A technique is presented, using the example of an Intel Pentium4 Willamette 1.9 GHz processor and a B66-1A cooler manufactured by ADDACorporation, describing the procedure for calculating ribbed heatsinks designed to cool REA fuel elements with forced convection and flat thermal contact surfaces with a power of up to 100 W. The technique allows to make a practical calculation of modern highly efficient small-sized devices for heat removal and apply them to the entire range of radio electronic devices that need cooling.

Parameters set in the initial data:

P= 67 W, power released by the cooled element;

q with\u003d 296 °K, temperature of the medium (air) in degrees Kelvin;

q before= 348 °K, limiting crystal temperature;

q R= nn °K , average temperature of the radiator base (calculated during calculation);

H\u003d 3 10 -2 m, height of the radiator fin in meters;

d\u003d 0.8 10 -3 m, rib thickness in meters;

b= 1.5 10 -3 m, the distance between the ribs;

l m\u003d 380 W / (m ° K), coefficient of thermal conductivity of the radiator material;

L\u003d 8.3 10 -2 m, radiator size along the fin in meters;

B= 6.9 10 -2 m, the size of the radiator across the fins;

BUT= 8 10 -3 m, the thickness of the radiator base;

V³ 2 m/s, air velocity in the radiator channels;

Z= 27, the number of radiator fins;

u R= nn K , the overheating temperature of the radiator base, is calculated during the calculation;

e R= 0.7, the degree of emissivity of the radiator.

It is assumed that the heat source is located in the center of the radiator.

All linear dimensions are measured in meters, temperature in degrees Kelvin, power in watts, and time in seconds.

The design of the radiator and the parameters required for calculations are shown in Fig.1.

Picture 1.

Calculation procedure.

1. Determine the total cross-sectional area of ​​the channels between the ribs by the formula:

S k \u003d (Z - 1) b H

For the accepted initial data - S k \u003d (Z - 1) b H \u003d (27-1) 1.5 10 -3 3 10 -2 \u003d 1.1 10 -3 m 2

For a central fan installation, the air flow exits through two end surfaces and the cross-sectional area of ​​the channels is doubled and equals 2.2 10 -3 m 2 .

2. We set two values ​​for the temperature of the radiator base and carry out a calculation for each value:

q р = ( 353 (+80°С) and 313 (+40°С))

From here, the overheating temperature of the radiator base is determined u R regarding the environment.

u p = q p - q with

For the first point u p = 57°K, for the second u p = 17°K.

3. Determine the temperature q required to calculate the Nusselt (Nu) and Reynolds (Re) criteria:

q = q c + P / (2 V S c r C p)

where: q with – ambient temperature, environment,

V is the air velocity in the channels between the fins, in m/s;

S to- the total cross-sectional area of ​​the channels between the ribs, in m 2;

r - air density at temperature q cf, in kg / m 3,

q cp = 0.5 ( q p+q with);

C R is the heat capacity of air at temperature q cf, in J/(kg x °K);

P is the power dissipated by the radiator.

For the accepted initial data - q \u003d q c + P / (2 V S to r C p) \u003d 296 K + 67 / (2 2m / s 1.1 10 -3 m 2 1.21 1005) \u003d 302.3 ° K (29.3 ° C)

* Value, for a given finned heatsink with central fan mounting, V from calculations 1.5 - 2.5 m/s (See Appendix 2), from publications [L.3] about 2 m/s. For short, expanding channels, such as the Golden Orb cooler, the cooling air velocity can reach 5 m/s.

4. Determine the values ​​of the Reynolds and Nusselt criteria necessary to calculate the heat transfer coefficient of the radiator fins:

Re = V L / n

where: n - coefficient of kinematic viscosity of air at q with, m 2 /with from Annex 1, table 1.

For the accepted initial data - Re = VL/ n\u003d 2 8.3 10 -2 / 15.8 10 -6 \u003d 1.05 10 4

Nu = 0.032 Re 0.8

For the accepted initial data - Nu = 0.032 Re 0.8 = 0.032 (2.62 10 4) 0.8 = 52.8

5. Determine the coefficient of convective heat transfer of the radiator fins:

a to = Nu · l in / L W / (m 2 TO)

where, l - coefficient of thermal conductivity of air (W / (m deg)), at q with from Appendix 1, table1.

