Which radiator is needed for the 50W LED. Thermocles for LEDs are an aluminum radiator with their own hands. We solve the cooling problem

LEDs are considered one of the most efficient light sources, their light stream comes to fantastic values, about 100 lm / W. Fluorescent lamps are two times less, namely 50-70 LM / W. However, for the long operation of the LED, you need to withstand their thermal modes. For this, branded or homemade radiators for LEDs are applied.

Why do diodes need cooling?

Despite the high indicators of light output, the LEDs emit light about a third of the power consumed, and the rest is highlighted in heat. If the diode overheats the structure of its crystal is broken, begins to degrade, the light stream decreases, and the degree of heating is avalanche-like.

Causes of overheating LEDs:

Too long current;
bad stabilization of the supply voltage;
bad cooling.

The first two reasons are solved by using a high-quality power source for LEDs. Such sources are often called. Their feature is not in stabilization of the voltage, namely in the stabilization of the output current.

The fact is that when overheating, the resistance of the LED decreases and the current flowing through it increases. If the voltage stabilizer will be used as a power supply unit as a power unit: more heating is more current, and a larger current is greater heating and so in a circle.

Stabilizing the current, you will partially stabilize the crystal temperature. The third reason is poor cooling for LEDs. Consider this question more.

We solve the cooling problem

Low-power LEDs, for example: 3528, 5050 and their likes are given warmly due to their contacts, and the power of such copies is much less. When the power of the device increases, the issue of removal of excess heat appears. This uses passive or active cooling systems.

Passive cooling - This is an ordinary radiator made of copper or aluminum. The advantages of materials for cooling are disputes. The advantage of this type of cooling is - no noise and the almost complete absence of the need for its service.

Installing LED with passive cooling into a point lamp

Active cooling system - This is a cooling method with the use of external force to improve heat removal. As the simplest system, you can consider the bundle of the radiator + cooler. The advantage is that such a system can be significantly more compact than passive, up to 10 times. The disadvantage is noise from the cooler and the need for lubrication.

How to choose a radiator?

Calculation of the radiator for the LED The process is not quite simple, especially for a beginner. To fulfill it, you need to know the thermal resistance of the crystal, as well as the transition of the crystal substrate, the substrate-radiator, radiator-air. To simplify the solution, many use the ratio of 20-30 cm 2 / W.

This means that each Watt LED lights need to use a radiator with an area of \u200b\u200babout 30 cm 2.

Naturally, such a decision is not unique. If your lighting design is used in the basement cool room, you can take a smaller area, but make sure that the LED temperature is within the norm.

Previous generations LED comfortably felt at a crystal temperature of 50-70 degrees, new LEDs can tolerate up to 100 degrees. It is easiest to determine - touch with hand if the hand barely tolerates - everything is in order, and if the crystal can burn you - make a decision to improve its working conditions.

We believe Square

Suppose we have a luminaire with a power of 3W. The radiator area for the 3W LED, according to the rule described above will be 70-100 cm 2. At first glance, it may seem big.

But consider the calculation of the radiator area for the LED. For a flat lamellar radiator, the area is considered:

a * B * 2 \u003d s

Where a., B. - the length of the sides of the plate, S. - Full square of the radiator.

Where did the coefficient 2 come from? The fact is that such a radiator has two sides and they are tantamount to the heat of the environment, so the complete useful area of \u200b\u200bthe radiator is equal to the area of \u200b\u200beach of its sides. Those. In our case, you need a plate with sizes of 5 * 10cm.

For the ribbed radiator, the total area is equal to the base area and the squares of each of the Röber.

Cooling with your own hands

The simplest example of the radiator will be a "sun" carved from a fuster or aluminum sheet. Such a radiator can cool 1-3W LEDs. Two such sheets between themselves through the thermal paste, it is possible to increase the area of \u200b\u200bheat transfer.

This is a banal radiator from the primary means, it turns out quite thin and it is impossible to use it for more serious luminaires.

Make your own hands the radiator for the LED on 10W will be impossible in this way. Therefore, you can apply a radiator from the central processor of the computer for such powerful light sources.

If you leave the cooler, the active cooling of the LED will allow you to use more powerful LED. Such a solution will create additional noise from the fan and will require additional nutrition, plus a periodic then the cooler.

