Re: AW: [EVGRAY] Re: buck plug ...

Hi Sven,

It must be the power that is infinite at zero resistance. If the energy is infinite it would probably explode like a super nova or at least result in a lightning flash. Steven Mark and Bob Boyce experienced a lightning flash from their devices when getting too close to pure resonance or whatever they reached. At least it's quite dangerous to one's health especially if not surviving the discharge.

If remembering correct the switching should take place at the peak voltage. This is where the current is zero and just before the current changes direction or where the voltage starts declining. A peak or a zero crossing detector can find that point. This signal then trigs the SCR.

Regards
Ole
 

---In [email protected], <s.friedrich@...> wrote :

 

 Hi Ole, I meant that with the Scope different I'm looking for the point exactly when unloading and no energy in the LC from Pushpull trafo can be delivered. I just read in the re ou 6.1 pdf which theoretically at 0 ohms the energy is infinite. 0 ohm does not convert any power and my thyristors break down. You have to recognize the LC curve where you have to engage or switch at the wrong time? Greetings Sven
 

 Von Samsung Mobile gesendet



-------- Ursprüngliche Nachricht --------
Von: "onielsen@... [EVGRAY]" 
Datum:13.02.2018 01:11 (GMT+01:00) 
An: [email protected] 
Betreff: [EVGRAY] Re: buck plug ... 

   Hi Sven,

"Is there a way to visualize the energy consumed by scope in the primary winding so that you can see that the energy needs to be consumed faster."
Yes if the scope can calculate the power and then integrate the power. Alternatively the energy in a an inductor is given by this formula: 

where E is the energy in Joule (or Ws). L is the inductance in Henry and U is the voltage in Volt. Knowing this the energy can be calculated at any given instant by knowing the current through the inductor at that particular moment..

If calculating the power by multiplying the current with the tension and the power is constant the energy can be calculated by multiplying the time with the constant power. As power is the rate of energy transfer energy equals power multiplied by time when the power is held constant or E = P_constant x T [Joule = Watt_constant x second]. No change in Watt is allowed if using this linear formula.

Transferring the energy to a capacitor through a rectifier is also a way to calculate the energy stored. The energy stored in a capacitor can be calculated by knowing its capacitance and the voltage across it.

Regards
Ole

 

---In [email protected], <s.friedrich@...> wrote :

 Hello to all interested,

I've tried a large load on the push pull transformer to lower the impedance even a DC motor I plugged in, you could really see how the recorded power diminished on the meter. About 30 watts less, but still I needed additional active power than with pure resonance when I operate the diode plug.

It is really difficult, even the negative diode plug makes more and more like the positive diode plug, I have already tested other capacitors, discharge capacitors coupling capacitors, the values ​​reversed it always stays there.

I measure with the scope on both capacitors simultaneously.

Is there a way to visualize the energy consumed by scope in the primary winding so that you can see that the energy needs to be consumed faster.

I had a new drawing uploaded to lower the primary coil impedance moderately directly at least at idle. Maybe you can also bring the thyristor for quicker shutdown, the capacitors are then not fully discharged but the energy can then be discharged depending on the impedance behavior of the secondary coil longer or shorter. If the primary coil is fully charged and no power is dissipated, a capacitor could be charged via a freewheeling diode so the capacitor would form an antipole and cancel the potential difference and the thyristor will turn off immediately. If I did not make a mistake.

regards

Sven

https://youtu.be/UY8pceAjm_E https://youtu.be/UY8pceAjm_E

Hi Ole, I meant that with the Scope different I'm looking for the point exactly when unloading and no energy in the LC from Pushpull trafo can be delivered. I just read in the re ou 6.1 pdf which theoretically at 0 ohms the energy is infinite. 0 ohm does not convert any power and my thyristors break down. You have to recognize the LC curve where you have to engage or switch at the wrong time? Greetings Sven

Von Samsung Mobile gesendet

-------- Ursprüngliche Nachricht --------
Von: "[email protected] [EVGRAY]" <[email protected]> 
Datum:13.02.2018  01:11  (GMT+01:00) 
An: [email protected] 
Betreff: [EVGRAY] Re: buck plug ... 

