So I know that providing a Pi with too little power can potentially cause physical damage to the circuitry (brownout). I don't 100% understand how or why this happens.

I know that it seems a lot of people are able to run the Pi on portable batteries, but I'm wondering what happens when the battery dies. Will it start to provide less and less power until it provides so little that the Pi gets damaged?

So this question also involves the behavior of a battery as it dies. As a portable battery is dying, what is happening to the current? Does the voltage start dropping or is it the amperage? Or both? My understanding is that a very low amperage is what can damage the Pi.

I just need a little explanation about how these different things are working and how to prevent issues, because I'm just worried about bricking my Pi.

  • raspberry won't function properly without sufficient current. Commented Oct 21, 2019 at 9:44

3 Answers 3


What happens during a brownout is not "physical damage" in itself, rather, digital devices start to behave erratically when underpowered. This can indeed result in physical damage too: if the SoC accidentally reconfigures an input pin as output, it can create an internal short circuit and fry the pin, the SoC itself, or an external device connected to that pin. Or it could send a command to the power management chip to increase the core voltage from 1.2V to 2V, which will kill the CPU cores.

Needless to say, the probability of such events is very low, as it requires to execute several erroneous commands in a defined sequence, and running random commands quickly results in a segmentation fault or a hard reset.

What usually happens during a brownout is RAM flipping a few bits here and there, and if that RAM was used as disk cache which is due to be written to the SD card, the cached data may get corrupted.


There are a lot of old-wives tales about the Pi and power supplies.

You definitely will NOT damage the Pi if it runs out of power, whether gradually or suddenly.

There is a possibility that the data on the SD Card can be damaged by uncontrolled power down, principally if it is writing at the time, although normal SD Card processes can cause problems if interrupted. (I have 6 Pi and 10 SD Cards, and despite many power failures have not experienced loss.)

If you want to be safe it is not difficult to detect low voltage and arrange an orderly shutdown.

Unfortunately your comment "Does the voltage start dropping or is it the amperage" indicates little understanding of electrical engineering, but is another of the old-wives tales circulating in Pi circles, which can be remedied by further study.

  • Please advice on a simple way to detect powerloss or low voltage? The less it requires building vs. programming the better. Commented Jul 8, 2017 at 12:05
  • 1
    For the Pi3 raspberrypi.stackexchange.com/a/54328/8697 shows how to detect low voltage as indicated by the PWR LED, (it is presumably possible with B+ and Pi2, but NOT Pi Zero, although I have never done this). There may be a kernel dev but I am unaware of any documentation. A simple hardware solution requires a comparator e.g. LM339.
    – Milliways
    Commented Jul 8, 2017 at 12:23
  • Indeed, I'm not very familiar with electrical engineering. That would be the reason I asked about how the current provided by the battery changes as it is dying.
    – Brooks
    Commented Jul 9, 2017 at 15:54

First of all I have to say about battery. Battery store energy as voltage. The term of current comes when we are trying to put a load across it. The voltage reduces in a rechargeable battery after using it for a while. A dead battery has no voltage,neither it can provide any flow of electricity.

If a battery dies, and that battery has some abnormal chemistry going in, it can damage the device(0.1% probability). Otherwise there will be no flow of electron in there. The amperage drops with the voltage. So I guess it won't affect the raspberry pi. (Caution: Do not leave a dead battery around any types of circuit, the discharge from the circuit can damage the circuit)

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