For the accepted initial data - a to = Nu l c / L \u003d 52.8 2.72 10 -2 / 8.3 10 -2 \u003d 17.3

6. Determine the auxiliary coefficients:

m = (2 a k / l m d) 1/2

we determine the value of mh and the tangent of the hyperbolic th (mh).

For the accepted initial data - m \u003d (2 a k / l m d) 1/2 \u003d (2 17.3 / (380 0.8 10 -3)) 1/2 \u003d 10.6

For the accepted initial data - m H = 10.6 3 10 -2 = 0.32; th (mH) = 0.31

7. Determine the amount of heat given off by convection from the radiator fins:

P pk = Z l m m S p u p th(m H)

where: Z is the number of ribs;

l m= coefficient of thermal conductivity of the radiator metal, W / (m · °K);

m- see formula 7;

S R- cross-sectional area of ​​\u200b\u200bthe radiator fin, m 2,

S p \u003d L d

u R– overheating temperature of the radiator base.

S p \u003d L d \u003d 8.3 10 -2 0.8 10 -3 \u003d 6.6 10 -5 m 2

P pk = Z l m m S p u p th (m H) \u003d 27 380 10.6 6.6 10 -5 57 0.31 \u003d 127 W.

8. Determine the average temperature of the radiator fin:

q cf = (q p / 2) [ 1 + 1 / ch (m H )]

where: ch (mH ) is the hyperbolic cosine.

For the accepted initial data - q cf = (q p / 2) [ 1 + 1 / ch (m H )] = (353/2) = 344 ° K (71 ° С)

* The value of the tangent and cosine of the hyperbolic is calculated on an engineering calculator by sequentially performing the operations “hyp” and “tg” or “cos”.

9. Determine the radiant heat transfer coefficient:

a l \u003d e p f (q cf, q c) j

f (q cf, q c) = 0.23 [ 5 10 -3 (q cf + q c)] 3

For the accepted initial data - f (q cf, q c) = 0.23 [ 5 10 -3 (q cf + q c)] 3 = 0.23 3 = 7.54

Irradiance coefficient:

j = b / (b + 2h)

j \u003d b / (b + 2H) \u003d 1.5 10 -3 / (1.5 10 -3 + 3 10 -2) \u003d 0.048

a l \u003d e p f (q cf, q c) j \u003d 0.7 x 7.54 x 0.048 \u003d 0.25 W / m 2 K

10. Determine the surface area of ​​the radiating heat flux:

S l \u003d 2 L [ (Z -1) (b + d) + d] +2 H L Z (m 2)

For the accepted initial data - S l \u003d 2 L [(Z -1) (b + d) + d] +2 H L Z \u003d 0.1445 m 2

11. Determine the amount of heat given off through radiation:

P l \u003d a l S l (q cf - q c)

For the accepted initial data - P l \u003d a l S l (q cf - q c) \u003d 0.25 0.1445 (344 - 296) \u003d 1.73 W

12. The total amount of heat given off by the radiator at a given temperature of the radiator q p \u003d 353K:

P \u003d P pk + P l

For the accepted initial data - P \u003d P pk + P l \u003d 127 + 1.73 \u003d 128.7 W.

13. Repeat Calculations for Heatsink Temperature q p \u003d 313K, and we build the thermal characteristic of the calculated radiator from two points. For this point P=38W. Here, on the vertical axis, the amount of heat given off by the radiator is plotted P R, while the horizontal temperature of the radiator q R .

Figure 2

From the resulting graph, we determine for a given power of 67W, q R= 328 °K or 55 °C.

14. According to the thermal characteristic of the radiator, we determine that for a given power P R=67W, heatsink temperature q R=328.5°C. Radiator overheating temperature u R can be determined by formula 2.

She is equal u p \u003d q p - q c \u003d 328 - 296 \u003d 32 ° K.

15. We determine the temperature of the crystal and compare it with the limit value set by the manufacturer

q to = q p + p ( r pc + r pr) °K \u003d 328 + 67 (0.003 + 0.1) \u003d 335 (62 ° C),

q R – temperature of the radiator base for a given design point,

R- the result of calculation according to formula 14,

r pc - thermal resistance processor case - crystal, for a given heat source is 0.003 K / W

r pr - thermal resistance of the body-radiator, for a given heat source is 0.1K / W (with heat-conducting paste).

The result obtained is below the limit temperature determined by the manufacturer, and is close to the data of [L.2] (about 57°C). In this case, the overheating temperature of the crystal relative to the surrounding air in the above calculations is 32°C, and in [L.2] 34°C.