Hi Sven,

"Is there a way to visualize the energy consumed by scope in the primary winding so that you can see that the energy needs to be consumed faster."
Yes if the scope can calculate the power and then integrate the power. Alternatively the energy in a an inductor is given by this formula: 

where E is the energy in Joule (or Ws). L is the inductance in Henry and U is the voltage in Volt. Knowing this the energy can be calculated at any given instant by knowing the current through the inductor at that particular moment..

If calculating the power by multiplying the current with the tension and the power is constant the energy can be calculated by multiplying the time with the constant power. As power is the rate of energy transfer energy equals power multiplied by time when the power is held constant or E = P_constant x T [Joule = Watt_constant x second]. No change in Watt is allowed if using this linear formula.

Transferring the energy to a capacitor through a rectifier is also a way to calculate the energy stored. The energy stored in a capacitor can be calculated by knowing its capacitance and the voltage across it.

Regards
Ole



---In [email protected], <s.friedrich@...> wrote :

Hello to all interested,

I've tried a large load on the push pull transformer to lower the impedance even a DC motor I plugged in, you could really see how the recorded power diminished on the meter. About 30 watts less, but still I needed additional active power than with pure resonance when I operate the diode plug.

It is really difficult, even the negative diode plug makes more and more like the positive diode plug, I have already tested other capacitors, discharge capacitors coupling capacitors, the values ​​reversed it always stays there.

I measure with the scope on both capacitors simultaneously.

Is there a way to visualize the energy consumed by scope in the primary winding so that you can see that the energy needs to be consumed faster.

I had a new drawing uploaded to lower the primary coil impedance moderately directly at least at idle. Maybe you can also bring the thyristor for quicker shutdown, the capacitors are then not fully discharged but the energy can then be discharged depending on the impedance behavior of the secondary coil longer or shorter. If the primary coil is fully charged and no power is dissipated, a capacitor could be charged via a freewheeling diode so the capacitor would form an antipole and cancel the potential difference and the thyristor will turn off immediately. If I did not make a mistake.

regards

Sven

https://youtu.be/UY8pceAjm_E

Hi Ole,

From a conventional point of view, it makes no sense for me to switch to 0 ohms, it's a simple short circuit, destroying the switches and delivering no usable energy.


We have a main LC (three-phase transformer) that generates reactive power, which we transfer via coupling capacitors to the diode plug. Two diodes split the positive and the negative half-wave into a half-LC.

So far so good. The diodes are not the problem because I can also switch reactive power through a FWBR without noticeable losses, that's how I see it.

Further, when this half LC is charged, it must perfectly re-enter the main LC so that it consumes no additional active power.

What does it look like based on a reactive window in which time does the energy have to be transferred? It is not enough to discharge the capacitor in time, the coil of the push-pull transformer must not deliver any magnetic energy in the direction.

Unfortunately, the fact is that I apply energy for reactive power and again the same when switching the thyristors and the later I discharge the thyristors the more active power I use in addition. If I increase the load secondary, the input energy decreases slightly.

If you reach the point where only the energy for the reactive power must be applied and a few watts for the switches then we are a long way further.

The transformer is quite big with 6.3 kva but it also has a fairly small inductance, you still have room on your thighs, you could now apply additional windings that further negate the inductance, but then I also transfer less energy or see that wrong. The high-inductive 230 / 24V transformer I do not need to connect directly because the inductance is so high, there is only a single switch and the energy dies, even if I short-circuit it on the secondary side.

The only element I know with an extremely low resistance are batteries that I have only 11 milli ohms although they are not very big, but I can not pop such discharges directly into a 24 volt battery right?!?

I try on different contacts such as diodes, push-pull-transformer, thyristor, etc. to make a scope picture maybe even a video if it's worth it.

regards

Sven

Hi Ole, I have here again created a picture as I imagine. The discharge must be done in this window and also have decayed, even if the capacitor is discharged in time, the transformer can deliver even more energy on the magnetic energy and move outside of this switching window, I think.

regards

Sven

Hi Sven
 I think (maybe) for your system, to fill capacitors all by themselves, and ONLY capactors.  (stage one)
 

 Then discharge ONLY capacitors to the load. (stage two)
 

 This is similar to diode plug, but there is a difference because 
 what might go wrong, is during discharge, the capacitors have been disconnected from the inductors and diodes that filled them up....so there is  a sudden moment when the load of the capacitors being filled has vanished..(vanished during the cap discharge to load moment) and this could cause many problems including "perhaps" reflection to primary from the very sudden change in impedance of the "load" of filling up capacitors.
 

 So solution is to have two parallel cap and diode systems ( A and B)  working filling up the caps during stage one, and so the caps A will always fill whenever caps B discharge and caps B always fill when caps A discharge.;....
 Difference with diode plug then, (I think) is there will be some moments in time when BOTH caps A and B fill at the same time too....so this description is not exactly like the diode plug but rather,  I would call it a
  "parallel dual two-stage cap discharge output circuit"
 

 So, there will changes of impedance to the impedance seen by system as the caps fill and discharge, but no so "severe" of changes....it will be the impedance of parallel caps A and B for some time, then there will be the impedance of filling just one cap A or just one cap B for the time chosen for whenever a cap A or a cap B discharges to a load....
 

 Advantages of this are you are not "restricted" to only filling caps for that sinewave portion "allowed" such in diode plug, or for discharging only during that sinwave portion allowed....
 

 Instead, such as if you have a farily high frequency system, you could for example  be filling caps for lets say 10 cycles, or 100 cycles, then discharge just one of the caps A or B after those cycles to the load and discharge caps in alternating fashion and alternating polarity, if you choose, if you want AC output....
 

 So it could be possible to have for an example,  a 500hz system, but there is chosen to be a cap discharge output of 50hz and you could even have the cap discharges mock the normal grid sinewave too if you want...
 

 One more thing, some caps are "storage" type capacitors, made simply to store joules of energy for some time, while there are more specialized capacitors made especially to be pulsed and certain frequencies, and to only store the energy for brief moments of time so maybe choice of capacitors will be very crucial to successful OU system, sorry I cannot recommend particular capactiors but look up and study "pulse-type capacitors".
 ciao
 Kone

Hi Sven,

I think that zero Ohm is meant as the resonant frequency of an LC-tank where the capacitive reactance cancels the inductive reactance. Only the resistive part remains which is the resistance of the conductors. This resistive part can be made smaller by using thicker wires and capacitors having thicker foils or copper foil instead of aluminum foil. But first make it work before going to the extreme which is also quite expensive. I don't think anybody needs that high efficiency except perhaps military researchers or other state financed researchers.

If not returning the power back to the source it is no longer reactive. Reactive power is when the energy sloshes back and forth between two points. This way the energy isn't consumed or dissipated to somewhere else like heating the ambient. Active power travels in one direction by not giving back the energy to its source.

Regards
Ole

 

---In [email protected], <s.friedrich@...> wrote :

 Hi Ole,

From a conventional point of view, it makes no sense for me to switch to 0 ohms, it's a simple short circuit, destroying the switches and delivering no usable energy.


We have a main LC (three-phase transformer) that generates reactive power, which we transfer via coupling capacitors to the diode plug. Two diodes split the positive and the negative half-wave into a half-LC.

So far so good. The diodes are not the problem because I can also switch reactive power through a FWBR without noticeable losses, that's how I see it.

Further, when this half LC is charged, it must perfectly re-enter the main LC so that it consumes no additional active power.

What does it look like based on a reactive window in which time does the energy have to be transferred? It is not enough to discharge the capacitor in time, the coil of the push-pull transformer must not deliver any magnetic energy in the direction.

Unfortunately, the fact is that I apply energy for reactive power and again the same when switching the thyristors and the later I discharge the thyristors the more active power I use in addition. If I increase the load secondary, the input energy decreases slightly.

