basic capacitor question?

[dyarker]

Sorry I asked. I was hoping for a water analogy. I guess mechanical will have to do.

BTW, You do NOT get more voltage out than voltage in. Always a little less. With perfect components, at resonance a parallel tank has infinite impedance. A perfect series LC circuit would have zero impedance at resonance. No such thing as perfect, so there is some loss. The farther from resonant frequency, the greater the loss. If there is an apparent voltage increase, then the inductor is actually a transformer with a step up secondary, or the input is to a tap part way up a single winding.

[Chris Smith]

Actually you get more volts out than you put in. [Smoke and mirrors]

Same or similar principle to the buck/ boost pump in that the timing is the key.
[but more naturally]

The current and voltage are different and accumulative in phase, thus increasing the voltage, not decreasing it.

Kind of like forcing things until they bunch up and get a small push up the rear, increasing things ahead of you.

But there are no free lunches in physics so the amperage drops slightly with this voltage increase.

However X+Y still = exactly Z [minus friction]

If you examine the fly wheel you also get this exact gain, for the same loss. We call that inertia.

The water analogy is the mass of the weight of water exactly like the electrons being more massive and traveling in a smooth and uniform direction and then using a smart and natural phase called the speed of the electron, light, etc. Timing is everything.

At certain points you get the bunch up in timing and voltage which gives the reading a slightly higher number.

As with all physics, the SUM is equal and always the same, but at certain points in time [the peaks] the voltage rises giving the over all voltage [the over all impression] a higher number. [RMS]

[MrAl]

Hi again,

Water analogy? To what?

Not sure if this helps or not, but...

You have a large pump with a 24 inch diameter outlet pipe that
pumps water into a huge reservoir from a lake. Somewhere at the
bottom of the reservoir you have a 1 inch diameter pipe that runs to
your kitchen sink.

A very large amount of water is quickly pumped into the reservoir
but only a small amount can leave the reservoir because the diameter
of the small pipe is so tiny. Thus, the reservoir acts as a capacitor
(or inductor depending on how you look at it). The thing is, even when
the big pump is turned off you can still get water to the kitchen because
there is much stored in the reservoir.
Another way of looking at this is...
if you keep the kitchen faucet turned 'on' so it's constantly pouring
out water down the drain, the water flow is constant even when the
huge pump at the reservoir is turned off. This is the way an inductor
or capacitor works...the inductor tends to keep the current flow constant
while the capacitor tends to keep the voltage across it constant. In this
way they both can be used as filters, to help keep a DC current constant
rather than have ups and downs in it's value.
Instead of the inductor or capacitor storing water, they store energy...
the inductor stores it (from a pure electrical view) in the form of a
current, while the capacitor stores it as a voltage.
Of course there is a little more to it, as the inductor works via a magnetic
field and usually orientation of macroscopic domains in it's core, while
the capacitor works on electrostatic principles, but for many many
problems you need not consider these physics but rather take the
more simple approach of looking at these circuit elements solely by
their electrical properties.

One thing i think beginning electrical students overlook is that the
electical properties are not hard to master. Even easier than trying
to visualize water flow through a pipe and trying to measure pressure
differences...which is much harder than using a volt meter

[dyarker]

There is NO voltage increase with tank circuits, even with smoke and mirrors.

The principle similar to buck/boost do not apply. In buck/boost a DC current is turned on and off. As current flows though an inductor a magnetic field is created. When the current is interupted, the field colapses and tends to keep the current the same as before the interuption. If the impedance is higher than during current on, the voltage will be higher (kick back). By using diodes this voltage and resulting current is subtracted (buck) or added (boost) from/to the supply voltage. The capacitor is a low pass filter. The inductor and capacitor in buck/boost circuits are NOT an LC tank.

[Externet]

Hello Dale.

..." Sorry I asked. I was hoping for a water analogy. I guess mechanical will have to do. "...
Perhaps you missed the suggestion replied on previous page.

As a diaphragm insertion may work for the capacitor; a propeller inserted in the pipe would be the inductor analogy. They are both mechanical elements inserted into the hydraulic flow for the water analogy.

The parallel and series resonance is still hard for me to visualize as analogy. Need a brainwash...

Miguel

[Chris Smith]

Nope, .....first year electronics and first year physics show you a tank circuit and why the voltage can be higher at the output than the input.

RMS is like a series of waves going in circles in the tank circuit.

When the wave behind catches up it shores up the wave front, in front of it, raising the peak to peak value, thus raising the RMS value.

Fact. Higher voltage reading.