In general, the thermal resistance between two flat surfaces when using solders, pastes and adhesives:

r = d to · l to -1 · S cont -1

where: d k is the thickness of the gap between the radiator and the body of the cooled unit filled with heat-conducting material, m,

l to- coefficient of thermal conductivity of the heat-conducting material in the gap W / (m K),

S cont is the area of ​​the contact surface in m2.

The approximate value of r cr with sufficient tightening and without gaskets and lubricants is

r cr = 2.2 / S cont

When using pastes, thermal resistance drops by about 2 times.

16. Compare q to with q before, we received a radiator providing q to= 325°K less q before = 348°K, - the specified radiator provides the thermal mode of the node with a margin.

17. Determine the thermal resistance of the calculated radiator:

r = u R/ P (°K/W)

r= u p / P (°/W) = 32/67 = 0.47°/W

Findings:

The calculated heat exchanger provides heat power removal of 67W at an ambient temperature of up to 23°C, while the crystal temperature of 325°K (62°C) does not exceed 348°K (75°C) allowed for this processor.

The use of a special surface treatment to increase the heat output through radiation at temperatures up to 50°C turned out to be ineffective and cannot be recommended, because does not cover costs.

I would like this material to help you not only calculate and manufacture a modern small-sized highly efficient heat exchanger, similar to those that are widely used in computer technology, but also make competent decisions on the use of such devices in relation to your tasks.

Appendix 1.

Constants for calculating the heat exchanger.

Table 1

q s, K(°C)	l* 10 -2 W/(m K)	n* 10 6 m 2 /sec	Wed J/(kg*K)	r , kg/m 2
273 (0)td>	2,44	13,3	1005	1,29
293 (20)	2,59	15,1	1005	1,21
373 (100)	3,21	23,1	1009	0,95

The values ​​of constants for intermediate temperatures, in the first approximation, can be obtained by plotting functions for the temperatures indicated in the first column.

Appendix 2

Calculation of the speed of movement of air cooling the radiator.

The speed of the coolant with forced convection in gases:

V \u003d Gv / S to

Where: Gv is the volumetric flow rate of the coolant, (for a 70x70 fan, S pr \u003d 30 cm 2, 7 blades, P em \u003d 2.3W, w \u003d 3500 rpm, Gv \u003d 0.6-0.8 m 3 / min .or actually 0.2-0.3 or V = 2m/s),

S to - free for the passage of the cross-sectional area of ​​the channel.

Taking into account that the area of ​​the fan flow section is 30 cm 2, and the area of ​​the radiator channels is 22 cm 2, the air purge rate is determined by the lower one, and will be equal to:

V = Gv /S = 0.3 m 3 /min / 2.2 10 -3 m 2 \u003d 136 m / min \u003d 2.2 m / s.

For calculations, we accept 2 m / s.

Literature:

Handbook of the REA designer, edited by R.G. Varlamov, M, Soviet radio, 1972;

Handbook of the REA designer, edited by R.G. Varlamov, M, Soviet radio, 1980;

http://www.ixbt.com/cpu/ , Coolers for Socket 478, spring-summer 2002, Vitaly Krinitsin, Published - July 29, 2002;

http://www.ixbt.com/cpu/ , Measuring Air Velocities Behind Cooling Fans and Coolers, Alexander Tsikulin, Alexey Rameikin, Published - August 30, 2002

Prepared in 2003 based on the materials of L.1 and 2

Often, when designing a powerful device on power transistors, or resorting to using a powerful rectifier in a circuit, we are faced with a situation where it is necessary to dissipate a lot of thermal power, measured in units, and sometimes tens of watts.

For example, the FGA25N120ANTD IGBT transistor from Fairchild Semiconductor, if properly mounted, is theoretically capable of delivering about 300 watts of thermal power through its package at a package temperature of 25 ° C! And if the temperature of its case is 100 ° C, then the transistor will be able to give 120 watts, which is also quite a lot. But in order for the transistor body to be able to give off this heat in principle, it is necessary to provide it with proper operating conditions so that it does not burn out ahead of time.

All power switches are produced in cases that can be easily installed on an external heat sink - a radiator. Moreover, in most cases, the metal surface of a key or other device in an output housing is electrically connected to one of the outputs of this device, for example, to a collector or to a transistor drain.

