Yes, I had an error.
I originally figured that for 3 MOSFETs in parallel, I would dissipate 113W each, and that was too much, so I did proportions from there to find the wattage dissipation for 4 FETs (84W), 5 FETs(67W), 6 FETS (56W). But it doesn't work proportionally like that. You need to do all the calculations for each combination of FETs, which comes out to 4FETs (63W), 5 FETs (40W), and 6 FETs (28W).
My math to find the wattage dissipation is as follows (I have made an excel spreadsheet that does all the work for me now):
Rds(on) max is 4.5mΩ (with worst case scenario) @ 25c junction temp. The temp VS. Rds(on) graph shows that at 120C, the Rds(on) will be multiplied by 1.75, which comes out to
7.87mΩ
Current limiting will be set at
360A
For 3 paralleled MOSFETs, the voltage drop across the MOSFET is going to be 120A (Package limit, worst case scenario) * .00787 = .94V. If passing 120A through it and it is dropping .77V, then P=IE, .77V * 120A =
113W (plus switching losses, negligible due to extremely fast switching times, calculated slightly above 1W)
For 6 paralleled MOSFETs, the voltage drop across the MOSFET is going to be 60A * .00787 = .47V. If passing 60A through it and it is dropping .47V, then P=IE, .47V * 60A =
28W (plus switching losses, negligible due to extremely fast switching times, calculated <1W)
Calculation for Switching loss:
When the MOSFET is in the middle of switching it is passing half of the current limit and half of the applied voltage is dropped across it. So, if it has 24V across it and 60A through it, then it is dissipating 1440W.
t = Q / A.
Qg = 210nC, Amp output from FAN3121 MOSFET driver @ 18V & 1.5ohm gate resistor is 11.4A.
210nC * 11.4A = 18.4nS
40,000 (2khz switching time) X 18.4nS = 737µS (the total time the MOSFET will be dissipating 1440W in one second). Since 737µS corresponds to about .000737% of 1 second, multiply 1440W by .0737 and get
1.06W
Calculation for heat sink:
Where
Tj = Maximum semiconductor junction temperature
θjc = Thermal resistance, junction to case.
Tc = Case temperature
θcs = Thermal resistance, case to heatsink (see Hot Tips)
Ts = Heatsink temperature
θsa = Thermal resistance, heatsink to ambient.
Ta = Ambient temperature
values:
Tj = 120C
Ta = 50C
Q = 56W
θjc = .402
θcs = .10 (estimate, based on seeing 1C/W when no thermal paste used, and seeing .09 when good thermal paste used)
When I was figuring 113W per MOSFET would get I get 0.137C/W
When I was figuring 56W per MOSFET would get I get 0.768C/W
Now that I am figuring 28W per MOSFET would get I get
1.99C/W, which is MUCH more manageable.