Application of Ceramic Substrate in LED Heat Dissipation

[Source: "High-tech led-technology and application" January issue / 甄海威] LED is a new type of semiconductor device, its light-emitting principle is that electrons and holes are combined to emit photons directly, this process does not need to emit heat, but LED The illuminating needs current driving, and only about 15% of the input LED energy is effectively converted into light energy, so the heat is generated during the LED illuminating process, so the LED is a cold light source that generates heat.

LED lamps are recognized as the next lighting technology due to power saving, environmental protection and long life, which will replace the existing lighting technologies. However, LEDs have more than 80% of their electrical energy converted into heat. LED heat dissipation is the biggest problem and has become the bottleneck for popularizing LED lighting. As the LED power increases, the influence of thermal characteristics on the LED becomes more and more significant, which not only causes the device life to be abruptly attenuated, but also seriously affects the wavelength, power, luminous flux and other parameters of the LED, which restricts the development of the LED industry.

First, the problem of aluminum substrate:

Since power devices such as LEDs generate large amounts of heat, ordinary PCB boards cannot withstand heat dissipation, but are soldered or glued to an aluminum substrate. The aluminum substrate is the carrier of the chip and is a bridge connecting the LED and the heat sink. At present, most of the mainstream LED lamps use aluminum substrates. The structure of the aluminum substrate is divided into three layers, which are composed of a copper foil (circuit carrier), a thermally conductive insulating layer and an aluminum plate. The upper layer of the aluminum substrate is used for the copper foil soldering device, and the lower layer of the aluminum substrate serves as a heat sink. The insulating layer is a thin insulating material used to insulate copper and aluminum. The sandwich insulating layer not only requires good insulation, but also has good thermal conductivity. Currently, epoxy or epoxy glass bonded sheets are used. The thermal resistance is still relatively large. The thermal resistance of the general aluminum substrate in China is about 1.7 ° C / W -3 ° C / W. It is better to use a thermal conductive aluminum substrate doped with nano ceramics. Due to the thermal conduction of nanotechnology, the thermal resistance can be 0.5 ° C / W, the insulation breakdown voltage is greater than 3KV.

Compared with PCB board, it has good thermal conductivity and electrical insulation. Therefore, aluminum substrates currently occupy a large market in the field of LED heat dissipation, but as the demand for heat dissipation of LEDs increases, the defects of aluminum substrates are fully exposed, which are summarized as follows:

1. Poor thermal conductivity: The heat generated by the LED is conducted to the metal substrate through the insulating layer and then transmitted to the heat sink through the thermal interface material, so that most of the heat generated by the LED can be diffused to the surrounding air by convection. in. The basic formula for heat conduction is Q = K × A × ΔT / ΔL, also known as the Fourier formula. Where Q represents heat, that is, heat generated or conducted by heat conduction; K is the thermal conductivity of the material, the thermal conductivity is similar to specific heat, but there is some difference from the specific heat, the thermal conductivity is inversely proportional to the specific heat, and the thermal conductivity is higher. The lower the specific heat value. For example, pure copper has a heat transfer coefficient of 396 and a specific heat of 0.39; in the formula, A represents the heat transfer area (or the contact area of ​​the two objects), ΔT represents the temperature difference between the two ends, and ΔL represents the distance between the two ends. . Therefore, we can find from the formula that the heat transfer is proportional to the heat transfer coefficient and the heat transfer area, and inversely proportional to the distance. The higher the thermal conductivity, the larger the heat transfer area, and the shorter the distance traveled, the higher the heat transfer energy and the easier it is to remove heat. However, most aluminum substrate insulation layers have little thermal conductivity or even thermal conductivity, so that heat cannot be conducted from the LEDs to the aluminum substrate, and the entire heat dissipation channel cannot be made clear. Since the aluminum substrate has an insulating layer separating the copper and the aluminum plate, the thermal resistance is the insulating material. This layer of material was previously made of epoxy resin and has a thermal conductivity of about 0.2-0.5 W/m.°C. Recently, the nano-ceramic insulating layer has a thermal conductivity of about 0.8-1.2 W/m.°C. When we actually tested the 5W LED spotlight, we found that the solder joint temperature and the lamp case temperature even reached 10 degrees or more, showing the difference in thermal conductivity, as shown in Table 1-1.
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For more information, please refer to the "High-tech LED - Technology and Applications" January issue

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