What is heatsink?
Actually the purpose of every heatsink is iden cal, which is to prevent accumulation of heat in a certain location, not to allow any temperature increase and to instantly disseminate the accumulated heat to the surrounding. The semiconductor parts of the silisium based materials used in electronics are deteriorated in 200˚C. Therefore, in order to instantly expel the accumulated heat on the chip, the chip part of the material is generally adhered on an aluminum plate using an adhesive with a high level of heat permeability. Then it is appropriately covered with and aluminum cover or plastic. However such material is not sufficient for expelling or disseminating to the outside in respect of the heat accumulated on the cases of the power elements such as transistor diode, triac or alike. It is required that these should be placed on a further heatsink.
Before picking a heatsink, it is required that the heat flow on the semi-conductor should be understood and the Thermal Circuit (Thermal Resistance) and the required parameters should be analyzed within this framework.
How to choose a heatsink?
Figure 1 indicates the heat relationship of the semi-conductor and heatsink with the surrounding. The definitions of the expressions seen on the Figure and some parameters required by us are as follows:
Q : Power
Tj : Maximum Junc on Temperature
Tc : Maximum Case Temperature
Ts : Maximum Cooler Temperature
Rsa : Thermal Resistance between heatsink-surrounding
Rcs : Thermal Resistance between case-cooler
Rjs : Thermal Resistance between cooler junction-case
TAmb: Temperature of surrounding
R=ΔT/Q
Wherein;
ΔT : is the temperature difference between two points,
R : Thermal Resistance (°C/W)
When all the thermal resistance seen on the above resistances are summed up and if we call them as Rja (Junc on-Surrounding Thermal Resistance);
Rja = Rjc+Rcs+Rsa therefore;
Tja = Tj-TAmb Tja = Q.Rja
Which results to;
Namely the highest value of the chip of the transistor given in the catalog should not be
exceeded. In this case, the equation should be as follows:
Rja = (Tj-TAmb)/Q
The two points to be noted in here is that; Tj = Tjmax
In case an insulator is used between the transistor body and the heatsink, for instance a flexglass pellet, then Rcs value changes between 0,2 ˚C/W and 0,5 ˚C/W. At the same me, the heatsinks made of various substances indicate different features based on the material they are made of. Aluminum is the cheapest solution as compared to the others.
The last phase is as follows;
The area of the heatsink is obtained using the above features. The calculated area changes based on its type of use namely based on being vertical or horizontal. Vertical usage areas are less than horizontal usage areas. The reason for this is the fact that air flow is better in vertical heatsinks.
An example calculation is as follows;
On an output circuit of the transistor;
Rjs = 1°C/W
Rcs = 0,5 "C/W (Insulator Mica Pellet)
Rsa = 2,5 “C/W (for Ver cal Posi ons)
Tj(max) = 175°C
Tamb = 30°C
In case the above equations are valid,
what should be the maximum power
Ptot in the surrounding of output?
Ptot≤(T(j(max))-TAmb)Rja
Rja = R(jc(msx))+Rcs+Rsa
Rja = 4 “C/W
Ptot = (175-30)/4
P(tot) = 36,25
Next, using the following table, the area of a surface of the cooler can be found:
The area of a surface corresponding to 2,5 ˚C/W' becomes;
60 inch2 or 387 cm2 in perpendicular position;
70 inch2 or 452 cm2 in vertical position;
Since the above values are the area of a surface of the heatsink, the total area becomes;
120 inch2 or 774 cm2 in perpendicular position
140 inch2 or 903 cm2 in perpendicular position
The heatsink obtained in here is an aluminum plate in the form of a heatsink plate.
This can be shaped in different forms and can be formed in compliance with the circuit.
