I'm working on an ultra-efficient fixed output half-bridge DC/DC converter. (Where the input is from variable sources of voltage and current like batteries, solar, wind, and also from fixed sources of voltage/current like ac grid)

So I've speced out my mosfets that I need to be able to switch the half bridge effectively and get excellent efficiency numbers and I've decided to use an ESP32 so that I can control other things as well.

And I'm looking at this GaNFET driver for my purposes (I will also be using this to control BLDC motors in this application): https://www.infineon.com/dgdl/Infineon-2EDF7275F-DataSheet-v02_07-EN.pdf?fileId=5546d462636cc8fb0163b08fd9203057

My issue is to try and find the most efficient way to power this. The ESP32 (or whatever simpler design without wifi, etc.) will be in modem sleep so it should use about 20mA at high speed from the data sheet for it. The GPIOs/PWMs will be used as the interupts can be triggered. It needs 5VDC of course.

The driver needs 3.3VDC which I can get from the ESP32, and it also needs the VOut power which is 17-20VDC depending on which GanFET I choose.

My understanding is that there is VERY little amperage on the driver and GanFET gate in this scenario so the draw and loss is negligable.

I will be putting a hot swap controller in at the VIN of the device for the requisite protections.

But my primary issue is I need to be able to have the ESP32 and Driver powered by the source voltage (i.e. not external) and the source voltage will be either 384VDC (+/- variations of state of charge etc.) and the output will be a fixed 380VDC.

So I've considered:

  1. A Voltage divider that gets me close from the bottom end of the 384VDC (320VDC) to 5VDC + a regulator and then another voltage divider that gets me close to the Gate Driver's 20VDC + a regulator.
  2. A less efficient buck converter that is self-powered from the source voltage in a single chip. (But I can't find anything that can step down from 440VDC max to 5/20VDC)
  3. ???

What's the best way to get the ESP32 or equivalent powered and power the GaNFET driver also powered off of the (variable) source voltage in this scenario?


  • \$\begingroup\$ Seems like that gets 5V @ 10mA, and 15V @ 10mA. The problem is that none of those will work for what I need. I tried to find one that would do what I wanted directly but none of them would drive 20mA at 5V nor 18-20V @ 10ma. I could find ones at lower voltages (40) that could regulate to what I need, but I'd have to use a voltage divider I guess? For reference I'm looking at: NJM78L05UA/EA and NJM78L20UA You'd think that they would just make stuff like this because VCC is almost always 5VDC with up to say an amp and Mosfet gate almost always works best between 15-20VDC. \$\endgroup\$ Oct 12, 2021 at 15:26

1 Answer 1


I've never done this so this is theoretical.

Apparently, hysteric converters or variations on them (such as constant on-time) excel at very large step down ratios are are relatively simple so you can scratch build one. They are about as simple as switching voltage regulators get but were avoided for the more conventional buck converters due to variable frequency characteristics but are becoming more popular again.

You could use a very high voltage linear regulator to bootstrap your hysteretic converter momentarily on startup until the convereter supply can take over to power itself:


EDIT: Looks like you can actually just buy them too. 7.5V-500V @ 300mA has to be enough, right?!


  • \$\begingroup\$ I saw your second link, but again it requires a 5V VCC and 230mV to power it which the ncp785a can't push. From your comments, I was thinking about something like this: mouser.com/datasheet/2/268/… but I'm not sure how to use it? Maybe this momentarily drives the buck IC you linked to and then it takes over for itself somehow? \$\endgroup\$ Oct 12, 2021 at 15:43
  • \$\begingroup\$ @JamesHancock Oh that does require a 5V Vcc. That's fine. They probably did that to give you options. No need for fancy switching. I would just use a couple of diodes and feed a slightly higher than 5V derived downstream from the hysteretic converter back to Vcc so that once the primary 5V was up and running it would exceed the 5V from the HV linear regulator thereby reverse biasing the diode on the series of the HV linear regulator. You can add other stuff to if you want like a 600V pass transistor to permanently disconnect the HV linear regulator afterwards so it's not stressed if you want. \$\endgroup\$
    – DKNguyen
    Oct 12, 2021 at 15:49
  • 1
    \$\begingroup\$ The feedback pin is just the standard divider that is also seen on adjustable linear regulators. Steps down the output voltage so it is comparable to the internal voltage reference and they are compared. The regulator keeps the stepped down output equal to the internal reference and in that way the divider ratio combined with internal reference voltage sets the output voltage. \$\endgroup\$
    – DKNguyen
    Oct 12, 2021 at 16:14
  • 1
    \$\begingroup\$ The VCC regulator on and off hysteresis 230 mV is a VOLTAGE, not a current supply. It's just the hysteresis to prevent the regulator from operating halfway or dancing on and off at a low supply voltage. As in, it must rise 230mV above the minimum nominal operating point to turn on a it must fall 230mV below the minimum nominal operating point to turn off. Different turn-on and turn-off thresholds. No stuttering about a single on/off threshold voltage. \$\endgroup\$
    – DKNguyen
    Oct 12, 2021 at 16:15
  • 1
    \$\begingroup\$ @j If you read page 11. Vcc seems to be the output of an internal HV linear regulator, much like the discrete one I linked earlier for boostraping, which itself is fed from Vin. Connecting an external low voltage 5V supply to Vcc effectively bypasses this internal regulator and reduces heat dissipation in the IC. So instead of relying on the internal linear regulator the entire time, you can use the more efficient switching output to override the internal regulator once everything has started up. \$\endgroup\$
    – DKNguyen
    Oct 12, 2021 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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