How to calculate the radiator. Scheme, description. Radiators for LEDs: area calculation, material selection, DIY manufacturing Calculation of the thermal resistance of a transistor radiator

All electronic components generate heat, so the ability to calculate heatsinks so as not to fly by a couple of orders of magnitude is very useful for any electronics engineer.

Thermal calculations are very simple and have a lot in common with the calculations of electronic circuits. Here, look at a typical thermal design problem I just ran into.

Task

You need to choose a radiator for a 5-volt linear regulator, which is powered by 12 volts maximum and produces 0.5A. The maximum allocated power is (12-5) * 0.5 = 3.5W

Dive into theory

In order not to produce entities, people scratched a pumpkin and realized that heat is very similar to electric current, and for thermal calculations you can use the usual Ohm's law, only

Voltage (U) is replaced by temperature (T)

Current (I) replaced by power (P)

Resistance is replaced by thermal resistance. Conventional resistance is measured in Volts/Amps, while thermal resistance is measured in °C/Watt.

As a result, Ohm's law is replaced by its thermal counterpart:

A small note - in order to indicate that thermal (rather than electrical) resistance is meant, the letter theta is added to the letter R: I don’t have such a letter on the keyboard, and I’m too lazy to copy from the symbol table, so I’ll just use the letter R.

We continue

Heat is generated in the stabilizer crystal, and our goal is to prevent it from overheating (to prevent overheating of the crystal, not the case, this is important!).

To what temperature the crystal can be heated, it is written in the datasheet:

Usually, the limiting temperature of the crystal is called Tj (j = junction = junction - the temperature-sensitive insides of microcircuits mainly consist of pn junctions. We can assume that the temperature of the junctions is equal to the temperature of the crystal)

without heatsink

The thermal diagram looks very simple:

Especially for cases of using a case without a radiator, the datasheets write the thermal resistance of the crystal-atmosphere (Rj-a) (you already know what j is, a = ambient = environment)

Note that the temperature of the "ground" is not zero, but equal to the temperature of the surrounding air (Ta). The air temperature depends on the conditions in which the radiator is located. If it is in the open air, then Ta = 40 ° C can be set, but if it is in a closed box, then the temperature can be much higher!

We write Ohm's thermal law: Tj = P*Rj-a + Ta. We substitute P = 3.5, Rj-a = 65, we get Tj = 227.5 + 40 = 267.5 °C. Too much, though!

We cling to the radiator

The thermal scheme of our example with a stabilizer on a radiator becomes like this:

• Rj-c– resistance from die to case heatsink (c = case = case). Given in the datasheet. In our case - 5 °C / W - from the datasheet

Rc-r– body-radiator resistance. It's not all that simple. This resistance depends on what is between the case and the heatsink. For example, a silicone gasket has a thermal conductivity coefficient of 1-2 W/(m*°C), while KPT-8 paste has a thermal conductivity of 0.75 W/(m*°C). Thermal resistance can be obtained from the thermal conductivity coefficient using the formula:

R = gasket thickness / (thermal conductivity coefficient * area of ​​one side of the gasket)

Often Rc-r can be ignored altogether. For example, in our case (we use the TO220 case, with KPT-8 paste, the average depth of the paste taken from the ceiling is 0.05mm). Total, Rc-r = 0.5 °C/W. With a power of 3.5W, the temperature difference between the stabilizer body and the radiator is 1.75 degrees. That's not a lot. For our example, let's take Rc-r = 2 °C/W

Rr-a- thermal resistance between the radiator and the atmosphere. It is determined by the geometry of the radiator, the presence of airflow, and a bunch of other factors. This parameter is much easier to measure than to calculate (see at the end of the article). For example - Rr-c = 12.5 °C / W

Ta\u003d 40 ° C - here we figured that the atmospheric temperature is rarely higher, you can take 50 degrees to be sure.

We substitute all these data into Ohm's law, and we get Tj = 3.5*(5+2+12.5) + 40 = 108.25 °C

This is significantly less than the limit of 150 °C. Such a radiator can be used. In this case, the radiator case will heat up to Tc = 3.5*12.5 + 40 = 83.75 °C. This temperature is already able to soften some plastics, so you need to be careful.