So, the task of the radiator is precisely to keep the transistor, and mainly its working transitions, at a temperature not exceeding the maximum allowable.

Andrey Povny

Radiators for semiconductor devices

During operation, powerful semiconductor devices emit a certain amount of heat into the environment. If you do not take care of their cooling, transistors and diodes can fail due to overheating of the working crystal. Ensuring the normal thermal regime of transistors (and diodes) is one of the important tasks. To properly solve this problem, you need to have an idea about the operation of the radiator and its technically competent design.

As you know, any heated object, cooling, gives off heat to the environment. As long as the amount of heat released in the transistor is greater than that given off by it to the environment, the temperature of the transistor case will continuously increase. At a certain value, the so-called thermal balance occurs, that is, the equality of the amounts of dissipated and released heat. If the temperature of the heat balance is less than the maximum allowable for the transistor, it will work reliably. If this temperature is higher than the allowable maximum temperature, the transistor will fail. In order for the heat balance to occur at a lower temperature, it is necessary to increase the heat transfer of the transistor.

Three methods of heat transfer are known: Thermal conduction, Radiation and Convection. The thermal conductivity of air is usually low - this value can be neglected when calculating the radiator. The proportion of heat dissipated by radiation is significant only at high temperatures (several hundred degrees Celsius), so this value can also be neglected at relatively low operating temperatures of transistors (no more than 60-80 degrees). Convection is the movement of air in the zone of a heated body, due to the difference in temperature between the air and the body. The amount of heat given off by a heated object is proportional to the temperature difference between the object and air, the surface area and the speed of the air flow washing the body.

In my youth, I came across an original solution for removing heat from powerful output transistors. Transistors (then transistors of the P210 type were used to build amplifiers) on long wires were outside the case. Two plastic jars of water were screwed to the case, and the transistors were in them. Thus, "water" effective cooling was provided. When the water in the jars was heated, it was simply replaced with cold water ... Instead of water, you can use mineral (liquid) or transformer oil ... Now the industry has begun to mass-produce water cooling systems for processors and video cards of computers - on the principle of car radiators (but this is already , in my opinion, exotic ...).

To ensure efficient heat removal from the semiconductor crystal, heat sinks (radiators) are used. Let's get acquainted with some of the designs of radiators.

The following figures show four types of heat sinks.

The simplest of these is the plate radiator. Its surface area is equal to the sum of the areas of the two sides. The ideal shape of such a heat sink is a circle, followed by a square and a rectangle. It is advisable to use a plate radiator for low power dissipations. Such a radiator must be installed vertically, otherwise the effective scattering area is reduced.

An advanced plate heat sink is a set of several plates bent in different directions. This radiator, with a surface area equal to the simplest lamellar one, has smaller dimensions. Such a heat sink is installed similarly to a plate heat sink. The number of plates can be different - depending on the required surface. The dissipation area of ​​such a radiator is equal to the sum of the areas of all the bent sections of the plates, plus the surface area of ​​the central part. This type of radiator also has disadvantages: a reduced efficiency of heat removal from all plates, as well as the impossibility of obtaining a perfectly straight surface at the junctions of the plates with each other.

For the manufacture of plate radiators, plates with a thickness of at least 1.5 (preferably 3) millimeters should be used.

A finned radiator - usually solid or milled - can be with one or two-sided fins. Double-sided finning allows for increased surface area. The surface area of ​​such a heat sink is equal to the sum of the surface areas of all the plates and the sum of the surface area of ​​the main radiator body.

The most effective of all of these is the pin (or needle) radiator. With a minimum volume, such a radiator has a maximum effective dissipation area. The surface area of ​​such a heat sink is equal to the sum of the areas of each pin and the area of ​​the main body.

There are also forced air heatsinks (for example, the CPU cooler in your computer). These heat sinks, with a small surface area of ​​the radiator, are capable of dissipating significant power into the environment (for example, the P-1000 medium-speed processor releases 30-70 watts of thermal energy, depending on the load). The disadvantage of such heat sinks is increased noise during operation and a limited period of operation (mechanical wear of the fan).

The material for radiators is usually aluminum and its alloys. Heat sinks made of copper have the best efficiency, but the weight and cost of such radiators are higher than those of aluminum.

