So.. I was preparing debouce basic circuit for the button based on single ceramic capacitor and resistor.
I found some old 1nF capacitors and 1M ohm resistor to see how it will be charging with 3,3V rectified power supply.
It is charging as expected: voltage on the capacitor 'leg' is increasing to flat out in several milliseconds.
The part that was a suprise was that it topped out to ~2,8V. Far away from 3,3V of the supply.
So I started to test the influence of resistor on the voltage and when changed to 86KOhm it topped out to 3,1V. Of course in shorter time.
That would mean that there must be a current flow, which is not really expected as the voltage is really rectified and capacitor should load and block any current. There is nothing that would be the load on it.
How should I understand it? What did I miss?
Just an idea and sounds like others have said similar (I'm just a beginner so not sure I'm comprehending all the comments), but using the same R value, try a different size cap, something bigger in the μF range, see if the voltage drop is less or reduced compared to the V drop of the 1nF. Maybe your multi is flowing cap current through itself during measuring, and enough to reduce cap V on a 1nF cap. When you use a smaller R (86kΩ vs 1MΩ), the possible current in the RC cct increases, so it's able to replenish the charge (that's being lost across your multi or leaking out the cap) at a higher rate than if a large R were having to be flowed through to refill your cap?
EDIT-
Idk if I'm close at all so just keep that in mind... Maybe leakage is adding to multimeter current theft, perhaps even more of a factor than multimeter current theft..
0.5V ÷ 1,000,000 Ω = 500nA
0.2V ÷ 86,000 Ω = 2,325nA
Rate in = rate out, but capacitor V levels off, so like a bucket with a hole in the bottom the more coulombs filling the capacitor the more pressure of water leaking out the bottom until there's enough of the bucket filled up to generate enough pressure at the hole in which the rate in = the rate out.