Measurement of radiator-atmosphere resistance.

Most likely, you already have a bunch of radiators lying around that can be used. Thermal resistance is very easy to measure. It needs resistance and power supply.

We sculpt the resistance on the radiator using thermal paste:

We connect the power source, and set the voltage so that some power is released on the resistance. It is better, of course, to heat the radiator with the power that it will dissipate in the final device (and in the position in which it will be, this is important!). I usually leave this design for half an hour so that it warms up well.

Once the temperature has been measured, the thermal resistance can be calculated.

Rr-a = (T-Ta)/P. For example, my radiator has warmed up to 81 degrees, and the air temperature is 31 degrees. thus Rr-a = 50/4 = 12.5 °C/W.

Estimated area of ​​the radiator

In the ancient handbook of a radio amateur, a graph was given, according to which you can estimate the area of ​​\u200b\u200bthe radiator. There he is:

It is very easy to work with him. We select the overheating that we want to get and see what area corresponds to the required power with such overheating.

For example, with a power of 4W and overheating of 20 degrees, you will need 250 cm ^ 2 radiators. This graph gives an overestimation of the area, and does not take into account a bunch of factors such as forced airflow, fin geometry, etc.

The claimed service life of LEDs is estimated at tens of thousands of hours. To achieve such a high figure without compromising optical performance, high-power LEDs must be used in tandem with a heatsink. This article will allow the reader to find answers to questions related to the calculation and selection of a radiator, their modifications and factors affecting heat dissipation.

And why is he needed?

Along with other semiconductor devices, the LED is not an ideal element with a 100% coefficient of performance (COP). Most of the energy it consumes is dissipated into heat. The exact value of the efficiency depends on the type of emitting diode and its manufacturing technology. The efficiency of low-current LEDs is 10-15%, and for modern white LEDs with a power of more than 1 W, its value reaches 30%, which means that the remaining 70% is spent in heat.

Whatever the LED, for stable and long-term operation, it needs a constant removal of thermal energy from the crystal, that is, a radiator. In low-current led, the outputs (anode and cathode) perform the function of a radiator. For example, in SMD 2835, the anode lead occupies almost half of the bottom of the element. In high-power LEDs, the absolute value of dissipated power is several orders of magnitude greater. Therefore, they cannot function normally without an additional heat sink. Constant overheating of the light-emitting crystal significantly reduces the service life of the semiconductor device, contributes to a smooth loss of brightness with a shift in the operating wavelength.

Kinds

Structurally, all radiators can be divided into three large groups: lamellar, rod and ribbed. In all cases, the base may be in the form of a circle, square or rectangle. The thickness of the base is of fundamental importance when choosing, since it is this area that is responsible for receiving and evenly distributing heat over the entire surface of the radiator.

The form factor of the radiator is influenced by the future mode of operation:

• with natural ventilation;
• with forced ventilation.

An LED heatsink to be used without a fan must have a fin spacing of at least 4mm. Otherwise, natural convection will not be enough to successfully remove heat. A striking example is the cooling systems of computer processors, where, due to a powerful fan, the distance between the fins is reduced to 1 mm.

When designing LED luminaires, great attention is paid to their appearance, which has a huge impact on the shape of the heat sink. For example, the heat dissipation system of an 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 to the heatsink case using hot glue.

Materials for the manufacture of radiators

Currently, high-power LEDs are cooled mainly on aluminum radiators. This choice is due to the lightness, low cost, flexibility in processing and good heat-conducting properties of this metal. Mounting a copper heatsink for an LED is justified in a luminaire where size is paramount, as copper dissipates heat twice as well as aluminum. The properties of materials that are most often used to cool high-power LEDs will be considered in more detail.

Aluminum

The thermal conductivity coefficient of aluminum is in the range of 202–236 W/m*K and depends on the purity of the alloy. According to this indicator, it is 2.5 times superior to iron and brass. In addition, aluminum lends itself to various types of mechanical processing. To increase the heat dissipation properties, the aluminum radiator is anodized (coated in black).

