After years of programming, I'm trying to do some electronics again.

So I wanted to build an amplifier. I think I now have most of the information, but I'm stuck with the heat sink.

So, basically, I figured out that the calculation goes something like:

temperature increase = power dissipation * total temperature resistance

This particular amplifier IC has the following characteristics (datasheet):

max output power:           40W
Rth (junction-case):        1.8°C/W
max junction temperature:   150°C
max operative ambient temp: 105°C

So, when I have a room temperature that shouldn't go over 30°C, my guess would be that I have 120°C of 'room'. But an average heatsink adds a few temperature resistance units, easily going over the maximum temperature. So I started wondering what the actual power is I need to dissipate.

I am using 4 ohm speakers. I don't intend to drive the volume too high (the speakers themselves are 35W, and while I only have a vague idea of how loud that is it seems much louder than I would ever listen to).

This is a graph from the datasheet.

So I do have multiple questions actually:

  • Are my assumptions and calculations correct?
  • What is the maximum temperature I should use for the junction, 150°C or 105°C?
  • How much warmth does the chip produce? I have trouble interpreting the graph.
  • And what kind of heat sink would be suitable for that? It appears this is measured in W/°C.


I've decided to go with a different IC, as this one produces a bit too much warmth. Either one that is much lower in power (and has a less confusing datasheet), or with this class D amplifier which doesn't even need a heat sink and is thus much smaller. Thanks for the help and explanation, I think I have a better grasp on calculating heat sinks now.

  • \$\begingroup\$ You might want to check that the power rating of the speakers is 35W RMS. If it is 35W PMPO ("peak maximum power output") then that is equivalent to a power-handling capacity of maybe 8W RMS. \$\endgroup\$ Aug 16, 2016 at 18:06

4 Answers 4


There are a few issues here you should understand to properly apply this amplifier IC to your project.

First, there is a distinction between the power delivered to the speakers and the power dissipated in the IC package.

Second, there are two channels. The data sheet doesn't seem terribly clear to me on this point, but Absolute Maximum Ratings Table 2 in Section 3.1 gives a Maximum Power Dissipation of 36 Watts. This implies 18 watts of dissipation per channel. That's 18 watts of power/channel dissipated in the IC, not power delivered to the speakers.

Third, looking at the graph, which is labeled "Dissipated Power & Efficiency vs Output Power, I take it that the "Ptot" vertical axis and the plot line are dissipated power. Whereas the X-axis "Po" is output power - power delivered only to the speakers. So that, for 28 watts delivered to one speaker (the maximum value shown on the "Po" axis, the Ptot plot line shows approx. 18 watts of dissipation (within the IC) on the left-hand Ptot axis. This seems to jive up with my point above that these power figures are "per channel". However, I'm not sure how the efficiency plot is calculated as it doesn't seem to agree with my interpretation of the power plot and axes. So, for two speakers run to the max capabilities of this IC you will be delivering 56 watts to the speakers, and you will need to dissipate 36 watts from the IC package.

Even though the package is rated at "36 Watts" in Table 1, from a practical standpoint getting 36 watts of heat out of package this size ( basically a 3/4" square footprint) while maintaining a die temperature of less than 150 deg-C (the "Maximum Junction Temperature") will require a pretty remarkable heat sink! You will need an immaculate assembly technique to ensure thorough and intimate contact between the back of the IC package and the mating surface of the heat sink.

To answer one of your questions about the distinction between the specified 150 degC and 105 degC values, you should interpret these as follows. 150 degC is the absolute maximum temperature the die of the IC can withstand without permanent damage. In practice you want to stay well away from this unmeasureable temperature. You should design for 130 or 135 max.

The 105 degC value means that if the IC is ideally attached to an ideal heat sink, and the die temperature is on the verge of destruction at 150 degC, the IC package surface will be at a temperature of 105 degC ( another temperature which is difficult or impossible to measure or verify because the package surface in question is bolted against the heat sink). Nonetheless, you want to select/design a heat sink which can theoretically maintain the package surface at 105 degC under maximum dissipation power conditions (28 watts into each speaker). However, you will want to derate this target temperature to 85 or 90 degC so that the die temperature stays at the targeted 130 to 135 degC temperature described above.

