CSD95495QVM: Thermal information

Daniel,

Thanks for posting.

See below for the thermal table for this part:

Daniel,

I am not sure what you mean by Pc - do you mean how much power the device can dissipate? That would just be the Tmax - Tj divided by the thermal impedance. 

The RthJA thermal impedance is just as much if not more so a function of the board and the thermal environment the part is being used on than it is the package or die inside. For instance, on 1x1" 2oz Cu, this RthJA would be around ~40deg C / W. For a min Cu board, the RthJA could be as high as 100 - 125deg C / W. 

The max case temperature is the same as the max junction temperature if both are in equilibrium. However, if you are at this case temperature, you can not dissipate any power in the junction as it will heat up beyond the 150 abs max. 

Does that answer your question? 

Daniel,
Sorry for the delay.

The amount of power you are dissipating in your application depends on many conditions - in the above example, they calculate using the input current, supply voltage, output voltage and output current.

There are other ways to estimate the power dissipation using just the output parameters, but if you want to do that, we need to take the conversation offline. I will send you a friend request just in case so we can have a private conversation.

The max power (Pd) the customer can dissipate in the package is determined by the following : Pd = (Tj,max-Ta)/RthJA. As stated above, the RthJA is going to be specific to you customer's application. It could be as low as 40deg C / W or as high as 100deg C / W depending on the thermal environement of the end equipment.

If I assume the Ta is 25deg, we would say Pd could vary from (150-125)/125 = 1W to (150-125)/40 = 3.1W.