Electrical – Mosfet Power Dissipation

mosfetpowerpower-dissipation

I'm trying to find power dissipation of my mosfet when I pull 17.5A. I'm using CSD18542ktt. Can anyone help me what equations can i use.

I have calculated the max power using the following equation.

P = (Tj(max) – Ta)/ RQja = 2.42Watts

Is this what i can dissipate without heat sinks?

So in my application, gate voltage is 10V and Rds(on) = 3.5mohms.

Power = 17.5 * 17.5 * 0.0035 = 1.07watts.

In order to check if i need a heatsink, i will multiply 1.07 with RQja. That gives me 66.34 which is way below my maximum junction temperature. Does that mean i do not need a heatsink?

Best Answer

$$P=IV$$

This is true for any circuit branch. For a MOSFET, if you aren't switching it extremely rapidly, \$I\$ is the current through the MOSFET channel. \$V\$ is the difference between the source and drain voltages.

If you are using the FET fully switched, this will be approximately \$I^2R\$, where \$R\$ is given by the \$r_{ds(\mathrm{on})}\$ parameter from the datasheet. And "fully switched" means with the gate voltage in the range given in the specified conditions for \$r_{ds(\mathrm{on})}\$

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