How do you choose a heat sink for this application? Here's my practical method. Start with one that will give you twice the dissipation capability you will need. You have 80 degC on the surface of the IC and a 30 degC ambient, that's a 50 degC drop the heat sink has to provide. You are dissipating 36 watts. So, you nominally need a heat sink with a 36/50 dissipation factor. 0.72 watts per degC. Start with one that can provide 72/50 = 0.36 watts per degC. When you get the whole amplifier circuit humming along with a 28 watt per speaker load, measure the surface temperature of the chip right where the package's side meets the heat sink. (Use a digital thermocouple type thermometer to do this. The thermocouple tip is very small and you can get a pinpoint temperature with it.) If you find that you have some temperature margin to play with (e.g. you measure 70 degC at the IC), you can cut away some of the purposefully oversized heat sink. Take away symmetrical sections of the heat sink, cutting off 10 % total mass per trial.

Another way to judge the amount of trimming needed is to get the amplifier humming under full load. Then, take multiple temperature measurements across the length and width of the heat sink. You will notice that this measured temperature drops as you move the probe further away from the IC in all directions. At a far distance you will find the surface temperature of the heat sink is very close to the ambient air temperature. This tells you the distance from the IC at which the heat sink is ineffective. This is simply due to the heat sink's ability to conduct heat over the increasing distance from the IC heat source. You can safely cut away any parts of the heat sink lying beyond this rather fuzzy boundary.

Simple rule of thumb: From a thermal perspective you can never make a heat sink too large! From a packaging perspective there are always size limits. So, use as large a heat sink as you can accommodate within your packaging constraints. The cooler the IC runs, the longer will be its life.

  • \$\begingroup\$ I see the graph is measured at 14.4V, at which point the maximum output power is 2x25W according to the 'Features' list at the top. Hence the graph doesn't go beyond 28W of output power, more evidence that it is about one channel. \$\endgroup\$
    – ayke
    Aug 19, 2016 at 13:29
  • \$\begingroup\$ But, at 28W of output power (per channel), with an efficiency of 76%, there would be a total power usage per channel of 28/0.76 ≈ 36W. That would be a loss of 8W per channel, 16W total, which I can't explain. \$\endgroup\$
    – ayke
    Aug 19, 2016 at 13:47
  • \$\begingroup\$ ayke: Chalk it up to just one more mediocre and poorly-crafted data sheet. \$\endgroup\$
    – FiddyOhm
    Aug 21, 2016 at 15:11

There appears to be a mistake in the graph, which is why you might be having trouble interpreting it. The efficiency \$\eta\$ and Ptot annotations are swapped. Say you want 10W out, the power Ptot is 24W and the efficiency is about 45%. From either of those numbers, you know the amplifier IC will be dissipating ~13W.

So if you allow a maximum Tj of 130°C and allow a maximum air temperature of 45°C at the heatsink you would need 8.5°C/W J-A. The device itself eats up 1.8 leaving 6.7°C/W for the heatsink. You can look up suitable heatsinks- the amplifier package is relatively wide.

Whether 10W (continuous, much higher for brief bursts) is enough or way too much is going to depend on you, your speakers, the source material etc.

The absolute maximum operating junction temperature is 150°C but that is very, very hot for a plastic packaged device. It's better to be more conservative.


Maximum junction temperature defines maximum temperature the component will survive to. So give yourself some headroom: say 30 degrees. Room temperature will be say 25±10 degrees at operating use. If in case, use a larger value.

Tj = 150°C

Theadroom = 30°C

Troom(max) = 35°C

Tdelta=Tj-Theadroom-Troom(max) = 150°C-30°C-35°C = 85°C

So, 85°C is the maximum temperature increase that your system will tolerate given design parameters.

Now, you use thermal resistance of your package to find out thermal resistance of your heatsink. Here, Rch will be thermal resistance of the link between the case and the heatsink (mica sheet, thermal paste, air, etc.) I assume 0.25°C/W for the example, but you have to find out the real value in spec sheets.

P = 40W

Rjc = 1.8°C/W

Rch = 0.25°C/W

Rhs = (Thermal resistance of heatsink: unknown)

Tdelta/P = Rjc+Rch+Rhs

85°C/40W = 1.8°C/W+1°C/W+Rhs

Rhs = 0.075°C/W

You need a heatsink with specified thermal resistance. If such an heatsink is not practical, you need to reduce your headroom, so that maximum allowed temperature increase is larger.

Note: I use arbitrary numbers for temperature and thermal resistance, so you will have to perform computations on your side with your numbers.


Look at it this way 40W*1.8C/W = 72C+30C = 102C, which is right around 105C for the max operating temperature. The key is to find out how much power your amplifier is going to dissipate, which is can be done with spice or a ballpark estimation with the current sourced and the output impedance.


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