So I'm using the LMR16006 DC/DC converter which is suggesting to place a Schottky diode from the SW pin to ground with a voltage rating 25% higher than my input voltage, which is 50V. Hence, the voltage rating should be at least 62.5V. Datasheet: https://www.ti.com/lit/ds/symlink/lmr16006.pdf

How much would it matter if I would take, for instance, a 60V diode?

Then, I'm struggling with the current rating. For instance, I'm looking at components at DigiKey at this moment and I've made a list like this: https://www.digikey.nl/products/en/discrete-semiconductor-products/diodes-rectifiers-single/280?k=schottky+diod&k=&pkeyword=schottky+diod&sv=0&pv914=119750&pv914=167594&sf=0&FV=-8%7C280%2Cmu70V%7C2071&quantity=&ColumnSort=0&page=1&stock=1&cad=1&pageSize=25

The datasheet tells me: ", the current rating for the diode should be equal or greater to the maximum output current for best reliability in most applications". Does this comply with the "Current - Average Rectified (Io)" in the list I obtained above? What does a higher or lower current mean in the application?

I assume an output current of 700mA in my system.

As I'm learning, I truly like some explanations on value choices

I've included a schematic in which D1 is the diode I'm talking about. enter image description here

  • \$\begingroup\$ What output voltage? Is it 5 volts? \$\endgroup\$
    – Andy aka
    Jan 11, 2021 at 14:50
  • \$\begingroup\$ Ah oops, forgot to mention... 5V \$\endgroup\$
    – Mart
    Jan 11, 2021 at 14:51
  • 2
    \$\begingroup\$ If the maximum load current is 700 mA, given the low duty cycle (60 volts input) then the diode should be continuously rated in excess of 700 mA by at least 30" for reliable long term operation. So you should be aiming for at least a 1 amp diode. \$\endgroup\$
    – Andy aka
    Jan 11, 2021 at 14:54
  • \$\begingroup\$ what's the meaning of ", seconds? How do you obtain the 1Amp if the output current is .7 amps? \$\endgroup\$
    – Mart
    Jan 11, 2021 at 15:04
  • 1
    \$\begingroup\$ I meant to write % and not " \$\endgroup\$
    – Andy aka
    Jan 11, 2021 at 15:18

1 Answer 1


From your search results I just picked this one to go off in my explanation. https://www.diodes.com/assets/Datasheets/ds30018.pdf (B170/B)

Its all about how the system is designed. If your system is designed to work at 50V and there are limiters in place to keep max voltage under 60V then a 60V diode would be fine. Therefore, you should pick a diode with a reverse voltage that is higher than a voltage you expect to see. 25% higher than the max voltage is a good metric to go by, but higher is for the most part normally better.

As for output current it pretty simple. 700mA is your average rectified current. Pick a diode with a higher average rectified current, like 1A.

For more information. The diode only conducts current when the mosfet is off. Therefore, your average current is actually less than 700mA because of the Ton time. These two images here show what happens during the off state of your mosfet. As you can see current through the diode starts at Imax and falls over time to Imin. These values can be found by using the " Output Inductor Selection" section of your technical documentation. Assuming a 40% ripple current will show that your peak current will be .98 A. Which is under even the average rectified current of the diode I picked.

Diode Current Visual Current in the system

Sources: https://en.wikipedia.org/wiki/Buck_converter

  • 3
    \$\begingroup\$ Please add a link or citation for the two graphics that you copied into your answer. We want to give credit to the original creator. \$\endgroup\$ Jan 11, 2021 at 20:34

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