Copper

The thermal conductivity of copper is 401 W / m * K, second only to silver among other metals. Nevertheless, copper radiators are much less common than aluminum ones, due to a number of disadvantages:

• high cost of copper;
• complex machining;
• big mass.

The use of a copper cooling structure leads to an increase in the cost of the lamp, which is unacceptable in a highly competitive environment.

Ceramic

Aluminum nitride ceramics, whose thermal conductivity is 170–230 W/m*K, has become a new solution in the creation of highly efficient heat sinks. This material is characterized by low roughness and high dielectric properties.

Using thermoplastic

Despite the fact that the properties of thermally conductive plastics (3-40 W / m * K) are worse than those of aluminum, their main advantages are low cost and lightness. Many manufacturers of LED lamps use thermoplastic to make the body. However, thermoplastics are outperforming metal heatsinks in designing LED fixtures above 10W.

Features of cooling high-power LEDs

As mentioned earlier, it is possible to ensure efficient heat removal from the LED by organizing passive or active cooling. It is advisable to install LEDs with a power consumption of up to 10 W on aluminum (copper) radiators, since their weight and size indicators will have acceptable values.

The use of passive cooling for LED arrays with a power of 50 W or more becomes difficult; the dimensions of the radiator will be tens of centimeters, and the weight will increase to 200-500 grams. In this case, you should think about using a compact heatsink along with a small fan. This tandem will reduce the weight and size of the cooling system, but will create additional difficulties. The fan must be provided with the appropriate supply voltage, and also take care of the protective shutdown of the LED lamp in case of a cooler breakdown.

There is another way to cool high-power LED matrices. It consists in using a ready-made SynJet module, which looks like a cooler for a medium performance video card. The SynJet module features high performance, thermal resistance less than 2°C/W, and weight up to 150g. Exact dimensions and weight may vary by model. The disadvantages include the need for a power source and the high cost. As a result, it turns out that a 50 W LED matrix must be mounted either on a bulky but cheap heatsink, or on a small heatsink with a fan, power supply and protection system.

Whatever the heat sink, it is able to provide good, but not the best, thermal contact with the LED substrate. To reduce thermal resistance, a heat-conducting paste is applied to the contact surface. The effectiveness of its impact has been proven by its widespread use in computer processor cooling systems. High-quality thermal paste is resistant to hardening and has a low viscosity. When applied to a radiator (substrate), one thin, even layer is sufficient over the entire contact area. After pressing and fixing, the layer thickness will be about 0.1 mm.

Radiator area calculation

There are two methods for calculating the heatsink for an LED:

• design, the essence of which is to determine the geometric dimensions of the structure at a given temperature regime;
• verification, which involves acting in reverse order, that is, with known parameters of the radiator, you can calculate the maximum amount of heat that it can effectively dissipate.

The use of one or another option depends on the available initial data. In any case, an accurate calculation is a complex mathematical problem with many parameters. In addition to the ability to use reference literature, take the necessary data from graphs and substitute them into the appropriate formulas, one should take into account the configuration of the rods or fins of the radiator, their orientation, as well as the influence of external factors. It is also worth considering the quality of the LEDs themselves. Often, in Chinese-made LEDs, the real characteristics diverge from the declared ones.

Accurate calculation

Before moving on to formulas and calculations, it is necessary to familiarize yourself with the basic terms in the field of distribution of thermal energy. Thermal conductivity is the process of transferring thermal energy from a more heated physical body to a less heated one. Quantitatively, thermal conductivity is expressed as a coefficient that shows how much heat a material is able to transfer through a unit area when the temperature changes by 1°K. In LED lamps, all parts involved in the exchange of energy must have high thermal conductivity. In particular, this concerns the transfer of energy from the crystal to the case, and then to the heatsink and air.

Convection is also a 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. This can be natural convection, which occurs due to the natural movement of the air flow, or forced, organized by installing a fan.

