Projects often require a lot of heat to be conducted away from a device generating it. Constructors often guess at the size of heatsink then build. I used to be that way myself, but in the late 70s I began using power transistors and the tears began. I had to educate myself. After a few experiments I developed 'rough' formulas for estimating the size of heatsink needed.
A silicon transistor should not normally be allowed to become much hotter than about 100°C. Most transistors will happily operate hotter, but do not count upon it.
Let us take the case of the simple PSU described earlier. The PSU may deliver 3 amperes at up to 15 volts. W=VA (Watts = Volts x Amperes). The load is therefore dissipating 3 x 15 = 45 watts. With a pre-regulated voltage of 24 volts, the regulator transistor has the remaining 9 volts accross it, also at 3 amperes. The transistor is therefore turning 27 watts (9v x 3A) into heat.
If you now turn the output voltage of the PSU DOWN to 3 volts, the problem gets much worse. The load is now taking 3 x 3 = 9 watts and the transistor is dissipating (24v - 3v) x 3A = 63 watts as heat. What do we do with all this heat? - use a heatsink, of course, but how big?
Heatsinks are given a characteristic ' °C/Watt '. A TO5 transistor (for example) without a heatsink has a case heatsink characteristic of somewhere in the region of 220°C per watt. So for every watt you make it dissipate it will increase its temperature by 220°C, or about 240 - 250°C at normal room temperatures. That's hot enough to melt solder!
A clip-on heatsink with a characteristic of 40°C/W will limit that device to 40°C + 30°C room temperature = 70°C. With the heatsink the transistor will therefore survive. The heatsink simply conducts heat away from the device and dissipates it into the room, so the larger the surface area, the lower (better) the °C/W rating that heatsink has.
Our transistor in the PSU is dissipating 62 watts, so to keep the temperature below 100°C we should calculate the smallest heatsink required. The room temperature is assumed to be 30°C, so the transistor can increase its temperature by only 70°C. 70°C/62W = 1.13 so we will need a heatsink rated at 1.13°C/Watt, or less. Most component suppliers catalogues give a heatsinks °C/W characteristic.
You can evaluate home-made heatsinks by using the following 'ROUGH' formula:
As an example, let us make a heatsink with 18 SWG aluminium sheet, folded as shown (viewed from above):
The heatsink is going to be 20cm wide (W), 10cm deep (D) and 12cm high (not shown). Each 'fin' is 10cm x 12cm = 120cm. Each fin also has two sides, area = 240 square cm. There are 10 fins so we have a total of 2400 square cm. The rear plate is also 2 sides x 20cm x 12cm = 480 square cm. Total area = 480 + 2400 = 2880 square cm.
Since and the square root of 2880 =
53.66 we therefore have 50/53.66 = 0.932°C/watt. That would be an
expensive heatsink to buy! It may not look pretty home-made, but
it works just as well.
It is common practice to use the chassis of a unit as the heatsink, or part of it. An aluminium box 5cm x 10cm x 20cm would therefore have an OUTSIDE surface area of 5 x 10 x 20 = 1000 square cm. Square root of 1000 = 31.623 so our equipment case has a thermal dissipation of (50/31.623)°C/W = 1.58°C/W Since we need 1.13 we do not have enough, but we can use the case to cut down the size of the heatsink we need. Notice that the inside of the case was not taken into account. The case is a closed box so without ventilation the inside of the case will dissipate nothing.
Ok, now let us work backwards using our PSU regulator transistor that is still dissipating over 60 watts without a heatsink. Let us mount it in the 5cm x 10cm x 20cm cabinet we found above. Transposing the formula we have:
We know we need a heatsink of 1.13°C/W so putting this into the formula we get 1957.9 square cm of heatsink required. We already have 1000 in the case, so we must provide the other 960 as fins on the back. 10 fins of 96 square cm (48 per side) will do the job. Each fin can therefore be 5cm x 10cm. Forget the backplane, that is part of the case. 5cm x 10cm is still quite big, but the case is only 5cm deep, so it is reasonable. Including heatsink, the case will now be 5cm x 20cm x 20cm. Note that only the fins were considered.
In reality, a 5cm tall case would be a little small so the case would not be able to accomodate the mains transformer, even if you used three 6.3v 3A fillament transformers. Increasing the height of the case by 1cm would reduce the size of the fins by 2cm; 6cm x 8cm (as opposed to 5cm x 10cm earlier).
Almost all devices that we need to mount on a heatsink have a metalic surface from which the heat is to be conducted. This surface is usually electrically connected to one of the terminals of the device. We therefore need to place an insulator between the device and the heat sink. In the case of transistors this is in the form of a mica washer or disk. The disk is lubricated with a heat conductive oil to help the heat to be transfered. The oil and the washer should never be omitted if the device is to be effectively protected.
Since aluminium is not the worlds best conductor of heat, I usually lay a strip of copper the full length of the heatsink. The device is mounted on the copper and the copper distributes the heat to the rest of the heatsink.
There are such things a temperature gradients, junction to case thermal resistance and many other parameters that could be investigated. If the subject of heatsinks were an English country mansion set in 110 acres of garden maintained by over 50 servants, I have just shown you the doorbell, or just enough to get you in the door (and out of trouble).
So, that is about all I want to say about heatsinking. Do have fun, but don't get your fingers burned! Best regards Harry, Lunda, Sweden.