0
\$\begingroup\$

I have two different MOSFETs controlling an actuator that is specified to have a current draw of 13A. There is also a note on the actuator datasheet about inrush current stating:

Typically, the inrush current will last between 75 – 150 milliseconds and can be up to three times higher than the current for the actuator and load

When choosing a heatsink to keep the component cool, do I need to consider the inrush current or is it okay to use the max current of 13A for the heat calculations? What is the method of determining how much the MOSFET increases in temperature as a function of time?

Here are the datasheets for the MOSFETs under question:

https://assets.nexperia.com/documents/data-sheet/PHP29N08T.pdf

https://www.infineon.com/dgdl/irf4905pbf.pdf?fileId=5546d462533600a4015355e329b1197e

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

You'll need to look at the safe operating area and transient thermal impedance graphs. Your peak surge current is 39 A for 150 ms, which puts you outside the SOA for the PHP29N08T (Vds of ~2 V at 50 mΩ). You're kind of off the edge of the SOA for the IRF4905 but looking at fig. 11 shows you that the single pulse thermal impedance is about 0.75. The maximum power dissipated during the pulse is ~30 W, so you have a junction temperature rise of 23 C relative to the case. The maximum junction temperature is 175 C so the maximum case temperature is 152 C.

Let's assume you've been operating the system at 13 A for a long time (steady state) and want to make sure that you will stay within limits if you turn the actuator off and immediately back on. How much heatsink do you need to keep the steady state case temp below 152 C? 13 A through 20 mΩ generates 3.38 W of heat. We will assume an ambient temperature of 50 C (sealed enclosure, in the sun, in summer, probably higher than 50 C). That means that the maximum case to ambient temperature difference is 102 C. 102 C/3.38 W = 30.1 C/W for case-to-ambient thermal resistance. The case-to-heatsink thermal resistance is 0.5 C/W, so the heatsink-to-ambient thermal resistance is 29.6 C/W, which is very achievable. Note that the junction-to-ambient (non-heatsinked) thermal resistance is 62 C/W which puts you at a 210 C rise above ambient in steady state so although your heatsinking needs are modest, some sort of heatsink is absolutely necessary.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.