[SOLVED] How should I calculate the PCB dissipating area?

How to calculate thermal dissipation size of PCB ground area?

LM22676 (https://www.ti.com/lit/ds/symlink/lm22676.pdf)
It says " The value of θJA for the TO-263THIN(TJ) package of 22°C/W is valid if package is mounted to 1 square inch of copper. The θJA value can range from 20 to 30°C/W depending on the amount of PCB copper dedicated to heat transfer. See application note AN-1797 for more information. "

I need θJA to be set at 80°C/W, I don't have too much copper area on the PCB board. How can I calculate the PCB dissipating area here?

AN-1797 (**broken link removed**)
It tells me nothing more than the θJA is 22°C/W under "Test Condition: No Air Flow, JEDEC 4Layer Test Board", which has thermal vias and full ground area on the bottom layer.

AN-2020 (**broken link removed**)
It tells you two formulas, if I use the second formula, A = 77.5 / (ThetaJA - ThetaJC), the copper area is 0.99 square inch when ThetaJA is 80°C/W, which is too large because 1 square inch of copper can get θJA of 22°C/W according to LM22676 datasheet.
(the ThetaJC of TO-PMOD-7 or TO-263 is 1.9°C/W according to AN-2020)

So how can I calculate the PCB dissipating area here?

It seems that no one can answer this question. I don't want to ask a difficult question. I just want to know what everybody does about thermal consideration.

Do you calculate the ThetaJA in your design? Do you calculate the PCB dissipating area?

I only want to know what a schematic designer does when he need to make his board work properly at 60 ℃

Thanks.

I think the calculations are fairly complicated; I've never been able to find a definitive answer to your question. I believe the information you see in data sheets is actually determined by experiment or expensive software simulation.