Calculating greatest current drawn from cap

[Peter_wadley]

Howdy

How do you calculate the peak current draw from a capacitor?


I have a 33000 uf cap..

I am charging it with 22v..

No resistor is used..

The explanations on wiki involve integrals .. my enemy.

 

[dknguyen]

It's just V=IR if you assume no inductance anywhere.

An ideal capacitor has zero parasitic resistance and inductance. Ideal connections have zero resistance and inductance. Ideal voltage sources have zero resistance and inductance and instantaneous transient response.

So with all this in mind, the current is infinite. You have to take into account some resistance or some inductance. In reality if you have no added resistance it is the resistance and inductance of the traces that come into play. THis is because an ideal capacitor when empty appears as zero impedance to an AC signal of infinitely high frequency (and we would probably assume this since we don't know the rise time/bandwidth or inductance without really REALLY getting into some nasty measurements and calculations.

So you can do a few things:
-make sure your voltage supply can provide way way way more current and hope that it can supply that initial current pulse...
-or use a resistor in parallel with a voltage controlled switch and when the cap charges up enough then close the switch to bypass the resistor
-guestimate the resistance (you can "probably" assume zero inductance) and use it in your calculations and go from there.
-you could also add an inductor which will control the rise of current and thus the peak level of the current, and it's inherent resistance will also help to limit the current pulse (and this resistance is probably listed for that inductor as well).

 

[Hero999]

If the ESR dominates the current will be very high which is the worst case.

However, my gut feeling tells me that the inductance will dominate, especially if the ESR is very low. The problem with this is that the capacitor will charge up in the wrong direction for a short length of time. You can get round this by adding a reverse polarity diode in parallel with the capacitor.

 

[Boncuk]

The max current to be drawn of a capacitor depends on many factors as already posted here.

As a rule of thumb you can use: 1F is the capacity which allows a current flow of 1A for a duration of 1 second.

A fully charged 330.000uF capacitor can deliver very high current for a very, very short time.

Bosco

 

[Roff]

The max current to be drawn of a capacitor depends on many factors as already posted here.

As a rule of thumb you can use: 1F is the capacity which allows a current flow of 1A for a duration of 1 second.

A fully charged 330.000uF capacitor can deliver very high current for a very, very short time.

Bosco

You left voltage out of your "equation". It would be accurate to say that if you charge 1 Farad with 1 Amp for 1 second, the capacitor will charge to 1 Volt. For constant current charging (and discharging), V=I*T/C. Given any two, you can solve for the third.
However, I don't think any of this is very relevant to the OP's question.