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Hi
What is your supply voltage?
You can only charging a capasitor 2/3 of the supply volatge, and discharging 1/3 of the supply volatge.
Hi
What is your supply voltage?
You can only charging a capasitor 2/3 of the supply volatge, and discharging 1/3 of the supply volatge.
i'm using 9 volts as a supply voltage and the maximum reading was 8.9v when discharging is stopped at 0.01v.
If you leave the capacitor connected to the voltage long enough, it will eventually charge completely to the supply voltage.
It may not discharge completely because of something called dielectric absorption. A small amount of charge gets trapped in the dielectric and slowly leaks out, which adds a small charge to the capacitor. If you short the capacitor long enough it will eventually discharge to zero (may take many minutes).
The capacitor voltage is asymptotic to the supply rails.hello
what are the reasons why the capacitor won't reach the supply voltage? and when discharging the capacitor it didn't get fully discharged it stopped at 0.01v.
thanx
Eventually it will get to within the thermal noise level of the final value, which is the practical limit. For example for a 1µF capacitor at room temperature this occurs after about 17 time constants (thermal noise = 64nV) for a 1V supply.The capacitor voltage is asymptotic to the supply rails.
define: asymptotic - Google Search
In principle, it never gets there.
Fer' sure, e to the -17th is a small number. . .Eventually it will get to within the thermal noise level of the final value, which is the practical limit. For example for a 1µF capacitor at room temperature this occurs after about 17 time constants (thermal noise = 64nV) for a 1V supply.