Mosfet Power Dissipation

Question

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?

Answer 1

$$P=IV$$

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})}\$

Answer 2

1) RQja is usually based on the part being soldered to a PCB with a certain area (perhaps double-sided, 3 x 3 cm^2), and the ambient of the PCB being at 25 C.

2) You appear to have used the 'typical' RDSON at 25 C. If the device is heating up, RDSON will increase -- at the max. 175 C TJ, RDSON will increase by nearly 100 %. You should also start with the maximum 25 C datasheet value. Thus the correct (worst case) RDSON to use in your calculations is 4mohm*2 (if you have 10 VGS), or 5.1mohm*2 (with 4.5VGS). So 17.5 A in 10.2 mohm dissipates 3.1 W. At 60 deg/W, this would exceed the max. TJ (when starting from ambient = 25). You will probably need some small amount of heatsinking.

3) Be aware that unless you are very careful with busses on your PCB, the resistance of those and/or connectors will be significant (mohm range) and will increase the overall power dissipation in your application/board -- so it may be difficult to keep the TA for the PCB & enclosure equal to the external 'ambient' temperature.

4) Don't forget to include switching losses if you are switching the FET frequently (> 10 Hz perhaps) (and it can take many amperes of gate current to switch this size device quickly). Also if your load is not DC, the correct value to use for the current is not the average value, but the RMS value.

Answer 3

since I can't use P = IIR

Sure you can. You look up Rds(on) and plug it in. Of course, in order to do that you need to know the gate drive voltage. For a gate voltage of 4.5 volts, the maximum Rds(on) is 5.1 mohm. So, for that gate drive, your worst-case power is $$P = 17.5\times 17.5\times .0051 = 1.56\text{ watts} $$

And yes, for a bare MOSFET, your calculation is correct. Although it probably applies to a FET which is in still air but with convection possible. If you put the circuit in a space which does not permit free convection, your maximum power will be less.