Mosfet Power calculation for maximum load

Hi

Am using 80NF10 mosfet its parameters are
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Vdss=100V
Ids=80A
Rds[on]=0.015 Ω
Pd=300W
Rthj-case=0.5°C/W
Rthj-amb=62.5°C/W
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i want to know for single Mosfet how can i calculate the maximum power it can be load [see picture].




i think Maximum power thorugh Mosfet =Vdss*Ids =8000W. is it currect ?

When the transistor is conducting 80A because of its resistance, it will be dissipating ~100W, so it needs to be put on a heat sink.
Frank

Totaly it can disspate 300W , so even its not reached that value [80*80*Rds[on]=96W ] , So after what power dissipation we need to keep heatsink? , means >96W or >300w ?

The 300W figure relates to 300 X Rj-c, i.e. you can't get the heat away from the junction fast enough, As Rj-c is .5, this says that at 300W, the junction is at 150 Degs C when the case is at 0 degs C. So I guess if you want to refrigerate the case, you can go over 300 W, but its cheaper to use a second device in parallel.
The point is always the junction temperature. You really want it to be as low as possible to give the device a long life. So look up the max junc. temp and work backwards for your expected power to find the case temperature, then what the max ambient temperature is going to be, then you can find a heat sink that has a thermal resistance of less then (Tc - Tamb)/power. Leave a margin for dirt and dust. . .
Frank

so i think , without heat sink we can use 300W as maximum load , then if i want to load 7500W , then can u just give me a currspdg heat sink spec, [pls add simple calculation with junction- case-heatsink-ambient ] . i know practicaly it is not possible... any way can u just me caln..?

No, no, NO without a heatsink the thermal resistance from the junction to ambient temperature is 62.5 degs/watt, so with 300 W the junction temperature will be 300 X 62.5 above ambient = 1675 degrees above the ambient (room) temperature.
Given an ambient temperature of 40 degs and a max junction temp of 150 degs, this gives about 110 degs to get rid off at a power of 100 W. So the temperature drop from junction to case is 100 X .5 = 50 degs, so this gives us 60 degs to loose in the heat sink at 100 W, so you need .6 deg/W heat sink, such as:- https://uk.farnell.com/h-s-marston/cp464-080-080-04-s-2/heat-sink-multifin-0-4-c-w/dp/936911
Frank