At the beginning of the article it was indicated that about 70% of the power consumed by the LED is spent in heat. To calculate the heat sink 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 PR, where:

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

It's time to count the number of obstacles located on the path of the heat flow from the crystal to the air. Each obstacle represents a thermal resistance (termal resistance), denoted by the symbol (Rθ, degrees / W). For clarity, the entire cooling system is represented as an equivalent circuit from a series-parallel connection of thermal resistances

Rθ ja = Rθ jc + Rθ cs + Rθ sa , where:

Rθ jc - thermal resistance p-n-junction-case (junction-case);
Rθ cs is the thermal resistance of the case-surface radiator;
Rθ sa is the thermal resistance of the radiator-air (surfase radiator-air).

If you intend to install the LED on a printed circuit board or use thermal paste, then you also need to take into account their thermal resistances. In practice, the value of Rθsa can be determined in two ways.

Rθ ja – p-n-junction-air resistance;
T j - maximum temperature of p-n-junction (reference parameter), °C;
T a is the air temperature near the radiator, °C.

Rθ sa = Rθ ja -Rθ jc -Rθ cs , where Rθ jc and Rθ cs are reference parameters.

Find from the graph "the dependence of the maximum thermal resistance on the direct current."

According to the known Rθ sa, a 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 over from old electronic equipment in their homemade products. At the same time, they do not want to delve into complex calculations and buy expensive imported novelties. As a rule, they are only interested in one question: “How much power can be dissipated by the available aluminum heatsink for LEDs?”

We suggest using a simple empirical formula that allows you to get an acceptable calculation result: Rθ sa \u003d 50 / √S, where S is the surface area of ​​\u200b\u200bthe radiator in cm 2.

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

We find the permissible dissipation power from the formula: P t \u003d (T j -T a) / Rθ ja.

The above calculation does not take into account many nuances that affect the quality of the entire cooling system (radiator orientation, temperature characteristics of the LED, etc.). Therefore, it is recommended to multiply the result obtained by the safety factor - 0.7.

Do-it-yourself LED radiator

It is not difficult to make an aluminum radiator for 1, 3 or 10 W LEDs with your own hands. First, consider a simple design, the manufacture of which will take about half an hour of time and a round plate with a thickness of 1-3 mm. In a circle, every 5 mm, cuts are made to the center, and the resulting sectors are slightly bent so that the finished structure resembles an impeller. To attach the radiator to the body, holes are made in several sectors. It's a bit more difficult to make a homemade heatsink for a 10 watt LED. To do this, you need 1 meter of aluminum strip 20 mm wide and 2 mm thick. First, the strip is sawn with a hacksaw into 8 equal parts, which are then stacked, drilled through and tightened with a bolt and nut. One of the side faces is ground for mounting the LED matrix. With the help of a chisel, the strips are unfolded in different directions. Holes are drilled at the mounting points of the LED module. Hot melt adhesive is applied to the polished surface, a matrix is ​​applied on top, fixing it with self-tapping screws.

Cheap heat sinks for amateur crafts

Especially for radio amateurs who like to experiment with different materials for heat dissipation and at the same time do not want to spend money on expensive finished products, we will give some recommendations on finding and making radiators with our own hands. For cooling LED strips and rulers, an aluminum furniture profile is perfect. These can be guides for wardrobes or kitchen fittings, the remains of which can be bought at cost in a furniture store.

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

Homemade cooling for a 50 W LED can be made from a radiator from a faulty chainsaw, lawn mower, sawing it into several parts. You can buy such spare parts in repair shops at the price of scrap. Of course, you can forget about the aesthetic qualities of the LED lamp in this case.

