For designing a low side non-sync buck regulator I am looking for MOSFET(s) that can provide low overall power loss. Conditions would be 60V to 48V, 10A, 600 kHz (worst case), 1 Ohm source and 0.35 Ohm sink gate driver. Schematic would be as shown below, although without a LED load.

enter image description here [Reference image from https://shadyelectronics.com/low-side-buck-step-down-converter/]

For this I was using the Cross reference comparison tool by EPC that provides the various power loss for a MOSFET for switching conditions specified. If we take a potential MOSFET such as FDD86110, we see that the diode reverse recovery loss is much higher than other losses. This is validated by calculation of loss by P=Vin×Qrr×Fsw.

enter image description here

My questions would be

  1. If this loss is such a big part of power loss at higher voltage, why don't most manufacturers or sellers have this as a criteria to sort or filter MOSFETS? And what is a good way to find MOSFETs with low Qrr along with other filter criteria.
  2. Schottky diode in parallel to the MOSFET would reduce power loss? If yes, how to calculate the power loss with this?
  3. Is there any other way to reduce this power loss?
  • \$\begingroup\$ The DFLS2100-7 diode will smoke when working with a load of 20 amps. I'm trying to figure out why you posted that schematic image given that it clearly has a known problem with the output being connected to an LED. \$\endgroup\$
    – Andy aka
    Aug 1 at 8:33
  • \$\begingroup\$ I have mentioned in the question that it is a reference schematic and the load isn't a LED. \$\endgroup\$
    – EarthLord
    Aug 1 at 9:19
  • 2
    \$\begingroup\$ In what way is the MOSFET being reverse biased? \$\endgroup\$ Aug 1 at 11:14
  • \$\begingroup\$ Oh, the MOSFET here won't have diode reverse recovery loss as its similar to the top MOSFET in a normal buck. The current through the MOSFET won't flow in the same direction as the body diode's forward direction. Thanks Tim. \$\endgroup\$
    – EarthLord
    Aug 2 at 6:07


Your Answer

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