Capacitor recharging .

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Jagau posted this 2 weeks ago

Capacitor recharging
How to recharge a capacitor after having used the starting voltage?

We need to understand the behavior of a coil in the presence of a charged capacitor, I start the thread with an image of Vladimir Utkin which I was inspired by for my research and he has all the credit for it.
Image I focused on.

Important quote:
You need to charge the capacitor using the electric component of the electromagnetic field of the inductor.
The best way to understand what is going on inside the capacitor is to do this very simple little experiment.
Here is a very simple little diagram:

We charge a capacitor at +9 volts, and I discharge it in a coil via a diode (schottky) connected to an inductor. Very simple and uncomplicated instead of using switches you can do it manually it works great too.
We charge the C by closing Sw1 first to charge the Capacitor and we open SW1, then we close SW2 and importantly, we let it close because that's where the magic happens.
On the oscilloscope what we observe:

At the moment when the capacitor is connected to the diode it discharges from the maximum of its charged voltage and the positive curve appears, we note the starting hysteresis point followed by the discharge with time constant of +9 up to 0 volts (1.47 millisecond).

The negative curve is the interesting part and this is where we agree on most of our research especially when it comes to capacitor recharging.
When the capacitor is completely empty, the coil produces, depending on the inductance used, a reverse voltage (BEMF, electric field) from the magnetic field stored by the inductor.
We have here in a first phase, the discharge time (time constant) of the capacitor produces in the coil a rise time of the magnetic field in the inductance. And in a second phase, the decay time of the inductance's magnetic field produces a reverse voltage that recharges the capacitor in reverse polarity.

At zero crossing, the magnetic field stored in the inductor returns to the capacitor as a voltage (volt). This voltage is now negative from 0 to -7.22 volts and its maximum is almost equal to the original voltage minus the resistive losses of the circuit components. Viewing on the oscilloscope is a breathtaking detail of the behavior of two passive elements with at the end an onset of resonance from the diode and parasitic elements of the circuit.

We can say that time is doing things well for us and that this simple little circuit is close to unity.
So after understanding what is going on we could better try to understand and learn to work with Lenz's Law and  use it to our advantage. As Chris has often demonstrated, it can now be visualized.

For those who would like to repeat the very simple experiment your oscilloscope must be in SINGLE mode, most oscilloscopes today have this option, it takes an instant depending on the programmed trigger point at extremely fast speeds. If you need help asked, I will accompany you.

Taking advantage of it in our research allows us to visualize this effect and helps us better understand the very important effect of capacitor recharging in a circuit using a C and an L.
It should be remembered that a capacitor is a passive source and it only produces when charged. Let's keep it simple for now, just trying to figure out what's going on with the image the oscilloscope produces.

Other observations to follow.

ùjagau

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Chris posted this 2 weeks ago

Hey Jagau,

Just incase you have not seen it:

Itsu is a good experimenter, and I am pleased he gives you Credit for your work!

I do wish those guys would expand, just a little, the Envelope that they allow themselves to be Enveloped by! Thinking in Terms of Onesies is not sufficient as Jagau has pointed out.

I was a little shocked to read this comment:

It is important to remember that a diode blocks reverse current - not reverse voltage.

Ref: Verpies

Some may ask why?

Well, Voltage does not "Charge" a Capacitor!

Perhaps I have misunderstood the post?

For some readers, the Thread: Parametric Excitations of Electric Oscillations may be of further assistance in understanding this Simple and Very Valuable Experiment!

Humanity has SO MUCH POTENTIAL! It just needs to be Realised and Accepted.

Jagau, your Experiment is Exceptional! Thank You for Sharing!

Best Wishes My Friend,

Chris

Jagau posted this 2 weeks ago

Hi Chris

It is an experience which may seem trivial but which when studied corectly gives yo us an answer on the behavior of a capacitor discharging through a coil and a capacitor recharge.
Do not forget that I used an unpolarized capacitor as I represented in my schematic and also a schottky diode for better voltage return.
The reproduced image is in real time and it is taken as I explained in my post.

