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})}\$