/\/\ thermal calculations problem (solved)

[3roomlab]

hi folks, so i have been mucking around components/PCB thermal thingies ...

say this situation :

junction --> plastic case --> ambient

if ambient is 30^oC
if R-ja = 50^oC/w
and the plastic case measures in at 50^oC

to get to actual T-junction, we find the watts going thru the plastic case( = 20/50 = 0.4watts)
and the junction temp is = (0.4 x Rja) + Tcase ? = 70^oC

am i right? i have the feeling that i am missing something ...
does R-ja actually means the resistance going thru the plastic package? or the entire item ?

[kxenos]

Rja is specified as the path of least resistance. If there is a tab in the chip then it's the resistance from junction to that tab. Else it's from junction to case

[3roomlab]

yes it is, so the tab itself R-jc and the R-ja are infact parrallel "resistors" am i right? they both do dissipate simultaneously once there is a higher heat potential generated at junction?

[kxenos]

No, they're not parallel because you can either have the hole case area to dissipate heat to ambient so you use Rja or you have a heatsink so you use Rjc but not both at the same time.
You could write that Pd = (Tj - Tc) / Rjc
And also that Tj = (Pd x Rja) + Ta
Therefore Tj = ((Ta x Rjc) - (Tc x Rja)) / (Rjc - Rja)
But I don't know if this is valid or not because Rjc and Rja are used in different use cases aka with and without heatsink respectively.

[IanB]

hi folks, so i have been mucking around components/PCB thermal thingies ...

say this situation :

junction --> plastic case --> ambient

if ambient is 30^oC
if R-ja = 50^oC/w
and the plastic case measures in at 50^oC

to get to actual T-junction, we find the watts going thru the plastic case( = 20/50 = 0.4watts)
and the junction temp is = (0.4 x Rja) + Tcase ? = 70^oC

am i right? i have the feeling that i am missing something ...
does R-ja actually means the resistance going thru the plastic package? or the entire item ?

You don't really have enough information to say anything here.

Rja is from junction to ambient (from junction to surrounding air). It says nothing about the case temperature, so any measurement of the case temperature must be thrown away. Now you are left with the air temperature alone. With only the air temperature you can say nothing at all about the junction temperature. It remains a complete unknown.

On the other hand, suppose you also know the power dissipated by the silicon (= voltage drop x current), then you have two pieces of information, and now you can do a calculation. You have:

  Power = (Junction temperature - Ambient temperature) / Rja

Therefore:

  Junction temperature = (Power x Rja) + Ambient temperature

Any time you have one equation in four unknowns you need to specify three of the unknowns in order to calculate the fourth.

[3roomlab]

ok i get it now thanks heaps