High-power LED light radiator introduction

The so-called LED (short for Light-emitting Diode) lamp refers to a new type of environmentally friendly lighting product using semiconductor light-emitting diode technology as a light source. The LED lamp has the characteristics of energy saving, easy control, maintenance-free, safe and environmentally friendly, and long service life. Its service life can reach 50,000-100,000 hours, far exceeding 1,000 hours of traditional tungsten light bulbs and 10,000 hours of fluorescent tubes. As a new type of energy-saving and environmentally-friendly green light source, LED is bound to be the future development trend.

However, LED lighting products are still insufficient. In addition to the higher price, the luminous efficiency is lower, and more heat is generated during operation. The heat sink is required to conduct heat and distribute it to the environment, otherwise it will affect the life of the LED or even jobs.

I. Introduction to general lighting high-power LED lamps (1W or more)

1. Common types - bulbs, spotlights, candle lights, tube lights, ceiling lamps, street lights, lights, etc., and depending on the interface specifications, there are many models (some of the lamps are as follows):

2. Composition - usually includes a lampshade, LED light board, heat sink, drive power supply, interface, insulating glue or sleeve (usually used for metal heat sinks) (as shown below):

2. The role and working principle of LED radiator

The main function of the heat sink is to continuously derive and dissipate the heat generated by the operation of the LED chip to the environment, so that the temperature of the chip is kept within the required range, thereby ensuring that the LED lamp can work normally. The quality of the heat sink depends mainly on the thermal resistance of the heat sink. The smaller the thermal resistance, the lower the junction temperature of the LED light under the same conditions, and the lower the junction temperature of the LED, the longer the life of the chip will be. The thermal resistance of the heat sink should include the thermal resistance and the thermal resistance. For a certain shape of the heat sink, the thermal resistance of the heat conduction is mainly related to the thermal conductivity of the heat sink material. The greater the thermal conductivity, the smaller the thermal conductivity and the better the thermal conductivity; under certain environmental conditions, the thermal resistance of the heat dissipation depends mainly on the heat sink. The heat dissipation area and the radiation coefficient of the surface material of the heat sink, the larger the heat dissipation area, the higher the radiation coefficient, the smaller the heat dissipation resistance, and the better the heat dissipation effect. Therefore, the LED lamp radiator must have a certain heat dissipation area, and the material of the heat sink must have a certain thermal conductivity and a high heat radiation coefficient; in addition, the heat conductive material itself should also have light weight, easy processing, and low price. Features.

III. Introduction of different materials for common LED heat sinks

The so-called LED heat sink is a component that can quickly derive and dissipate heat generated in the operation of the LED chip into the environment. Therefore, it is impossible to have only a certain heat dissipation area as a heat sink, and the material for manufacturing the heat sink must have a certain heat conduction. Sex, that is, to have a higher thermal conductivity, the heat generated by the chip can be continuously exported and finally dispersed into the environment; of course, for the thermal conductive material, in addition to thermal conductivity, it should have a small specific gravity, low price, and high strength. Easy to process and other features.

Commonly used materials usually include metal materials, inorganic non-metal materials and polymer materials. The polymer materials include plastics, rubber, and chemical fibers. The thermally conductive material is typically a metal and a partially inorganic non-metallic material. Among the common metals, the thermal conductivity of aluminum and copper is relatively high, but the price of copper is high, the ratio is large, the processability is not as good as aluminum, and the aluminum heat sink can fully meet the requirements of LED heat dissipation, so aluminum in the LED metal heat sink The main heat sink is the main one, and the copper heat sink is rare. In addition, a small number of radiators are iron. Inorganic non-metallic materials with good thermal conductivity are generally powdery before processing, and need to be processed by special processes. After processing into a heat sink, they are basically ceramic; some inorganic non-metallic materials have high thermal conductivity and are very insulating, but the price is very high. Such as diamond, boron nitride, aluminum nitride, etc.; although some have high thermal conductivity but are not insulated, similar to metals such as graphite, carbon fiber, etc.; and it is very difficult to process inorganic non-metallic powder into a ceramic radiator of complex shape. Therefore, although ceramic LED heat sinks are used, they have great limitations and need to be further improved and improved. The thermal conductivity of the polymer material itself is relatively poor. However, if a metal powder or a non-metal powder having good thermal conductivity is added to a plastic or a rubber to form a heat conductive plastic or rubber, the thermal conductivity thereof is greatly improved. Because of its unique elasticity, thermal conductive rubber is irreplaceable by other thermal conductive materials, so it has been widely used in some fields, but because of its poor rigidity, it may not be suitable as a heat sink material. At present, no relevant reports have been reported; Or the LED heat sink made of metal insert is a new generation of LED heat sink which has been successfully developed recently. If properly designed, it can achieve the heat dissipation effect of the die-cast aluminum heat sink, and the outer surface is insulated and safer to use. The direction of development. The following is a brief introduction to several common LED heat sinks of different materials.

