Steve

Hi,

I found a good schematic of the auto ignitioncoil.
As you can see, it is actually 1 coil.

An ignition coil is essentially an autotransformer with a high ratio of secondary to primary windings. By "Autotransformer", I mean that the primary and secondary windings are not actually separated - they share a few of the windings.
The ratio of secondary to primary turns in an ignition coil is somewhere around 100:1. The ignition coil is operated directly off a 12 volt source. However, the ignition coil does not work like an ordinary transformer. An ordinary transformer will produce output current at the same time that input current is applied. An ignition coil actually does most of its work acting as an inductor. When the ignition coil is connected to the battery, the inductor is 'charged' with current. It takes a few milliseconds for the current to build up the magnetic field - this on account of reverse voltage caused by the increase in magnetic field. During this short charging period, maybe a thousand volts are produced at the high voltage terminal, not enough to actually cause a spark.

The actual spark is generated when the breaker contacts open.
For an ideal inductor, the current and voltage relate by:

V = L dI
Where V is voltage, L is inductance (in henrys) and dI is the rate of change of the current.


Thus, seeing that L is constant for the inductor, the abrupt change in current will cause a very large voltage to be produced. This produces a very short, very high voltage spike. Of course, the change in current is on the primary side, but because the primary and secondary coils have a large mutual inductance (this is where the transformer part comes in), you get a spike on the order of 100 or more volts on the primary, and 10000 volts on the secondary. Even the primary side of the coil can give you a bit of a jolt if you hold the wires wile disconnecting power. Also note that any contacts you will use will get a lot of sparks, also on account of this.
A less theoretical way of looking at it is that disconnecting the current source will cause the magnetic field in the coil to collapse (because nothing is forcing it magnetic anymore), and the rapid change of magnetic field in turn induces a large voltage in the windings.

In an engine with four or more cylinders, the high voltage terminal of the coil is connected to the distributor, which is just a fancy high voltage rotating switch, for selecting which of the spark plugs needs to be fired. This is much cheaper than having one ignition coil for each cylinder.

Modern cars of course all use electronic ignitions instead of the unreliable mechanical breaker points points for interrupting the current to the ignition coil. However, the ignition coil still works the same way, so you should be able to use an ignition coil even from a modern car. The main thing that has changed is that the 'points' have been replaced with fancy computer gadgetry and power electronics.

One thing that cars usually have that I didn't include in the circuit is a small capacitor across the breaker points, often called the "ignitor". This capacitor in cars before electronic ignition had a tendency to fail over time, so you may have heard people refer to it. The capacitor helps to protect the points by liming voltage across the contacts, so the contacts themselves won't spark. It also forms a bit of a resonant circuit with the coil, and that way enhances the spark, although it probably cuts down on the output voltage a little bit.


br
Steve

Ok,

point by point:
1. chokes...in Stans circuits they can do 2 things. 1 = become magnetically charged and then discharging into a HV negative spike, or 2 = filter of some noices and spikes
The theory that fits by explanation nr 2 is that Stan was charging his wfc like a capacitor and the chokes were there to soften the raising voltage pulses of the circuit.

2. Nathan Stubblefields coils had to do with Scalar waves and not with tesla's longitudinal waves.
3. and yes, if you have 10Kv and 1ma, you have automatically reduced amps..

I hope that i didnt forget any questions.......

br
Steve

Thats also nice stuff, Najman!
Thanks for posting.

br
Steve

Hi,

Well, Stan said that his hydrogen generator worked on a system that was 180 degrees out of phase.
Sofar i can follow Dankie. The question is what was 180 out of phase?
If you hit voltage against voltage, you get zero.
Same frequency against same frequency = noice
In a RC setup you can shift voltage against amps.
What would it bring if you cut of the signal before amp kick in and after voltage raised?

Again, i just wrote some food for thoughts

br
Steve

more:

In physics, interference is the addition (superposition) of two or more waves that result in a new wave pattern.

Food for thought:

Two oscillators that have the same frequency and different phases have a phase difference, and the oscillators are said to be out of phase with each other. The amount by which such oscillators are out of step with each other can be expressed in degrees from 0° to 360°, or in radians from 0 to 2?. If the phase difference is 180 degrees (? radians), then the two oscillators are said to be in antiphase. If two interacting waves meet at a point where they are in antiphase, then destructive interference will occur. It is common for waves of electromagnetic (light, RF), acoustic (sound) or other energy to become superposed in their transmission medium. When that happens, the phase difference determines whether they reinforce or weaken each other. Complete cancellation is possible for waves with equal amplitudes.

br
Steve

Hi,

Here an explanation of RLC resonance, in serie or in parallel.
Its good and readable.

