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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.

Dark Patate
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1 Answers1

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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.

Dmitry Grigoryev
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