1. Aluminum heat sink - At present, the more common LED heat sink is mainly aluminum heat sink, generally including die-cast aluminum and tensile aluminum heat sink. The main reason is that metal aluminum has a high thermal conductivity, a small specific gravity, easy processing, and low price. Among them, die-cast aluminum radiator is the most commonly used, its thermal conductivity is 70-90W/mK, the appearance is more beautiful, the shape can be changed, the price is moderate; the disadvantage is that it is not insulated, for the safety needs to increase the insulation rubber or sleeve, and the insulation glue Or the sleeve is generally poorly heat-conducting, which is unfavorable for heat dissipation, and the cost is also increased. In addition, the cost of the die-cast aluminum radiator mold is high, which is not conducive to the development of new products; the production process needs to consume more energy, and the secondary processing cost is too large. The thermal conductivity of the stretched aluminum radiator is about 200W/mK, the heat conduction is very good, the processing cost is low at one time, but the secondary processing is often required, the secondary processing cost is high, and the form is relatively simple, and the insulation treatment is also required. For street lights and high power indoor LED lighting.

2. Thermally conductive plastic radiator - The heat-conducting plastic LED lamp radiator developed by several large foreign companies not only solves the problem that the aluminum radiator itself is not insulated, but also the weight of the radiator is also reduced. At present, the company producing LED heat-dissipating plastics mainly includes DSM in the Netherlands, SABIC in the US, and Toray in Japan. The heat conductive plastic is mainly composed of high heat conductive inorganic filler such as nylon and PBT, such as aluminum nitride and boron nitride; its thermal conductivity is not very high, generally only about 1-6W/mK. The LED heat sink made of heat-conductive plastic alone is not very good, and can only be used for LED products with lower power. Therefore, the heat-conducting plastic heat sink is generally composed of two parts: a heat-conductive plastic and an in-line aluminum. The aluminum parts are made of stamped aluminum with a thermal conductivity of 200W/mK or more. The heat dissipation of the aluminum-plastic composite heat sink can be close to or equivalent to that of the die-cast aluminum heat sink, and the processing is relatively easy. However, the price of the existing thermal plastics is quite high, about 200-300 yuan / Kg, the price of the radiator is also higher than the die-cast aluminum radiator, although in some large companies producing LED lamps, such as Philips There are certain applications, but in the near future it is impossible to replace aluminum radiators in large quantities.

3.Thertrans heat-conducting plastic radiator - The thermal transmission plastic radiator Thertrans, which was recently developed by Shanghai Hefu New Material Technology Co., Ltd., is a thermal plastic plastic Therpoxy (patent applied) and embedded aluminum parts (aluminum parts) developed by the company. The material is composed of a thermoplastic LED heat sink. Therpoxy heat-conductive plastic is a kind of heat-conductive plastic which is made by mixing some heat-conductive fillers and other related materials. Its thermal conductivity is 3.6W/mK, and the price is about 100 yuan per kilogram. The material has the characteristics of heat conduction, insulation, flame retardant, heat resistance, environmental protection, aging resistance, easy molding, light weight and low price. Among them, the flame retardant can reach UL94V01.3mm, the environmental protection has completed ROHS certification, and the surface breakdown voltage is above 4000V. Thertrans thermal plastic radiator has beautiful appearance and variable sample. It can be processed by injection molding machine. Its weight is about one-third lower than that of die-cast aluminum radiator of the same shape, and it does not need to add additional insulation treatment. Forming together, effectively reducing the assembly process and cost of LED lamp manufacturers. From the test results, the heat dissipation effect of Thertrans thermal plastic radiator can be close to, or even slightly better than, the same shape die-cast aluminum heat sink, indicating that the thermal conductivity of Thertrans heat sink is not much different from that of die-cast aluminum, or the same shape of Thertrans heat conduction. The thermal resistance of the plastic heat sink and the die-cast aluminum heat sink is basically the same.

The thermal conduction mechanism of Thertrans thermal plastic radiator is that the heat generated by the LED is first transferred to the inlaid aluminum through the aluminum substrate, and then the aluminum is conducted to the outer surface of the radiator through the thermal plastic. Finally, the heat on the outer surface of the radiator passes through convection and heat. Radiation and other methods are distributed to the environment. In addition, the thermal radiation coefficient of general plastics is higher than that of metal aluminum. Therefore, the surface of the heat-conducting plastic heat sink should be better than the aluminum heat sink of the same shape, that is, the radiation heat dissipation effect of the heat-conductive plastic heat sink should be slightly better than the same. Shaped metal radiator.

