Comparison of LED thermal plastic and aluminum heat dissipation

Regarding the heat dissipation capability of thermal plastics, some netizens have recently smashed an article, "Plastic Radiator for LEDs". A chart is used to illustrate that a thermally conductive plastic heat sink can be better than an aluminum heat sink. I can't agree with this.

In order to be targeted, the original text and the chart are as follows (the italic part):

So why can thermal plastics be made into heat sinks? Can its heat dissipation be compared to traditional aluminum radiators? Perfect, even better! Can you believe it?

We know that any heat sink, in addition to being able to quickly transfer heat from the heat source to the surface of the heat sink, still relies on convection and radiation to dissipate heat into the air. The high thermal conductivity solves the problem of fast heat transfer, and the heat dissipation is mainly determined by the heat dissipation area, shape, natural convection and heat radiation, which are almost independent of the thermal conductivity of the material. So as long as there is a certain heat transfer capacity, the plastic radiator can still be a good radiator!

Generally speaking, if the distance from the heat source to the surface of the heat sink is less than 5 mm [Question 1], as long as the thermal conductivity is greater than 5, the heat dissipation is dominated by convection, and at this time, conduction heat dissipation has no effect. This can be seen from the curve below.

Figure 1. Relationship between thermal conductivity and conduction convection heat dissipation

The relationship between thermal conductivity and conduction convection heat dissipation is given in Figure 1. In the figure, the abscissa is the thermal conductivity and the ordinate is the temperature difference between the heat source and the environment. The curves for the various colors are the distance between the heat source and the surface of the heat sink. If the distance is less than 5mm (green curve), as long as the thermal conductivity is greater than 5, its heat dissipation capacity is completely determined by convection. The smaller the temperature difference and the smaller the distance, the less important the thermal conductivity is.

And for an ideal good heat sink, about 70% of the heat is convective heat dissipation, and 30% of the heat is radiated by radiation [Question 2]. The heat dissipation capability of the heat-conductive plastic is not inferior.

In the above, I identified the key points with [Question 1] and [Question 2]. Let's analyze separately.

Second, the discussion of question 1

It is said that the distance from the heat source to the heat sink is less than 5 mm, and if the thermal conductivity is greater than 5, the heat dissipation is completely determined by the convection. Can not help but ask, can the actual application situation ensure that the heat source to the surface of the radiator is kept less than 5mm? In this case, the spacing of each LED should not be greater than 10mm. Does this situation satisfy the area required for heat dissipation? In this case, the heat dissipation area of ​​each LED is only 1 square centimeter. According to some people's experience, the heat dissipation per watt requires 50 square centimeters, and the area of ​​1 square centimeter is only suitable for 0.02W LED. Even an ordinary 20mA LED heat dissipation is not enough (about 0.06W). What is the practical significance of this? If the heat sink is a conventional finned heat sink, and the LED is an integrated light source, the distance of the heat source to the surface of the heat sink will be greater than 5 mm, then it is obvious that the heat conductive plastic heat sink will fail. When the LED is actually used, it is often impossible to ensure that the surface distance of the heat source to the heat sink is less than 5 mm. If this is basically impossible to achieve, it is of no practical significance. However, thermally conductive plastics are not completely unusable. In some cases, thermally conductive plastics can be used due to certain conditions and requirements, sacrificing some parameters and performance. For example, for small bulbs, the insulation of the thermally conductive plastic is easier to solve with the use of aluminum due to the need for insulation. Relative to the same power of the heat source, the same structure of the aluminum radiator, using heat-conducting plastic to make the radiator, it is necessary to sacrifice the life of the LED. Because of the use of a thermally conductive plastic heat sink of the same construction size, the junction temperature of the LED is higher than in the case of an aluminum heat sink. The above are qualitative explanations. The results of the test test and the simulation calculation are explained below.

(1) Test test situation

The two thermally conductive plastic radiators and aluminum radiators for comparison are shown in Figure 2. From the specification of such a thermally conductive plastic, the thermal conductivity is 15 W/(m ° C).

In Figure 2, the 1# plastic cup is a nominal 6W heat sink, and the 2# plastic cup is a nominal 3W heat sink. Therefore, the nominal 3W aluminum heat sink is also used for test comparison. (Because there is no aluminum heat sink corresponding to the 1# plastic cup at hand, no comparison test is performed.) The test junction temperature is calculated by the voltage method and the voltage temperature coefficient of the packaged LED.

The test data is shown in Table 1.

From this comparison test, it can be seen that the heat-conducting plastic heat sink has the same heat sink effect as the aluminum heat sink, and the volume is nearly doubled.