The area of \u200b\u200bthe radiator for 10W LED will be quite large - about 300 cm 2. A good solution will be the use of ready-made aluminum products. In a construction or economic store, you can purchase an aluminum profile and use it for cooling powerful LEDs.

Having made the assembly of the desired area from such profiles, you can get a good cooling, do not forget all the joints to miss at least a thin layer of thermal paste. It is worth saying that there is a special profile for cooling, which is produced industrially a wide variety of species.

If you do not have the ability to make a radiator of the cooling of the LED with your own hands, you can search for suitable instances in the old electronic equipment, even in the computer. There are several on the motherboard. They are needed to cool the chipsets and power key power keys. An excellent example of such a solution is depicted in the photo below. Their area is usually from 20 to 60 cm 2. Which allows you to cool the LED with a power of 1-3 W.

Another interesting embodiment of the radiator from aluminum sheets. This method will allow you to score almost any necessary cooling area. Watch the video:

How to fix the LED

There are two main fastening methods, consider both of them.

First method - It is mechanical. It is to fasten the LED with self-draws or other fasteners to the radiator, for this you need a special substrate type "Star" (see Star). A diode is soldered to it, pre-blurred thermocons.

On the "Bureau" of the LED there is a special contact patch with a diameter as a SLIM cigarette. After that, the supply wires are soldered to this substrate, and it is screwed to the radiator. Some LEDs go on sale already fixed on the transition plate as in the photo.

Second way - This is adhesive. It is suitable for both mounting through the plate, and without it. But the metal to the metal is not always fixed, how to glue the LED to the radiator? To do this, you need to purchase a special thermal conducting glue. He can meet both in the economic and in the radio component store.

The result of such an attachment as follows.

conclusions

As you might be convinced by the radiator for the LED, you can find both in the store, and rushing in your old devices, or just in the deposits of all sorts of little things. Do not be used special cooling.

The area of \u200b\u200bthe radiator depends on a number of conditions, such as humidity, ambient air temperature and radiator material, but at home solutions they neglect.

Always pay special attention to checking the thermal modes of your devices. Thus, you will provide their reliability and durability. You can determine the temperature hand, but it is better to purchase a multimeter with the possibility of its measurement.

The stated lifetime of the LEDs is calculated with tens of thousands of hours. To achieve such a high indicator, without worsening the optical characteristics, powerful LEDs must be used in a pair with a radiator. This article will allow the reader to find answers to questions related to the calculation and choice of radiator, their modifications and factors affecting heat dissipation.

Why is it needed?

Along with other semiconductor devices, the LED is not an ideal element with a 100% efficiency (efficiency). Most of the energy consumed is dissipated in heat. The exact value of the efficiency depends on the type of the emitting diode and the technology of its manufacture. The effectiveness of low-current LEDs is 10-15%, and in modern white temperatures more than 1 W, its value reaches 30%, and therefore the remaining 70% is spent in heat.

Whatever the LED, for stable and long work, he needs a constant removal of thermal energy from the crystal, that is, the radiator. In low-current LED, the radiator function (anode and cathode) is performed. For example, in SMD 2835, the output of the anode takes almost half of the lower part of the element. In powerful LEDs, the absolute magnitude of the power dissipated is several orders of magnitude. Therefore, they cannot function normally without additional heat sink. Permanent overheating of the light-emitting crystal at times reduces the service life of the semiconductor device, contributes to the smooth loss of brightness with the displacement of the working wavelength.

Views

Structurally, all radiators can be divided into three large groups: plate, rod and ribbed. In all cases, the base may have a circle form, a square or rectangle. The base thickness is of fundamental importance when choosing, since it is this site that is responsible for the reception and a uniform distribution of heat over the entire surface of the radiator.

The influence of the radiator is influenced by the future mode of operation:

with natural ventilation;
with forced ventilation.

The cooling radiator for LEDs, which will be used without a fan must have a distance between ribs at least 4 mm. Otherwise, the natural convection is not enough to successfully remove heat. A bright example is the cooling system for computer processors, where, due to the powerful fan, the distance between the ribs is reduced to 1 mm.

When designing LED lamps, great importance is given to their appearance, which has a huge effect on the shape of the heat sink. For example, the heat energy of the LED lamp should not go beyond the standard pear shape. This fact forces developers to resort to various tricks: use printed circuit boards with an aluminum base, connecting them with a radiator body with thermoclause.