The Therpoxy thermal plastic can also quickly package the LED driving power supply through the processing equipment, or integrally form the heat sink and the driving power supply, so that the heat dissipation problem of the power supply itself can be effectively improved on the one hand, and the packaged driving power source can be easily disassembled on the other hand. ,replace. Thermally conductive plastic Therpoxy can also be used in other applications where heat dissipation is required.

Four. LED heat sink thermal resistance analysis

The so-called thermal resistance generally refers to a comprehensive parameter reflecting the ability to block heat transfer. The size of the thermal resistance is the key to the heat dissipation of the LED heat sink. The smaller the thermal resistance, the better the heat conduction and heat dissipation effect of the heat sink.

The thermal resistance of the heat sink refers to the total thermal resistance of the heat sink to the environment. It consists of the heat conduction heat resistance in the heat conduction process of the heat sink and the heat dissipation resistance of the heat sink from the outer surface of the heat sink to the surrounding environment through convection and radiation. . The following discusses the thermal resistance of each part of the LED heat sink and its influencing factors:

1. The heat conduction thermal resistance of the heat sink itself

The thermal resistance of the heat sink can be calculated by the following formula:

R derivative = (T1-T2) / W (thermal resistance unit is ° C / W) (1)

In the middle

R guide - heat conduction thermal resistance of the radiator itself

T1 - the temperature of the heat sink at the point of contact with the aluminum substrate

T2——the average temperature of the outer surface of the radiator

W——LED lamp power

In addition, according to the law of conservation of energy, the heat generated by the LED lamp after heat balance is equal to the heat derived from the heat sink itself, and is expressed by the formula: Q production = Q guide + Q cover (the heat is rarely emitted from the lamp cover, for convenience) Analysis, ignored here, where

Q production = aW (2)

Q derivative=bs(T1-T2)/L (3)

In the middle

Q production - the heat generated when the LED works

Q guide - the heat derived from the radiator itself

T1 - the temperature of the heat sink at the point of contact with the aluminum substrate

T2——the average temperature of the outer surface of the radiator

a——LED heat production coefficient

W——the actual power of the LED lamp

B——heat conductivity of the radiator material

S——the average heat transfer area of ​​the radiator

L——heat transfer average distance of the radiator

For a specific heat sink b, s, L is certain, so the formula (3) can be simplified to Q derivative = m. (T1-T2), where m = bs / L, after derivation, m. (T1-T2) = aW, Therefore, (T1-T2)=aW/m, brought into the formula (1), we can know that R is = a/m. From this formula, it can be seen that for a specific heat sink, the thermal resistance of the heat sink is one when the LED light source is fixed. Value. In addition, in the thermal resistance calculation formula, W represents the total power of the LED, and a part of the power of the LED is used for illumination, and a part of the power is converted into thermal energy. Therefore, since the thermal resistance is calculated, the W in the formula is replaced with the heat generation power. (aW) is more scientific, so R is =1/m=L/(bs), which means that the thermal resistance of the heat sink itself is the same as the resistance. It is a constant related only to the parameters of the heat sink itself. The thermal distance is proportional to the average heat transfer area of ​​the heat sink and the thermal conductivity of the heat sink material.

2. Radiator surface to ambient heat dissipation

The heat dissipation from the surface of the heat sink to the ambient air can be calculated by:

R 散 = (T2-T3) / aW (4)

In the middle

R dispersion - heat dissipation from the surface of the heat sink to the environment

T2——the average temperature of the outer surface of the radiator

T3 - ambient temperature

W——LED lamp power

a——LED heat production coefficient

The way to dissipate the outer surface of the radiator to the environment is mainly convection, followed by heat radiation, and the heat conduction is negligible. The convection and radiative heat dissipation formula of the radiator is as follows:

Q flow = c. (T2-T3). S scattered (5)

Q spoke = d. (T42 - T43). S (where T is the absolute temperature) (6)

In the middle

Q flow - convection heat dissipation from the surface of the radiator to the environment

Q-radiation - radiated heat from the surface of the radiator to the environment

C——the natural convection coefficient of the environment around the radiator

D——the emissivity of the surface material of the radiator

S scattered - the outer surface area of ​​the radiator

T2 - the average surface temperature of the radiator

T3 - ambient temperature

According to the law of conservation of energy, after the thermal balance of the LED lamp Q production = Q guide + Q cover = Q dispersion + Q cover = Q flow + Q spoke + Q cover (the heat is rarely emitted from the lamp cover, for the sake of analysis, it is ignored here ), ie aW = c. (T2-T3). S scattered + d. (T42-T43). S scattered.

