Heatsink and BD140 maths

[jmx66]

Hi all,

I only want someone check, and explain the right way to get the result.

Aim: BD140 heatsink in an enclosure, with fan cooling.

Total power dissipation for T Ambiant 25° 1,25° / W
Derate above 25°C 10mW / °C

Total power dissipation up to Tc=25°C 12,5 W
Derate above 25 °C 100 mW / °C

Junction temperature Maxi 150 °C

Thermal resistance R junction to case 10 °C / W

R junction to ambiant 100 °C / W


My choices:

T ambiant: 40 °C Junction temperature: 110 °C


Some elementary maths:

- First case: P max = (Tj - Ta) / Rth

1,25 - ( 15 x 10 x 0,001) = (110 - 40 )/ Rth

Rth= 70 / 1,10 = 63 °C/W

Rth= Rth junction to case + Rth heatsink + Rth heatsink to case ( will use of paste )

63 = 10 + Rth heatsink + 0,5 So Rth for this heatsink must be 53 maxi ? Is it right, or mistake of my own?


- Second case: If i use Pmax = 12,5 Watts my Rth result is about 6,36.....

So 6,36 = 10 + Rth heatsink + 0,5
Rth heatsink = -4 I don't think , it's a good result!!!!!!!!!!!!!


Would you kind enough to tell me, why there is 2 power - 1,25 and 12,5 Watt -in this datasheet?????

Thanks a lot.

jm

 

[FvM]

Isn't it obvious that a maximum power disspation for 25°C case temperature is mainly a theoretical value (for calculation purposes) and the actually achieveable power dissipation will be considerably smaller?

Instead of calculating a heatsink for Pmax = 12.5 W (ending up in negative Rth numbers) you'll want to calculate the actual total Rth and determine the achievable power disspation. If it's too low, choose a larger transistor with lower internal Rth.

 

[chuckey]

I think the second power/temperature rating (12.5W @25 degC) refers to the device being fixed to an infinite heat sink. As you have not given us the power input to the device, or the thermal resistance of your heat sink I am not sure what to calculate.
A more typical question might be :- if the device was handling 5W what is the minimum thermal resistance of the heat sink for Tamb = 40 and Tj max of 40?
Frank

 

[godfreyl]

Would you kind enough to tell me, why there is 2 power - 1,25 and 12,5 Watt -in this datasheet?????

1.25W is the maximum power with no heatsink, and ambient air temperature = 25°C.

12.5W is the maximum power with the case kept at 25°C.
That information isn't very useful on it's own, but it allows us to quickly compare the power ratings of different transistors.

Let's do a proper heatsink calculation:

Let's say you want to dissipate 5W.
According to the datasheet, Rth(junction-case) = 10°C/W.
It also says max junction temperature = 150°C, but you said you want to limit junction temperature = 110°C (not a bad idea), so we'll use that number instead.

Let's also allow for ambient temperature up to 40°C, as in your example, and assume Rth(case-heatsink) = 0.5°C/W for your thermal paste and insulating washer.

Now, as you pointed out:
Junction temp - ambient temp = Rth(tot) * power dissipated
i.e.: Rth(tot) = (junction temp - ambient temp) / power dissipated
So: Rth(tot) = (110 - 40) / 5 = 14°C/W

And, as you also pointed out: Rth(tot) = Rth(junction-case) + Rth(case-heatsink) + Rth(heatsink)
i.e.: Rth(heatsink) = Rth(tot) - Rth(junction-case) - Rth(case-heatsink)
So: Rth(heatsink) 14.5 - 10 - 0.5 = 4°C/W

So you'd need a heatsink rated at 4°C/W (or lower).

Another example:
Let's say you have a heatsink rated at 10°C/W and want to know how much power you can safely dissipate.

In this case, Rth(tot) = 10 +10 + 0.5 = 20.5°C/W
Maximum power = (110° - 40°) / 20.5°C/W = 3.4W


Hope that helps :smile:

 

[jmx66]

Hi,

Thanks a lot for your advices.

jm