The semiconductor device is attached to the heat sink using special flanges. If it is necessary to isolate the device from the radiator, various insulating gaskets are used. The use of spacers reduces the efficiency of heat transfer from the crystal, therefore, if possible, it is better to isolate the heat sink from the chassis of the structure. For more efficient heat dissipation, the surface that comes into contact with the semiconductor device must be even and smooth. To increase the efficiency, special thermal pastes are used (for example, "KPT-8"). The use of thermal paste helps to reduce the thermal resistance of the section "case - heat sink" and allows you to slightly lower the temperature of the crystal. Mica, various plastic films, and ceramics are used as gaskets. At one time, I received a copyright certificate on the method of isolating the transistor case from the heat sink. The essence of this method is as follows: The surface of the heat sink is covered with a thin layer of thermal paste (for example, type KPT-8), a layer of quartz sand is applied (by pouring) to the surface of the paste (I used sand from a fuse), then the excess sand is removed by shaking off and the transistor is pressed tightly using a clamp made of insulating material. During factory tests of this method, the "gasket" withstood a short voltage supply of 1000 volts (from a megger).

Some foreign high-power transistors are produced in an insulated case - such a transistor can be attached directly to a heat sink without the use of any gaskets (but this does not exclude the use of thermal paste!).

The heat source in the transistor-radiator-environment system is the collector P-N junction. The entire heat path in this system can be divided into three sections: junction - transistor case, transistor case - heat sink, heat sink - environment. Due to the imperfection of heat transfer, the temperatures of the junction, the transistor case and the environment differ significantly. This is because heat encounters some resistance along its path, called thermal resistance. This resistance is equal to the ratio of the temperature difference at the boundaries of the section to the dissipated power. The above can be illustrated with an example: according to the reference book, the thermal resistance of the P214 junction-case is 4 degrees Celsius per watt. This means that in case of power dissipation of 10 watts on the junction, the junction will be "warmer" by 4 * 10 = 40 degrees! If we take into account the fact that the maximum junction temperature is 85 degrees, it becomes clear that the case temperature at the specified power should not exceed 85-40 = 45 degrees Celsius. The presence of the thermal resistance of the radiator is the reason for the significant difference in the temperature of its sections, which are at different distances from the place where the transistor is installed. This means that not the entire surface of the radiator is involved in the active heat transfer, but only a part of it, which has the highest temperature and therefore is best washed by air. This part is called the effective surface of the radiator. It will be the greater, the higher the thermal conductivity of the radiator. The thermal conductivity of a radiator depends on the properties of the material from which the heat sink is made and its thickness. That is why copper or aluminum is used to make heat sinks.

A complete calculation of the radiator is a very time-consuming process. For a rough calculation, the following data can be used: To dissipate 1 watt of heat generated by a semiconductor device, it is sufficient to use a heat sink area of ​​30 square centimeters.

Diode designation		Max. temp. env. environments	Radiator area
KD202A,KD202V			WITHOUT RADIATOR
KD202D,KD202Zh
KD202K,KD202M
KD202B,KD202G
KD202E,KD202I
KD202L,KD202N

In the journal "Radioamator-Konstruktor" an article by an unknown author was published on the method of simplified calculation of radiators. .

Literature

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). The larger the surface area of ​​the heatsink, the more intensely it will transfer heat from the LED to the air.

Read more about the principles of operation of LEDs.

The amount of heat absorbed by the air mass per unit area does not depend on the material of the radiator: the efficiency of a 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, the following condition should be taken into account: its thermal conductivity must be at least 5-10 W. 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. In general, this material is inferior to aluminum in terms of lightness of construction and manufacturability (copper is a less pliable metal).

It is impossible to manufacture a copper radiator by pressing - the most economical - method. And cutting gives a large percentage of waste of expensive material.

Copper radiators

Ceramic

One of the most successful options for a 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 ribbed 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 can be square, rectangular or round in cross section.

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.

You can get the most reliable answer about the optimal way to 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 simplest design of a homemade radiator is a circle cut out of an aluminum sheet with notches 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

The radiator for the LED can be made with your own hands from a piece of rectangular pipe and an 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. To do this, a drop of adhesive mass is applied to the metal surface, 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. Special panels are now being produced with which you can quickly and reliably mount the LED.

Some models have clips 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.

= ([Temperature at the hot spot, °C] - [Temperature at the cold point, °C]) / [Dissipated power, W]

This means that if a thermal power of X W is supplied from a hot spot to a cold one, and the thermal resistance is Y cg / W, then the temperature difference will be X * Y cg.

