I need to power an ESP32 board from a 54 V battery. For that purpose, I understand that I need to step it down to 5 V so I need to design a step-down buck converter.

I tried the following circuit using an LM2576HV as its input range is up to 60 V DC:

enter image description here

I have tried following values of inductor L:

  1. 100 μH: when I used a 100 μH inductor and tried to measure output voltage using a multimeter, the IC burned as soon as I supplied 54 V.
  2. For an L value of 1000 and 2200 μH the output was 5 V according to the multimeter connected at the output, but when I connected the ESP32 across the output again the IC burned.
  • Why did the IC burn?
  • Does the capacitor have anything to do with the output current? I understand it is only for output voltage ripple.

I used this diode.

  • 1
    \$\begingroup\$ What the saturation current of the inductors you tried? Post switching waveforms. \$\endgroup\$
    – winny
    Feb 12 at 17:26
  • \$\begingroup\$ There's a lot more to inductor selection -- even in a "simple switcher" than the inductance. Please edit your question to include the exact part numbers of the inductors you tried, as well as the diode and capacitor. Note that as a rule of thumb, a buck (or boost) regulator with a 10:1 or higher voltage ratio is considered the highest reasonable limit -- you're slightly past that. I'd consider bucking down to some intermediate voltage (12V to 18V), then again to 5V, or maybe a flyback topology if a transformer could be sourced reasonably. \$\endgroup\$
    – TimWescott
    Feb 12 at 18:36
  • \$\begingroup\$ @winny sorry but i do not have access to that facility \$\endgroup\$ Feb 14 at 5:23
  • 2
    \$\begingroup\$ Then you’re driving blind. I suggest you buy a premade and pretested module which fits your needs. \$\endgroup\$
    – winny
    Feb 14 at 7:03
  • 1
    \$\begingroup\$ Where did you get your LM2576HV? The ones available on eBay and AliExpress are often fakes that are really the lower voltage rated parts. And you might need to add some low impedance ceramic or film capacitors. \$\endgroup\$
    – PStechPaul
    Feb 14 at 18:25

2 Answers 2


Do you want to design your own step-down converter?

As others pointed out creating huge step-down ratios with the mentioned topology can have numerous pitfalls.

Of course you would learn a lot by designing your own converter. But I wanted to point out you could also focus on the rest of the project instead ;-)

You don't really have to do your own.

Small and proven modules such as R3M or R5M series from Recom might solve the problem. No, I'm not a sales rep for anybody.

Such parts are available from multiple manufacurers, also as SIP and DIP modules. Some also have enable-inputs that may be useful as well if you need to shut-down for any reason.


If you look at the typical application in section 8.2 of the LM2576HV datasheet, where it shows a 5V output design that matches yours, the maximum input voltage for 5V out at 3A is just 15V.

If you scroll down further, to Application Curves, you'll see the operating ranges of input voltage vs. output current and the part codes that are required. Figure 8.5 for the 5V fixed output variant of the device shows that for anything above 15V input you're going to need inductor code H1000 (a 1000uH inductor) and even then you may not exceed around 0.45A draw.

An ESP32 can draw far more than 0.45A when the WiFi transmitter is powered, so this buck converter part is wholly unsuitable for your needs. This isn't really all that surprising - this chip was designed nearly 25 years ago. A more modern buck converter IC should be able to handle this with far greater ease.

EDIT: I misread the datasheet, the above is wrong.

My suspicion is that the efficiency is what's killing your chips. Figure 6-9 shows the efficiency curves vs. input voltage. The efficiency is not very good for 5V out at high input voltages. The efficiency increases with load (this is typical for buck converters) but it doesn't outpace power dissipation. With 54V input, 5V output, and 3A current, you get about 76% efficiency. Your load power is 15W, which means you're putting 19.74W in and dissipating 4.74W in your converter. That's enough to immolate it.

It's possible that your inductor is not suited for the job either - it's not enough to just pick any old 100uH inductor. It needs to at least have appropriate DCR and saturation current ratings.

  • 1
    \$\begingroup\$ By looking figure 8.5, I understand it can be used for 3A current with L value of 100uH at 60 V. How are you saying that it is not designed for 3A at 60V \$\endgroup\$ Feb 14 at 5:21
  • \$\begingroup\$ @NikhilPandya Sorry, yes, I did completely misread that graph. My answer is wrong. I suspect the problem is actually the efficiency - I'll edit to fix. \$\endgroup\$
    – Polynomial
    Feb 14 at 16:12

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.

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