(2) Case of simulation calculation

The model of the simulation is shown in Figure 3. In Fig. 3, the heat source has a diameter of 10 mm and a thickness of 1 mm, assuming a copper material, and a heat source is provided on the surface. The thermal power is set to 1W. The diameter and thickness of the heat sink are simulated in different sizes to simulate. When simulating, the aluminum material was selected 6061 type.

First, when the thickness of the heat sink is 1.5 mm, the heat dissipation of the heat conductive plastic and the aluminum material is compared when the diameter is changed. The distance from the bottom of the heat source to the longitudinal direction of the heat sink is less than 5 mm, and the distance from the heat source to the edge of the edge to the heat sink increases from 5 mm.

Table 2 is the result of the simulation calculation.

First explain the surface condition of aluminum. Oxidation and blackening treatment, this needless to say. The high oxidation condition is generally oxidized without filler and the oxide layer is thin. Light oxidation but oxidation of aluminum under natural conditions. Polishing, everyone knows this. In the actual design of the heat dissipation, the polished surface is rarely used. This is just for comparison to calculate this type of surface.

It can be seen from the calculation results that the diameter of the heat sink is 20 mm, that is, when the edge of the heat source is 5 mm to the lateral edge of the heat sink, the maximum temperature is higher than 100 degrees when the heat conductive plastic heat sink is used. Not to mention the comparison with aluminum, such temperature is far from meeting the requirement that the LED junction temperature is less than 70 degrees. This kind of heat dissipation design is very bad, there is no need to talk about the material superiority! When the diameter of the radiator is 40mm, it can be seen that after the anodization of the aluminum material, the heat sink capability is stronger than that of the heat conductive plastic. Especially in the case of oxidative blackening, there is no way to compare heat-conductive plastics. If the heat sink is increased, even the aluminum heat sink on the polished surface will have a heat dissipation capacity that exceeds that of the heat conductive plastic.

Let's take a look at the effect of the thickness of the heat sink. Table 3 shows the simulation results of the constant diameter of the radiator and the thickness change.

As can be seen from Table 3, for the thermally conductive plastic, the thickness is increased and the improvement in heat dissipation is very significant. It is indicated that the thickness is increased, and in the case where the thermal conductivity is not high, the effect of reducing the thermal resistance by increasing the cross-sectional area is very significant. At the same time, it is also seen that for light oxidation and polished aluminum, the heat dissipation effect after the thickness of the heat sink is increased is also very obvious. This shows that a higher surface emissivity of the thermally conductive plastic and a lower surface emissivity of the aluminum, in order to achieve good heat dissipation, requires good, low thermal resistance conditions to achieve. That is, when there is insufficient heat dissipation from a place near the heat source, it needs to be conducted to a distant surface as soon as possible to dissipate heat.

As the heat dissipation area is further increased, it can be seen that the heat dissipation capability of the heat conductive plastic is the worst. The simulation data is no longer given here.

The actual heat dissipation, heat flow is not only transmitted in one direction, the heat from the heat source can never rely on the area of ​​only 5mm in the upper, lower, left, right, front and rear directions to dissipate heat. The size of the heat sink must break through 5mm in some directions, otherwise it will not be enough to dissipate heat. Therefore, the content of Figure 1 illustrates the advantages of thermally conductive plastics, which actually has no advantages at all!

Third, the discussion of question 2

It is said that about 70% of the heat of the radiator is cooled by convection, and 30% of the heat is radiated by radiation. Generally speaking, it is misleading. The average person has not systematically studied heat transfer, that is, he has learned that it is also very troublesome to calculate the heat dissipation of convection and radiation for a specific heat sink structure. Now with the simulation software, the readers who use it can use the software to calculate the amount of radiation and understand whether the statement is correct. The ratio of flow to radiation is very much related to the structural design of the heat sink. For radiative heat dissipation, the angular coefficient has a large effect. For example, two parallel fins, when the height is high, the relative surface radiation heat dissipation is very small, that is, for the radiation, the opposite two surfaces do not substantially radiate heat dissipation. Therefore, how to design the heat sink structure requires a little effort. The design is good, the amount of radiation and the flow rate can be divided equally.

Fourth, the conclusion

Just from a good heat dissipation design, the heat dissipation capability of the heat conductive plastic has no advantage over the aluminum material. Unless the luminaire is required for insulation, consider using a thermally conductive plastic as the outer casing and heat sink, but this will sacrifice heat dissipation. To achieve the heat dissipation effect of aluminum, the volume must be large. To sum up, it is a sentence: when the insulation requirements are not met, consider insulating plastics with insulation. The result is of course a loss of thermal performance.