Electronics > Beginners
RC circuit basics
tester43:
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?
RoGeorge:
Good observation! :-+
The instrument has internal resistance. In order to measure, some small current will go through the instrument, and together with the 1Mohm resistor, will make a voltage divider.
(Offtopic: 'So' is for drawing a conclusion, not good to use as a first word.)
thinkfat:
--- Quote from: RoGeorge on January 21, 2020, 01:16:16 pm ---Good observation! :-+
The instrument has internal resistance. In order to measure, some small current will go through the instrument, and together with the 1Mohm resistor, will make a voltage divider.
(Offtopic: 'So' is for drawing a conclusion, not good to use as a first word.)
--- End quote ---
Also, the capacitor will have a leakage current.
tester43:
Declared oscilloscope input impedance is 1 MΩ+-1%.... I am guessing that in terms of DC it will be simply 1 MΩ resistance (?) and then it will allow some small current to go through the probe as we have voltage divider there leading to ... 1/2 voltage (because 1M resistor + 1M probe?)?
mikerj:
Are you sure you don't have a 10x probe? That would provide a 10M impedance which would put the final voltage closer to your observation.
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