Formula for calculating the cooling of a force element

For the case of calculating the heat removal of an electronic power element, the same can be formulated as follows:

[Power element crystal temperature, GC] = [Ambient temperature, °C] + [Dissipated power, W] *

where [ Total thermal resistance, Hz / W] = + [Thermal resistance between the case and the radiator, Hz / W] + (for the case with a radiator),

or [ Total thermal resistance, Hz / W] = [Thermal resistance between the crystal and the case, Hz / W] + [Thermal resistance between the case and the environment, Hz / W] (for case without heatsink).

As a result of the calculation, we must obtain such a crystal temperature that it is less than the maximum allowable value indicated in the reference book.

Where can I get the data for the calculation?

Thermal resistance between die and package for power elements is usually given in the reference book. And it is marked like this:

Do not be confused by the fact that the units of measurement K / W or K / W are written in the reference book. This means that this value is given in Kelvin per Watt, in Hz per W it will be exactly the same, that is, X K / W \u003d X Hz / W.

Usually, reference books give the maximum possible value of this value, taking into account the technological spread. We need it, since we must carry out the calculation for the worst case. For example, the maximum possible thermal resistance between the crystal and the case of the power field effect transistor SPW11N80C3 is 0.8 c/W,

Thermal resistance between case and heatsink depends on the case type. Typical maximum values ​​are shown in the table:

TO-3	1.56
TO-3P	1.00
TO-218	1.00
TO-218FP	3.20
TO-220	4.10
TO-225	10.00
TO-247	1.00
DPACK	8.33

Insulating pad. In our experience, a properly selected and installed insulating pad doubles the thermal resistance.

Thermal resistance between case/heatsink and environment. This thermal resistance, with an accuracy acceptable for most devices, is quite simple to calculate.

[Thermal resistance, Hz / W] = [120, (gC * sq. cm) / W] / [The area of ​​the radiator or the metal part of the element body, sq. cm].

This calculation is suitable for conditions where elements and radiators are installed without creating special conditions for natural (convection) or artificial airflow. The coefficient itself is chosen from our practical experience.

The specification of most heatsinks contains the thermal resistance between the heatsink and the environment. So in the calculation it is necessary to use this value. This value should be calculated only if tabular data on the radiator cannot be found. We often use used heatsinks to assemble debug samples, so this formula helps us a lot.

For the case when heat is removed through the contacts of the printed circuit board, the contact area can also be used in the calculation.

For the case when heat is removed through the leads of an electronic element (typically diodes and zener diodes of relatively low power), the area of ​​the leads is calculated based on the diameter and length of the lead.

[Lead area, sq. cm.] = Pi * ([ Length of the right output, see] * [Right outlet diameter, see] + [Length of the left output, see] * [Left outlet diameter, see])

An example of calculating heat removal from a zener diode without a radiator

Let the zener diode have two terminals with a diameter of 1 mm and a length of 1 cm. Let it dissipate 0.5 watts. Then:

The output area will be about 0.6 sq. cm.

The thermal resistance between the case (terminals) and the environment will be 120 / 0.6 = 200.

The thermal resistance between the crystal and the case (terminals) in this case can be neglected, since it is much less than 200.

Let us assume that the maximum temperature at which the device will be operated will be 40 °C. Then the temperature of the crystal = 40 + 200 * 0.5 = 140 °C, which is acceptable for most zener diodes.

Online calculation of heat sink - radiator

Please note that for plate radiators, the area of ​​\u200b\u200bboth sides of the plate must be calculated. For PCB tracks used for heat dissipation, only one side needs to be taken, since the other does not come into contact with the environment. For needle radiators, it is necessary to approximately estimate the area of ​​\u200b\u200bone needle and multiply this area by the number of needles.

Online calculation of heat dissipation without a radiator

Several elements on one radiator.

If several elements are installed on one heat sink, then the calculation looks like this. First, we calculate the temperature of the radiator using the formula:

[Radiator temperature, gc] = [Ambient temperature, °C] + [Thermal resistance between the radiator and the environment, Hz / W] * [Total power, W]

[Crystal temperature, c] = [Radiator temperature, gc] + ([Thermal resistance between the crystal and the body of the element, Hz / W] + [Thermal resistance between the body of the element and the radiator, Hz / W]) * [Power dissipated by the element, W]