Somebody explain capacitors

[forbin80]

Ok, I've read explanations about capacitors over and over, and maybe it just hasn't sunk in.  I understand that they build up charge on one of the plates, but I don't really understand how they affect the flow of a circuit.  For instance, in an effect circuit, why does a larger capacitor enable lower frequencies?

What I would like is for the people who have a good understanding of this to explain in their own words what capacitors do exactly.

Thanks!
Whit

[R.G.]

First, read the first part of How It Works at GEO,

http://geofex.com/Article_Folders/How_It_Works/hiw.htm

That gets you to the "charge on the plates" level.

Think of two capacitors, one only 1" Square (25.4mm square) and one 12" square (305mm square). Imagine that the plates are not metal, but hollow water containers, each flat and with the iner sides glued together and both filled with water. The leads are hoses in and out.

Lets say I want to pull water through that water-capacitor. I suck on one hose, the hose on the other side being tossed into a bucket of water. At first slurch, I get water as I pull it from the hollow in the plate I'm sucking on, and as water fills up the other side. At some point, the flow to me quits because I've sucked out all I can and the other side is as full as it can get. So flow stops.

I try this on both capacitors, and I find that the big, wide one can supply water much longer than the little one, but still cannot supply continuous water through it.

If I reverse the pressure and try to press water into the side I have hold of, it initially fills with water, but then fills up and the other side empties.

The pressure I apply to suck/fill the hose on one side is like voltage - the higher the pressure/voltage, the more water/charge I can pull out/force in before flow stops. And the bigger one has it's 144-times bigger area to fill, so it lets in/out 144 times as much water/charge.

So a continuous flow can't get through (unless the pressure is so high it ruptures the chambers, exactly analogous to high voltage destroying an electrical cap). But if I suck, blow, suck, blow repeatedly, I can keep some water moving into and out of the capacitor all the time. That is, both water and electrical capacitors can pass **alternating** currents.

And I notice that the big cap can be counted on to pass alternating currents to a much lower frequency than littler ones, because the littler ones fill up and empty faster, and so block the alternating flow at a much higher frequency than the big one.

You could, in fact calculate the "resistance" of the water-capacitor to water flow. At DC, there is a transient, and then the resistance to flow is infinite. AT some alternation frequency, you could measure the amount of pressure that you had to exert to get Q amount of water to move back and forth per second.

We could in fact calculate that the "capacitance" of the two water plates is C = Qp, Q being the amount of water "forced in" and p being the pressure forcing it.

For an electrical cap, that's C = QV, where Q is the charge and Vis the voltage.

Then the "resistance".

There's not a simple handwaving derivation I know of for this one. The amount which our water-capacitor "impedes" water flow is

Z = 1/(2*pi*F*C) where F is the alternation frequency, C is the water-capacitance as defined above, and 2*pi is to make the numbers come out right.

Electrically Zc=1/(2*pi*f*c), same form of relationship, different units.

But that's how they work.

[niftydog]

A simple way to think of it is this; caps block DC, but pass AC. The amount of AC they pass is dependent on the capacitance.

A low pass filter shunts the high frequencys to ground while passing the low frequencys.

A high pass filter blocks low frequency signals while passing the high frequencys.

why does a larger capacitor enable lower frequencies?

Thus, this must then be in relation to a high pass filter. Adjust the capacitance in the filter and you effect how much of the low frequencys are passed/blocked by the filter.


there's plenty of more in depth explanations out there, but here's my abriged version;

caps store charge. (not voltage, not current, charge!)

Charge a cap with a DC voltage. As the charge builds up, so does the voltage differential between the two pins. It quickly reaches a point where the VDC across the cap is very, very close to the VDC that is providing the charge to the cap. This essentially stops any further DC current flowing through the cap. Hence, we can assume that caps block DC.

However, a cap cannot remain fully charged when being fed with an AC signal. Everytime the signal voltage is going down, the cap is discharging. The smaller the capacitance, the faster the rate of discharge and vice versa.

A large enough cap being fed with an audio signal will charge up but will NOT discharge quickly enough, resulting in a reduced audio signal amplitude. (The troughs of the AC signal are altered by the discharging cap, causing the peak to peak amplitude to drop)

Even larger caps will not discharge hardly at all, causing the audio signal to be "converted" to a pure DC voltage. Due to the different rates of charging/discharging, different caps will affect different frequencies in different ways.

Again, smaller caps charge/discharge quickly, meaning they have less effect on low frequency signals. Large caps charge/discharge slowly, meaning they have a greater effect on low frequency signals. The cap response can be calculated and used to create filters at specific frequencies.

Google "RC time constant", "low pass RC filter" "cutoff (or -3dB) frequency" "reactance" and "impedance" etc etc for more detailed explanations, otherwise, visit my favorite http://www.tpub.com/electronics.htm and search on "capacitors" and "filters" etc...

[forbin80]

Ok, after reading your answers and looking at a couple of other things I think I see where my confusion has been all these years.  I always read that charge builds up on the plates (+ on one plate, - on the other), but that electrons don't flow between the plates.  So I thought, doesn't this break the circuit, if the electrons can't flow?  But I think I see now that with an alternating current, the buildup/decay of charge creates an electric field within the atoms between the plates and that in a sense allows current to flow "through" the capacitor.  Is this right?

Thank you for taking the time to write those explanations, both were extremely helpful.

[forbin80]

So in the case of the audio probe (just the 1/4 inch phone jack w/ a grounded sleeve and a capacitor connected to the tip) what is the purpose of the capacitor?  Just as protection to make sure only so much current can flow to the amp?

[Paul Perry (Frostwave)]

So in the case of the audio probe (just the 1/4 inch phone jack w/ a grounded sleeve and a capacitor connected to the tip) what is the purpose of the capacitor?  Just as protection to make sure only so much current can flow to the amp?

Yes, you want the signal (AC) to go through, but not the DC (especially, if the thing you are probing is a valve amp and there is 400V DC on the other side of the cap :shock:

And here is a way to think about why higher frequencies go thru caps easier: since the AC 'goes through' the cap (or appears to) by alternatively charging and discharging it, a signal at twice the frequency does so twice as often, so there is twice as much current flow :wink: