A.D.J
Junior Member

Offline
Posts: 14
|
I was trying to use constant I to charge one capacitor. The charging process is based on the equation: I*t=CV. If C=1pF, t=8us, I=10nA, so V=I*t/C=80mV. The schematic simulation is very closer to 80mV. However, if changing the C=100fF, the expected V=8us*10nA/100fF=800mV. The simulation result is less than 800mV. I have no idea what results in the voltage reduction. Thank you.
-------------------------- Is it related to capacitor leakage current issue? I read a article talked about the leakage current I is proportional to the charges on the capacitor plate. For smaller capacitance, we got higher charge, so the leakage current is bigger. As a result, the final charged voltage could be smaller than expected. Is that reasonable? Thank you.
|