For a laugh when I get back Friday I will put a 220 on the heat sink and see how much current I put through it and still keep it at 25C.
Afraid the answer is no current, in theory that is. The thermal budget illustrates that:
MOSFET Maximum Junction Temperature (170DegC or 150 degC) → Thermal Resistance Junction to Case (0.5 degCW) → Thermal Resistance Case to Insulating Washer & Insulating Washer to Heatsink (0.75 DegCW) →Th Resistance Heatsink to Ambient Air (2 degCW)→ Ambient Air Temperature (50 Deg C).
Taking the typical thermal resistance values in brackets, the total thermal resistance from junction to ambient is: 0.5 + 0.75 + 2 = 3.25 Degrees C W
Taking the maximum junction temperature to be 175 Deg C, the differential temperature is 175- 50= 125 Deg C
From this it follows that the maximum allowable power dissipation of the MOSFET is 125/3.25 =
38.46W
Suppose you had a 100A, 120W MOSFET and in your circuit you had 24V between its drain and source, in an amplifier for example, the maximum current you could allow is 38.46W/20V= 1.923 Amps.
On the other hand suppose you were using the MOSFET as a switch:
Say the resistance between drain and source of the MOSFET were 10mili Ohms, at 25 Deg C junction temperature. At a guess the resistance would increase to 50m Ohms at a junction temperature of 175 Deg C .
Thus, the maximum safe drain current would be found from, I * I * 50 mili Ohms = 38.4W. Thus I =
27.71A
It would make no difference how high the maximum current or power is shown on the MOSFET data sheet, it could only handle 38.4W and as a result it can only ever switch 27.71A.