Author Topic: Quick Question on MOSFET power losses  (Read 449 times)

0 Members and 1 Guest are viewing this topic.

Offline augustolTopic starter

  • Contributor
  • Posts: 41
  • Country: br
Quick Question on MOSFET power losses
« on: October 16, 2021, 07:08:12 pm »
Hi, everyone,

I was reading an "Application Note" from the International Rectifer and studying it yesterday. I like to do a first read along the whatever-I-am-studying and than come back re-reading it paying more attention to finer detail.

So I was re-reading it today and taking a closer look the formulas than I noticed some terms in the power conduction losses that I couldn't identify.

Pcond=(RDS(on)/RL).P0

The P0 term could be PO (for power output maybe?) Since the note doesn't explain anything about the formulas, I suppose it's something dumb and basic, but it's new to me.

So, the RL and P0 terms are reffering to what?

Here's the link to the Application Note if anyone want to check it (page 6): https://www.infineon.com/dgdl/an-1071.pdf?fileId=5546d462533600a40153559538eb0ff1
 

Offline Manul

  • Super Contributor
  • ***
  • Posts: 1108
  • Country: lt
Re: Quick Question on MOSFET power losses
« Reply #1 on: October 16, 2021, 08:15:52 pm »
RL should be load resistance and PO should be output power (power dissipated in RL).
 
The following users thanked this post: augustol

Online magic

  • Super Contributor
  • ***
  • Posts: 6747
  • Country: pl
Re: Quick Question on MOSFET power losses
« Reply #2 on: October 16, 2021, 08:38:15 pm »
P=IRMS²·RDS(on), simple as that.

Your formula works exactly when PO is the total power output of the PSU and RL is the combined resistance of the load and MOSFET. Do the math (I²·R) and see yourself. If RDS(on) is much less than load resistance, then the approximation given above also gets close enough.
 
The following users thanked this post: augustol

Offline Picuino

  • Frequent Contributor
  • **
  • Posts: 724
  • Country: 00
    • Picuino web
« Last Edit: October 16, 2021, 09:28:13 pm by Picuino »
 
The following users thanked this post: augustol


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf