0

I have a Raspberry Pi 4 that's is powered on the USB port by a circuit I made. The key component is a Buck Regulator (LM2576-5.0WT), the input is a AC adapter (12 V - 8 A). The regulator is able to provide the 3 A required by the Pi (in 5 V).

But when I plug the RPi, the power LED turns on, but not the ACT one. When I check the tension, it's 5 V without the Pi and 6.9 V with it. And, when I check the current, it is 0.3 A, of course not enough to switch on the Pi.

So I thought it was my power circuit the problem, but when I plugged a motor instead of the Pi, it could go to 2 A (I have to find other loads to go up the 3 A). But, is there something that could limit the Pi to 0.3 A?

Thanks in advance for any clue.

  • 2
    If you have fed 6.9V into the Pi you have destroyed the Pi. – joan Oct 12 at 10:04
  • 1
    If you have a AC adapter and connected that to the Pi via a DC/DC regulator you have probably broken it. The DC/DC reg. is a DC component and the Raspberry Pi also. – Mats Karlsson Oct 12 at 11:20
  • The Pi is still working. I could use my USB adaptor and it still works. After I didn't get why when there are no devices, I have 5 V, and when I checked with the Pi I got 6.9. I don't get why the DC/DC regulator would broke the PI ? – Dark Patate Oct 13 at 6:22
  • I tested with an additional load, my circuit can provide, at least, 2 A. – Dark Patate Oct 13 at 7:14
  • 1
    This is a question about whatever (undisclosed components) you have connected - nothing to do with the Pi – Milliways Oct 13 at 8:51
1

The key component of a buck converter is the inductor, not the silicon. If you pick the wrong one, it won't be able to generate enough current for your load, or won't lower the voltage to the value you expect. If the inductor saturates at 1 A, you won't get 2 A out of it no matter what switching IC you're using.

You can check the picture here to get an idea of a converter capable of powering a Pi 4.

If you measure 6.9V on the Pi side, it's likely game over. Let the Pi cool down 10 minutes or so, then try powering it with a compatible power supply. It it doesn't work then, order yourself a new one. If the Pi is working, I would double-check your measurements: it's likely that you never applied 6.9V to it after all.

| improve this answer | |
  • For the value of the inductance I chose according the datasheet, and one that can manage 5 A (I chose this one). But maybe I should check something else? Moreover, my Pi still works, even after the 6.9 V. I checked again, and still same voltage. But now, when I check without charge, I have 10 V instead of 5 V yesterday... I'm lost... Maybe my multimeter is broken... – Dark Patate Oct 14 at 6:00
  • @DarkPatate Yes, such an inductor (with 5A rating) should be sufficient. You should get trustworthy measurements in order to make progress in this project. I also urge you to use a dummy load in your tests (a bunch of car lightbulbs or something similar) which won't blow up as easily as a Pi and wouldn't cost that much to replace. – Dmitry Grigoryev Oct 14 at 12:04
  • Thanks for your advices. I'll use a DC motor to test it. Actually, after some tests, I think there are probably some problems in the soldering of that PCB. Because I just tried with a test PCB I made with only the Buck regulator, and here I have my 5 V. When I plug the Pi, the ACT led switches on (but the Pi doesn't seem to work, no screen and no ssh). I also checked with an oscilloscope to verify the stability of the tension, it looks ok (about 100 mV of oscillation). I'll pursue my effort :) – Dark Patate Oct 14 at 14:06

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