Why do plugged in chargers, which aren’t connected to anything, still draw power?

Question

e.g. If a phone charger is plugged into an outlet yet has no phone connected to be charged, why does it draw power?

In: Technology

Answer 1

Are you saying that because you have measured it, or because you believe that to be true? The no-load consumption of a modern switch-mode power supply (and all chargers are switch-mode) is very low, and below the ability of the typical domestic power meter to measure.

A test here

[https://www.howtogeek.com/231886/tested-should-you-unplug-chargers-when-youre-not-using-them/](https://www.howtogeek.com/231886/tested-should-you-unplug-chargers-when-youre-not-using-them/)

managed to find 0.3W when connecting six chargers, three laptop, three phone/tablet, so an average of 0.05W each.

This is ELI5, so…

Imagine it’s a hot day, and you want to keep a water bowl for your dog filled up, so there is always something him to drink. But the only water supply you have is a fire hydrant. You can fill the bowl by flashing the hydrant open for a fraction of a second. If your dog is drinking a lot, then you need to do that quite often to keep the bowl full. But if your dog is off doing something else, you will only need to let water in from the main very occasionally, as the bowl dries out on a sunny day.

That’s how a switch-mode power-supply works, It lets mains electricity in for a short period, to fill a bowl (a “capacitor”). If something is plugged in (“the dog is drinking”) it has to repeat that process frequently. But if nothing is plugged in, then it only has to make up the losses caused by the capacitor leaking.

Most of the “vampire power” nonsense stems from people who think that power supplies are transformers. They were, thirty years ago, but today they’re all switch-modes (the “flash the mains, charge a capacitor” process). I don’t know about the US, but in Europe there _is_ a transformer in a power supply, because you have to have galvanic isolation between input and output for safety reasons, but it’s not being driven while the capacitor is full, so its efficiency is almost irrelevant when there is no load.