Materials manufacturing radiators

Currently, the cooling of powerful LEDs produce mainly on aluminum radiators. Such a choice is due to lightness, low cost, support in processing and good thermal conducting properties of this metal. Installation of the copper radiator for the LED is justified in the lamp, where the dimensions are of paramount importance, as copper is twice as much as heat dissipates than aluminum. Properties of materials that are most commonly used to cool powerfish LEDs, consider in more detail.

Aluminum

The thermal conductivity coefficient of aluminum is within 202-236 W / m * to and depends on the cleanliness of the alloy. According to this indicator, it is 2.5 times greater than iron and brass. In addition, aluminum is amenable to different types of machining. To increase the heat sink properties, the aluminum radiator is anodized (covered in black).

Copper

Thermal conductivity of copper is 401 W / m * K, yielding only silver among other metals. Nevertheless, copper radiators occur much less than aluminum, due to the presence of a number of flaws:

high cost of copper;
complex mechanical processing;
large mass.

The use of copper coolant leads to an increase in the cost of the lamp, which is unacceptable in the conditions of rigid competition.

Ceramic

A new solution in the creation of highly efficient heat sinks was alumonitride ceramics, the thermal conductivity of which is 170-230 W / m * to. This material is distinguished by low roughness and high dielectric properties.

Using thermoplastic

Despite the fact that the properties of heat-conducting plastics (3-40 W / m * K) are worse than that of aluminum, their main advantages are low cost and ease. Many manufacturers of LED lamps use thermoplastic for the manufacture of the housing. However, thermoplastic loses the competition with metal radiators in the design of LED lamps with a capacity of more than 10 W.

Features of cooling powerful LEDs

As mentioned earlier, it is possible to ensure an effective heat removal from the LED with the organization of passive or active cooling. LEDs with power consumption up to 10 W it is advisable to install on aluminum (copper) radiators, since their mass-dump indicators will have acceptable values.

The use of passive cooling for LED matrices with a power of 50 W and more becomes difficult; The size of the radiator will be tens of centimeters, and the mass will increase to 200-500 grams. In this case, it is worth thinking about the use of a compact radiator along with a small fan. This tandem will reduce the mass and size of the cooling system, but will create additional difficulties. The fan must be ensured by the appropriate supply voltage, as well as take care of the protective disconnection of the LED lamp in the event of a cooler breakdown.

There is another way of cooling powerful LED matrices. It consists in applying the finished Synjet module, which externally resembles a cooler for the medium performance video card. The Synjet module is characterized by high performance, thermal resistance is no more than 2 ° C / W and weighing up to 150 g. Its accurate dimensions and weight depend on the specific model. The disadvantages should be assigning the need for a power source and a high cost. As a result, it turns out that the LED matrix of 50 W must be fixed either on a bulky, but a cheap radiator or a small radiator with a fan, power supply and protection system.

Whatever the radiator, it is capable of providing good, but not the best thermal contact with the LED substrate. To reduce heat resistance, heat-conducting paste is applied to the contacted surface. The effectiveness of its impact is proven by widespread use in computer processors cooling systems. The quality thermalcase is resistant to solidification and has a low viscosity. When applied to the radiator (substrate), there is enough one fine level layer on the entire area of \u200b\u200bcontact. After pressing and fixing, the layer thickness will be about 0.1 mm.

Calculation of the Radiator Square

There are two methods for calculating the radiator for the LED:

the project, the essence of which consists in determining the geometric dimensions of the structure at a given temperature mode;
the caller, which involves acting in the reverse order, that is, with the known parameters of the radiator, you can calculate the maximum amount of heat that it is able to effectively dispel.

The use of a particular option depends on the available source data. In any case, the accurate calculation is a complex mathematical task with a variety of parameters. In addition to using reference literature, take the necessary data from the charts and substitute them to the appropriate formulas, the configuration of the rods or the radiator ride should be taken into account, their focus, and the influence of external factors. It is also worth considering the quality of the LEDs themselves. Often in Chinese production LEDs, real characteristics disagree with the declared.

Accurate calculation

Before switching to formulas and calculations, it is necessary to familiarize themselves with the main terms in the field of thermal energy propagation. The thermal conductivity is the process of transmission of thermal energy from a more heated physical body to less heated. Quantitative thermal conductivity is expressed as a coefficient, which shows how much heat is capable of transferring the material through a unit of the area with a change in temperature by 1 ° K. In LED lamps, all parts involved in the exchange of energy should have a high thermal conductivity. In particular, this concerns the transmission of energy from the crystal to the housing, and then to the radiator and air.

