# another stupid capactior question

• posted

I would like to know in common sense terms what the hell a capacitor does. is this accurate?:

a flaky low voltage charge comes through a wire and hits a capacitor, the capacitor will accept the charge and "charge up", when a certain amount (specified by the farad rating) of charge is reached the capacitor (slowly?) dissipates the charge in a more uniform and often higher voltage.

so a capacitor can "clean up" a flaky or unpredictable line of current/charge?

thanks.

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• posted

The simplest way to describe what a capacitor does is as follows:

A capacitor is a device which stores electrical energy in the form of an electric field.

That's it; that's all there is to it. Any time you have an electric field between two points/electrodes/whatever, it represents a certain amount of stored energy, and the "quality" of the space between those two points which enables this energy storage is "capacitance."

For a more concrete visualization, imagine it this way. If I have two flat conductive plates, separated by some distance (the important thing being that the plates are NOT in contact, and you can't get a conductive path between them - including via an arc), then I can put opposite charges on the two plates (one "+" and one "-"), and those charges will be held there by the simple fact that opposite charges attract.

But how much charge can you put on the plates? If you remember that *like* charges repel, you might also think that at some point, there will be so much "like" charge on a given plate that you can't force any more on there. Or, more correctly - it will take a little bit MORE "force" to put this next bit of charge onto that plate than it did to put the last one on. That "force" is voltage - in other words, the more voltage (which can be viewed as "electrical pressure") I have, the more charge I can force onto the plates before they won't accept any more. The relationship between the amount of charge you store and the voltage it takes to store that charge, for a given physical situation, is the "capacitance" of that particular situation. It is determined by the arrangement and size of the plates, the properties of the physical material separating them, etc.. The relationship between charge, capacitance, and voltage is given as

Q = CV or C = Q/V

where C is the capacitance (given in a unit called "farads"), V is the voltage, and Q is the charge in coulombs.

Clearer?

Bob M.

• posted

It doesn't need to be low voltage.

It may also come thru a resistor or inductor or active device.

Electric charge is measured in coulombs. It is one of the *fundamental* units of measurement. Capacitance is measured in farads which, as the equation Bob showed, includes a voltage variable.

What you described is an integrator.

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An integrator includes a series resistor at its input and is one of the many ways a capacitor can be used.

You have described a power supply filter. It is closely related to the integrator and is another of the many ways a capacitor can be used.

A capacitor does NOT have gain. You can't get more voltage out of it than you put in (without adding more devices).

Feeding a tech term into Google thusly is usually useful:

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A good place to start learning about a device or phenomena is:

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Notably, that article has a DC Circuits section and an AC Circuits section.

• posted

One good way to understand how a capacitor works is to look up "Leyden jars", which were the original capacitors. And they worked with high voltage, not low. Charge itself does not have a voltage.

The capacitor will present a voltage when it is charged. The charge is flowing in the wire because of an applied voltage. The flow will stop when the charge on the capacitor equals the applied voltage. There is then no potential difference to drive a current (flowing charge). The basic equation is Q= C*V, so that tells you how much charge (Q) must be stored in any given capacitor, C, to get a given voltage.

• posted

If you look at a capacitor in terms of voltage and current in a graphical way, set the current to vertical access momentarilly. Assumming the capacitor is at zero charge, initially, when the switch is closed the voltage is applied, but momentarilly short circuit current flows through the capacitor. This current reduces exponetually by the natural logarithm or 2 over time until the current aproches a very small static value almost zero depending upon the capacitor. This is how Capacitors behave in electric circuits.

• posted

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*-could-care-less+*-couldn't-care-less&num=100

• posted

the plumbing equivalent of a capacitor is a pipe with a rubber diaphragm across the middle, blocking it. No net steady current could flow through, obviously; but a short pulse of pressure (voltage) would go through, as a little bump of current. If a pressure is applied to it then current flow blocked, it will retain the pressure internally until it is allowed to release it with a small movement of current. The wider the pipe, the bigger the capacitance, and the more current it will soak up until it reaches an equal back pressure.

That above describes a capacitor in series; I suspect you are talking about a shunt capacitor, in parallel; the same analogy still holds. It won't affect steady current flow, but short pulses of pressure or random noise will have a path through the capacitor that bypasses the circuit further downstream. Same idea as a muffler on a car exhaust.

• posted

I know I'm little late on this thread, but it piqued my interest.

Capacitors behave differently on AC than they do on DC. Applying DC power directly through a capacitor will initially pass decreasing current while it charges up to capacity. Once its "full" for the given voltage it will block the current. Applying AC power directly through a capacitor can "smooth out" "noisy/spiky" power, but the resulting voltage waveform is not a smooth sine wave. These simplistic statements regard the capacitor as an individual device with power fed directly through it. When arranged with other devices such as resistors or transistors, capacitors are very versatile.

A previously thread mentioned wiring capacitors to obtain the sum of their capacitance. Capacitors in parallel do sum their capacitance. 10mF +10mF =

20mF

When wired in series, the capacitance is reduced. (the inverse of the sum of their inverses)

1/(1/10mF+1/10mF) = 5mF

Scott

• posted

```--
That's not true.

For instance, if you supply a nice sine wave to a load through a```
• posted

Thank you for that clarification. I did not even consider the behavior of a high pass filter. Doesn't a capacitor in series smooth out voltage spikes in the amplitude of a sine wave, if not the frequency?

Actually I was referring the the "stepping" that occurs to the attenuated waveform. But because it lets high frequency junk through, no it does not do much for noise without other of filters.

I appreciate that alternate formula. When pairing capacitors up in series like this, I seem to recall the effect of doubling the voltage rating? Scott

• posted

The short form answer to this is to note that a capacitor represents a component whose impedance decreases with frequency. "Spikes" and other such noise are high-frequency components, so a capacitor which appears "across" such signals will decrease their amplitudes more than they affect the low-frequency content (which is what you're trying to preserve).

Note that that is not how "capacitor in series" would normally be interpreted, though.

Yes, but don't push your luck with that. "Doubling the voltage rating" assumes that the voltage across the pair is exactly split between them, which is generally not the case. And it's definitely not the case if the capacitors are of different values.

Bob M.

• posted

```--
No.

What a capacitor is is an energy storage device with a reactance```
• posted

Ah, I see.

I thought of that just after posting. I was considering a special case where two identical caps were used to obtain a given capacitance at a higher voltage than a single cap would allow.

Thanks! Scott

• posted

I think so.

Not fully at this point, but I am still mulling your detailed formulae example over. Thank you for all of the time taken to explain this.

The "stepping" I remember may have been a voltage harmonic(s) in a simulated circuit. We had a oscilloscope hooked to a ceiling fans induction motor at one time too, but the induction of the windings may have contributed to that profile. I am very fuzzy in my knowledge of inductors.

I am working with split pole induction motors with a single running capacitor (4µF 350V) to phase shift the directional winding. Lower speeds are provided by another pair of capacitors in series. (Medium is the two speed capacitors in parallel, in series with the rest of the circuit. Low is one of the speed caps in series with the rest of the circuit.) This decrease in voltage seems to relate directly to your formulae, no?

We have encountered using an incorrect size speed capacitor can cause a motor hum, and I had thought it was the capacitors introducing the step in the sine wave. Now I am thinking a voltage harmonic (still perhaps from the caps?) may be a better description of the problem. The type of hum we are encountering does have a slow harmonic repetitiveness, instead of a steady

60 cycle hum or even the buzz found with spiky/noisy voltage. I wish we still had that o-scope.

Yes, I thought of that only after I had posted. It was a special case.

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