Steve

Hey Stevie... I would measure before the power supply.

how about using a small fully charged battery... testing how long it will take to drop to a predetermined voltage. like 10 volts.

Hi Jolt,

Well, i thought the same. Can you do me a favour and run a test on your PS?
Just measure amps and volts on the 110V, and monitor that when you add some load to the ps. It totally doesnt match here.
The PS pulls like 60watts, but on the PS outlet it does like 10watts.
Seems to me like a lot of losses.

Its pretty hard to explain like this. Can we meet on Skype or msn one of coming days?

Steve

Hi members,


I have a case here. Please try to help me out here.
The story is as follows.
I got an energymeter from Cresta which was tested a Best Buy in a consumertest. This meter was most correct in his measurements.

I put the meter between the 230V wandoutlet and my lab variable powersupply. On this PS are an ampmeter and a voltmeter.

After doing tests with Strait dc and also with a setup with 50% pulses and new coils it was very difficult to say what the powerconsumtion was in these tests.

What i did was keeping the watts on the energymeter on 60watts.
The reading from my PS display where between the 2.5v till 9v and most of the time on 2A.

Question: is the energy meter right because it includes the whole powersupply or should i use the PS readings?

My first impression is to choose for the energy meter.
If thats the right way of measuring then i have good news for you 
Strait dc vs 50% duty vs 50% with diodes and coils.....
All tests did 60watts on the energy meter.
The coils setup was a winner....

Steve

In did it is but that controller is hard to find.

I did find one source that has them stocked.

http://www.alliedelec.com/search/productdetail.aspx?SKU=2734242

www.farnell.com has those IRF smartrectifiers....
Do they switch max 200VAC an meed 20V for Vcc?
 
Steve
 
 
 

Nice pdf, Electrojolt!
 
 
Steve

@Donaldwfc: Thanks 
 
Here is something more
 
 
A major cause of losses in a conventional power supply using a 50/60-Hz transformer is the bridge rectifier. This article shows how to build a “greener” rectifier, substantially reducing losses by eliminating the diodes in the bridge rectifier and substituting modern low-RDS(ON) power MOSFETs. The MOSFETs used are typically employed in high-frequency switch-mode power supplies. Aside from the power MOSFETs, the circuit uses only two comparators and a few inexpensive transistors, diodes, capacitors, and resistors.
Four IRF2804 n-channel power MOSFETs, T1-T4, replace the bridge diodes (Fig. 1). The remaining components are needed to steer the gates of the MOSFETs. The power MOSFETs’ body diodes (shown by dashed lines) make up a diode bridge rectifier in the usual way.
During the first half cycle after power-up, this “parasitic” bridge rectifier charges load capacitor C3. When VOUT becomes higher than 2.7 V, comparators U1 and U2 get into the act. In addition, driver stages T9-T12 on the right side, which are also powered by VOUT, now have enough supply voltage to switch on the gates of the T3 and T4.
After the second half cycle, the two boost capacitors on the left side, C1 and C2, have charged to the peak value of the input voltage, and supply driver stages T5-T8 for the power MOSFETs T1 and T2. The voltages across C1 (VB1) and C2 (VB2) are always positive with respect to the source connections of T1 and T2, respectively.
Comparator U2’s inputs are connected to T4’s source and drain connections, so it also senses the voltage polarity of this transistor’s body diode. Whenever the polarity across T4 becomes negativethat is, when a forward current could flow through T4’s body diodethe power MOSFET is switched on via U2’s output and the driver stage T11/T12. (The gate voltage, VG, is shown as R_A in Fig. 2.) The drain-source voltage VDS (the blue trace, Ch. 2, in Fig. 2) now becomes very small, since VDS = ID × RDS(ON), and the transistor’s RDS(ON) is only 2 m?.
Virtually all the current now flows from source to drain and almost no current is flowing through the body diode. Notice that VDS remains negative, so the comparator can keep T4’s gate high. At the same time, T1 is also switched on, with the help of T14 (trace R_B in Fig. 2) and driver stage T5/T6.
Later in the cycle, when the current through T1 and T4 drops to zero (that is, when the transformer output voltage dives below VOUT), T4’s VDS also becomes zero, and the comparator cuts off both T4 and its leftside partner, T1. While T1 is conducting, boost capacitor C2 amasses charge that’s needed one-half period later for dumping into T2’s gate.
After that one-half period, similar things happen to the other power MOSFET pair. Comparator U1 senses T3’s VDS and switches on this transistor and its cousin T2 on the left side just at the moment before a current begins to flow through the respective body diodes.
The values of C1 and C2 must be high enough to ensure that the gate-source voltage at the end of the gate-charging process is high enough to switch on the respective power MOSFET completely. For a gate charge of QG = 160 nC (the data-sheet value for the IRF2804) and an allowed voltage drop of, say, ?UG = 100 mV, the minimum capacitance would be CMIN = QG/? UG = 1.6 F. Therefore, 10 ?F is high enough. Multilayer chip capacitors can be used, but beware of the voltage dependency of dielectrics like Y5V.
The two comparators are LT1716 low-power devices in small SOT-23 packages. They are particularly suitable for this application because they can cope with negative voltages on their inputs, even when running from a single supply. That’s important because the drains of T3 and T4 become negative with respect to ground.
Another advantage of this comparator is its wide operating voltage rangefrom 2.7 V to 44 V. Unfortunately, the device’s output drive is too low to drive T3 and T4 directly. That’s why the need arises for driver stages T9-T12. They are small p- and n-channel MOSFETs that put a maximum voltage swing on the gates of the power MOSFETs.
With a 5-A load, the circuit worked with transformer voltages of 2.8 V rms to 14 V rms. The lower limit is determined by the gate threshold voltages of the MOSFETs, and the upper limit is determined by the maximum allowed gate voltages. If the circuit must run at higher transformer voltages, the supply voltages for the driver stages should be limited by resistors/Zeners or voltage regulators.
 
