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series LC resonance in a DC circuit

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Steve:

--- Quote from: X-Blade on December 29, 2018, 15:29:01 pm ---This is a misunderstanding.
The way you get the double of voltage if when a real capacitor fully charges to voltage supplied and there will be no current flow anymore, so the inductor collapses producing the pulse BUT THE POWER SUPPLY IS STILL CONNECTED, and since that, it charges to Vsupplied + inductor pulse in series.
In a pulsed system as Meyer with non-ideal "capacitor" there will be a second pulse and not the voltage doubling effect.

--- End quote ---

Hi XB,

I know that.
Meyer is pulsing.
This video is proofing a serie LC with dc.
If our cells would act as a capacitor, then the voltage across the cell would be higher then the 12v of my battery that i use...
Can you try the same test?
See what you will measure?

Cheers

Steve our big Steve.
Look closely to the circuit in the video and the description of the guy.
In that example, you hear the guy saying that inductor collapses because there is no current flowing after the capacitor is fully charged, there was not the power supply interrupted, and because of that, the inductor collapses and the voltage sums to the power supply, then the capacitor has a lower voltage than of the inductor and the power supply in series, that is the only reason why capacitor charges to the double of the charge.
If you interrupt the circuit like Meyer does, it has only one power source: the inductor(s).
Thats why you see pulse doubling effect on Meyer circuit.
I aready showed where and how to read the doubled pulse across the cell:
(https://i.postimg.cc/sgrXDqpt/pic-105-4-onoff.gif)

The falling part of the waveform is produced by the inductor(s) collapsing.
This is not a real capacitor in my opinion.
I only see the water behaving like a resistive non-linear load.

sebosfato:

--- Quote from: sebosfato on December 29, 2018, 16:24:41 pm ---if you connect the inductor in such way that it can collapse freely than you get more than double actually... it will win any barrier to finish the discharge

--- End quote ---

What I mean is when you pulse a transformer and open the switch the current will want to keep flowing and will burn the switch if no other dissipative or regenerative means exist!

Im thinking about this and i'm kind of inclined to think that the water below 1.23 v is a capacitor and this is the big problem.. What im thinking is that when water resonance is achieved the cell will behave like a short circuit allowing huge current flow...

Something tellme Steve that your idea is right but at low voltage there is going to be lower power per electron

Fabio, why dont you go to your lab, test and see if it is right or wrong according to your theories?
There are lot of "theorists" and nothing proved right or wrong.
There will be no "lc resonance" the way you thing.
There is pulsed D.C. not A.C. so there will be no current magnification due to the LC resonance,

Where are the bench people? Where is the builders?

People are comparing and bring theories to the level of a competition?
Please people, go to your lab / bench and test, prove right or wrong, document it, and share if you like.
There is a lot of "smoke and mirrors" on Meyer technology.
No one saw voltage alone performing work on that stuff.
It is time to wake up, Xmas is gone.

Aussepom:
Ok guys can I put my 2bob’s worth in?
I will put some test in that were done it may help.

What was not stated in this series circuit was the value of the coil, its DC resistance and impedance XL and the same for the Capacitor.
To get things correct you will need to know them or measure them, calculate whatever.

On a stable 12V Dc supply with a coil and a cap in series, that is it is switched on and left on.
The voltage will show a high peak across the coil then will be 12v the current will rise as shown extensionally, opposing the input, and it will lag.
At the T point this is one time constant, it will get to 63%  after 5time contents it will be fully ‘charged’ yes it will hold a charge, once it gets there it will allow full current to flow in the circuit to charge up the cap to again 5time contents. Then current will stop if the capacitor is a good one and does not ‘leak’. The voltage will equal the supply voltage of 12VDC.
Now if you disconnect the power supply, extremely fast, if it is a switch there will be an arc as the coil tries to discharge across the gap, the voltage on the cap cannot as the coil will oppose it as if it was the supply like the battery.
There will now be some lost charge in the coil, its voltage will be lower than the cap, so now the cap take the place of the battery and tries to charge up the coil, again it will take time to do this, another 5 time constants of what is left of the voltage.
Now we have a problem, the coil is charged, so what now, the coil has a DC resistance, this will try to discharge, ‘leak’ the capacitor, now the reverse and it starts again, this is known as ringing.
Here are some test circuits if they come out .
T is calculated by in a coil L/R  L in henneries and R in DC resistance this is T and one time content for the coil to get to 63%  of the max current, 5 T and it is fully charged.
Remember if XL=XC then the current in the circuit will be V/R  dc resistance
there is one for Bob Boyce test will put in BB

Have fun
Brian