Say I am operating a MOSFET 20A DC and the voltage drop across the drain and source (VDS) is 2V. The operating point is red as shown in the safe operating area (SOA) of the MOSFET as shown below. Obviously, the operating point is located within the "safe area". If we do the calculation, however, the power dissipation of the MOSFET is 40W (20A x 2V,) which wil definitely kill the MOSFET. I am confused. Can anyone solve my confusion?

enter image description here

  • 1
    \$\begingroup\$ Are you using the mosfet in the ohmic (linear) region? \$\endgroup\$
    – Jeroen3
    Nov 30, 2023 at 7:01
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    \$\begingroup\$ the mosfet will be fine at 2 volts and 20 amps if the heatsink can remove the 40 watts \$\endgroup\$
    – Autistic
    Nov 30, 2023 at 7:34
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    \$\begingroup\$ Please link to the data sheet that shows the graph you embedded. You need to name/link picture/image sources. \$\endgroup\$
    – Andy aka
    Nov 30, 2023 at 8:42
  • \$\begingroup\$ "which wil definitely kill the MOSFET" -- citation needed ;) \$\endgroup\$ Nov 30, 2023 at 10:43

1 Answer 1


Please note the test conditions in the upper right corner. As long as you have a cooling mechanism that manages to keep the case temperature at 25 °C, the die will not go above 175 °C, and it will survive.

The SOA allows 10 V × 30 A, so the thermal resistance between junction and case can be computed as RθJC = (175 °C − 25 °C) / (10 V × 30 A) = 0.5 °C/W. (There should be another graph with more detailed values.)

So for 40 W, the maximum case temperature (at which the die will die) is 175 °C − 40 W × 0.5 °C/W = 155 °C. So assuming a worst-case ambient temperature of 40 °C, you need a heat sink with a total thermal resistance of less than (155 °C − 40 °C) / 40 W = 2.875 °C/W. (The heat sink datasheet will tell you this value. Such a small value requires lots of metal with a fan.)


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