Not all tank circuits are designed to “raiseâ€

[rshayes]

The electrons don't actually pass through the capacitor. A excess of electrons on one plate creates an electrostatic field which forces electrons off of the other plate. The field can propagate through the dielectric layer without requiring the electrons to actually move through the dielectric layer. On the other side, the field can in turn move other electrons.

An easier way to view a capacitor is to look at the relationship between voltage and charge. The voltage across a capacitor is equal to the charge in the capacitor divided by the capacitance. If the voltage changes, the charge changes, and this is interpreted as an electric current. The current is proportional to the rate of change of voltage.

When the voltage applied does not change (DC) there is no current flow.

A fluid analog to a grounded capacitor is a cylindrical tank of water. Here the pressure at the bottom of the tank is proportional to the amount of water in the tank. The pressure is the analog of voltage, the volume is analogous to charge, and the cross section of the tank is analogous to the capacitance.

An ungrounded capacitor may be a bit trickier to visualize using a fluid model.

[MrAl]

Hi there,

Im not sure how we got on the question of whether or not the output
voltage from a 'tank' circuit can be higher than the input, but here
is something to think about from a real example...

Take a circuit: an AC voltage of 1v peak and put it in series with
an inductor also in series with a cap. Measure the output voltage
peak AC that develops arcoss the cap.
Using an inductor of 1 Henry and a cap of 10,000uf and set the AC
frequency to exactly 1.59 Hertz the voltage across the cap measures
very close to 513.84 volts peak.
Thus, a sine wave voltage of 1v peak produces a voltage over
500 volts, also a sine wave.
Why the increase? Because in the equation for this network the
denominator tends toward zero as the input frequency approaches
"fo", where fo is usually called the 'center frequency', or
'resonate frequency'.
Tune the ac off center frequency and the voltage drops considerably.
In theory, using perfect components (no losses) the voltage is
infinite when the frequency is EXACTLY at the center frequency
of the network, even if the input voltage is very small like 0.001 volts.
How is this possible? Because at the exact center frequency of
the network the denominator of the equation is zero, and a zero in the
denominator means even with a tiny tiny voltage (which is in the
numerator) the output (result) is infinite.

Here's the equation in case anyone wishs to fool around with it...

Vo=v*(1/sC)/(sL+1/sC)
where
Vo is the peak voltage across the cap
v is the input voltage ac peak
C is the value of the cap in farads
L is the value of the inductor in Henrys
s=j*w
w=2*pi*frequency
j is the complex imaginary operator


Simplified (to show how the denominator might tend to zero):

Vo=v/(1-w*w*L*C)

Note in the above simplified equation with some values of w,L, and C,
the denominator will become equal to or very close to zero, which
will, even with a very small value for 'v', make the output voltage Vo
very very large.

Using a frequency sweep of say from 0 Hertz to 2 Hertz there would be
a large blip around 1.59 Hertz.

[Chris Smith]

If you have ever surfed in your life, you know the first wave behind yours is re-enforced by the front wave picking its front higher [your wave] ,, thus all the analogy in physics and electrons are really the same.....

First rule of thumb, all of life is the same, physics, doctorates, mechanics, etc,.......doesn’t matter what you seek?

The rules are the same.

You cant beat it, you cant cheat it.....

[MrAl]

Hi again,

Chris:
i just wanted to show some solid proof that what you said was true
about the output voltage could be higher than the input. I thus used
pure circuit analysis, and to keep it simple i left out the resistance
(even though it is always there in real life). I found mathematically
that all the resistance does is to eat up some of the energy but not
all (depending of course on size) so the output voltage isnt as high
as it would be without the reistance, but nonetheless still higher
than the input for the proper choice of frequency.
Of course not all frequencies will show a gain such as this, only
a narrow band of frequencies...all others will produces less output
voltage than input voltage...but i guess we all knew this part.

Next time i'll include the resistive part in the equation just for fun

Of course everyone is certainly welcome to verify all this using
their favorite circuit analysis program. All you have to do is
put a inductor, resistor, and cap in series and excite it by
an AC generator, and measure the output voltage across the cap.

One little side note:
Note that since the voltage across the cap is very high (could be 500v
or more) and the input generator is only 1v, that puts a high voltage
across the inductor also.

[Chris Smith]

Al
My point of the smoke and mirrors was that it appears simply because of RMS, or a given point in time gives us a peak value, and then is measured over all and it Appears higher.

But by the same token this higher value is used in radios to lower the impedance going to the Antenna.

What we measure in this tank is the peak value, but the accumulative value always remains the same. There is no actual gain, just less of one for the gain in the other.

Physics is simple, electrons and waves act the same.

But for some,... this seems impossible because they never include time which is always the biggest factor.

When ever you deal with a cap, time is always the key to what happens.