Can a capacitor let DC current through?

Precisely! Or, as a sign in the men's room at a local drinking establishment succinctly puts it:

"We don't sell beer, we only rent it."

Bob M.

But on their trip to your house and back, they do work, like lighting light bulbs, cooking, heating water, etc. - so what you're paying for is the power you dissipate, or, over the course of a month, the energy that the power company sends to you, you use for awhile, and throw away. ;-)

Cheers! Rich

Ick! Who wants to be the second cusomer to rent a particular pint?

Ewww!

Thanks, Rich

We've noticed. Thanks for the reminder, Michael. :-)

Cheers! Rich

On Tue, 21 Aug 2007 23:02:53 -0700, vorange wrote: ...

OK. ;-)

Cheers! Rich

SNIP

SNIP

Part of the problem is that AC and DC are, as John said, extremely vague terms. They have no fundamental basis in theory other than as a shorthand in those cases where the context makes their meaning clear.

AC and DC are descriptive terms that can cause confusion whenever applied to anything other than a pure, time-varying sinusoidal voltage or a time-invariant voltage, respectively. That these are quasi-bogus terms becomes clear when we recall that they have no units! They don't appear in formulas. We can't measure the AC-ness or DC-ness of a circuit or device. Would electronics theory even miss AC and DC?

I wonder whether there are other descriptive terms used in electronics the way AC and DC are used. Offhand, I can't think of any at all!

As someone else pointed out, the confusion disappears when we consider the basic definitions of capacitors and inductors (which do not include AC or DC) rather than "capacitors pass AC but not DC".

Chuck

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Hi,

could you take a look at the first animation on this page (below). Is that what happens when (steady state) DC encounters a capacitor for the first time. Does the capacitor pass a small "blip" of a charge INITIALLY on startup and then block all further DC current.

If so then my mental model is right. I liken it to blowing air out of my lungs (in one breath) onto a paper fan. The fan turns at first (a blip) however I eventually run out of air in my lungs to blow out (just like the opposite plate of the cap runs out of free electrons to push away). And that's why the load (the paper fan) stops turning.

But there is that initial blip is there not? If so one should be careful when using caps.... what if the input pin of a microcontroller registers that 'blip' as a HIGH input signal and does something in response to it!? Would it be wise to place a resistor ahead of the cap if its going into some low impedance input pin of some digital/ logic chip?

Thanks to all for your help.

Yep. Once the voltage stops changing, the current fades toward zero.

Yes, there is. And if you look further down, it shows that you can extract the reverse of that blip by connecting a load across the capacitor.

The main capacitor you have to worry about is your body. If you walk across a rug, you may mechanically charge up your body capacitance and when you touch the processor, you can dump that charge into a pin. It can be a large enough pulse to damage the device.

A pseudo-science TV programme over here in Britland suggested that statistically in every glass of water that you drink is at least one molecule that was urinated by Isaac Newton!

SNIP

Well, the body is a capacitor in the sense that distributed somewhere in the universe is the "other plate" containing the opposite charge. The resulting electric field created by that capacitor is usually not strong enough to overcome the repulsive effects of the charges on the body when the body comes into contact with another object.

The problem with describing the body as a capacitor is that one might then imagine that touching one plate of a "conventional capacitor" will discharge it.

Pedagogically, it is probably better to describe the body as a charged object, which is not the way we normally describe a capacitor.

Chuck

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Think of it more like a chamber with a rubber diaphragm stretched across the middle:

---------------------- | / | | \\ | | / | ------- \\ ------- ------- / ------- | \\ | | / | | \\ | ----------------------

When you blow in the left, at first the diaphragm moves to the right, and displaces the air on its right. Eventually, the diaphragm "bottoms out", and the flow stops (or, you're blowing as hard as you can, even though the diaphragm is only part-way, but the back-pressure stops you). When you release the pressure, it blows the air back in your face. That's like the cap discharging, except that for the charge to flow like the air does here, you need wires and a complete circuit. :-)

That's the flaw I see in the fluid models of electricity - the fluid needs to be in a pipe. If you break the pipe, all the fluid spills out. If you break a wire, the flow stops. So they're kind of opposites in that respect, but otherwise it seems to work well.

Hope This Helps! Rich

Actually it's not so much a fluid, more a case of blue smoke.

When the circuit goes wrong, something breaks and you see the blue smoke all over the place.

It's actually a little more complicated. it helps if you understand a little calculus. The capacitor passes the derivative of the signal, i.e. the rate of change. So, with DC, there is no change, there is no current throughput. With your square wave, the signal changes where it rises and falls, so at those points you will get spikes through, on the flat parts, nothing, i.e. a spike waveform plus and minus around a baseline of zero. The reason "capacitors pass AC" is that AC as people generally think of it is made of sine waves, and the derivative of a sine wave turns out to be the same sine wave, just shifted a bit in time.