If you reach the point where only the energy for the reactive power must be applied and a few watts for the switches then we are a long way further.

The transformer is quite big with 6.3 kva but it also has a fairly small inductance, you still have room on your thighs, you could now apply additional windings that further negate the inductance, but then I also transfer less energy or see that wrong. The high-inductive 230 / 24V transformer I do not need to connect directly because the inductance is so high, there is only a single switch and the energy dies, even if I short-circuit it on the secondary side.

The only element I know with an extremely low resistance are batteries that I have only 11 milli ohms although they are not very big, but I can not pop such discharges directly into a 24 volt battery right?!?

I try on different contacts such as diodes, push-pull-transformer, thyristor, etc. to make a scope picture maybe even a video if it's worth it.

regards

Sven

Hi Sven,

Your triggering circuit doesn't detect the peak. It may actually trig before reaching the highest voltage of the capacitors which is on the red curve.

Regards
Ole
 

---In [email protected], <s.friedrich@...> wrote :

 Hi Ole, I have here again created a picture as I imagine. The discharge must be done in this window and also have decayed, even if the capacitor is discharged in time, the transformer can deliver even more energy on the magnetic energy and move outside of this switching window, I think.

regards

Sven

Hi Ole,

yes, the energy spills back and forth like a wave. A wave power station is also there because the waves work through the up and down waves.

I think we divert some of the reactive power to let them do some work and then bring them back so the main LC does not notice what's missing.

If it were just a mere sloshing of energy, voltage and current would not rise so much, but would have to stay at the same level as the input energy.

Actually, the flyback in an induction is the boosting component. If there were no stored magnetic field energy, the energy would not increase in a resonant circuit even if it is blind.

Really hard to grasp this topic.

Maybe I should try the resonance experiment with my 3 spotlights, now I have 1200W as a light bulb in 3 lamps, if they shine bright with resonance energy that would be a confirmation of the energy, of course, only if the energy in the input source does not rise sharply.

regards

Sven

SvenDon't fotget, you may have to adjust your tuning for a new load perhapsCheers Warren

Sent from Yahoo Mail on Android 
 
  On Tue, Feb 13, 2018 at 9:29 AM, [email protected] [EVGRAY]<[email protected]> wrote:       
Hi Ole,

yes, the energy spills back and forth like a wave. A wave power station is also there because the waves work through the up and down waves.

I think we divert some of the reactive power to let them do some work and then bring them back so the main LC does not notice what's missing.

If it were just a mere sloshing of energy, voltage and current would not rise so much, but would have to stay at the same level as the input energy.

Actually, the flyback in an induction is the boosting component. If there were no stored magnetic field energy, the energy would not increase in a resonant circuit even if it is blind.

Really hard to grasp this topic.

Maybe I should try the resonance experiment with my 3 spotlights, now I have 1200W as a light bulb in 3 lamps, if they shine bright with resonance energy that would be a confirmation of the energy, of course, only if the energy in the input source does not rise sharply.

regards

Sven
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Hi Sven,

"If it were just a mere sloshing of energy, voltage and current would not rise so much, but would have to stay at the same level as the input energy."
The voltage and current must increase when energy is increased. The energy can be directly calculated from knowing the LC-tank size and the peak voltage or peak current.

"Actually, the flyback in an induction is the boosting component. If there were no stored magnetic field energy, the energy would not increase in a resonant circuit even if it is blind."
Again knowing the stored energy of the magnetic field (i.e. calculated from the inductance and the current) is what determines the peak of the flyback voltage depending on what impedance the coil looks into. A high impedance flyback load must give a high flyback voltage peak as the current must and will stay continuous at the change in impedance by opening the circuit. The voltage to drive the same current at increased impedance makes the voltage step up at the rate of impedance changing. E.g. doubling the impedance requires doubling the flyback voltage to drive the same current through it. Thus the voltage is doubled but the time it lasts is halved.

