So I have this MOSFET: IRF4110

I am switching on for 2ms and off for 18ms (Making it a 50Hz 10% Duty Cycle) and current in on time is 100A at 10V.

So the datasheet says 4.5mohm and thus I calculate the power dissipation as: P = (100*10%)^2 * 4.5mohm P = 0.45W

Thus the power derating says 2.5 W/C and from the looks it says that the temperature rise will be 0.45/2.5 = 0.18 C ?

I find my answer highly wrong and would appreciate if someone can give a nudge in the right direction.

Edit: Assume ambient temperature as 30 Degree Celsius


2 Answers 2


(100*10%)^2 * 4.5mohm P = 0.45W is wrong because the square of a mean value is not equal to the mean of the squares.

You are interested in the mean power that's why you first have to calculate the power for continuous current and then multiply by 0.1 for the duty-cycle, i.e. you have to calculate

\$(100A)^2 \cdot 4.5m\Omega \cdot 0.1 = 4.5W\$

in order to get the mean power dissipation in the MOSFET.

For calculating the temperature rise you must not use the derating factor (units [\$W/°C\$]; it just gives you an estimate how much the max. power dissipation has to be derated per each °C above 25°C) like you have done, but the thermal resistance \$R_\theta\$ (units [\$°C/W\$]; it tells you how much the temperature rises per Watt).

If you don't use a eat sink the datasheet tells you that thermal resistance from junction to ambience is

\$R_{\theta JA} = 62°C/W\$

If you use a heat sink the datasheet tells you that thermal resistance from junction to heat sink is

\$R_{\theta JS} = R_{\theta JC} + R_{\theta CS} = 0.402°C/W + 0.5°C/W \approx 1 °C/W\$

So without heat sink junction temperature rise with respect to ambience would be

\$4.5W \cdot 62°C/W = 279°C\$ (which is much too high; i.e. you need a heat sink).

With heat sink junction temperature rise with respect to heat sink would be

\$4.5W \cdot 1°C/W=4.5°C\$.

Of course it depends on the thermal resistance of the heat sink what you will get as total temperature rise with respect to ambience.

Still you have to take care that instantaneous power (=45W) is within limits during the on-time.

  • \$\begingroup\$ Agreed in those 2msec he will have 45W in there. Datasheet says it can handle that, but additional power disipation is still required. :) \$\endgroup\$
    – Dimitri
    Nov 29, 2019 at 13:34
  • \$\begingroup\$ So let's just say a safe 10 C increase? \$\endgroup\$ Nov 29, 2019 at 16:17
  • \$\begingroup\$ It depends on the heat sink. You need to get the thermal resistance value for heat sink to ambience (from heat sink datasheet) and include it in the calculation, i.e. add it to \$R_{\theta JS}\$. \$\endgroup\$
    – Curd
    Dec 3, 2019 at 20:00

I read your calculations and hope mine will be correct.

But first you as you are saying you have 100A and 10 V which is basicly 1000W on input. Which is a lot, so 0.45W is something I wouldn´t expect. You will need good cooling for that.

From what it seems you are making theorical calculations. I think you are mixing here a little bit. You will have at least:

P = 100A^2 * Rin = 100*100*4.5/1000 = 45W You will have this every 20msec.

So each cycle of the clock you will have this in your rectifier, not 10% of the current like you calculated. So in ideal situation I would expect almost 20°C more

difT = 45/2.5.

Now you apply Duty cycle Power Increase = 20°C * 0.1 = 2°C

But as I said, Ideal. As for your question, I assume that you want someone to check the calculations

  • \$\begingroup\$ So it would be safe to say like the temperature of mosfet will increase by 10 C \$\endgroup\$ Nov 29, 2019 at 16:16

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