The circuit’s efficiency is quite good. At a 10-A dc output (7 V rms ac input), none of the components require a heatsink. The power MOSFET case temperatures stay well below 50°C.
Due to a lack of equipment, I could not test the circuit at higher currents. But beyond 10 A, it may be worthwhile to connect two MOSFETs in parallel to reduce RDS(ON) even further. But pay attention to the resistances of the PCB traces, since they could be higher than the MOSFETs’ RDS(ON)!
The circuit was compared to a popular KBU8B silicon diode rectifier. At an input voltage of 5 V rms at 50 Hz and a constant load of 5 A dc, the KBU8B’s output was 4.45 V dc, average, measured across C3 (15,000 ?F). Under the same conditions, the “greener” rectifier produced an output of 5.9 V dc, average.
Another comparison that may be even more meaningful involves determining what rectifier input voltage is needed for a given dc output voltage. For this measurement, a transformer with several output windings (Ultron ULT2) was connected to the mains via a Variac. The desired output was 5 V dc, average. Measurements were done at two constant load currents: 5 A and 10 A.
For the KBU8B rectifier and a 5-A load, the transformer’s 6-V output winding was used. The Variac had to be adjusted for a transformer output (secondary) voltage of VSEC = 5.55 V rms, which had to be corrected slightly to 5.48 V when the rectifier got hot. The measured input power was 47 W. With a 10-A load, which is already beyond the specs of the KBU8B, the 8-V output winding had to be used. The Variac was adjusted to 5.97 V rms (5.87 V rms when hot). Under these conditions, the real input power of the Variac was 88 W.
Using the “greener” rectifier with a load current of 5 A, the Variac had to be tuned back to a transformer output voltage of VSEC = 4.34 V rms (off the 6-V winding). The Variac’s real input power was only 36 W. At 10 A, the 6-V winding could still be used, with the Variac tuned to 4.82 V rms. The real input power was 69 W. Thus, the power MOSFET rectifier circuit saved roughly 10 W at 5 A and 20 W at 10 A.
At high currents and low voltages, and especially when the output ripple voltage increases, the two power MOSFETs on the right get a little warmer than those on the left. The reason is because the driver stages on the left have their own filter capacitors (C1 and C2) that provide a smooth dc voltage, while the driver stages on the right are directly supplied from the high-ripple output voltage. Unfortunately, right at the moment when the gates of the right-side MOSFETs should be taken high, the available output voltage is rather low, since output capacitor C3 has discharged to its minimum value (traces CH1 and R_A in Fig. 2).
The cure for this problem is simple. Add a diode and a capacitor to supply the right half of the circuit (Fig. 3).
The whole circuit fits into roughly the same volume as a conventional bridge rectifier. Considering that there’s usually no need for heatsinks, the circuit should pay off quickly. Also, in many cases, a smaller and cheaper transformer can be employed.
 
