I'm currently working with some WS2812B LEDs and tried out adding a capacitor to be safe from voltage spikes (was recommended by Adafruit). I noticed that a soon as I connect a capacitor between 5V and ground that the USB is disconnect from the PC (Windows Usb disconnect sound plays). I wondered why and tried another setup. This time just the capacitor between 5V and ground. Nothing else. According to my knownledge this isn't a short (capacitor is intact) and therefor this setup should be valid. Again the same thing happens: USB disconnects...
I really don't know why this happens but want to understand it. The capacitor I use is rated 16V 1000 mF (electrolytic in case this is relevant). The capacitor is also connected the right polarity. What forces the USB to disconnect (and let Windows play the USB disconnect sound). Am I may damaging the USB Port somehow and therefor a disconnect is forced as a security measurement?
Oh yeah... Using an Arduino Uno R3 made by Elegoo.
Hope you can help me out, Tmirror
Yes, connecting a capacitor is a dead short.
Because it is not charged at the moment you connect it.
There is a lesson here which everyone will tell you. (Quite soon!)
Unless something is very carefully designed (USB) to be "hot-plugged", you must switch off the power to make any change in connections.
I'm currently working with some WS2812B LEDs...
How many are 'some'?
Seven addressable LEDs (and the Arduino) could already be the limit for USB.
Therefore we usually power 'some' LEDs from an external supply,
which also eliminates the problem you're having.
Bypass capacitors are in the sub uF range. As described above, a meaty electrolytic looks like a dead short until it charges, then the charge is held unless there is a load of some sort. That is what makes it useful for stabilisation. However, the computer is seeing the inrush current and trying to protect itself.
Forget USB power, sort a decent 5V supply capable of a few Amps and your issue will go away. It will also be low enough internal resistance to cope with the electrolytics.