We can just imagine what we can do with `` DC resosant charging '' the topology for recharging the 2 capacitors on which many engineers have looked has just gone up in smoke using that method.

The rest of the experiment will surely be of interest to more than one, it follows the use of switchable inductance in real time who will be next experiment.

For those who have questions about what charges a capacitor:
When charging a capacitor from a source, initially the capacitor has 0 volts so the voltage is zero. When connected to a source, current flows through the capacitor and the capacitor voltage increases until it is equal to the voltage source. At this point, the charge current ceases and the stored charge is proportional to the applied voltage (q = cv), with a constant of proportionality called capacitance.

P.S to Itsu, If you do not want to see the parasitic resonance at the end of the waveform, place your scope probe on the anode of the diode, you will have a cleaner waveform.

Jagau

scalarpotential posted this 2 weeks ago

Is there a 0V ground  ref in the real experiment when SW2 is closed and SW1 is open?

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Jagau posted this 2 weeks ago

No  there is no other reference to the earth ground than that of the capacitor charged

by the battery or otherwise.

Jagau

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scalarpotential posted this 2 weeks ago

What I  am seeing is a half sine-wave for the current as measured by Itsu, this current goes through the inductor, the voltage on the scope is the derivative of the current, which is shown as the first half of a cosine, when the current stops because the capacitor is discharged, then the magnetic field collapses and it is forced into a natural oscillation until it is dissipated as shown on the scope. What is the function of the diode?

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Jagau posted this 2 weeks ago

Ok you it seem you are very confused.

First, I did not measure the current in my waveform, find out whoever tested it.

Only the return voltage is the BEMF of the inductance which returns in a single oscillation, the only oscillation is parasitic there is no other oscillation other than the unwanted one at the moment and that only oscillation is determined by the LC values.I have already written everything and explain this in my post at the beginning, please reread it carefully.

Did you try the circuit ???
Seriously you ask me what a diode is in such a basic circuit, wow
Try it with a diode and without the schottky diode you will understand how a diode works and what it is used for.
Try it out and show me your waveform?

Jagau

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scalarpotential posted this 2 weeks ago

I didn't build it, but I trust your waveform. Looking at the Utkin pdf, an H-field in the capacitor is confusing, it's probably the series inductance of the cap.

Negative voltage on the capacitor, what is the expected behavior?

I believe it is normal: when capacitor current decays (ITSU shows the current), it is a negative I change for the inductor, so the voltage on the inductor becomes negative, or positive when the probes on L are swapped, then the capacitor is charged with + on the lower side, and because the current remains positive, the diode remains forward biased, so the capacitor gets a negative polarity when the + probe is on the top.

Doesn't the diode turn it into DC? It allows only half of a single LC oscillation period of a discharge-charge cycle in one direction.

VL=L.di/dt -> decaying current -> di/dt = negative -> VL=  negative.

What point am I missing?

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Chris posted this 2 weeks ago

Hey Scalarpotential,

One must understand where the Charge is, in Time t, and where is is moving.

I should point out, there is an inaccuracy in the second video, compared to Jagau's experiment, but this does depend on ones point of view, and what is being observed! Capacitor Peak Voltage, is Peak Sinusoidal Wave!

Not Zero, which was shown in the video.

NOTE: When Capacitor Voltage is Zero, the Inductors Magnetic Field is at Maximum.

Like Jagau said, the Capacitor was Charged to 9 Volts, so we start the Cycle, with the Capacitor fully Charged, Peak Voltage, and watch the Capacitor empty into the Inductor, and then from the Inductor, negatively back into the Capacitor.

At no time, does the Charge change its Polarity!

The Inductor Terminals Change Polarity, between the Conversion from Electric Field to Magnetic Field, but the Charge does not.