Regards
Ole
 

---In [email protected], <s.friedrich@...> wrote :

 Hi Ole,

yes, the energy spills back and forth like a wave. A wave power station is also there because the waves work through the up and down waves.

I think we divert some of the reactive power to let them do some work and then bring them back so the main LC does not notice what's missing.

If it were just a mere sloshing of energy, voltage and current would not rise so much, but would have to stay at the same level as the input energy.

Actually, the flyback in an induction is the boosting component. If there were no stored magnetic field energy, the energy would not increase in a resonant circuit even if it is blind.

Really hard to grasp this topic.

Maybe I should try the resonance experiment with my 3 spotlights, now I have 1200W as a light bulb in 3 lamps, if they shine bright with resonance energy that would be a confirmation of the energy, of course, only if the energy in the input source does not rise sharply.

regards

Sven

Hi Sven,

By the way in an LC-tank the capacitor does to the current what the inductor does to the voltage and vice versa. The inductor gives a voltage kickback when opening the circuit to high impedance while the capacitor gives a current kickback.when closing the circuit to low impedance. The inductor conserves the current during opening the circuit (continuous current and discontinuous voltage) while the capacitor conserves the voltage during closing the circuit (continuous voltage and discontinuous current).

The inductor stores the current as a magnetic field and thus becomes a permanent magnet with the circuit closed and no resistance to dissipate the stored energy.

The capacitor stores the voltage as an electric field and thus becomes a permanent electrified dielectric (electret) with the circuit open and no resistance to dissipate the stored energy.

Regards
Ole
 

---In [email protected], <onielsen@...> wrote :

 Hi Sven,

"If it were just a mere sloshing of energy, voltage and current would not rise so much, but would have to stay at the same level as the input energy."
The voltage and current must increase when energy is increased. The energy can be directly calculated from knowing the LC-tank size and the peak voltage or peak current.

"Actually, the flyback in an induction is the boosting component. If there were no stored magnetic field energy, the energy would not increase in a resonant circuit even if it is blind."
Again knowing the stored energy of the magnetic field (i.e. calculated from the inductance and the current) is what determines the peak of the flyback voltage depending on what impedance the coil looks into. A high impedance flyback load must give a high flyback voltage peak as the current must and will stay continuous at the change in impedance by opening the circuit. The voltage to drive the same current at increased impedance makes the voltage step up at the rate of impedance changing. E.g. doubling the impedance requires doubling the flyback voltage to drive the same current through it. Thus the voltage is doubled but the time it lasts is halved.

Regards
Ole
 

---In [email protected], <s.friedrich@...> wrote :

 Hi Ole,

yes, the energy spills back and forth like a wave. A wave power station is also there because the waves work through the up and down waves.

I think we divert some of the reactive power to let them do some work and then bring them back so the main LC does not notice what's missing.

If it were just a mere sloshing of energy, voltage and current would not rise so much, but would have to stay at the same level as the input energy.

Actually, the flyback in an induction is the boosting component. If there were no stored magnetic field energy, the energy would not increase in a resonant circuit even if it is blind.

Really hard to grasp this topic.

Maybe I should try the resonance experiment with my 3 spotlights, now I have 1200W as a light bulb in 3 lamps, if they shine bright with resonance energy that would be a confirmation of the energy, of course, only if the energy in the input source does not rise sharply.

regards

Sven

Hello Ole,

I try again with simple words to give it back.
I think the magnetic energy adds to the added energy. Practically, we only use one kind of energy at a time.

A flyback converter uses only the stored energy in the magnetic field, the energy used to build up the magnetic field disappears as a loss. Maybe I think that's too easy.

When I read through the relevant Pdfs by Hector or Dan Combine, it sounds like a dream of the future.

I will try to tune in the 1200W spotlights today and see if this happens at power factor 0.
I can measure the current across the shunt and see the shift. If that succeeds at power factor 0 and the lights are on, there is usable energy in it.

regards

Sven

Hi Ole,

I do not understand that completely, finally we have an alternating field and the polarity is constantly changing.

