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skibum 9th May 2012 03:12 PM

Maximum DC power for a Transistor before heatsink necesssary?
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I have a NPN transistor that has 60mA of emitter current and Vce = 3.18VDC for a power of 0.18 watts.

The only specifications I can find on the data sheet are:

PD Total Device Dissipation(TC=25C) = 50 Watts

Derate above 25C = 0.4W/C

RθJC Thermal Resistance, Junction to Case = 2.5 C/W

It is a T-220 case.

I have attached a Power derating curve that is on the data sheet also.

For this example I feel that I am OK with out a heatsink because when I press my finger against the transistor it is just barely warm.

I just want to know how to do this for in the future.

How can I calculate at what power should I be using a heatsink?

Loudthud 9th May 2012 06:25 PM

You need to know the thermal resistance junction to ambient. For a TO-220 it's probably 50 to 60 degrees C per Watt. If it's not on the manufacturers' spec sheet, look for other parts made by the same manufacturer, or a similar part by another company. Then just figure the junction temperature based on power dissipation.

nigel pearson 9th May 2012 06:47 PM

A rule of thumb I use is 1.5 watts for TO220 . 1 watt for TO126 and 2 W for TO3P . Seems to work if there is ventilation . The printed circuit can dissipate heat if plenty of copper is connected with the transistor . I ran some BD 135 in pairs at 1.13 Watts for days with no problems . Another rule of thumb . If you can not hold on to them for 5 seconds they are too hot . Small clip on heat sinks cost very little money .

AndrewT 10th May 2012 01:07 PM

I have pulled up an ONsemi MJE15032 because I knew it was rated at 50W when Tc=25C
The datasheet shows 2.0W when Ta=25C
Sometimes the manufacturer shows two plots on the P vs Tc/Ta graph. ONsemi show both.
Again some manufacturers show Rth j-c and Rth j-a for the two operating conditions of infinite heatsink or no heatsink.

All these forms can be used to determine the same de-rated SOAR.

50W @ Tc=25C gives Rth j-c = {150-Tc} / P = 125 / 50 = 2.5C/W for the infinite heatsink.
2.0W @ Ta=25C gives Rth j-a = {150-Ta} / P = 125 / 2 = 62.5C/W for the no heatsink.

You have 180mW and no sink.
Tj = {Rth j-a * P} + Tambient = {62.5 *0.18} + 40C = 51C. 51<<100 so OK.

You can re-arrange the formula to give Pmax = {Tjlimit - Tambient} / Rth j-a.
If you set your Tjlimit to 100C and Tambient = 50C then,
Pmax = {100-50} / 62.5 = 0.8W compared to absolute maximum of 2W when Ta= 25C

nigel pearson 10th May 2012 01:36 PM

I remember doing all those calculations for a design once and getting nowhere . I didn't realize that the heat transfer to the heat sink is not ideal and doesn't happen in real life . I say when real power comes in double your estimate . We did this at work recently . The guys I work with are no idiots . The design was to be air cooled . Now we have a small fan as it didn't quite work out . This cost the project months because the boss didn't want a fan . It is the quietest fan we could find and comes on when needed . Some customers in Asia turn of the air con . It blows our devices up if 45 C in the room .

skibum 10th May 2012 02:07 PM

Thanks Andrew

I will still do the finger test but it's nice to have an equation.

nigel pearson 10th May 2012 02:11 PM

I know if we get into trouble we have Andrew . I have many at work and I am lazy . I dig a hole and they get me out . I am the boss you know . It was me who didn't want fans .

Boss of projects that is . Not money boss thank goodness .

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