0
\$\begingroup\$

I refer to a 2N3055 power transistor, the spec. sheet gives the following:

Max. current 15A, Max. Voltage 50V, Power rating is 115 Watts.

  • What is the power rating calculated on?
  • If it is E*I then the power would be 750 Watts, this is of course incorrect, so how is it calculated please.?
\$\endgroup\$
0
\$\begingroup\$

You can have maximum current through the transistor and you can have maximum voltage across it but you can't have both at the same time. This is specified in the "safe operating area" power curve.

enter image description here

Figure 1. The safe operating area of a BDV66C transistor. Source: Wikipedia Safe operating area.

In Figure 1 we can see the following:

  1. 16 A at about 12 V. Pmax = 192 W.
  2. 5 A at 50 V. Pmax = 250 W.
  3. 0.2 A at 120 V. Pmax = 30 W.

Taking the two maxima from the graph one might have expected that the device could handle 16 A at 120 V = 1,920 W which is not the case.

SOA is defined as the voltage and current conditions over which the device can be expected to operate without self-damage. The SOA figures are, of course, subject to other constraints such as maximum junction temperature and so the power dissipated as heat, even when inside the SOA, must be conducted away to maintain a safe operating temperature. There is more information in the linked wiki article.

\$\endgroup\$
  • \$\begingroup\$ Thank you all that took the trouble of answering. This has served to bring me up to date on this subject. It's a pity I couldn't give a best answer to all replies. This has had me puzzled for more years than I care to tell. Again I thank you all for your answers \$\endgroup\$ – Rodney Jackson Feb 17 '19 at 9:00
1
\$\begingroup\$

The power rating is based on the thermal design of the TO-3 transistor package, how easy or hard it is to extract heat from the transistor die inside. The MJE3055 is basically the same die in a smaller TO-220 package, and is rated for only 75 W. The thermal path from the die to the mounting surface is better than the old TO-3 package, but that mounting surface is significantly smaller.

\$\endgroup\$
0
\$\begingroup\$

Total Power Dissipation @ Tc = 25°C Derate PDmax=115W Above 25°C by 0.657 W/°C for case temp, Tc so Rjc= 1/0.657= 1.52'C/W

This implies the max junction temp, Tj = 175'C+25'C=200'C but this accelerates failure rate significantly, so a good design has Tj<< 100'C.

That assumes an infinite heatsink which is not possible.

So to compute Tc , you need Tja heatsink rating like 0.1 'C/W for a super CPU cooler with Fan or 1'C/W for a large passive heatsink then add internal ambient rise above room temp (or outside temp)

Then Temp rise of junction is Rja*W= Rjc+Rca(total) * Watts = Tj rise

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.