mosfetpower electronicsswitch-mode-power-supply
I decided to use IRFP460 in a two-switch forward converter. This Mosfet can dissipate 280 watts. Can I get 1000 Watt from my converter?
As a general rule how much power a Mosfet can transfer with respect to its maximum dissipation loss?
I concluded that the mosfet can dissipate a derated loss power at a specified temparature if thermal resistance from junction to ambient be a required value; eg 2 c/w for 20 watts at 100 deg. Celcuis of case temperature.
Is it possible?
You have to change the design of your heat sink or increase the capacity of the fan. Using the dimensions provided and the dimensions of a IRFP460, Tj = 163C in a 45C ambient based on calculations using the heat sink calculator (http://www.heatsinkcalculator.com). The Rcs (resistance case to heat sink) was estimated using a film of thermal grease of 0.1mm thick with a thermal conductivity of 1.35W/(m.K) with the contact area equal to the width X length of the IRFP460. This gives a Rcs value of 0.24 C/W.
Keep in mind that the maximum flow rating of a fan is based on no obstructions being in front of the fan. Once the fan is place behind the heat sink the flow provided will be less than the maximum flow. The exact value is dependent on the design of the heat sink and the fan flow curve.
I understand that they are at different case & junction temperature but what do they actually mean & how to use this data?
Using the example of case temperature being controlled/limited to 25 degC, the spec says you can dissipate 71 watts. Now look at maximum junction to case thermal resistance - it says 2.1 degC/watt.
So, for 71 watts I would expect the junction to rise to 149 degC above the case. This ties nicely with the case being fixed at 25 degC because 25 + 149 = 174 degC - ok a slight discrepency with the stated max junction temperature.
8 watts would raise the junction by nearly 17 degC above the case. However using the "Maximum Junction-to-Ambient Steady-State" value, 8 watts will raise the junction up to 8x50 degC because now, the case isn't "held" at some value. This is why you need a heatsink.
If you went for a heatsink that was rated at 3 degC per watt (case physical connection to ambient), you use that figue (3 degC/watt) plus junction to case (2.1 degC/watt) to get a combined number of 5.1 degC/watt. Thus, for a junction temperature of not more that 175 degC in a controlled ambient of 25 degC, the temperature rise of 150 degC divided by 5.1 degC gives you the max continuous power i.e. 29.4 watts.
A heatsink of 16.65 degC/watt would be on the cusp of tolerating an 8 watt internal dissipation because 16.65 (external) + 2.1 (junction-to-case) = 18.75 and 150/8 = 18.75.
Does this help? Maybe these simplified pictures will: -
Best Answer
The 280 watt figure is really optimistic .you would get this on an infinite heatsink that was held by some means at 25celcius.Now the fet would be running at Tj max of 150 celcius .This is all totally unreasonable .For good reliability 100 celcius is more acceptable and its reasonable that your heatsink could be at say 60 celcius .If the heatsink were to be at 25 celcius it would have to be infinitely large.So we only have a differential of 40 between the junction and the heatsink now.If we use some insulater like a mica washer or silpad etc then this adds to the total thermal resistance between the junction and the heatsink.Well 1 degree celcius per watt seems a reasonable total figure.Therefore 40 watts of dissipation is acceptable and your SMPS if its worth its salt will waste less than this per device .In fact a good SMPS that has 2 switches should waste 10watts per switch.