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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?

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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.

https://www.daenotes.com/electronics/basic-electronics/diode-in-parallel

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

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  • \$\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
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    \$\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

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