I have built quite some buck converters for ~3A@3-5V (~1MHz, 24V Input) and I always used schottky diodes due to their low forward voltage drop (and therefore low power dissipation during free wheeling). Now I naively replaced the converter‘s feedback divider to regulate an output voltage of 12V. After a short time the regulator failed and soon I found out that it was the shottky diode that was destroyed.

I think I understand the Problem: schottky diodes don‘t only have a low forward drop (which is desirable), but they also have a relevant reverse leakage current which rises significantly with die temperature. While that wasn’t a problem in the 3-5V versions, at a reverse voltage of 12V the power dissipation (12V*100mA=1.2W) is too high for the device. I replaced the diode with a common silicon diode for experiment and it failed as well, but (as I assume) for a different reason: the ~0.7V forward drop at 2A causes at least 1.4W of dissipation.

  • How do I choose a suitable diode for such a buck converter (high current and high voltage) where schottky diodes don‘t appear to be a good candidate?
  • Is there a class of diodes which has a low forward drop and low reverse leakage or should I use a bunch of schottky diodes to share the Power?
  • Someone told me I could use a transistor instead of a diode, but I have no idea about that; does it make sense?

Edit: this is the schottky diode that failed in my case

Edit2: I'm using the TPS54240 regulator and I need the converter to provide at least 2A. circuit layout

  • 1
    \$\begingroup\$ Please link to the data sheet for the Schottky diode that drew 100 mA at 12 volts. Yes, using a synchronous buck converter is better in terms of efficiency. \$\endgroup\$
    – Andy aka
    Jan 12, 2020 at 15:27
  • \$\begingroup\$ It could be you have wrong passive components causing this and wrong freq. try 500kHz \$\endgroup\$ Jan 12, 2020 at 16:10
  • \$\begingroup\$ Was that schottky diode an ultrafast recovery type? You need that for 1MHz. \$\endgroup\$
    – Janka
    Jan 12, 2020 at 16:40
  • \$\begingroup\$ @Andyaka added the datasheet. I ran it at 3A, not as I wrote at 4A. \$\endgroup\$
    – Sim Son
    Jan 12, 2020 at 16:50
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 I think I stuck pretty close the converter‘s recommendations. Its datasheet provided lots of formulas to determine suitable components. Which component do you consider most critical in my case? \$\endgroup\$
    – Sim Son
    Jan 12, 2020 at 16:55

2 Answers 2


When power loss is square of current at 4A, the differences between low RdsOn FETs and diodes becomes more significant if you want up to 94% efficiency while up to 80% @ 0.5A.

20W out with 6% loss best case = 1200 mW heat that must be dissipated.

This requires a Synchronous Buck converter with 50A Pch/Nch FETs and around 500 mm² 2x Cu fill PCB with about isolated 5 nodes to dissipate losses.

Your problems could be any component incl. PCB area and convection air. I can suggest LM3150MH with dual FETs using TI's design webench power designer or AD's similar tool but this depends on your investment of existing design and other demands.


with added schematic & layout , but still missing critical parameters for Ptot = Pcon + Psw + Pgd + Pq and Rth*Ptot= ΔT 'C rise , DCR for each part, I don't see any obvious design electrical errors for 2.5A out a very nice compact layout, but a thermodynamic failure.

Consider new Diode retrofit with force air cooling with an 18mm 24V fan

or rip up and retry , with some useful experience gained to pay attention to Rth*Ptot +Tamb. (max) next time with better design specs.

Diode should look more like this. https://www.digikey.ca/product-detail/en/toshiba-semiconductor-and-storage/CMS04-TE12LQM/CMS04-TE12LQM-TR-ND/871534

And future board design should look more like this, using TI Webbench. Note > 500 sq mm with above specs

enter image description here

This is a more efficient Buck converter for reducing diode losses.

enter image description here

Heat can cause thermal runaway issues with ferrite and higher peak currents results in thermal issues with diodes from lack of copper to release heat. Higher RdsOn also causes higher Coss and higher T causes more leakage in FETs and Sch. diodes

So SMD is nice and small but switches need to be chosen carefully to be efficient and resonate at zero valley current when turning off.

  • \$\begingroup\$ Thank you for your answer! I think I'll try to provide the diode with a better heatsink. Also the diode you proposed looks interesting, it seems like there are more suitable diodes (wasn't aware of the fast-recovery type) and I'll look for a different one. As I'm pretty close to my target and am also limited in space I hope I don't have to change my approach completely. Beside the free-wheeling and reverse losses, what are those terms in the power estimation "Pcon + Psw + Pgd + Pq"? \$\endgroup\$
    – Sim Son
    Jan 12, 2020 at 19:25
  • \$\begingroup\$ things you need to learn (re-search) \$\endgroup\$ Jan 12, 2020 at 19:27

In your case, I wonder whether your diode actually had a reverse current of 100 mA. That seems humongous.

Let's start from the bottom:

When currents become large, the forward voltage, even that of Schottky rectifiers, becomes a significant contributor to loss.

Thus, one usually uses synchronous switch mode power supplies:

Someone told me I could use a transistor instead of a diode, but I have no idea about that; does it make sense?

Yes, it makes very much sense! You know exactly when that diode should be conducting, and when not. Instead of a diode, you could have another switch like the transistor feeding input voltage into your inductor.

Typically, MOSFETs (for medium currents, like yours) are used, which can have excellent r_ds,on, meaning there's very low voltage drop.

You can buy, for your power regime, ICs that integrate both the high-side switch and the low side switch (examples, esp. this).

  • \$\begingroup\$ Thanks for your answer! I wonder if the gate capacity would be a problem when switching with 1Mhz? \$\endgroup\$
    – Sim Son
    Jan 12, 2020 at 17:01
  • \$\begingroup\$ I added a datasheet, 100mA of reverse current appear to be quite realistic as it raises pretty much with temperature. I did tests at lower currents and using a head bed to nail it down (destroyed quite few diodes) \$\endgroup\$
    – Sim Son
    Jan 12, 2020 at 17:07

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