How to calculate for selecting a heat sink of a semiconductor device (2)

A calculation example is shown with the following products.

＜Target＞
Target temperature is less than 90 deg.C in consideration of reliability.

＜Condition＞
Product: TTD1409B
Ambient temperature: 60 [deg.C]
Heat sink: No use
Power dissipation: 1 [W]
R_th(j-a): 62.5 [deg.C/W] refer to datasheet (= (150 – 25 ) [deg.C] / 2 [W])
R_th(j-c): 5 [deg.C/W] refer to datasheet = (150 – 25 ) [deg.C] / 25 [W])
(Absolute maximum rating Tj(max) = 150 [deg.C], P_C= 2 [W] at Ta = 25 [deg.C], P_C= 25 [W] at Tc = 25 [deg.C])
This situation: T_j = 122.5 [deg.C](= 62.5 [deg.C/W] × 1 [W] + 60 [deg.C])

＜Calculation＞
When the contact resistance between a semiconductor device and a heat sink is disregarded, “Thermal resistance, Junction to Case R_th(j-c)” and “Thermal resistance, Junction to heat sink R_th(j-cf)” are considered to be the same.
R_th(cf-a) = (( 90 – 60 ) / 1 ) – 5 [ deg.C / W ] = 25 [ deg.C / W ]
To use the heat sink with a thermal resistance value lower than 25 deg.C / W, you can obtain the result to be expected.

When using the above-mentioned technical content such as product data, figures, and tables, please judge whether the application is appropriate at your own responsibility based on sufficient evaluation of the product alone and of your whole system.