 
 
 

Here another one....
 
 
Rectifier Bridge Has No 2Vf drop! 
 
 
The venerable full-wave rectifier bridge (Fig. 1) is a common, familiar circuit for converting an AC input voltage to a DC output voltage.  It is also useful for translating a DC input of arbitrary polarity into a DC output of known polarity, as is commonly required in electronic telephones or other telephony devices, and has application in protecting against battery reversal in battery-powered circuits.
(http://www.thetaeng.com/images/Bridge.GIF)
 Fig. 1  A drawback of the classic four-diode rectifier bridge is the unavoidable forward voltage drop (Vf) of two diodes when current is flowing.  With conventional silicon diodes, this could typically amount to 1.5 volts or more.  The result of this is wasted power and reduced efficiency in power supply applications, or loss of working voltage in telephony or battery-powered applications.
In telephony applications in particular, it is possible for a device to have as little as 4 volts available to it under worst case conditions of loop current and line length.  Since most integrated circuits, telephony or otherwise, are decidedly unfriendly about power supply reversals, it is common practice for the line-powered electronics to be surrounded by a full-wave rectifier bridge in order to guarantee power supply polarity.  But with only 4 volts of line voltage, a 1.5-volt drop in the rectifier would leave only 2.5 volts for the electronics!
Similarly, in battery powered circuits, it is often the case that the loss of efficiency caused by  series diodes to protect against inadvertent battery reversal is unacceptable.
The circuit shown in Fig. 2 eliminates these drawback by replacing the diodes with MOSFETS.  The four MOSFETs are connected in such a way as to conduct in opposing pairs.  Which pair conducts is a function of the polarity of the applied voltage.  The conducting pair is such as to steer the applied voltage to the appropriate output terminals so as to always maintain the same polarity at the output.  In other words, the circuit rectifies.
 Fig. 2  Interestingly, if one looks at the intrinsic drain-to-source body diodes of the MOSFETs, ignoring the MOSFETs themselves, they form the conventional rectifier bridge configuration.  Indeed, when voltage is first applied, the circuit acts the same as a conventional rectifier bridge in that the forward voltage drop of two diodes (2Vf) appears between the input and the output.  But as soon as the applied voltage exceeds the turn-on threshold of two MOSFETs (or more precisely, the sum of an N-channel threshold and a P-channel threshold), the appropriate pair of MOSFETs turns on, effectively bypassing the pair of diodes that is conducting.  The voltage-drop performance of the bridge is now a function of drain-to-source resistance (RDS(on)), which, with modern MOSFETs, is pretty darn good!  In telephone line applications, a voltage drop in the millivolt range can easily be achieved.  Also, with low-threshold MOSFETs achieving thresholds in the 1-volt range these days, it is possible to construct a bridge where the MOSFET turn-on occurs not long after the diode turn-on as the applied voltage ramps up.
A limitation of the circuit, as shown, is that the applied voltage cannot exceed the gate-to-source voltage (VGS) rating of the MOSFETs.  Typically, this is 20 volts.  For higher voltage applications, it is possible to put a resistor in series with each gate and use a zener clamp between the gate and source of each MOSFET to limit the VGS experienced by any individual MOSFET, as shown in Fig. 3.  With such a provision, the primary limitation on applied voltage then becomes the drain-to-source breakdown (BVDS) rating of the MOSFETs.  Fig. 3  One caveat of the FET bridge circuit: do not use it as the rectifier in front of a capacitor-input power supply!  In a conventional rectifier bridge, the diodes prevent the backflow of current from the power supply input capacitor as the applied voltage drops below the voltage on the capacitor.  With this design, the MOSFETs act like switches rather than one-way valves for current flow.  They don’t care which way current flows, hence the input capacitor of the power supply will be discharged to near zero volts with each half-cycle of the applied AC power!  This limits the power supply applications for this circuit to inductive- or resistive-input designs.
However, it would be possible to use this circuit with a polarized capacitor in power-factor correction applications.  Correction of an inductive power factor would normally require a non-polarized capacitor directly across the AC line.  By putting the FET bridge circuit in front of the capacitor, a polarized capacitor could be used instead which may be advantageous in terms of size and cost.  I haven't tried this particular application, so I can't vouch for it, but if you have success with the idea, please let me know.

