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Charging and discharging a capacitor???

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snakegaer

New Member
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 :)
 
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Divtech

New Member
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.
 

snakegaer

New Member
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.
 

rainman1

New Member
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.
That is incorrect.
You can charge the capacitor to a level thats pretty close to the supply voltage.

Afet RC sec, you'll reach 2/3 of the supply voltage when charging,
and 1/3 of its initial voltage when discharging.
 

snakegaer

New Member
i know that it won't reach supply voltage
but my question was what is the reason why it won't reach the supply voltage? and why there is some voltage left in the capacitor when fully discharged???
 

crutschow

Well-Known Member
Most Helpful Member
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).
 

snakegaer

New Member
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).
thanks that helped a lot :)
 

tresca

Member
think of voltage as 'pushing power'
Your supply as a certain voltage (ie. pushing power). When the capacitor is completely discharged, the power supply will have to resistance to its pushing power. But as time increases, the capacitor charges (develops some type of resistance to the pushing power) and begins to slow its charge down. At some point, both voltage of the power supply and capacitor would be relatively equal, and so its harder for the power supply to 'push' the capacitor.

Pretend you are getting bullied. A big bully is pushing you around, and you take fall and cry. Then you tell a buddy about it, and he comes to help you out, now there are 2 of you and one bully, but the bully is strong enough to knock both you down. And so you tell another friend and so on. At one point the bully is going to have a hard time pushing you down because you got your buddies to help out.

Thats just a super dumb example, but it makes the point clear. At least I think so.
 

crutschow

Well-Known Member
Most Helpful Member
The capacitor voltage is asymptotic to the supply rails.
define: asymptotic - Google Search
In principle, it never gets there.
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.
 

Willbe

New Member
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.
Fer' sure, e to the -17th is a small number. . .:D
 
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