Under certain environmental conditions, the convection coefficient and the radiation coefficient are constant for a specific heat sink, and the heat dissipation area is also constant; for a constant power source, the heat production power is constant, then the above formula shows that (T2-T3) is Certainly, the heat dissipation resistance of the heat sink to the surface environment is certain. However, if the heat production power (aW) of the light source increases (or decreases), the temperature of the outer surface of the heat sink will also increase (or decrease). By analyzing the radiation heat dissipation formula, it can be known that the temperature T2 of the outer surface of the heat sink rises ( The ratio of the reduction or decrease is less than the ratio of the increase (or decrease) of the heat generation power; therefore, the thermal resistance of the outer surface of the radiator to the ambient air is related to the heat production power, the heat generation power is increased, and the heat resistance of the surface of the heat sink to the environment should be slightly Decrease, on the contrary, the heat production power is reduced, and the heat resistance of the surface of the heat sink to the environment is slightly increased.

For different heat sinks, in the case of a certain heat generation, it can be known from equations (5) and (6) that the larger the heat dissipation area, the convection coefficient, and the radiation coefficient, the smaller (T2-T3), and the smaller the heat resistance of the heat sink. That is to say, the heat dissipation resistance of the heat sink is not only related to its own heat dissipation area and the radiation coefficient of the surface material of the heat sink, but also related to the ventilation condition of the environment in which the LED lamp is located, that is, the natural convection coefficient of the surrounding air, which is a variable. .

3. Total thermal resistance of the heat sink (to the environment)

The total thermal resistance of the heat sink is equal to the thermal resistance of the heat sink itself plus the thermal resistance of the heat sink surface to the environment. The heat resistance of the heat sink to the environment can be calculated by the following formula:

R3=R1+R2=(T1-T3)/aW (7)

According to the previous analysis, the factors affecting the total thermal resistance of the heat sink can be summarized as follows:

(1). Influence of the parameters of the radiator itself: The shorter the average heat transfer distance of the radiator, the smaller the thermal resistance of the radiator; the larger the average heat transfer area of ​​the radiator, the smaller the thermal resistance of the radiator; the greater the thermal conductivity of the radiator material, the heat dissipation The smaller the thermal resistance of the heat sink is, the larger the heat dissipation area of ​​the heat sink is, the smaller the heat resistance of the heat sink is. The larger the radiation coefficient of the surface material of the heat sink is, the smaller the heat resistance of the heat sink is.

(2). The influence of convection coefficient: the better the ventilation around the radiator, the larger the natural convection coefficient, the smaller the thermal resistance of the radiator

(3). The effect of heat production: the same heat sink, in the same environment, the actual heat production power is larger, the heat resistance of the heat sink is slightly reduced.

Therefore, the total thermal resistance of the heat sink is not only related to the heat dissipation area of ​​the heat sink, the geometric size, the radiation coefficient of the surface material, but also the external factors such as the heat production power of the LED and the convection coefficient of the surrounding environment. Constant value. However, in general, the convection coefficient does not change much under natural convection. Under normal circumstances, the change of LED heat production power will not be too large, and the impact on thermal resistance should be small. For ease of analysis and calculation, we can approximate that the total thermal resistance of the heat sink is certain.

Five. Introduction to the heat sink test method

Here, two heat sinks of die-cast aluminum and Thertrans heat-conductive plastic with similar shapes were selected for testing, and the test results were analyzed and compared.

Experimental equipment

Four-channel temperature tester - accuracy 0.1 ° C

Temperature line

Digital thermometer

computer

2. Test sample

(1) A19 die-cast aluminum radiator - commercially available (surface white paint)

(2) TR-E101W Thertrans thermal plastic radiator (built-in aluminum sleeve) - Shanghai Hefu Composite Materials Technology Co., Ltd. (white paint on the surface)

The above two heat sinks are basically the same size, and the heat dissipation area is equivalent, as shown below.

