsa, Tax, Tah, and Thc. I additionally have up to 5 watts of power sunk into the case from other sources directly attached to the case. So, I have 6 watts of power for TcH and THA.

Can I then say that my external ambient maximum is:

Tjs + Tsa + Tax + Tah + Thc + TcH + THA, or 0.5 + 9.5 + 3.0 + 3.0 + 0.25 + 0.6 + 6.0 = 22.85 °C/watt?

With Tj of 105, my maximum external ambient would be 82.15 °C?


Yes, thermal resistance/conductance is a two-way street, for passive conductive cooling devices, with equal speed limits posted for both directions.

OK, now for the caveats: If your heat sink is rated for forced air flow, and you have no forced air flow, then the heat flow through the heat sink will be different. Your heat sink is rated for certain conditions at both ends. If you meet those conditions (at both ends), except for reversal of temperature conditions, then you can expect equal but opposite heat flow.

Say that your 10C/W heat sink is rated 10C/W for conduction of heat from a 1W source to still air, with the contact area with air being fins. Now, you put those fins INSIDE your enclosure, in contact with still air, and you place the outside end of your heat sink in contact with a device (say, a cold plate) that will keep that end of the heat sink 10C cooler than the inside air. In that case you will get 1W of energy flow from the fins of the heat sink to the cooling device (cold plate).

You would want to pay attention to such things as: Warm air rises and cooler air falls. Air fins are most effective when hot air can rise from them and allow cooler air to come in contact with the fins. Cooling fins, on the other hand, would be more efficient when placed such that cooled air can fall away from them.

  • 1
    \$\begingroup\$ the outside of an enclosure is likely to experience some air movement, even if only due to the convection effect you mention, but inside an enclosure, I wouldn't want to rely on simple convection. definitely would want a fan on the inside. \$\endgroup\$ – JustJeff Jun 20 '11 at 22:03

How much heat is drained to the environment is determined by two factors: the difference in temperature between the heat source and the environment and the total thermal resistance between those.
One of the elements in this thermal resistance chain is the resistance between the air inside your enclosure and the enclosure's wall. If you place a heatsink inside the enclosure this thermal resistance is replaced by the sum of two new resistances: that between air and heatsink and that between heatsink and enclosure wall. Intuition says that the former will be lower because of the larger contact area. And the thermal resistance between two metal object in close contact is also much lower than that between air and an object. Thermal paste between heatsink and enclosure reduce this resistance further.
All in all, the total thermal resistance will be reduced by adding a heatsink on the inside, but you can't just calculate it as a mirror image from the one on the outside.

OK, I see what you mean by reverse, and the answer is yes. A difference in temperature will cause heat to flow from the higher to the lower temperature. Peltier coolers use this to force heat away from the source of the heat by creating a much lower temperature. If you can't draw no more heat away the temperature difference will decrease and reach an equilibrium state when the temperature is equal everywhere.

  • \$\begingroup\$ I guess I'm asking if a heat sink's thermal resistance value goes both ways... both in the temperature differential caused per watt, and a wattage of heat removed from ambient given a specific temperature differential. I extended the question. \$\endgroup\$ – darron Jun 20 '11 at 18:38

I do not have any numbers to offer, but can tell you that there are systems that do exactly what you propose. A fan and heat sink inside the case, coupled to a fan and heat sink outside the case appears to be a viable design.


I wonder if a peltier cooler bolted inside the case might help by reducing internal temperature.

  • \$\begingroup\$ It'd be highly inefficient. The peltier would heat the case, and the rest of the case's internal surface area would be transferring much of that heat right back into the internal air. A peltier on the CPU itself might work, but at 1W power dissipation for my CPU the peltier would be a large percentage of my total product's power usage. \$\endgroup\$ – darron Jun 21 '11 at 2:01

Thermal resistance is not a fixed value: it depends, amongst other things, on the air temperature.

So technically, the heatsink inside the box will not have the same thermal resistance as the heatsink out side the box, unless the temperature inside the box is the same as the temperature outside the box. Which it cannot be, to get any heat flow at all.

This is a second order effect. Thermal resistance will depend mostly on air flow.

  • \$\begingroup\$ "Thermal resistance is not a fixed value: it depends, amongst other things, on the air temperature." - Uhm, no. Thermal resistance is the fixed number in the equation. That's why the thermal resistance can be and is specified by manufacturers for e.g. their heat sinks. It does not change with temperature, whereas of course the actual flow of heat/energy does depend on temperature difference and thermal resistance. \$\endgroup\$ – JimmyB Jan 15 '19 at 15:25
  • \$\begingroup\$ @JimmyB The "constant" value specified by manufacturers is only given at a specific air temperature, and you may calculate the effective resistance for other conditions. The thermal resistance of air is not a constant, because air changes density with temperature. So thermal resistance of an heatsink system with air is /not/ a fixed value. Fortunately, this is not a sharp optimum. \$\endgroup\$ – david Jan 17 '19 at 23:16
  • \$\begingroup\$ "The "constant" value specified by manufacturers is only given at a specific air temperature" - Any source for that? - So you are saying that the heat dissipation capacity in Watts of a given heat sink is different if it operates at 50°C/0°C from that in a 150°C/100°C situation? That's just not the case. That is, of course, assuming that you actually keep the temperature at the cold side at the given level, e.g. via forced air flow. \$\endgroup\$ – JimmyB Jan 18 '19 at 13:03
  • \$\begingroup\$ "The thermal resistance of air is not a constant" - Notice that the thermal resistance of air or whatever medium is in heat sinking cases irrelevant. We normally don't rely on heat propagating slowly through still air; which would be hard to achieve anyway due to convection which we would have to stop then. - When we want insulation, e.g. in buildings, we actually must take measures to prevent convection in the insulation layer to leverage the high thermal resistance of air. \$\endgroup\$ – JimmyB Jan 18 '19 at 13:13
  • \$\begingroup\$ As noted in the reply: "This is a second order effect". In reply to the original post "could I use an internal heat sink" And the density of air is as relevant to moving air as with static. Your comments address neither the special case of the original question nor the special case of the given reply -- as well as being technically incorrect. \$\endgroup\$ – david Jan 22 '19 at 4:34

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