Now it’s not a problem to buy a scattering of powerful LEDs, but the radiator is expensive for them, because. already has tangible dimensions and mass. I offer my solution to this problem. As you know, the main thing in the radiator is the surface area, so the needle ones are the most efficient. Knowing the golden formula of a radiator 1 W = 10-30 sq.cm. it can be estimated that for 10 W LED you will need approximately 200 sq.cm. area. It was decided to dial this area with an aluminum plate, which can be found in any large hardware store. Here's what happened to me.
Video instructions for making
And I got almost 400 sq.cm. radiator area from a strip of 1000x20x2 mm. This is enough for 20W and even 50W LED with a small fan.
Temperature
And for my 10 W, according to the well-known dependence (see figure), a delta of 30º is obtained.
Maximum allowable temperature LED +80º, so this radiator can be operated without forced cooling at ambient temperatures up to +50º. It is not surprising that in fact the radiator practically does not heat up, because. creates natural air circulation and you can safely take the plate narrower or put the LED more powerful, up to 50 watts. Already bought a few 1000x15x2 mm. If they sold 10 mm wide, you could also try. By the way, it is better to fasten with two bolts or rivets, which are easy to make from the same piece of aluminum strip.
Accessories
In addition to an aluminum strip from the nearest hardware store / market, you may also need:
In the latter case, pay attention to the input voltage of the driver. I use 24 V for my network, but you can find it immediately for 220 V. A pack of 10 pcs. will be cheaper.
LEDs are considered one of the most efficient light sources, their luminous flux reaches fantastic values, on the order of 100 Lm / W. Fluorescent lamps give out half as much, namely 50-70 Lm / W. However, for a long operation of the LED, it is necessary to withstand their thermal conditions. For this, branded or home-made radiators for LEDs are used.
Why do diodes need cooling?
Despite the high light output, LEDs emit light for about a third of the power consumed, and the rest is released into heat. If the diode overheats, the structure of its crystal is disturbed, begins to degrade, the luminous flux decreases, and the degree of heating increases like an avalanche.
Causes of LED overheating:
- Too much current;
- poor stabilization of the supply voltage;
- bad cooling.
The first two reasons are solved by using a quality power supply for LEDs. Such sources are often referred to as . Their feature is not in voltage stabilization, but in stabilization of the output current.
The fact is that when the LED is overheated, the resistance of the LED decreases and the current flowing through it increases. If you use a voltage stabilizer as a power supply, the process will turn out to be an avalanche: more heating - more current, and more current - this is more heating, and so on in a circle.
By stabilizing the current, you partially stabilize the temperature of the crystal. The third reason is poor cooling for LEDs. Let's consider this question in more detail.
Solving the cooling problem
Low-power LEDs, for example: 3528, 5050 and the like, give off heat due to their contacts, and such specimens have much less power. When the power of the device increases, the question of removing excess heat arises. For this, passive or active cooling systems are used.
Passive cooling- This is a conventional radiator made of copper or aluminum. The benefits of cooling materials are controversial. The advantage of this type of cooling is the absence of noise and the almost complete absence of the need for its maintenance.
Installation of LED with passive cooling in spotlight Active cooling system is a cooling method using external force to improve heat dissipation. As the simplest system you can consider a bunch of radiator + cooler. The advantage is that such a system can be much more compact than a passive one, up to 10 times. The disadvantage is the noise from the cooler and the need to lubricate it.
How to choose a radiator?
Calculating a radiator for an LED is not an easy process, especially for a beginner. To perform it, you need to know the thermal resistance of the crystal, as well as the crystal-substrate, substrate-radiator, radiator-air transitions. To simplify the decision, many use the ratio of 20-30 cm 2 /W.
This means that for every watt of LED light, you need to use a radiator with an area of about 30 cm2.
Naturally, this solution is not unique. If your lighting design will be used in a basement cool room, you can use a smaller area, but make sure that the temperature of the LED is within normal limits.
Previous generations of LEDs were comfortable at a crystal temperature of 50-70 degrees, new LEDs can tolerate temperatures up to 100 degrees. The easiest way to determine is to touch it with your hand, if the hand barely tolerates it, everything is in order, and if the crystal can burn you, make a decision to improve its working conditions.
