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How to calculate time required to charge a capacitor? Answered

I have a 1.1μ capacitor with voltage capacity of 440V. I am inserting a 1000Ω in series. My question is how much time will it take to almost completely charge if I'm applying 12v at 0.5A?



6 months ago

To fully charge that 440v capacitor to 440v, can you guess how much voltage you would need to apply to the capacitor? Yup. 440v minimum.

The physicists will tell you that even with 440v supply, the capacitor will never see the full 440v across it, because the closer it gets to being fully charged, the slower it charges. it only converges in the limit as time -> infinity.

If you derive the differential equation you see that the RC network will have an inverse/flipped exponential characteristic if you set your initial conditions (initial capacitor charge) to zero. Also from a frequency or S domain, you can see that an RC network is a low-pass filter. You can use a fourier or laplace transform to convert things over and see for yourself! That is a very helpful way to analyze RC circuits when feeding them AC sine signals (like 60hz ac mains).

same problem bro

There exist circuits for charging capacitors to 100s of volts DC (e.g. 300 VDC), from a much lower voltage source (e.g. 3 VDC, or 1.5 VDC), and the usual place to observe such a circuit in the wild, is in the circuit that charges the capacitor needed for a camera flash tube.

Essentially this circuit is a kind of DC-to-DC converter, and in general, something like this will be required if you want to charge a capacitor to a voltage much higher than that of the voltage source actually supplying the energy.

Just connecting the capacitor and low-voltage, voltage source together, through a resistor, will charge the capacitor to that low voltage; e.g. connecting a capacitor to a 12 volt battery, will charge the capacitor to 12 volts. Connecting a capacitor to a 3 volt battery will charge the capacitor to 3 volts.

Speaking of energy, the amount of energy stored in a charged capacitor, is U = 0.5*C*V^2, where U is energy in joules, C is capacitance in farads, and V is voltage in volts. Because of that factor of voltage squared, that is kind of where the energy is.

So to actually maximize the stored energy, you need to charge the capacitor to a voltage close to, but less than, its maximum rated voltage. I mean, this is kind of the meaning of, "full", or "fully charged", for a capacitor, in terms of energy.

Typical size of a photo flash capacitor is about 300 uF, at 300 volts, which corresponds to stored energy of,

U = 0.5*(300e-6 F)*(300 V)*(300 V) = 13.5 J

I think camera flash charging circuits are pretty easy to find, about as easy to find as a broken camera. Although if you just want to study the camera flash charger circuit, just from like looking at circuit diagrams, I usually recommend this page,


Also, I think this circuit is an example of a , "flyback coverter", and there is a Wikipedia page with that title, here,


As always plus one..

But Strong language to a newbie learning eelectronics..


7 months ago

RC = 1000*1.1*10-6=1.1*10^-3 = time to raise the voltage 66.7% of 12V.

The capacitor will reach 8V in 1.1 milliseconds..

And it will never reach 400 volts !

You cannot charge that cap with a half amp..

The current will start at 12 milliamps reducing to 0.67 ma after 1.1 milliseconds and eventually approaching Zero !

Thanks for the reply. By the way, what of I use input as 220V 15A? If not, then what should I do?

It has to be DC not AC and what resistor..

You can never exceed the supply voltage,

R is in ohms,

C is in Farads

Time is in seconds to reach 2/3 of supply DC voltage..

Current is a variable !

You can do that math for yourself :-)

I don't know, what you want to do...

I know I'm confusing you. But I just want high power. I can't use direct AC because it blows my circuit off no matter what I do. I have a transformer but it's step down. I just don't know what to do then I thought of using capacitor but I don't know when to stop giving capacitor power. I just want high currency.