Hi,
 
The losses in a bridge rectifier can easily become significant when low voltages are being rectified. The voltage drop across the bridge is a good 1.5 V, which is a hefty 25% with an input voltage of 6V. The loss can be reduced by around 50% by using Schottky diodes, but it would naturally be even nicer to reduce it to practically zero. Thats possible with a synchronous rectifier. What that means is using an active switching system instead of a passive bridge rectifier.The principle is simple: whenever the instantaneous value of the input AC voltage is greater than the rectified output voltage, a MOSFET is switched on to allow current to flow from the input to the output. As we want to have a full-wave rectifier, we need four FETs instead of four diodes, just as in a bridge rectifier. R1 R4 form a voltage divider for the rectified voltage, and R5 R8 do the same for the AC input voltage. As soon as the input voltage is a bit higher than the rectified voltage, IC1d switches on MOSFET T3. Just as in a normal bridge rectifier, the MOSFET diagonally opposite T3 must also be switched on at the same time. Thats taken care of by IC1b. The polarity of the AC voltage is reversed during the next half-wave, so IC1c and IC1a switch on T4 and T1, respectively. As you can see, the voltage dividers are not fully symmetrical. The input voltage is reduced slightly to cause a slight delay in switching on the FETs. That is better than switching them on too soon, which would increase the losses. Be sure to use 1% resistors for the dividers, or (if you can get them) even 0.1% resistors. The control circuit around the TL084 is powered from the rectified voltage, so an auxiliary supply is not necessary. Naturally, that raises the question of how that can work. At the beginning, there wont be any voltage, so the rectifier wont work and there never will be any voltage... Fortunately, we have a bit of luck here. Due to their internal structures, all FETs have internal diodes, which are shown in dashed outline here for clarity.They allow the circuit to start up (with losses). Theres not much that has to be said about the choice of FETs its not critical. You can use whatever you can put your hands on, but bear in mind that the loss depends on the internal resistance. Nowadays, a value of 20 to 50 mW is quite common. Such FETs can handle currents on the order of 50 A. That sounds like a lot, but an average current of 5 A can easily result in peak currents of 50 A in the FETs.The IRFZ48N (55 V @ 64 A, 16 mW) specified by the author is no longer made, but you might still be able to buy it, or you can use a different type. For instance, the IRF4905 can handle 55 V @ 74 A and has an internal resistance of 20 mR. At voltages above 6 V, it is recommended to increase the value of the 8.2-kR resistors, for example to 15 kR for 9V or 22 kR for 12 V.

Regards
Steve
 

Hello steve
 
Do you remember that youtube video with a kind of iron U  I  core with two coils, witch the guy energize and than become a permanent magnet and witch when he pull out the I part of the core the electricity was discharged and the lamp lights? Do you have any info about that principle and the link to that video please? I'm looking for it since some time and not finding...


Thanks


ps do you remember the my proposed resonant circuit with one coil 2 diodes two tubes?
now get the dan's circuit substitute the other capacitor by another tube and take out the transistor and drive with my circuit, using distilled water...


isn't the same circuit? forget about my modification on the last drawing , even if is not wrong it would only be needed if you want to run with only one tube.. but a little different than what i showed...


also forget about the cap in series with the tubes in my circuit...




