I am looking at buck-boost converters on the Texas Instruments design tool, and multiple of these topologies have parallel diodes on the output. I cannot see any reason to do this.

Can anyone explain the purpose of paralleling D4 and D2 on the output?



The parallel Schottky diodes are there to allow a higher forward current through the load. Usually done to prevent scenarios where the load current exceeds the maximum current rating of a single diode.


EDIT: You can refer to the datasheet, page 23 for a similar design example. The diode packaging is shipped as a single component (MBRD1035). http://www.ti.com/lit/ds/symlink/lm25118-q1.pdf

  • \$\begingroup\$ Nope. Notice in that article the two diodes have different curves? One diode will turn on first then runaway hogging all the current. They address this by adding resistors. \$\endgroup\$ – Fat Diode Sep 2 at 16:46
  • \$\begingroup\$ @FatDiode Pretty sure gusgus is right. That's not to say it's not bad design. \$\endgroup\$ – DKNguyen Sep 2 at 16:57
  • \$\begingroup\$ @FatDiode I don't understand. Where do you see the different curves? And where are the resistors? Do you mean Rfbt/Rfbb? \$\endgroup\$ – gusgus Sep 2 at 16:59
  • 3
    \$\begingroup\$ Current sharing can be a problem but usually these two diodes are on a single die so we can expect a decent sharing between the two dice: a) they are at the same operating temperature and b) their \$V_f\$ are probably quite close as they come from the same wafer lot. I've seen that in many hi-volume adapter designs and even paralleled diodes bridges sometime! \$\endgroup\$ – Verbal Kint Sep 2 at 17:03
  • \$\begingroup\$ those parallel diodes may actually be in one package \$\endgroup\$ – jsotola Sep 2 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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