Read also

= ([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 case 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:

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]

Let's say right away that there is no scientifically based method for calculating cooling radiators. On this occasion, you can write more than one dissertation or monograph (and many have been written), but it is worth changing the configuration of the cooling fins or rods, placing the radiator not vertically, but horizontally, bringing any other surface closer to it from below, above or from the side - everything will change, and sometimes dramatically. That is why manufacturers of microprocessors or processors for video cards prefer not to take risks, but to supply their products with radiators with a fan - forced airflow, even weak, increases the efficiency of heat dissipation by dozens of times, although for the most part this is not required at all (but they act according to the law “it’s better to overdo it, what not to do, and rightly so). Here we will only give a couple of empirical methods that have proven themselves in practice and are suitable for calculating passive (that is, without airflow) heatsinks for such amplifiers or for analog power supplies, which will be discussed in the next chapter.

Rice. 8.4. Typical plate radiator

First, let's look at how to calculate the area of ​​\u200b\u200bradiators based on their geometry. On fig. 8.4 schematically shows a typical plate radiator. To calculate its area, you need to add the total area of ​​\u200b\u200bits edges (also on each side) to the area of ​​​​its base. If the bottom side of the radiator is pressed against the board, then it is better to consider only one side of the base as working, but we will assume that the radiator "hangs" in the air (as often happens) and therefore the base area doubles: Socn-‘^-LyLi. The area of ​​one rib (also on both sides) Sp = 2-Lyh, but to this value you also need to add the lateral surfaces of the rib, the area of ​​​​which is equal to SQoK = 2'hd. There are only 6 ribs, so the total area of ​​the radiator will be equal to S = Soctt + 6-5r + b-b'bok. Let L1 \u003d 3 cm, I2 \u003d 5 cm, L \u003d 3 cm, s \u003d 0.2 cm, then the total area of ​​​​such a radiator will be 145 cm ^. Of course, this is an approximate calculation (mts did not take into account, say, the side surface of the base), but for our purposes, accuracy is not required.

Here are two empirical ways to calculate power dissipation as a function of surface area, and don't be too hard on me for not seeing any special scientific calculations here.

The first and simplest method: the area of ​​the cooling radiator should be 100 cm2 for each watt of power released. So the radiator shown in fig. 8.4 dimensions, according to this rule, it can dissipate 14.5 W of power - just under our amplifier with some margin. And if you are not pressed by the dimensions of the case, then you may well limit yourself to this estimated calculation.

Rice. 8.5. The effective heat transfer coefficient of a finned radiator under conditions of free convection with different fin lengths: 1 - /7 = 32 mm; 2 - /7 = 20 mm; 3 - /7 = 12.5 mm

To estimate the heat output of the radiator, you can use the formula W = azff-e.5, where:

W is the power dissipated by the radiator, W;

Aeff - effective heat transfer coefficient, W / m ^ ° С (see graph in Fig. 8.5);

0 - value of overheating of the heat-releasing surface, °С, Q = Т^- Tq^ (Гс - average temperature of the radiator surface, Gos - ambient temperature);

S is the total area of ​​the heat-releasing surface of the radiator, m1

Please note that the area in this formula is substituted in square meters, not centimeters.

So, let's get started: first, let's set the desired overheating of the surface, choosing a not too large value equal to 30 °C. Roughly speaking, we can assume that at an ambient temperature of 30 °C, the surface temperature of the radiator will be 60 °C. Considering that the difference between the temperature of the heatsink and the temperature of the transistor or microchip crystal with good thermal contact (which is discussed below) can be approximately 5 ° C, then this is acceptable for almost all semiconductor devices. The height of the ribs h is 30 mm, so we use the upper curve in the graph in Fig. 8.5, from where we learn that the value of the heat transfer coefficient will be approximately 50 W / m ^ ° C. After calculations, we get that W = 22 W. According to the simplest rule, earlier we got 14.5 W, that is, by making more accurate calculations, we can slightly reduce the area, thereby saving space in the case. However, we repeat, if the place does not press us, then it is better to always have a reserve.

The radiator should be placed vertically, and the fins should also be placed vertically (as in the figure), and its surface should be painted black. I want to remind you again that all these calculations are very approximate, and even the method itself can change if you put the radiator not vertically, but horizontally, or if you equip the radiator with needle fins instead of lamellar ones. In addition, we do not take into account the thermal resistance of the crystal-case and case-heatsink junctions in any way (simply assuming that the temperature difference is 5 °C).

