STM32 Power Configuration – Microcontroller Power Supply Guide

Why your regulator needs bypass capacitors

Those capacitors are usually there to provide stability to the output of the regulator itself. Linear regulators use a feedback loop to regulate under changing load conditions... the bypass capacitors helps stabilize the feedback loop to prevent oscillations.

Why bypass/decoupling capacitors are required

The general recommendation is to bypass at the point of load for all your ICs which, in this case, would be the Vdd pins on your microcontroller. A small ceramic cap, 0402 or 0603, close to each Vdd pin, with a short via to your ground plane (or to your ground track) of 0.1uF value should suffice nicely.

This is because the power draw from something like a microcontroller is pulsed... not steady. Think about it like this: your microcontroller is running a task very frequently... let's say a single task 20 times a second. If the regulator adjusts to deliver power when the microcontroller is idle, and then your task runs, the voltage is going to sag as the regulator tries to meet the current demand. The bypass capacitors are there to supply power when the regulator can't supply it fast enough. This is due simply to the fact they are capacitors and due to the fact they are placed physically close to the IC... closer than the voltage regulator is.

This is leaving out a lot of stuff, like the signal return path, but generally speaking, you're just trying to make sure that current demands can be met locally without tons of current having to travel all over the board. There's also the issue of reducing noise (I believe because you're reducing the di/dt traveling across your power traces by having current loops stay local to your bypass capacitors) but I don't want to try and explain that because my knowledge isn't that great in that area. :)

There is an excellent video by Dave Jones of the EEVBlog explaining why you can power almost every IC through almost every pin. This applies to your situation as well since you're supplying power through VDD. Vbat will be connected through the ESD protection diodes (which are present on virtually all pins on the IC) to the VDD pin and taking it's supply via this "detour".

It is just good practice not to leave the Vbat pin open but instead connecting it to VDD. Although leaving Vbat open works perfectly well as you have found, the circuits feeding from Vbat will operate at a lower supply voltage because they receive their supply voltage through the ESD diodes. In certain situations this could lead to unwanted effects. For example the minimum supply voltage at which the processor will work might be different.