#### BobaMosfet

- Joined Jul 1, 2009

- 2,123

You're not looking at this correctly. A capacitor NEVER passes current. If it does, it's failed. A capacitor is merely a device that enlarges the diameter of a conductor to outrageous proportions- upon whose surface electrons pile up unable to jump the gap (through the dielectric).Standard answer: iequalscdvdt, as the man behind the CapSite has named it. So there is the

capacitance to start with, and the ratio dv/dt to complete the product that ought to give you the

momentary charge current to the cap. But how ... ? Consulting textbooks and tutorials, you are

bound to find a phrase like: 'presents a near-short circuit', to describe the cap's behavior after

switch-on. Few if any specify the 'near'-part, however. The only thing to do then, is go look for an

R (i.e. in the cap itself, not the obvious resistances in source, wiring, etc.). A glance at one of the

many different equivalent circuit diagrams for an e-cap, reveals its series resistance ESR, which

indeed turns out to be less than one Ohm usually. And by applying this man's familiar law, we can

finally make more than an educated guess as to the value of the charge current surge. This is

because - while dv in the original formula is not easy to determine - V in Ohm's Law simply is the

full applied voltage, as there has not yet been generated a counter e.m.f. (where the 'm' stands for

'motive', not 'magnetic') directly after switch-on. So the inrush current surge equals the voltage

applied to the e-cap, divided by its ESR, or does it ? ....

I invite and await your comments and criticism as to method, reasoning and outcome of the above.

(Please try to apply as little math as possible - thank you).

When you put voltage across a capacitor you are dropping.losing volts (ie. potential for current to move), and it appears as a short simply because current is flowing off the leg into the plate as fast as can be allowed by physics (if not impeded by a serial resistance of any sort). For all intents and purposes Dv/Dt is only applicable if an impedance exists. This why in graphs you see current start at zero in time, and voltage start at max, and they change inversely to one another. Current stops flowing and no further voltage can be dropped when the capacitor is pushing back against the supply with equal voltage.

Max current discharged, if the power-supply removed, is solely a function (ideally) of coulombs on the plate- movement is near instantaneous, so is max based on charge quantity. In physics, There is only electrons moving and the impedance to that flow- volts are an imaginary value representing the reciprocal of that ratiometric relationship between movement of electrons (neutralization of charge) and the impedance to that movement.

In short, impedance determines max charge from capacitor and the time delta.