How to Size Heat Sinks for Semiconductors

First, lets start with the definition of terms:

qja - Thermal resistance required between the semiconductor junction and the ambient air
qjc - Thermal resistance between the semiconductor junction and its case
qcs - Thermal resistance of the heat sink compound
qsa - Thermal resistance the heat-sink
Ta - Maximum ambient temperature in ºC
Tc - Maximum case temperature in ºC
Tj - Maximum junction temperature in ºC
P - Maximum power dissipated in Watts

Lets take a part with the following parameters:

  1. There is about 50 linear ft/min airflow, which is insignificant enough to be ignored.
  2. The ambient temperature (Ta) is 40°C.
  3. The part data sheet specifies a maximum case temperature (Tc)of 110°C.
  4. The part data sheet specifies a maximum junction temperature (Tj) of 150°C.
Therefore, the heat sink must maintain the junction temperature at or below the maximum with 50 ft/min of airflow or less.

After the designer determines the heat sink requirements, the next step is to calculate the thermal resistance required between the semiconductor junction and the ambient air qja with the expression:

(Tj - Ta)
 (150 - 40)
qja
 =
 ---------
 =
 -----------
 =
 7.1ºC/W
P
 15.4
where Tj = maximum junction temperature in ºC; Ta = maximum ambient temperature in ºC; and P = maximum power dissipated in Watts.

The maximum junction temperature and power dissipated are found in the voltage-regulator data sheet, and the maximum ambient temperature is measured at the motherboard. The thermal resistance for the entire assembly is the sum of resistances of the regulator, heat sink, and the thermal interfaces between them. The resistance of the part is the thermal resistance between the semiconductor junction and its case, qjc, found by,

(Tj - Tc)
 (150 - 110)
qjc
 =
 ----------
 =
 ------------
 =
 2.6ºC/W
P
 15.4
where Tc is the maximum case temperature in ºC.

A high thermal-conductive material placed between the regulator and the heat sink improves mechanical contact at the interface and lowers the total thermal resistance. The thermal resistance of the material, qcs, is rated at 0.1°C/W.

After determining the required resistance for the entire assembly and the resistances of the regulator and thermal material, the heat-sink thermal resistance qsa may be solved by:

qja
 =
 qjc
 +
 qsa
 +
 qcs
  
therefore,
qsa
 =
 qja
 -
 qjc
 -
 qcs
=
 7.1
 -
 2.6
 -
 0.1
 =
 4.4ºC/W


The next step is to select a heat sink based on the known thermal resistance and available dimensions. Designers need not be stuck with a catalog part that only marginally fits their application. Sink manufacturers often modify standard heat-sink designs to provide the characteristics exactly needed.

Standard TO-220 heat sink selection

PART NUMBER
 THERMAL
RESISTANCE,
75ºC RISE ABOVE
AMBIENT
 DIMENSIONS, (IN.)
W — D — H
5297
 5.5ºC/W
 1.65 _ 1.00 _ 1.00
5298
 3.7ºC/W
 1.65 _ 1.00 _ 1.50
5299
 3.4ºC/W
 1.65 _ 1.00 _ 2.00