# Heatsink calculation with limited data

• posted

Hi,

I am using a modified version of the lighting controller circuit at...

=2E..and I'm not sure what heatsink I need for the PNP Darlington power transistor. As the specified model is no longer available, I am using a TIP125. (An aside: is this a suitable replacement?) The datasheet is available at...
=2E..but the only information it gives regarding thermal characteristics is the following:

Thermal resistance, junction-to-case, R(theta)JC =3D 1.92=B0C/W Thermal resistance, junction-to-ambient, R(theta)JA =3D 62.5=B0C/W Operating and storage junction, temperature range, Tj, Tstg:

-65--150=B0C

With regards to power dissipation, all it says is:

TC =3D 25=B0C =3D> PD =3D 65W derating at 0.52W/=B0C above 25=B0C TA =3D 25=B0C =3D> PD =3D 2.0W derating at 0.016W/=B0C above 25=B0C

Given this, how do I calculate the heatsink I need?

(The maximum current, ideally, is 1.67A, so presumably it's worth working to a maximum of at least 2A, and the maximum voltage across the transistor will be about 16V.)

• posted

I think the TA curve is for no heat sink, so it's the TCase curve that applies.

That's the voltage when turned off. For a switching circuit, you need the voltage when turned on and the current. You might need to consider the switching times, too. The circuit is running at 10 kHz, or 100 us period, and the switching times may be about 1 us each way. That would be 2% of the time in a much higher power dissipation area.* Anyway, the data sheet says the saturation voltage (Vce when turned fully on) is a maximum of 2 V at a collector current of 3 A; they don't give a figure for 2 A, so a safe figure for the dissipation would be 6 W. If you derate 65 W to 6 W, you have 59 W / (0.52 C/W), or a maximum case temperature of (25 C + 113.5 C) =3D 133.5 C. Then decide what the maximum ambient is - pretend it's 100 C; then you have

23.5 degrees C available to you to get rid of 6 W through the heat sink, so its maximum thermal resistance, case to ambient, is 23.5 C / 6 W, or about 3.9 degrees C / W. Then you need to look at the specs for heat sinks, to see what you need, and whether you need thermal compound, or forced air, or not, to get to that thermal resistance. You can visualize all this as a string of thermal resistors from one temperature to another, like voltage drops, with the power like a current flowing through all the resistors.

-- John (Offering the usual money-back guarantee on all guesses).

• Suppose all the switching time was spent at 16 V and 2 A, or 32 W. That times 2% is 0.64 W average additional, so maybe it's not so much. It also seems safe when looking at the safe operating area curve in the data sheet.
• posted

tp://

25

Sorry for my arithmetic above; if 100 C ambient, and 133.5 C max case temp, you have 33.5 C to play with; 33.5 / 6 =3D 5.6 C / W for the heat sink.

-- John

• posted