i =3D C*dV/dt. A passable estimate, therefore, is
2.2u*400/20u =3D 44A.

Mike H

On Wed, Nov 2, 2011 at 10:45 AM, Electron <electron2k4@infinito.it> wrote:

>
> Hi all,
> a 2.2uF capacitor charged to 400V discharges in 20uS: what is the formula
> to calculate the average discharge current? Huge precision is not
> necessary,
> it's just to size appropriately a MOSFET/BJT switch.
>
> Thank You very much,
> Mario
>
> PS: I wanna play in the game too! Here's my attempt, maybe correct, maybe
> not:
>
> I'm using two formulas. The first is:
>
> Coulomb =3D Farad * Volt
>
> thus I get 0.00088 Coulomb inside that charged capacitor.
>
> Second formula:
>
> Ampere =3D Coulomb / second
>
> thus, if I'm not wrong, the right answer should be:
>
> 44 Ampere
>
> Am I right or wrong?
>
> If I'm right, then (since this is a pulse application) I need a switch
> which
> can handle peak currents of more than 44A (say 60A to be safe).
>
> Did I do my calculation correctly?
>
> I'm 100% self-taught *blush* :o
>
> Cheers,
> Mario
>
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