The Capacitor Terminals also Change their Polarity, but the Charge does not. This is the same, this is as the Magnetic Fields is Converted back into the Electric Field.

This Conversion is due to Faradays Law of Electromagnetic Induction, The Magnetic Field Cutting the Turns of the Inductor, Creating a Voltage, and thus a Current.

A most efficient and seemless process of genius!

The Charge does Change its Polarity after the Peak at the end of this shown Cycle, at 270 Degrees, traveling back the way it came, to only repeat the process again, eventially Damping out to nothing unless a means of Sepperationg More Charge is achieved, to over come I2R Losses.

@Jagau or any other readers, please correct any errrors or inconsistencies.

Best Wishes,

Chris

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Jagau posted this 2 weeks ago

Alright Chris you have patience and a gift to teach that I don't have
As you said it depends on what you want to observe.
The schottky diode retains the charge in reverse polarity after its trip in the inductor in order to allow that one and only once the change of polarity, it becomes polarized in the reverse direction I believe that is the obvious even, It is the aim of the experiment to recharge the capacitor after using it only once.
thanks for your help, the simplest things are sometimes the most difficult to explain.
Jagau

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Jagau posted this 2 weeks ago

The image Itsu reproduced with a current probe (which I don't have), we can visually see that the theory is correct, it shows that in a capacitor the current (green trace) is ahead of the voltage (trace yellow) up to 90 degrees.
When the yellow trace (voltage) is at 0 volts, green line (current) is at its maximum and this is the reverse for an inductor, thanks Itsu,
a picture is worth a thousand words. simple things simple fact

Jagau

scalarpotential posted this 2 weeks ago

Yes, I agree it's a tank circuit, but with initial energy not from a battery but from the capacitor of E=0.5*C*V2, with V(0)=9V, so Q=C*V=7μ*9=36uC, that's pumped around from one plate towards the other minus losses and diode v-drop, the  charge travels through the inductor creating voltage proportional to how fast current varies or how fast charge accelerates (hence di/dt=dq/dt/dt=d2q/dt2).

The energy back in the cap comes from the Lenz effect when magnetic energy stored by self-induction 'returns' as voltage and current.

I thought this was helpful in showing how to add initial energy in circuit analysis:

?t=542

Correct me where I am missing the point.

Take care.

Jagau posted this 2 weeks ago

Hi all

For the following of the experiment we will focus on a practical example in order to improve the charge transfer from one capacitor to another.

Of course, the Lapalace transform can be useful to help us turn a differential equation into an algebraic equation, but the main goal here is to find a practical way of charge transfer from one C to another C.

Another demo on DC resonant charging is in the works with picture and expalnation and we should be able to discuss it shortly.

Jagau

Jagau posted this 2 days ago

DC resonant charging’’
Here I suggest this very simple circuit with only a discharge pulse in order to study the behavior of the circuit:

So the change here is that the coil is in series with the diode and charges a second capacitor C2 with a single pulse. The DC charge of C1 passes through the coil and the diode and charges C2 at a potential equivalent to C1 according to a time constant T=RC
As for the secondary oscillations produced by the BEMF, the oscillation is perpetuated by observing a gradual drop due to the impedance and resistance of the circuit following the well-known rule.

A LT spice simulation with probe ar B gives us this:

A LT spice simulation with probe ar C gives us this:

We should, according to the Lt spice software, have almost double the original voltage, very close to 18 volts.

The question where does this surplus come from, C1 from 9 volts to 0volt is now empty??

Is this really the case?
Real example to follow

P.S. The new series of lithium ion supercapacitors on the market and are starting to hear about them, yet they have been around since 2001, just like the graphite ones.

Jagau

scalarpotential posted this yesterday

Hi,

I remember this circuit:

http://www.richieburnett.co.uk/dcreschg.html#resonant

It would be interesting if you really manage to charge C2 to a steady double voltage on the bench, with only a charged C1.

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