Electronic – BJT power dissipation – Which value to use? (Ta vs Tc)


I'm using a TIP120 (Darlington pair BJT) brought to saturation for a project of mine. I have \$V_{CE(sat)}=1V\$, \$I_C=2A\$, \$V_{BE}=2.5V\$ and \$I_B=0.005\$A, which give me a total power dissipation of:

\$P_D=V_{CE(sat)}*I_C+V_{BE}*I_B \approx 2W\$

When I look up the component datasheet to check the absolute maximum ratings, there are two values given for power dissipation: one at 65W (@ \$T_C=25°C\$) and one at 2W (@ \$T_A=25°C\$), as seen on the image below:

Absolute Maximum Rating section of TIP120 datasheet

So my question is: what is the difference between the two values? What is the difference between \$T_A\$ and \$T_C\$?

Sorry if this is a common question, I've searched everywhere to try and answer that question, but search engines are not very helpful when I want to know the purpose of parameters found in electronic datasheets (if there exists a glossary for the most common parameters found in datasheets somewhere, and someone has a link, I'd be very happy to use it!).

I suspect that I should use the first value for some reason, but given that my calculated \$P_D\$ value is pratically the same as the second one, I don't want to take any chance and destroy my future setup, making all that magic smoke escape…


Best Answer

The ON datasheet is rather confusing (or rather doesn't explain its notations). The 65W refers to the [max] power dissipation if you manage to keep the case at 25C. The 2W refers to an ambient temp of 25C, but no restriction on the case temp. This is a bit more clear from the Bourns datasheet of their similar product.

enter image description here

What this means in practice is that 65W is the max you can hope for with an ideal [possibly very large] heatsink.

Both of these data are actually a rather convoluted way of saying the same thing, namely that the max junction temperature allowed is 150C. This can be verified using the following data:

enter image description here

  • 1.92*65 + 25 = 124.8 + 25 = ~ 150C
  • 62.5*2 + 25 = 125 + 25 = 150C.

Which is actually given as such in the datasheet:

enter image description here

Now for practical purposes, I would suggest using a small heatsink rather than betting you won't fry it at exactly the dissipation limit for use without one.

If you want to calculate the temp rise with a heatsink, say one which gives 13C/W, then you add the heatsink's thermal resistance to that of the case (1.92C/W) and the interface material, let's say 1C/W, which would give you about 16C/W total resistance. For 2W that translates into 32C temp rise over ambient, so at 25C you'd have 57C. That's pretty decent for not frying yourself when accidentally touching it.