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Okay, so I'm trying to figure out why use heat sinks at all.

Let's say if have a component that, by itself, is trying to dissipate (PD) 11W of power. The current TJ is at 50°C and the RJ-A of the component is 2.6°C/W. TA is 21.4°C.

TJ = TA + (RJ-A x PD)

But if I attach a heat sink with a RTh of 1.2°C/W that's adhered with TIM with a RTh of 0.8°C/W, then the TJ, based on the math, would rise to 72°C.

TJ = TA + ((RJ-C + RC-TIM + RTIM-A) • PD)

Obviously heat sinks work. So what am I missing? Where is the drop in TJ?

Thanks!

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    \$\begingroup\$ You are missing the energy transferred from the heatsink away to the environment \$\endgroup\$ – PlasmaHH Jan 22 '16 at 15:52
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    \$\begingroup\$ You haven't quoted a figure for R(j-c) but it is certainly NOT the same as R(j-a) - if it's (say) 0.1K/W then your heatsink thermal resistance will be 2.1K/W (0.1+0.8+1.2) and temperature rise 23.1K. \$\endgroup\$ – Brian Drummond Jan 22 '16 at 16:01
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    \$\begingroup\$ What component has 2.6C/W Rth j-a? Are you sure that is not the Rth j-c figure? \$\endgroup\$ – user1582568 Jan 22 '16 at 16:06
  • \$\begingroup\$ @user1582568 A high power SMD part with a very specific minimum land-pattern than cannot be further helped by a heat-sink to the plastic top would be an example of such Rth(j-a) figures. Some power A-brand LEDs on specifically prescribed metal base PCBs sport those number ranges as well. \$\endgroup\$ – Asmyldof Jan 22 '16 at 16:09
  • \$\begingroup\$ The heat sink is in parallel with the existing Rth j-a thereby reducing it. \$\endgroup\$ – user1582568 Jan 22 '16 at 16:18
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You are missing that your Junction to Ambient value depends on the surface emissivity and thermal resistance.

Your Juntion to Ambient is already very low, so to be honest I suspect you are using a value from a datasheet that includes a strict minimum PCB copper area with a soldered-on metal tab. But in cases, where say the Junction to Ambient is 10 degrees per watt, that is considering the surface of the device as the only emission surface.

When you then add a heatsink the total Junction to Ambient will change to a lower value.

What you need to know to correctly justify the case of the heat-sink is the Junction To Case thermal resistance for the surface that you will connect the heat-sink to.

For a large transistor die in a TO220 with full-metal back the junction to ambient may be in one or more dozens of degrees per watt region, but the junction to case (metal tab surface) thermal resistance can be 1 or 2 degree per watt or even less. If you then add a 4 degree per watt heat-sink + paste combination, you have already gone from 10 ~ 15 down to 6 degrees per watt, since 4 + 2 = 6 and for the heat-sink you don't use junction to ambient values.

Edit: Addition:

To clarify, I specifically say "Junction to Case for the surface you want to use", because a datasheet does say "Junction to Case", but they mean the part of the case where it makes sense to glue or screw a heat-sink on. So if a TO220 Datasheet says "Junction to Case = 2K/W", they mean to the metal tab, because that's where a sane person puts a heat-sink and it's usually assumed that the full metal surface is covered by the heat-sink.

If you use the 2K/W in the above as a value, but then glue a heat-sink to the plastic top of the case, you will be unhappily surprised by lots of smoke.

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