and


have you ever read this?
Frequency dependent capacitorsIf a capacitor is driven with a time-varying voltage that changes rapidly enough, then the polarization of the dielectric cannot follow the signal. As an example of the origin of this mechanism, the internal microscopic dipoles contributing to the dielectric constant cannot move instantly, and so as frequency of an applied alternating voltage increases, the dipole response is limited and the dielectric constant diminishes. A changing dielectric constant with frequency is referred to as dielectric dispersion, and is governed by dielectric relaxation processes, such as Debye relaxation. Under transient conditions, the displacement field can be expressed as (see electric susceptibility):indicating the lag in response by the time dependence of ?r, calculated in principle from an underlying microscopic analysis, for example, of the dipole behavior in the dielectric. See, for example, linear response function.[6][7] The integral extends over the entire past history up to the present time. A Fourier transform in time then results in:where ?r(?) is now a complex function, with an imaginary part related to absorption of energy from the field by the medium. See permittivity. The capacitance, being proportional to the dielectric constant, also exhibits this frequency behavior. Fourier transforming Gauss's law with this form for displacement field:where j is the imaginary unit, V(?) is the voltage component at angular frequency ?, G(?) is the real part of the current, called the conductance, and C(?) determines the imaginary part of the current and is the capacitance. Z(?) is the complex impedance.
When a parallel-plate capacitor is filled with a dielectric, the measurement of dielectric properties of the medium is based upon the relation:where a single prime denotes the real part and a double prime the imaginary part, Z(?) is the complex impedance with the dielectric present, C(?) is the so-called complex capacitance with the dielectric present, and C0 is the capacitance without the dielectric.[8][9] (Measurement "without the dielectric" in principle means measurement in free space, an unattainable goal inasmuch as even the quantum vacuum is predicted to exhibit nonideal behavior, such as dichroism. For practical purposes, when measurement errors are taken into account, often a measurement in terrestrial vacuum, or simply a calculation of C0, is sufficiently accurate.[10] )
Using this measurement method, the dielectric constant may exhibit a resonance at certain frequencies corresponding to characteristic response frequencies (excitation energies) of contributors to the dielectric constant. These resonances are the basis for a number of experimental techniques for detecting defects. The conductance method measures absorption as a function of frequency.[11] Alternatively, the time response of the capacitance can be used directly, as in deep-level transient spectroscopy.[12]
Another example of frequency dependent capacitance occurs with MOS capacitors, where the slow generation of minority carriers means that at high frequencies the capacitance measures only the majority carrier response, while at low frequencies both types of carrier respond.[13][14]
At optical frequencies, in semiconductors the dielectric constant exhibits structure related to the band structure of the solid. Sophisticated modulation spectroscopy measurement methods based upon modulating the crystal structure by pressure or by other stresses and observing the related changes in absorption or reflection of light have advanced our knowledge of these materials.[15]

That dude was showing bemf.....i remember. Sadly i dont have the links....

Of course is your circuit and Dans circuit similar with the exception of the 2 tubes and 2 diodes from you. Dan has a giant coil to create a massive spike of at least 1500v.
He first charge the water up with amps.
Amps create a path for the charge. Then the spike hits and does something...so it seems...

Hi,
 
Just found a simple proof of a better working coil because of the bifilar concept.:
 
 
This easy experiment demonstrates the extra power in Nikola Tesla's patent for electromagnets.
 Things you'll need:      Wind the first nail with 100 turns of magnet wire.  Leave about 3 inches of wire on both ends of the winding.
Wind the second nail with 100 turns of magnet wire, but in the following way.  Cut two equal length wires about 12" long each.  Holding the two wires together, begin turning 50 parallel turns of magnet wire around the nail.  When you have finished winding the coil trim off the excess wire so that there are 3" of wire on both ends of the coil.  Take the two inside leads from each end and twist them together.  Remember to clean the ends of the magnet wire so they can make an electrical connection.
This is what they should look like: (click on the image for a closer view)    Two Electromagnets
Now connect the battery to the end leads of the single wound nail.  This will energize the coil and cause the nail to become magnetic.  Now pick up as many paper clips with the nail as you can.
OK, connect the battery to the ends of the bifilar wound coil.  Now pick up as many paper clips as you can with this electromagnet.
The same amount of voltage, from the same battery, produces twice as much energy in the bifilar wound coil as in the single wound coil.  This is just one of the many techniques Nikola Tesla used to make his inventions highly efficient.
 
 

Here a parallel LC calculator.

http://www.whatcircuits.com/lc-resonance-frequency-calculator/

hey can some one point me to a direction of variable capacitors for the types of circuits we work with? pcb mount or what ever kind it is we would use on frequency generators.

Most of the time, we use variable resistors for that purpose or we change voltage.
Just drop the schematic and say what you want and ill help you out...


Steve

12v peak.
 

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.

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