Nevertheless, these methods give a good approximation to the truth, but if we do not provide good thermal contact, all our calculations may go down the drain. You can, of course, simply press the transistor tightly against the heatsink with a screw, but only if the surface of the heatsink at the point of pressing is perfectly flat and well polished. In practice, this never happens, so the radiator in the place of pressure is lubricated with a special heat-conducting paste. It can be bought in stores, and sometimes a tube with such a paste is applied to "coolers" for microprocessors. It is necessary to lubricate with a thin, but uniform layer, do not overdo it in quantity. If two devices are placed on one radiator, in which the collectors are under different voltages ^ then an insulating gasket must be laid under the case, insulating plastic washers under the fixing screws, and a piece of insulating cambric tube with a length equal to the thickness of the radiator at the hole ( Fig. 8.6).

Rice. 8.6. Fastening the transistor in the TO-220 case to the radiator if it is necessary to isolate it: 1 - radiator; 2 - a hole in the radiator; 3 - insulating washers; 4 - tightening screw; 5 - nut; 6 - insulating tube; 7 - mica gasket; 8 - plastic part of the transistor case; 9 - metal part of the transistor case; 10 - transistor outputs

The most convenient insulating pads are mica, anodized aluminum pads are very good (but they must be carefully monitored so as not to scratch a thin layer of insulating oxide) and ceramics (which, however, are quite fragile and can crack under too much pressure). By the way, in the absence of branded gaskets, you can use a thin fluoroplastic (but not polyethylene, of course!) Film, making sure that it does not break through. When installed on a gasket, a heat-conducting paste is applied in a thin layer on both surfaces - both on the transistor and on the radiator.

There is a well-known law in physics, electrical engineering and atomic thermodynamics - the current flowing through the wires heats them up. Joule and Lenz came up with it, and they turned out to be right - the way it is. Everything that runs on electricity, one way or another, part of the passing energy transfers to heat.

It just so happens in electronics that the most heat-affected object in our environment is air. It is to the air that the heating parts transfer heat, and from the air it is required to receive heat and put it somewhere. Lose, for example, or scatter on your own. We call the process of heat transfer cooling.

Our electronic designs also dissipate a lot of heat, some more than others. Voltage stabilizers are heated, amplifiers are heated, the transistor that controls the relay or even just a small LED is heated, except that it heats up quite a bit. Okay, if it gets a little warm. Well, if it is fried so that you can’t hold your hand? Let's take pity on him and try to help him somehow. So to speak, to alleviate his suffering.

Recall the device of the heating battery. Yes, yes, the same ordinary battery that heats the room in winter and on which we dry socks and T-shirts. The larger the battery, the more heat will be in the room, right? Hot water flows through the battery, it heats the battery. The battery has an important thing - the number of sections. Sections are in contact with air, transferring heat to it. So, the more sections, that is, the larger the area occupied by the battery, the more heat it can give us. By welding a couple more sections, we can make our room warmer. True, at the same time, the hot water in the battery can cool down, and there will be nothing left for the neighbors.

Consider the transistor device.

On a copper base (flange) 1 on a substrate 2 fixed crystal 3 . It connects to outputs 4 . The whole structure is filled with plastic compound 5 . The flange has a hole 6 for installation on a radiator.

This is essentially the same battery, look! The crystal heats up, it's like hot water. The copper flange is in contact with air, these are battery sections. The contact area of ​​the flange and air is the place where the air is heated. The heated air cools the crystal.

How to make a crystal colder? We cannot change the device of the transistor, this is understandable. The creators of the transistor also thought about this, and for us, the martyrs, they left the only path to the crystal - the flange. A flange is like a single section of a battery - frying is frying, but heat is not transferred to the air - a small contact area. This is where the scope for our actions is given! We can build up the flange, solder a couple more sections to it, that is, a large copper plate, since the flange itself is copper, or fix the flange on a metal blank called a radiator. Fortunately, the hole in the flange is prepared for a bolt with a nut.

