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Here is the problem I am working on, from pg105-106 of Fundamentals of Power Electronics 2nd edition. enter image description here

As stated in this book, "The ability of the MOSFET channel to conduct current in the reverse direction makes it possible to employ a MOSFET where a diode would otherwise be required". So I believe two MOSFETs' conducting resistance are source of conducting loss, and should be calculated with \$(D+D')I_L^2*R_{on}=I_L^2*R_{on}\$ where \$I_L\$ equals inductor or load current.

But I have no idea on how to solve switching loss. In question (a) and (b), I am not sure what is the source of switching loss. Though I have seen example on how to calculate diode switching loss in the book, but it is the case that diode are conducting current. Now the only thing open and close to conduct current is the MOSFET channel. So I totally do not know how to make use of parameters of body/external diodes and how to calculate switching loss.

What makes me most confused is (c), as the question gave me the feeling that on some range of frequency an external diode will worsen the efficiency!

So how to do the math correctly? Thanks.

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  • \$\begingroup\$ Take a look at this application note from TI: ti.com/lit/an/slpa009a/slpa009a.pdf Also keep in mind that they only want you to consider the switching loss from the FET's body diode. \$\endgroup\$
    – joribama
    Commented Sep 4, 2019 at 3:26
  • \$\begingroup\$ @joribama In question (a), conduction loss is included xD. \$\endgroup\$
    – Page David
    Commented Sep 4, 2019 at 3:51
  • \$\begingroup\$ what I meant was as far as switching losses, only care about the body diode loss. \$\endgroup\$
    – joribama
    Commented Sep 4, 2019 at 4:11
  • \$\begingroup\$ The conduction loss of the MOSFET \$r_{DS(on)}\$ depends on the duty ratio \$D\$ (or \$D'\$ depending on which MOSFET we talk about) minus the body diode conduction time at both ends of the conduction time (there is always a deadtime to avoid shoot-through currents). Then switching losses can be calculated using information given here:electronics.stackexchange.com/questions/133889/… \$\endgroup\$
    – Verbal Kint
    Commented Sep 4, 2019 at 6:19

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