We consider the area
Let's say we have a lamp with a power of 3W. The radiator area for a 3W LED, according to the rule described above, will be equal to 70-100cm 2. At first glance, it may seem large.
But consider the calculation of the radiator area for the LED. For a flat plate radiator, the area is considered:
a * b * 2 = S
Where a,b are the lengths of the sides of the plate, S – total area radiator.
Where did coefficient 2 come from? The fact is that such a radiator has two sides and they equally give off heat environment, so the total useful area of the radiator is equal to the area of each of its sides. Those. in our case, we need a plate with side dimensions of 5 * 10cm.
For a ribbed radiator, the total area is equal to - the area of the base and the areas of each of the ribs.
Do-it-yourself cooling
The simplest example of a radiator would be a "sun" cut out of tin or aluminum sheet. Such a radiator can cool 1-3W of LEDs. By twisting two such sheets together through thermal paste, you can increase the heat transfer area.
This is a banal radiator made from improvised means, it turns out to be quite thin and cannot be used for more serious lamps.
It will be impossible to make a radiator for a 10W LED with your own hands in this way. Therefore, it is possible to use a radiator from the computer's central processing unit for such powerful light sources.
If you leave the cooler, active cooling of the LEDs will allow you to use more powerful LEDs. Such a solution will create additional noise from the fan and require additional power, plus periodic maintenance of the cooler.
The radiator area for a 10W LED will be quite large - about 300 cm 2. good decision will use finished aluminum products. At a hardware store or hardware store, you can purchase an aluminum profile and use it to cool high power LEDs.
Having made the assembly of the required area from such profiles, you can get good cooling, do not forget to grease all the joints at least thin layer thermal paste. It is worth saying that there is a special profile for cooling, which is industrially produced in a wide variety of types.
If you do not have the opportunity to make a do-it-yourself LED cooling radiator, you can look for suitable items in old electronic equipment, even in a computer. There are several on the motherboard. They are needed to cool chipsets and power switches for power circuits. An excellent example of such a solution is shown in the photo below. Their area is usually from 20 to 60 cm 2 . That allows you to cool the LED with a power of 1-3 watts.
Another interesting option production of a radiator from aluminum sheets. This method will allow you to dial almost any required area cooling. Watch video:
How to fix the LED
There are two main ways of fastening, we will consider both of them.
First way- it's mechanical. It consists in screwing the LED with self-tapping screws or other fasteners to the radiator, for this you need a special star-type substrate (see star). A diode pre-lubricated with thermal paste is soldered to it.
On the "belly" of the LED there is a special contact patch with a diameter like 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 adapter plate, as in the photo.
Second way- it's glue. It is suitable both for mounting through the plate and without it. But it’s not always possible to attach metal to metal, how to glue an LED to a radiator? To do this, you need to purchase a special thermally conductive adhesive. It can be found both in the household and in the store of radio components.
The result of such a fastening looks like this.
conclusions
As you can see, the radiator for the LED can be found both in the store and by rummaging through your old appliances, or simply in the deposits of all sorts of little things. It is not necessary to use special cooling.
The radiator area depends on a number of conditions, such as humidity, ambient temperature and radiator material, but they are neglected in a household solution.
Always give Special attention checking the thermal conditions of your devices. Thus, you will ensure their reliability and durability. You can determine the temperature by hand, but it is better to purchase a multimeter with the ability to measure it.
There are approximate data from Taiwanese specialists for aluminum finned radiators:
- 1W 10-15kv/cm
- 3W 30-50kv/cm
- 6W 150-250kv/cm
- 15W 900-1000kv/cm
- 24W 2000-2200kv/cm
- 60W 7000-73000kv/cm
This data is for passive cooling.
But these data were calculated for their climatic conditions and yet they are approximate. the values are not exact, there is a run-up in the area.
To calculate, you need to know the following parameters:
1. You need to understand what type of radiator you are going to use:
plate, pin, ribbed
- lamellar
- Pin (needle)
- Ribbed
2. You also need to consider the material that makes up the radiator. Most often it is copper or aluminum, but hybrids have also appeared recently.
Hybrids have a built-in copper plate that is in contact with the working element (an element that requires cooling, in this case LED), then aluminum.