What is a radiator? I’ve been repeating for the third paragraph about him, but I haven’t really said anything! Okay, let's see:

As you can see, the design of radiators can be different, these are plates, and fins, and there are also needle-shaped radiators and various others, just go to the radio parts store and run through the shelf with radiators. Radiators are most often made of aluminum and its alloys (silumin and others). Copper radiators are better, but more expensive. Steel and iron radiators are used only at very low power, 1-5W, as they slowly dissipate heat.

The heat released in a crystal is determined by a very simple formulaP=U*I, where P is the power dissipated in the crystal, W, U = voltage on the crystal, V, I is the current through the crystal, A. This heat passes through the substrate to the flange, where it is transferred to the radiator. Further, the heated radiator comes into contact with air and heat is transferred to it, as the next participant in our cooling system.

Let's look at the complete transistor cooling circuit.

We have two pieces - this is a radiator 8 and a gasket between the heatsink and the transistor 7 . It may not be, which is both bad and good at the same time. Let's figure it out.

I'll tell you about two important parameters - these are thermal resistances between the crystal (or junction, as it is also called) and the transistor case - Rpc and between the transistor case and the radiator - Rcr. The first parameter indicates how well heat is transferred from the crystal to the flange of the transistor. For example, Rpc, equal to 1.5 degrees Celsius per watt, explains that with an increase in power by 1 W, the temperature difference between the flange and the radiator will be 1.5 degrees. In other words, the flange will always be colder than the crystal, and this parameter shows how much. The smaller it is, the better the heat is transferred to the flange. If we dissipate 10W of power, then the flange will be colder than the crystal by 1.5 * 10 = 15 degrees, and if 100W, then by all 150! And since the maximum temperature of the crystal is limited (it cannot be fried to a white heat!), The flange must be cooled. At the same 150 degrees.

For example:
The transistor dissipates 25W of power. Its Rpc is 1.3 degrees per watt. The maximum temperature of the crystal is 140 degrees. This means that there will be a difference of 1.3 * 25 = 32.5 degrees between the flange and the crystal. And since the crystal cannot be heated above 140 degrees, we are required to maintain the flange temperature not hotter than 140-32.5=107.5 degrees. Like this.
And the Rcr parameter shows the same thing, only the losses are obtained on the same notorious gasket 7. Its Rcr value can be much larger than Rpc, therefore, if we are designing a powerful unit, it is undesirable to put transistors on gaskets. But still, sometimes you have to. The only reason to use a spacer is if you need to isolate the heatsink from the transistor, because the flange is electrically connected to the middle terminal of the transistor case.

Let's look at another example here.
The transistor is fried at 100W. As usual, the temperature of the crystal is no more than 150 degrees. Rpk it has 1 degree per watt, and even on the gasket, which has Rkr 2 degrees per watt. The temperature difference between the crystal and the radiator will be 100*(1+2)=300 degrees. The radiator must be kept not hotter than 150-300 = minus 150 degrees: Yes, my dears, this is the very case that only liquid nitrogen will save: horror!

It is much easier to live on a radiator for transistors and microcircuits without gaskets. If they are absent, and the flanges are clean and smooth, and the radiator sparkles with brilliance, and even heat-conducting paste is put, then the Rcr parameter is so small that it is simply not taken into account.

There are two types of cooling - convection and forced. Convection, if we remember school physics, is the independent distribution of heat. The same goes for convection cooling - we installed a radiator, and he himself will somehow sort out the air there. Convection-type radiators are most often installed outside the devices, like in amplifiers, have you seen? On the sides are two metal plate gizmos. From the inside, transistors are screwed to them. Such radiators cannot be covered, air access is closed, otherwise the radiator will have nowhere to put the heat, it will overheat itself and refuse to receive heat from the transistor, which will not think for a long time, it will also overheat and: you yourself understand what will happen. Forced cooling is when we force the air to blow more actively around the radiator, making its way along its ribs, needles and holes. Here we use fans, various air cooling channels and other methods. Yes, by the way, instead of air, there can easily be water, oil, and even liquid nitrogen. Powerful generator tubes are often cooled by running water.