Convection is also the process of heat transfer, which occurs due to the movement of molecules of liquids and gases. With regard to LED lamps, it is customary to consider the exchange of energy between the radiator and air. It can be a natural convection, which is due to the natural movement of the air flow, or forced, organized by installing the fan.

At the beginning of the article it was indicated that about 70% of the capacity consumed by the LED was spent in heat. To calculate the radiator for LEDs, you need to know the exact amount of energy dissipated. To do this, we use the formula:

P T \u003d K * U PR * I, where:

P T - power secreted in the form of heat, W;
k is a coefficient that takes into account the percentage of energy turning into heat. This value for powerful LEDs is taken equal to 0.7-0.8;
U PR is a direct drop in the voltage on the LED when the rated current, B;
I PR - nominal current, A.

It was time to calculate the number of obstacles located on the way of passing the heat flux from the crystal to the air. Each obstacle is thermal resistance (Termal Resistance), denoted by the symbol (Rθ, degree / W). For clarity, the entire cooling system is represented in the form of a substitution scheme from the sequential-parallel inclusion of thermal resistances

Rθ ja \u003d rθ jc + rθ cs + rθ sa, where:

Rθ jc - thermal resistance P-N-Case-case (Junction-Case);
Rθ CS - thermal resistance Case-Radiator (Case-Surfase Radiator);
Rθ SA - heat resistance radiator-air (Surfase Radiator-AIR).

If it is supposed to install a LED on a printed circuit board or use a thermal paste, then you also need to take into account their thermal resistance. In practice, the value of RθSA can be defined in two ways.

Rθ ja - P-N-air resistance;
T j - maximum temperature of the P-N transition (Reference parameter), ° C;
T A - air temperature near the radiator, ° C.

Rθ sa \u003d rθ ja -rθ jc -rθ cs, where rθ jc and rθ cs are reference parameters.

Find from the "Dependence of maximum heat resistance from direct current".

According to the known Rθ SA, the standard radiator is chosen. In this case, the passport value of thermal resistance should be slightly less than the calculated one.

Approximate formula

Many radio amateurs are accustomed to using radiators left in their homemakes remaining from old electronic equipment. At the same time, they do not want to deepen into complex computing and buy expensive innovations of imported production. As a rule, they are only interested in the only question: "What power can disseminate the existing aluminum radiator for LEDs?"

We offer to take advantage of a simple empirical formula that allows you to obtain an acceptable result of the calculation: rθ sa \u003d 50 / √s, where S is the surface area of \u200b\u200bthe radiator in cm 2.

Substituting into this formula the known value of the total heat sink area, taking into account the surface of the roiber (rods) and side faces, we obtain its thermal resistance.

The permissible dispersion capacity is found from the formula: p T \u003d (t j -t a) / rθ ja.

The above calculation does not take into account many nuances affecting the quality of the work of the entire cooling system (the direction of the radiator, the temperature characteristics of the LED, etc.). Therefore, the result obtained is recommended to multiply to the reserve coefficient - 0.7.

Radiator for the LED with his own hands

Make an aluminum radiator for LEDs 1, 3 or 10 W with their own hands is easy. First, consider a simple design, for the manufacture of which will take about half an hour of time and the round plate with a thickness of 1-3 mm. During the circle every 5 mm, cuts are made to the center, and the resulting sectors are slightly bend, so that the finished design resembles the impeller. For the fastening of the radiator to the body in several sectors make holes. It is a little more difficult to make a homemade radiator for a 10 watt LED. To do this, you need 1 meter of aluminum strip 20 mm wide and a thickness of 2 mm. First, the strip is cut with a hacksaw on 8 equal parts, which are then stacks, drilled through and tighten with a bolt with a nut. One of the side faces is ground under the fastening of the LED matrix. With the help of the chisel, the strip is inflicted in different directions. In places of fastening the LED module, drill holes. Termocles are applied to the ground surface, the matrix is \u200b\u200bapplied on top, fixing it with self-draws.

Cheap heat sinks for amateur self-herinks

Especially for radio amateurs who love to experiment with different materials to remove heat and at the same time do not want to spend money on expensive finished products, we will give several recommendations for the search and manufacture of radiators with your own hands. To cool the LED tapes and lines, the furniture profile of aluminum is perfectly suitable. It can be guides for wardrobes or kitchen accessories, the remains of which can be bought at cost in a furniture store.