3. The radiator is not calculated by surface area, but by usable area scattering.
4. The next factor is how heat is removed from the working element to the radiator, i.e. applied thermal paste or thermal tape, or simply soldered.
5. It will be useful to know the resistance of the crystal - the housing of the LED
6. Will there be additional cooling of the radiator, and what will it be:
- With a cooler (small fan):
- Water cooling:
Certainly water cooling will be more efficient than just a cooler, but its cooling, depending on the power, will allow you to reduce the radiator area by 3-5 times. And with water, other problems may arise, such as the tightness of the system, for example.
7. It is also necessary to take into account the input power, i.e. if the LED works at the maximum of its capabilities, then it will need more cooling, excess power will completely turn into heat, but if the load is reduced, say, by half, then overheating will be much lower.
You should also consider the location of the device indoors or outdoors, it will be operated.
Also on the Internet there is a formula obtained experimentally, it may be useful:
S cooler = (22-(M x 1.5)) x W
S – radiator (cooler) area
W - power input in watts
M – remaining unused LED power
With the resulting area, no additional cooling radiator is required, cooling occurs naturally and will give good heat dissipation in any conditions.
The formula is applicable for an aluminum radiator. For copper, the area will be reduced by almost 2 times.
Thermal conductivity in W / m * °C of various materials
silver - 407
gold - 308
aluminum - 209
brass - 111
platinum - 70
gray cast iron - 50
bronze - 47-58
LEDs appeared only a few years ago. But they have already managed to consolidate their leadership positions in the lighting products market. They can be used not only in lighting systems, but also in various crafts or amateur circuits. When dealing with led, cooling options must be taken care of. One way to cool LEDs is to install a heatsink.
Radiators for cooling LEDs
Our article will reveal to you all the secrets of how to properly and at the same time assemble a cooling device with your own hands.
Why is a heat sink needed?
Before proceeding to self assembly heat sink for LEDs, you need to know the features of the light source itself.
LEDs are semiconductors that have two legs (“+” and “-”) i.e. they have polarity.
LEDs
In order to properly manufacture a radiator for them, it is necessary to carry out a certain calculation. First of all, this calculation should include voltage measurements, as well as current strength. In addition, it must be remembered that any electrically intensive device, including LEDs, has a tendency to heat up. Therefore, a cooling system is needed here.
When calculating, remember that only 1/3 of the indicated power of the light source will be converted into a luminous flux (for example, 3-3.5 out of 10w). Therefore, the main part will be heat losses. In order to minimize heat loss and use radiators.
Note! Overheating the LED leads to a decrease in its life. Therefore, the use of a radiator also allows you to extend the "life" of the light source.
Therefore, LED circuits have a cooling complex for all the main elements.
Today, to cool the elements of the electrical circuit, which includes LEDs, you can use three options for heat removal:
- through the body of the device (not always possible to implement);
- through printed circuit board. Cooling is carried out through minor conductive paths through which current flows;
- using a radiator. It is suitable for both boards and LEDs.
Note! In the latter situation, it is necessary to correctly calculate what area it should be.
Radiator LED
by the most effective way LED cooling is the use of a heatsink, which you can easily build yourself. The main thing to remember is that the shape and number of fins affect the operation of the heat sink.
Design features of heat sinks
Puzzled by their own hands to assemble a radiator suitable for LEDs, many ask the quite logical question “which one is better?”. Indeed, today there are two groups of heat sinks, which differ in their design features:
- needle. Most commonly used for cooling systems natural type. Such models are used for high-power LEDs;
needle radiator
- ribbed. Used in forced cooling systems. They are selected depending on the geometric parameters. However, they can also be used to cool high-power LEDs.
Finned radiator
When choosing the type of heat sink, it must be remembered that the needle passive device exceeds the efficiency of the ribbed model by 70%.
A radiator of any design (ribbed or needle-shaped) can have a different shape:
- square;
- round;
- rectangular.
The heatsink option suitable for LEDs should be selected depending on the needs of the cooling system.
Computing Features
The calculation of the circuit for creating a radiator with your own hands should always begin with the selection of the element base. Don't forget that the rating here must meet not only the potential of the heat sink being assembled, but also the prevention of creating additional losses. Otherwise homemade apparatus will have low efficiency. And first of all, for this it is necessary to calculate the area of \u200b\u200bthe radiator.
