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I am analyzing the thermal dissipation of the FDB33N25 MOSFET - datasheet

The procedure I usually follow is as stated:

$$I_{d} = 15~A (requirement)$$ $$R_{ds} = 92~m\Omega$$ $$\Theta_{ja} = 62,5~K/W$$ $$P = I^2 \cdot R_{ds} = 20,7~W$$ $$T = P \cdot \Theta_{ja} = 20,7 \cdot 62,5 = 1293~K$$

The MOSFET can sustain a max of 20 A. So 15 A should be fine. But the temperature generated is mind blowing.

Is this analysis correct? If yes, then I can never achive 15 A without extravagant cooling. Also, will this temp of 1293 K be in the ambient space (theoretically of course)?

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    \$\begingroup\$ Rthja = 62.5 Celsius/Watt tell us that for every 1 watt of power dissipation in the MOSFET the junction temperature will rise 62.5 celsius above the ambient temperature. And welcome in the real world. mcmanis.com/chuck/robotics/projects/esc2/FET-power.html And Rthja is a thermal resistance between junction and ambient without any additional heat sink \$\endgroup\$ – G36 Mar 12 '17 at 14:00
  • \$\begingroup\$ You used the Rth junction-ambient figure, which is applicable when you don't use any heatsinking at all. The max figures, especially the marketing-department-written ones on the first page of a datasheet, assume the most favourable situation, in this case an unlimited heatsink. \$\endgroup\$ – Wouter van Ooijen Mar 12 '17 at 14:05
  • \$\begingroup\$ @ThePhoton it's actually (i)x(i)xRds. Sorry about the typo \$\endgroup\$ – Board-Man Mar 12 '17 at 16:00
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Thermal Resistance, Junction-to-Case, Max. 0.53 °C/W

So, if you could stick a 2°C/W heat sink on your D2PAK, it would dissipate 20W without problems. I suspect you will find this rather difficult though, so here are other solutions:

  • FET with lower RdsON
  • FET with a more heat-sink friendly package
  • Several FETs in parallel

The last one allows both to lower the dissipation, and to spread it between several packages, where it will be easier to dissipate.

Important Note: Since you selected a 250V MOSFET, I'm assuming you need the voltage rating. If this is a low voltage application, you will get much better RdsON with an appropriate choice of lower-voltage MOSFET.

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What is your applied gate-drain voltage.

RdsON is a fickle number that has a number of dependent variables. Do not just rely on the front sheet marketing number.. Check the graphs.

Your gate voltage may be insufficient.

And yes.. 20W = LARGE HEATSINK. 62.5 Celcius/Watt is for the device package to air on it's own.

You should also consider looking for a better MOSFET. 92mOhs is not that great these days. It also increases with temperature, so dumping heat is very important.

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  • \$\begingroup\$ BTW I was going to show you a better MOSFET example, but you did not list your REAL voltage requirement. \$\endgroup\$ – Trevor_G Mar 12 '17 at 17:05

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