Radiators from Soviet tape recorders and amplifiers are suitable for cooling LED matrices 3-10 W and amplifiers, which are more than enough on the radio rolls of each city. You can also use spare parts from old office equipment.

Homemade cooling for 50 W LEDs can be made from the radiator from faulty chainsaws, lawn mowers, sawing it into several parts. You can buy such spare parts in repair shops for the price of scrap. Of course, about the aesthetic qualities of the LED lamp in this case can be forgotten.

Read the same way

LEDs appeared just a few years ago. But they have already managed to consolidate the leadership positions in the lighting market. They can be applied not only in lighting systems, but also in various crafts or amateur schemes. When you deal with LED, you need to take care of cooling options. One of the ways to cool the LED is the installation of the radiator.

Radiators for cooling LEDs

Our article will reveal all the secrets, as it is possible, and at the same time collect the device for cooling.

Why the heat sink is needed

Before proceeding to self-assembling of the heat sink for LEDs, it is necessary to know the features of the source itself.
LEDs are semiconductors that have two legs ("+" and "-") i.e. They possess polarity.

LEDs

To properly make a radiator for them, it is necessary to carry out a specific calculation. First of all, this calculation should include voltage measurements, as well as current strength. In addition, it is necessary to remember that any electrical device, including LEDs, features a tendency to heating. Therefore, the cooling system is needed here.
Conducting calculation, remember - only 1/3 of the specified power of the light source will be transformed into a light stream (for example, 3-3.5 of 10W). Therefore, the main part will be thermal losses. In order to minimize heat loss and use radiators.

Note! Overheating of the LED leads to a decrease in its life. Therefore, the use of the radiator allows you to extend the "life" of the light source.

Therefore, LED circuits have a cooling complex of all major elements.
Today, for cooling elements of the electrochmem, which includes LEDs, you can use three variants of heat loss:

through the body of the device (not always can be implemented);
via printed circuit board. Cooling is conducted through non-core conductors by which current flows;
using the radiator. It is suitable for both payments and LEDs.

Note! In the latter situation, it is necessary to correctly calculate what kind of area it should be.

Radiator on LEDs

The most efficient way to cool the LED is the use of a radiator, which can be easily constructed independently. The main thing remember that the form and number of ribs affect the operation of the heat sink.

Features of the construction of heat sinks

Purchased to properly assemble the radiator suitable for LEDs, many are asked a regular question "What better?". After all, today there are two groups of heat sinks, which differ in their structural features:

needle. More often used for the cooling system of natural type. Such models are used for powerful LEDs;

Needle radiator

ribbed. Used in compulsory cooling systems. They are chosen depending on the geometric parameters. At the same time, they can be used for cooling powerful LEDs.

Ribbed radiator

Choosing the type of heat sip, it is necessary to remember that the needle passive machine exceeds the efficiency of the ribbed model by 70%.
The radiator of any design (ribbed or needle) can have a different form:

square;
circular;
rectangular.

The variant of the radiator suitable for LEDs should be selected depending on the needs in the cooling system.

Features of calculations

Calculation of the scheme to create with your own radiator hands should always be started with the selection of the element base. Do not forget that the denomination here should answer not only the potential of the collected heat sip, but also preventing the creation of additional losses. Otherwise, the homemade apparatus will have low efficiency. And first of all, this requires calculating the radiator area.
What should include the calculation of this parameter as area:

modification of the device;
what is the dispersion area;
ambient air indicators;
the material from which the heat sink is manufactured.

Such nuances must be considered when a new radiator is designed, and the old one is reworked. The most important for independently assembling the heat sink will be the indicator of the maximum allowable dispersion of the power of the heat exchange element.
To calculate the area of \u200b\u200bthe radiator there are two ways.
The first method of calculation. In order to determine the required area, you need to use the F \u003d A X S (T1 - T2) formula, where:

F - thermal stream;
S - Surface Surfaces;
T1 - the temperature indicator of the medium that removes heat;
T2 - the temperature that has a heated surface;
a is a coefficient reflecting the heat transfer. This coefficient for unpolished surfaces is conditionally accepted equal to 6-8 W / (M2K).