Do you have any idea what I could try?

regards

Sven

Hi Sven,

"A flyback converter uses only the stored energy in the magnetic field, the energy used to build up the magnetic field disappears as a loss. Maybe I think that's too easy."
First energy is stored in the inductor (input cycle) and then it is transferred to the output load (output cycle). The load then dissipates or converts the energy into another form. As long as a current flows it has movement and a magnetic field surrounding it. Decreasing the current decreases the magnetic field of the current. How fast the current is changed determines the voltage that is induced. Or how fast the voltage is changed determines the current that it drives. This is if the inductor has only pure inductance. A real inductor also has resistance and capacitance which also affects the result.

Regards
Ole
 

---In [email protected], <s.friedrich@...> wrote :

 Hello Ole,

I try again with simple words to give it back.
I think the magnetic energy adds to the added energy. Practically, we only use one kind of energy at a time.

A flyback converter uses only the stored energy in the magnetic field, the energy used to build up the magnetic field disappears as a loss. Maybe I think that's too easy.

When I read through the relevant Pdfs by Hector or Dan Combine, it sounds like a dream of the future.

I will try to tune in the 1200W spotlights today and see if this happens at power factor 0.
I can measure the current across the shunt and see the shift. If that succeeds at power factor 0 and the lights are on, there is usable energy in it.

regards

Sven

Hi Sven,

"Do you have any idea what I could try?"
I lost you here. Idea to peak detector or stable output or what?

Regards
Ole
 

---In [email protected], <s.friedrich@...> wrote :

 Hi Ole,

I do not understand that completely, finally we have an alternating field and the polarity is constantly changing.

Do you have any idea what I could try?

regards

Sven

Empty body

SvenHeat would be good.So would refrigerationHave you read about the gravity function?We are likely fooling with time already, ha, ha.
It is just scratching at the door of quantum physics, the practice of which, we still don't have a clue about.If the great abstract thinkers have found it in the warm fuzzy world of mathamatics, it is now up to us to render concepts into reality. We have to build working apparatus.Cheers Warren
Sent from Yahoo Mail on Android 
 
  On Wed, Feb 14, 2018 at 9:13 AM, 'Sven Friedrich' [email protected] [EVGRAY]<[email protected]> wrote:       

Hello Ole,

are you sure the current continues to flow primarily through the inductor, he could still do extra work after the inductor.

I think there is something else. I would like to see the one who makes the inductor cool and freeze at work.
I have already seen pictures of it. But that would be the thermal path from heat to electrical energy.
Maybe just a dream.

regards

Sven Gesendet: Mittwoch, 14. Februar 2018 um 13:48 Uhr
Von: "[email protected] [EVGRAY]" <[email protected]>
An: [email protected]
Betreff: Re: AW: [EVGRAY] Re: buck plug ... 
Hi Sven,

"A flyback converter uses only the stored energy in the magnetic field, the energy used to build up the magnetic field disappears as a loss. Maybe I think that's too easy."
First energy is stored in the inductor (input cycle) and then it is transferred to the output load (output cycle). The load then dissipates or converts the energy into another form. As long as a current flows it has movement and a magnetic field surrounding it. Decreasing the current decreases the magnetic field of the current. How fast the current is changed determines the voltage that is induced. Or how fast the voltage is changed determines the current that it drives. This is if the inductor has only pure inductance. A real inductor also has resistance and capacitance which also affects the result.

Regards
Ole


---In [email protected], <s.friedrich@...> wrote :
 Hello Ole,

I try again with simple words to give it back.
I think the magnetic energy adds to the added energy. Practically, we only use one kind of energy at a time.

A flyback converter uses only the stored energy in the magnetic field, the energy used to build up the magnetic field disappears as a loss. Maybe I think that's too easy.

When I read through the relevant Pdfs by Hector or Dan Combine, it sounds like a dream of the future.

I will try to tune in the 1200W spotlights today and see if this happens at power factor 0.
I can measure the current across the shunt and see the shift. If that succeeds at power factor 0 and the lights are on, there is usable energy in it.

regards

Sven
 
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