Question about capicitors

Capacitors store charge; charge is a function of current and time.

For the purposes of the question you ask[*], you can think of a capacitor as a variable resistor - it starts at near-zero ohms, but as it charges, its resistance increases, tending towards (but never reaching) an open circuit.

But in real life the behavior of capacitors is more complex than that.

Capacitors pass current in proportion to the rate of change of voltage across them. The proportionality factor is the value of capacitance that describes the capacitor. The formula that relates rate of change of voltage across and current through a capacitor is I=C*(dv/dt), with I in amperes, C in farads and dv/dt in volts per second.

So, in answer to your initial question, yes, capacitors pass current as they charge up to some DC voltage across them. They act like short circuits only when you try to produce an infinite rate of change of voltage across them. i.e. attach them to a very stiff voltage source that tries to charge them up to that voltage, instantaneously.

On Mon, 2 Oct 2006 15:54:08 -0800, in message , "Dana" scribed:

The technically correct answer to the question is "no." An uncharged capacitor, when subjected to the application of a direct current source, such as a battery, will indeed allow current to pass while it accumulates charge. Technically, however, the charging current is a transient current, and subject to different mathematics than direct current. In the math model, one doesn't see direct current until the capacitor is charged and the transients settle.

yes, a spike of current will flow through it until it's charged up.

this is the capacitor passing the change in voltages as zero volts becomess some other fixed voltage.

Bye. Jasen

current,

Correct me if I'm wrong, but I would have thought that ALL you'd see IS direct current, while the cap is initially charging, and that from a source with a low resistance, at the instant that the voltage is applied, the capacitor "appears" to be a short.

... Johnny

They act 'like a short circuit' to brief impulses. No net DC flows though.

Graham

You mean if the capacitor is in parallel with the source and load? Yes, the current will be split between the load and charging capacitor. Then when the cap is charged it will tend to maintain stead flow to the load if the source fluctuates or is pulsating. It's called filtering. Current leads the voltage.

Indeed so. One of the reasons we use gated power for switchable power domains is to 'soft-start' the inrush current. For the OP - check out hotswap controllers; you can find them easily on all the major mfrs - Linear tech, AD, National, TI, Maxim to name just a few. If you read the datasheets carefully you'll see that the switch is slowly turned on to present some series resistance during that 'spike', as Graham puts it.

A test on one ceramic cap (4.7uF) showed that applying raw power at

Cheers

PeteS

I have in fact seen supply rail decoupling caps literally 'catch fire' due to switch-on transients.

Graham

just think of a CAP as a dead battery. what happens when you put the charger on it? it draws a lot of continuous current until the battery is saturated (charged), this current causes flow. flow is flow, so this means depending on the size of the cap, charging source it has to work with and load on the other side, will gauge the time it take's for it to get fully charged and not cause flow in the circuit any more. i hope that was a simple enough explanation.

I have seen some impressive pyrotechnics from caps in my time and certainly switch-on transients are up there for the root cause.

Cheers

PeteS

On Tue, 3 Oct 2006 18:36:44 +1000, in message , "Johnny Boy" scribed:

Yes to the last statement. Taking a simple series circuit containing a capacitor: at t = t0, applying a "DC" source, the capacitor drops zero voltage (axiom: the voltage cannot change instantaneously across a capacitor), and hence will at that instant draw a current equal to the voltage source divided by whatever resistances are present. The voltage across the cap will then rise - as the current falls - logarithmically until the voltage stabilizes at the "DC" calculated level (tn - in this example, that level would be zero). The current is a curve - a transient - between t0 and tn, and as such isn't calculated by steady-state DC equations. That's what I mean when I say that until tn, there is no direct current in the circuit, rather there is only transient current.

Some DC does flow through during the charge time. A sometimes useful "trick" is to put a large cap in series with a relay coil. When power is first applied, the relay "flicks" on briefly. But, as you indicated, no DC flows through following that (and ignoring leakage).

Ed through.