I have a PSU with 5V and 3A (see here https://www.aliexpress.com/item/32961533195.html?spm=a2g0o.order_list.order_list_main.25.6c841802TcDBSs) and I want to power my Wemos D1 mini with it. So I connected the power supply with the board and it powers on for a few milliseconds and then turns off again. When I do the exact same thing with my Arduino Uno, it just works fine. I have two Wemos D1 mini and it didn't work with either of them. I, however, bricked one of them by connecting +/- inverted and in an attempt to get it working which didn't do the board any good. When I use USB to power the board it works just fine. Is there anything obvious, that I might be missing here?
EDIT: I measured my PSU and it provides 5.4V instead of 5V. Is that a problem?
I would try either an Apple or Galaxy USB charger as the PSU. You will need a breakout board to get at the +5V and G lines of course. Measure that voltage in order to determine if your meter is off.
The VIN pin simply goes to the ME6211C33 chip which is a LDOVR that supplies the 3.3V the board uses.
It should be able to handle up to 6V so the 5.4 you measured isn't a problem.
Based on all that I suspect the charger is somehow flawed, so using a high quality charger would be my approach.
Coincidentally, I had one of those lieing around, just tried it and it's working. Thank you so much for that suggestion, I think I would never have thought of that (escpecially because I'm not really aware of what a diode does, but I'll read up on that).
A diode is a component which allows current to flow in one direction and blocks it from flowing in the opposite direction. But they are not perfect. Even when current is allowed to flow, some of the voltage is lost or "dropped". You are making use of that imperfection.
You might be thinking "why not use a resistor instead". Not a good idea. The voltage drop of of a resistor varies depending on the current that is drawn. The current drawn by any circuit varies depending on what's happening from moment to moment. But the variations in the current would cause the voltage dropped by the resistor to vary also. The circuit will not like that! Circuits are designed to use a steady voltage. If the voltage isn't steady, all kinds of weird problems start to happen.
But the voltage drop of a diode hardly varies at all when the current varies.