What should include the calculation of such a parameter as area:
- device modification;
- what is the scattering area;
- ambient air indicators;
- the material from which the heat sink is made.
Such nuances must be taken into account when a new radiator is being designed, and the old one is not being redone. The most important indicator for self-assembly of the heat sink will be the indicator of the maximum allowable power dissipation of the heat exchange element.
There are two ways to calculate the area of a radiator.
The first calculation method. In order to determine the required area, you need to use the formula F = a x S x (T1 - T2), where:
- F is the heat flux;
- S is the heat sink surface area;
- T1 - an indicator of the temperature of the medium that removes heat;
- T2 is the temperature that the heated surface has;
- a is the coefficient reflecting heat transfer. This coefficient for unpolished surfaces is conventionally assumed to be 6-8 W/(m2K).
Circumference
When using this calculation method, it must be remembered that a plate or fin has two surfaces for heat removal. In this case, the calculation of the surface of the needle is carried out using the circumference (π x D), which must be multiplied by the height indicator.
The second method of calculation. A somewhat simplified formula derived experimentally is used here. In this case, the formula S = x W is used, where:
- S is the heat exchanger area;
- M is the idle power of the LED;
- W - input power (W).
Moreover, if a ribbed aluminum apparatus is to be manufactured, the data obtained by Taiwanese specialists can be used in the calculations:
- 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 must be remembered that the above data is suitable for the climatic conditions of Taiwan. In our case, they should be taken only when carrying out preliminary calculations.
Heat sink material
The service life of LEDs directly depends on what material is used in the semiconductor, as well as on the quality of the cooling system.
When choosing a material for a heat sink, you must be guided by the following:
- the material must have a thermal conductivity of at least 5-10 W;
- the thermal conductivity level must be above 10 W.
In this regard, for the manufacture of a heat sink it is worth using the following materials:
- aluminum. Aluminum variant Today, LEDs are most often used for cooling. But at the same time, the aluminum heat sink has a significant disadvantage - it consists of a number of layers. As a result of this structure, the aluminum apparatus provokes thermal resistance. They can only be overcome with the help of additional heat-conducting materials, which can be insulating plates;
Note! aluminum radiator, despite its drawback, copes well with heat dissipation. It uses an aluminum plate, which is blown by a fan.
aluminum radiator
- ceramics. Ceramic heat sinks have special paths through which current is conducted. LEDs are soldered to the same tracks. Such products are able to divert twice more heat;
- copper. There is a copper plate here. It has a higher thermal conductivity than aluminum. But copper is inferior to aluminum in technical specifications and weight. At the same time, copper is not a malleable metal, and after processing there are a lot of scraps;
copper heatsink
- plastic. The advantages include affordable cost, as well as high level manufacturability. At the same time, in the minuses, there is less thermal conductivity.
As we see, the most the best option for the price and quality, there will be a do-it-yourself manufacturing of a radiator for aluminum LEDs. Consider several ways to make a heat sink for LEDs.
How are heat sinks made?
Not all radio amateurs willingly undertake the manufacture of such devices. After all, it will play a leading role. The service life of a lighting installation made of LEDs depends on how well the heat sink is made by hand. Therefore, many prefer not to take risks and buy devices for the cooling system in specialized stores.
Homemade radiator for diodes
But there are situations when it is not possible to buy, but it can be made from improvised means, which can be found without any problems in the home laboratory of any radio amateur. There are two manufacturing methods here.
The first way of self-assembly
The simplest design for a homemade radiator, of course, will be a circle. It can be cut like this:
- cut out a circle from aluminum sheet and make on it required amount incisions;
Sliced aluminum circle
- then we bend a little sector. The result is a kind of fan;
- 4 antennae must be bent along the axes. With their help, the device will be attached to the lamp body;
- LEDs on such a radiator can be fixed with thermal paste.
Ready-made radiator for round diodes
As you can see, this is a fairly simple manufacturing method.
The second way of self-assembly
The cooling apparatus, which will be connected to the LEDs, can be independently made from a piece of pipe, which has a rectangular cross section, as well as from aluminum profile. Here you will need:
- press washer with a diameter of 16 mm;
- pipe 30x15x1.5;
- thermal grease KTP 8;
- W-profile 265;
- hot glue;
- self-tapping screws.
