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Charging Time of a Capacitor

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tom_pay

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Hi All,

I have been searching the internet for a while now about calculating the time it takes for a capacitor to charge/discharge. Unfortunately I dont understand the information given.

All I wish to find out is, if there is a certain voltage source applied to a certain value capacitor via a certain value resistor, how long will it take to charge up to a certain voltage. As well as this, if a certain value capacitor at a certain initial voltage is discharged through a certain value resistor, how long will it take to reach a certain voltage.

Is anybody able to explain it, in simple terms :), or know of any formulas that may be helpful.

Thanks so much, I am totally confused and Dr. Google hasnt been helpful.

Tom
 
Thanks Mike for such a prompt reply.

I have read the article and, I think, I understand it. However, there must be a better way of determining the RC values other than the graphs, is there a proper formula for their first graph?

Thanks so much,

Tom
 
1 time constant = Resistance x Capacitance as was mentioned in article. Also recall, it takes 5 time constants for C to be fully charged. When using the TC=RxC be sure to use correct units, see link as it is explained.
 
Ahh, Thanks Eric, that link was quite helpful,

So For charging:

[LATEX]V_{c}=V_{b}[1-e^{-t/RC}][/LATEX]

And for discharging:

[LATEX]V_{c}=V_{0}e^{-t/RC}[/LATEX]

Thanks so much

Tom
 
Hey Eric... :p just kidding.
 
Hi Tom,


Solving the equation:
Vc=Vb*(1-e^(-t/RC))

for t gives us:
t=ln(Vc/Vb-1)*RC

and then we could solve for RC if we needed to determine an appropriate resistor or capacitor:
RC=t/(ln(Vc/Vb-1)

Also, if the capacitor has some initial voltage across it before it starts to charge, we can use this:
Vc=(Vb-Vi)*e^(-t/RC)+Vi

where Vi is that initial voltage, and solving for t we now get:
t=RC*ln((Vi-Vb)/(Vi-Vc))

There are really two formulas for this though...
For Vb>=Vi we have:
Vc=(Vb-Vi)*e^(-t/RC)+Vi

and for Vi>Vb we swap Vi and Vb so we have:
Vc=(Vi-Vb)*e^(-t/RC)+Vb
 
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