3. Test light source - six light beads (series) LED (luminous efficiency about 20%)

4. Test power supply - 5W DC drive power supply (located outside the heat sink during testing)

5. Temperature measurement point

* LED lamp angle temperature - temperature line and lamp angle solder (approximate to junction temperature)

* The lower end surface temperature of the aluminum substrate - the temperature sensing wire is screwed to the aluminum substrate (approximating the temperature of the heat sink at the point of contact with the aluminum substrate)

* Radiator outer surface blade temperature - temperature sensing line and blade iron clip fixed (position fixed)

* Ambient temperature - the temperature line is naturally placed

As shown below:

6. Measurement method

First, connect the thermometer, LED light board, power supply, heat sink, computer, temperature sensing line, etc. as required, turn on the power and start temperature measurement. After each set of experiments, wait until the temperature curve is relatively stable for a certain period of time, save the test. Curves and data, and make a record. At least three sets of data were measured for each set of experiments, and the average was taken.

7. Test results

A. Die-cast aluminum radiator test results and curves (vertical axis is temperature, horizontal axis is time):

B.Thertrans thermal conductive plastic test results and curves (vertical axis is temperature, horizontal axis is time):

8. Analysis of results

It can be seen from the above test results that the A19 die-cast aluminum radiator lamp has an angular temperature of 63.1 ° C and an ambient temperature of 27.1 ° C; the TR-E101W Thertrans thermal plastic radiator has an angular temperature of 62.4 ° C and an ambient temperature of 27.9 ° C. Next, we use equation (7) to calculate the thermal resistance of the two heat sinks. The thermal resistance of the die-cast aluminum heat sink R1=(63.1—27.1)/(5×0.8)=9°C/W, the thermal R2 of Thertrans thermal plastic radiator = (62.4 - 27.9) / (5 × 0.8) = 8.6 ° C / W (Because the average temperature of the surface of the heat sink is difficult to measure, it is impossible to accurately calculate the thermal resistance and heat dissipation resistance of the heat sink). Through the above analysis, it can be shown that the heat-dissipating effect of the same shape of the heat-conductive plastic Thertrans heat sink and the die-cast aluminum heat sink can be equal or slightly better. In addition, it can be seen from the heat sink picture that the die-cast aluminum heat sink needs to be additionally insulated plastic sleeve because it is not insulated by itself, and the heat-conductive plastic heat sink can be used more easily and safely because it can be insulated without isolation.

Six. LED radiator selection and design points

In order to design or select a suitable radiator product, not only must the radiator be aesthetically pleasing, easy to process, and the price is low, but more importantly, it must meet the heat dissipation requirements of the LED chip, so that the LED operating temperature is kept within the specified range. Inside.

First, let's take a look at the factors that affect the junction temperature of LED lights. According to our experience, the factors affecting the LED junction temperature are summarized as follows: for your reference:

1. LED light board -

LED lamp power: the higher the power, the higher the junction temperature

Chip luminous efficiency: the same power, the higher the luminous efficiency of the chip, the lower the heat generation, the lower the junction temperature

Number and distribution of lamp beads: same power, the larger the aluminum substrate, the more the lamp beads, the more uniform the distribution, the lower the junction temperature

2. The connection between the LED aluminum substrate and the heat sink (thermal resistance between the aluminum substrate and the heat sink) -

Aluminum plate: The aluminum substrate is directly connected to the heat sink than the aluminum substrate plus aluminum plate (if the aluminum substrate is too small to be directly connected to the heat sink, an aluminum plate is required) and the LED junction temperature connected to the heat sink is low; the flatness of the aluminum plate and the aluminum substrate surface The better the general junction temperature is lower

Connection method: Apply a small amount of thermal grease in the case of mechanical connection, the LED junction temperature is generally low

3. Heatsink - the smaller the thermal resistance of the heat sink, the lower the junction temperature of the LED. 4. Ambient temperature - the lower the ambient temperature, the lower the junction temperature of the LED

From the above influencing factors, it can be seen that in order to reduce the junction temperature of the LED lamp, the first choice is to select a suitable lamp panel. In the case of a certain power, the LED chip has higher luminous efficiency, and the less heat generated during operation, therefore, the light is selected as much as possible. High-efficiency chip-made light board. In the case of the determination of the lamp board and the driving power, in addition to optimizing the connection mode, the most important factor affecting the junction temperature of the LED is the thermal resistance of the selected heat sink. The smaller the thermal resistance of the heat sink, the lower the junction temperature of the LED. The longer the life of the luminaire. The influencing factors of the heat resistance of the radiator have been introduced in detail. The user can communicate with the radiator manufacturer according to their actual requirements. The knowledge of the heat resistance of the radiator is analyzed and their experience is best established. A reasonable mathematical model, choose or design a suitable LED radiator product.

references:
(1) Xu Guiping. Analysis of heat conduction and heat dissipation mechanism of high-power LED radiator. Gaogong LED, 2010-09 P72-76