We make a radiator as follows:
- drill three holes in the pipe;
Radiator pipe option
- Next, we drill the profile. With its help, fastening to the lamp will be carried out;
- we attach the LEDs to the pipe, which will act as the base of the heat sink, using hot glue;
- at the joints of the radiator elements, apply a layer of thermal paste KTP 8;
- it remains to assemble the structure using self-tapping screws equipped with a press washer.
This method will be somewhat more difficult to implement than the first option.
Conclusion
Knowing what a radiator connected to LEDs is, it is quite possible to make it yourself from improvised means. Its proper assembly will help you not only cool efficiently lighting installation, but also to avoid the situation of reducing the life of LEDs.
The device and principles of operation of the radiator for LEDs. Rules for choosing the material and area of the part. We make a radiator with our own hands quickly and easily.
The common belief that LEDs do not heat up is a misconception. It arose because low-power LEDs do not feel hot to the touch. The thing is that they are equipped with heat sinks - radiators.
The principle of operation of the heat sink
The main consumer of the heat generated by the LED is the ambient air. Its cold particles approach the heated surface of the heat exchanger (radiator), heat up and rush upward, making room for new cold masses.
When colliding with other molecules, heat is distributed (dissipated). How more area surface of the radiator, the more intense it will transfer heat from the LED to the air.
Read more about the principles of operation of LEDs.
Amount absorbed air mass heat per unit area does not depend on the material of the radiator: the efficiency of natural " heat pump» is limited by its physical properties.
Materials for manufacturing
Radiators for cooling LEDs vary in design and material.
Ambient air can take no more than 5-10 W from a single surface. When choosing a material for the manufacture of a radiator, one should take into account the performance next condition: its thermal conductivity should be at least 5-10 watts. Materials with a smaller parameter will not be able to transfer all the heat that air can take.
Thermal conductivity above 10 W will be technically excessive, which will entail unjustified financial costs without increasing the efficiency of the radiator.
For the manufacture of radiators, aluminum, copper or ceramics are traditionally used. Recently, products made of heat-dissipating plastics have appeared.
Aluminum
The main disadvantage of an aluminum radiator is the multi-layer design. This inevitably leads to the appearance of transient thermal resistances, which have to be overcome by using additional heat-conducting materials:
- adhesive substances;
- insulating plates;
- materials that fill air gaps, etc.
Aluminum radiators are the most common: they are well pressed and cope with heat dissipation quite tolerably.
Aluminum Heatsinks for 1W LEDs
Copper
Copper has a higher thermal conductivity than aluminum, so in some cases its use for the manufacture of radiators is justified. On the whole given material inferior to aluminum in terms of lightness of construction and manufacturability (copper is a less pliable metal).
Manufacturing copper radiator pressing method - the most economical - is impossible. And cutting gives a large percentage of waste of expensive material.
Copper radiators
Ceramic
One of the most good options The heat sink is a ceramic substrate on which current-carrying traces are preliminarily applied. LEDs are soldered directly to them. This design allows you to remove twice as much heat compared to metal radiators.
Bulb with ceramic heatsink
Heat-dissipating plastics
Increasingly, there is information about the prospects for replacing metal and ceramics with thermally dissipating plastic. Interest in this material is understandable: plastic costs much less than aluminum, and its manufacturability is much higher. However, the thermal conductivity of ordinary plastic does not exceed 0.1-0.2 W / m.K. It is possible to achieve acceptable thermal conductivity of plastics through the use of various fillers.
When replacing an aluminum radiator with a plastic one (of equal size), the temperature in the temperature supply zone increases by only 4-5%. Given that the thermal conductivity of heat-dissipating plastic is much less than aluminum (8 W/m.K versus 220-180 W/m.K), we can conclude that the plastic material is quite competitive.
Bulb with thermoplastic heatsink
Design features
Structural radiators are divided into two groups:
- needle;
- ribbed.
The first type is mainly used for natural cooling of LEDs, the second - for forced cooling. With equal overall dimensions a passive needle radiator is 70 percent more efficient than a finned one.
Needle type heatsinks for high power and smd LEDs
But this does not mean that plate (finned) radiators are only suitable for working in tandem with a fan. Depending on the geometric dimensions, they can also be used for passive cooling.