How to recognize a radiator - is it for convection or forced cooling? Its efficiency depends on this, that is, how quickly it can cool the hot crystal, what flow of thermal power it can pass through itself.

We look at the photos.

The first radiator is for convection cooling. Large fin spacing ensures free airflow and good heat dissipation. A fan is put on top of the second radiator and blows air through the fins. This is forced cooling. Of course, you can use both those and those radiators everywhere, but the whole question is their efficiency.
Radiators have 2 parameters - this is its area (in square centimeters) and the coefficient of thermal resistance of the radiator-environment Rrs (in Watts per degree Celsius). The area is calculated as the sum of the areas of all its elements: the area of ​​the base on both sides + the area of ​​the plates on both sides. The area of ​​\u200b\u200bthe ends of the base is not taken into account, so there will be very few square centimeters there.

Example:
the radiator from the example above for convection cooling.
Base dimensions: 70x80mm
Fin size: 30x80mm
Number of ribs: 8
Base area: 2х7х8=112 sq.cm
Fin area: 2х3х8=48 sq.cm.
Total area: 112+8x48=496 sq.cm.

The coefficient of thermal resistance radiator-environment Rpc shows how much the temperature of the air leaving the radiator will increase with an increase in power by 1W. For example, a Rpc of 0.5 degrees Celsius per watt tells us that the temperature will increase by half a degree for 1W of heat. This parameter is considered to be three-story formulas and our feline minds are by no means within the power: Rpc, like any thermal resistance in our system, the smaller the better. And you can reduce it in different ways - for this, radiators are blackened chemically (for example, aluminum is well darkened in ferric chloride - do not experiment at home, chlorine is released!), There is also the effect of orienting the radiator in the air for better passage along the plates (vertical radiator cools better than recumbent). It is not recommended to paint the radiator with paint: paint is an excess thermal resistance. If only slightly, so that it was dark, but not a thick layer!

The application has a small program in which you can calculate the approximate area of ​​\u200b\u200bthe radiator for some microcircuit or transistor. With it, let's calculate the radiator for some power supply.

Power supply circuit.

The power supply outputs 12 volts at a current of 1A. The same current flows through the transistor. At the input of the transistor is 18V, at the output is 12V, which means that a voltage of 18-12 \u003d 6V drops on it. Power dissipated from the transistor crystal is 6V * 1A \u003d 6W. The maximum temperature of the crystal in 2SC2335 is 150 degrees. Let's not use it in extreme conditions, let's choose a lower temperature, for example, 120 degrees. The thermal resistance of the junction-case Rpc for this transistor is 1.5 degrees Celsius per watt.

Since the transistor flange is connected to the collector, let's provide electrical insulation to the heatsink. To do this, we put an insulating gasket made of heat-conducting rubber between the transistor and the radiator. The thermal resistance of the gasket is 2 degrees Celsius per watt.

For good thermal contact, let's drop a little PMS-200 silicone oil. This is a thick oil with a maximum temperature of +180 degrees, it will fill the air gaps that are necessarily formed due to the unevenness of the flange and radiator and improve heat transfer. Many use KPT-8 paste, but many consider it not the best heat conductor.
We will bring the radiator to the back wall of the power supply, where it will be cooled by room air + 25 degrees.

We will substitute all these values ​​into the program and calculate the area of ​​the radiator. The resulting area of ​​113 sq. cm is the area of ​​the radiator, designed for long-term operation of the power supply in full power mode - more than 10 hours. If we do not need so much time to drive the power supply, we can get by with a smaller, but more massive radiator. And if we install a radiator inside the power supply, then there is no need for an insulating gasket, without it the radiator can be reduced to 100 sq.cm.

In general, my dears, the stock does not pull the pocket, do you all agree? Let's think about the margin, so that it is both in the area of ​​\u200b\u200bthe radiator and in the limiting temperatures of transistors. After all, not just anyone, but you yourself will have to repair devices and change overcooked transistors! Remember this!