Circumference

Using this method of calculation it is necessary to remember that the plate or edge has two surfaces for heat removal. At the same time, the calculation of the surface of the needle is carried out using the circumference length (π x D), which must be multiplied by the height rate.
The second method of calculation. It uses a somewhat simplified formula derived by experimentally. In this case, the formula S \u003d x W is used, where:

S is the heat exchanger area;
M is an unused power of the LED;
W - Power Supply (W).

At the same time, if the ribbed aluminum machine will be made, you can use the data on the calculations that Taiwan experts received:

60 W - from 7000 to 73000 cm2;
10 W - about 1000 cm2;
3 W - from 30 to 50 cm2;
1 W - from 10 to 15 cm2.

But in such a situation it is necessary to remember that the above data is suitable for the climatic conditions of Taiwan. In our case, they should be taken only when conducting preliminary calculations.

Material for the manufacture of heat sink

The life of the LEDs directly depends on which material is involved in the semiconductor, as well as on the quality of the cooling system.
When choosing a material for the heat sip, it is necessary to be guided by the following:

the material must have a thermal conductivity of at least 5-10 W;
the thermal conductivity level should be higher than 10 W.

In this regard, for the manufacture of the heat sink it is worth using such materials:

aluminum. The aluminum version today is used most often for cooling LEDs. But at the same time, the aluminum heat sink has a significant minus - consists of a number of layers. As a result of such a structure, the aluminum apparatus provokes thermal resistance. It is possible to overcome them only with the help of additional heat-conducting materials, in the role of which insulating plates can act;

Note! Aluminum radiator, despite its drawback, perfectly copes with a heat dissipation. It uses an aluminum plate that is blown by the fan.

Aluminum radiator

ceramics. Ceramic heat lines have special routes for which the current is carried out. LEDs sold to the same tracks. Such products are capable of falling twice as much heat;
copper. There is a copper plate here. It is distinguished by a higher thermal conductivity than aluminum. But copper is inferior to aluminum in technical characteristics and weight. At the same time, copper is not a malleable metal, and after processing there remains a lot of cropping;

Radiator from the medium

plastic. The advantages should be attributed to the available cost, as well as a high level of technological. At the same time, in the minuses there are smaller thermal conductivity.

As we see, the most optimal option for the price and quality will be the manufacture of radiator for the LEDs from aluminum. Consider several ways to make a heat sink for LEDs.

How heat sinks are made

Not all radio amateurs with hunting are taken for making such devices. After all, it will perform a leading role. From how qualitatively the heat sink will be made with his own hands, the life of the lighting installation, made of LEDs, depends. Therefore, many prefer not to risk and buying devices for the cooling system in specialized stores.

Homemade radiator for diodes

But there are situations when there is no possibility to buy, but it can be made of submitted means, which without problems will be delayed in the home laboratory of any radio amateur. And here two methods of manufacture are suitable.

The first way of self-assembly

The simplest design for the homemade radiator, of course, will be a circle. It can be cut as follows:

from the leaf of aluminum cut the circle and make the required number of incisions on it;

Cutting Circle of Aluminum

next, flex a little sector. As a result, it turns out some kind of fan;
on the axes it is necessary to bend 4 mustache. With their help, the device will be attached to the lamp housing;
lEDs on such a radiator can be fixed with the thermal paste.

Ready Radiator for Round Form Diodes

As you can see, this is a fairly simple way of manufacture.

Second way of self-assembly

The cooling machine that will connect to LEDs, you can independently make their piece of pipe, which has a rectangular cross section, as well as from an aluminum profile. Here you will need:

press-washer with a diameter of 16 mm;
pipe 30x15x1.5;
thermalcase KTP 8;
W-shaped profile 265;
termoklay;
saws.

We make the radiator as follows:

in the pipe we drill three holes;

Pipe option for radiator

next drill profile. With it, the lamp will be completed;
lEDs secure to a pipe that will act as the base of the heat sip, with the help of a thermoclaus;
in places of connection of the elements of the radiator, we apply the thermal palary of KTP 8;
it remains to assemble the design with the help of self-tapping screws equipped with a pole.

This method will be somewhat more complicated in the implementation than the first option.

Conclusion

Knowing what is the radiator connected to the LEDs, it is quite possible to make it with your hands from the remedies. Its correct assembly will help you not only cool the lighting installation effectively, but also avoid a situation of reducing the limits of exploitation of LEDs.