LED lamp with ribbed heatsink
Pay attention to the distance between the plates (or needles): if it is 4 mm - the product is designed for natural heat removal, if the gap between the radiator elements is only 2 mm - it must be equipped with a fan.
Both types of radiators cross section can be square, rectangular or round.
Radiator area calculation
Methods for accurately calculating the parameters of a radiator involve taking into account many factors:
- ambient air parameters;
- scattering area;
- radiator configuration;
- properties of the material from which the heat exchanger is made.
But all these subtleties are needed for a designer developing a heat sink. Radio amateurs most often use old radiators taken from end-of-life radio equipment. All they need to know is what is the maximum power dissipation of the heat exchanger.
F \u003d a x Sx (T1 - T2), where
- Ф – heat flux (W);
- S is the surface area of the radiator (the sum of the areas of all fins or needles and the substrate in sq. m). When calculating the area, it should be borne in mind that the fin or plate has two heat removal surfaces. That is, the heat sink area of a rectangle with an area of 1 cm2 will be 2 cm2. The surface of the needle is calculated as the circumference (π x D) multiplied by its height;
- T1 is the temperature of the heat-removing medium (boundary), K;
- T2 is the temperature of the heated surface, K;
- a is the heat transfer coefficient. For unpolished surfaces it is assumed to be 6-8 W/(m2K).
There is another simplified formula obtained experimentally, which can be used to calculate the required radiator area:
S = x W, where
- S is the heat exchanger area;
- W - input power (W);
- M is the unused power of the LED.
For finned radiators made of aluminum, you can use the approximate data provided by Taiwanese experts:
- 1 W - from 10 to 15 cm2;
- 3 W - from 30 to 50 cm2;
- 10 W - about 1000 cm2;
- 60 W - from 7000 to 73000 cm2.
However, it should be noted that the above data is inaccurate, since they are indicated in ranges with a fairly large range. In addition, these values are determined for the climate of Taiwan. They can only be used for preliminary calculations.
Get the most reliable answer about the best way You can calculate the radiator area in the following video:
DIY
Radio amateurs rarely take up the manufacture of radiators, since this element is a responsible thing that directly affects the durability of the LED. But in life there are different situations when you have to make a heat sink from improvised means.
Option 1
The most simple design homemade radiator - a circle cut out of an aluminum sheet with cuts made on it. The resulting sectors are slightly bent (it turns out something that looks like a fan impeller).
4 antennae are bent along the axes of the radiator to fasten the structure to the lamp body. The LED can be fixed through the thermal paste with self-tapping screws.
Option 1 - homemade aluminum radiator
Option 2
You can make a radiator for an LED with your own hands from a piece of pipe rectangular section and aluminum profile.
Necessary materials:
- pipe 30x15x1.5;
- press washer with a diameter of 16 mm;
- hot glue;
- thermal grease KTP 8;
- profile 265 (W-shaped);
- self-tapping screws.
To improve convection, three holes with a diameter of 8 mm are drilled in the pipe, and holes with a diameter of 3.8 mm are drilled in the profile for its fastening with self-tapping screws.
The LEDs are glued to the pipe - the base of the radiator - with hot glue.
At the joints of the radiator parts, a layer of KTP 8 thermal paste is applied. Then the structure is assembled using self-tapping screws with a press washer.
Methods for attaching LEDs to a radiator
LEDs are attached to heatsinks in two ways:
- mechanical;
- gluing.
You can glue the LED on hot glue. For this on metal surface a drop of adhesive mass is applied, then an LED sits on it.
To obtain a strong connection, the LED must be pressed down with a small load for several hours - until the glue is completely dry.
However, most radio amateurs prefer the mechanical fastening of LEDs. Now released special panels, with which you can quickly and reliably mount the LED.
Some models have clamps for secondary optics. Installation is simple: an LED is installed on the radiator, on it is a socket, which is attached to the base with self-tapping screws.
But not only radiators for the LED can be made independently. Fans of plants are advised to familiarize themselves with the LED.
High-quality cooling of the LED is the key to the durability of the LED. Therefore, the selection of a radiator should be approached with all seriousness. It is best to use ready-made heat exchangers: they are sold in radio stores. Radiators are not cheap, but they are easy to install and the LED protects more reliably from excess heat.
