AW: Re: AW: Re: Aw: Re: AW: Re: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug.

Hello Ole, you mean a voltage divider that ensures that the auxiliary thyristor is not so high voltage is applied. This resistance creates additional active power and the capacitor discharges. While the circuit is not reflective to the source, a lot of extra power is applied to charge and discharge the capacitors. I do not know how Hector did that without additional active power. The circuit works perfectly as you can see on the scope images. Somewhere the power factor shifts again direction 1 but how do I find out? The stray field energy is also a problem I can not get rid of them so easily because this converter works with alternating polarities for the supply. I have to keep the power reactive. Greetings Sven


Von Samsung Mobile gesendet

-------- Ursprüngliche Nachricht --------
Von: "[email protected] [EVGRAY]" <[email protected]> 
Datum:10.02.2018  21:36  (GMT+01:00) 
An: [email protected] 
Betreff: Re: AW: Re: Aw: Re: AW: Re:  Re: AW: Re:  Re:  Re: AW: [EVGRAY] Re: All information about Diode Plug. 

Hi Sven,

Reconsideration

The small thyristor (SCR) can be used as a non-variable triggering device that trig at its breakdown voltage if the cathode is connected to the gate of the main SCR. The anode is then fed through the current limiting resistor to the opposite side of the capacitor for monitoring the voltage across the capacitor. Perhaps it even can be made variable by some resistor voltage dividing network like shown here if the pasting gets through:

Regards
Ole



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


Hello Ole, you're right the series resistor only limits the current not the voltage. I'll probably stay with the 1200v thyristors. The thyristors in the pdf consume 60% less power but only 1000v. I have only measured the resonant coil windings in which I have measured current and voltage and then calculated the inductance via the no-load impedance, only the windings in the one leg. Unfortunately, the Variacs seem to be overwhelmed. One is broken complete insulation fault. Greetings Sven

Von Samsung Mobile gesendet


-------- Ursprüngliche Nachricht --------
Von: "onielsen@... [EVGRAY]" 
Datum:09.02.2018 18:23 (GMT+01:00) 
An: [email protected] 
Betreff: Re: Aw: Re: AW: Re: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug. 

 
Hi Sven,

"Actually, I do not need so high voltage-resistant auxiliary thyristors, the resistor is limited by the applied voltage and is only on the power thyristor."

Resistors don't limit the voltage. They limit the current. When current begins to go through a resistor the voltage across it increases proportionally with the current through it. Thus if the thyristor has a lower breakdown voltage than the voltage being applied to it it will still start conducting without any control.

The small thyristor does see the same voltage as the main thyristor minus the gate cathode voltage of the main thyristor.

"I have several options now:

      218.615 mH = 450V Urms max.
      882,037 mH = 800V Urms max.
    1010,000 mH = 800V Urms max.
    2274,000 mH = ??????"

Is this the inductance calculated from the impedance at different loads? At no load the full inductance is available. At heavy load most of the inductance is canceled. At least this is how ordinary transformers work.

Regards
Ole





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

Hello Ole,

I think you're right with the power diodes between inductor and capacitor, the circle is not closed or permeable in both directions, so no resonance can arise.

I have again made a part of my triggering for the thyristors. I had to use the 1.1 k resistor because the 2.2 K resistor was too large and the current was not enough to ignite the power thyristor. If I take out the diode and replace it with a resistor, I have to reduce the 1.1 K resistor or just remove the diode. Just as the diode plug circuit now works perfectly almost clean sine only slight distortions could risk it.

Unfortunately, I do not get the Tyn1225 auxiliary thyristors cheap anymore, so I do not want to break that much anymore.

Actually, I do not need so high voltage-resistant auxiliary thyristors, the resistor is limited by the applied voltage and is only on the power thyristor.

Last night I again measured the resonance coil of the 3-phase transformer via the impedance and of course got quite different values ​​as my LCR meter had indicated. I will not grow any more so that works very well.

I have several options now:

      218.615 mH = 450V Urms max.
      882,037 mH = 800V Urms max.
    1010,000 mH = 800V Urms max.
    2274,000 mH = ??????

My probe is only up to 2.5 kv, and I have at the penultimate measurement already 2.02 kv peak to peak. Although I think that the probe for the peak voltage is specified but one should not challenge his luck.

Anyway, I've driven with the higher voltage and smaller coupling capacitors and have a much better efficiency, so output to input energy.

I have the feeling that the resonance only up to the max. Anything above increases only the reactive current and the voltage does not increase any further, but this only costs input power.

regards

Sven
 
Gesendet: Freitag, 09. Februar 2018 um 03:00 Uhr
Von: "onielsen@... [EVGRAY]" <[email protected]>
An: [email protected]
Betreff: Re: AW: Re: Aw: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug.
 
Hi Sven,

If the gate was too sensitive an extra resistor could be placed between the gate and the cathode to form a voltage divider with the gate current limiting resistor.

The reason for the extra diode at the variable inductor could be some leftover from an earlier design from before adding the inductor or choke. At least it can be omitted with the two other diodes in place as the current will never go in the reverse direction. Without the inductor (choke) only one diode is needed. The bottom diode is for charging the capacitor and the one sharing its anode (or cathode) is for 'discharging' or demagnetizing the inductor. I.e. the flyback voltage from disrupting the current through the inductor makes the current 'discharge' to the capacitor which is then being charged.

Regards
Ole



---In [email protected], <s.friedrich@...> wrote :
 
Hi Ole, I'm afraid that over the Powerthyristor cathode on the gate a reverse voltage destroyed the auxiliary thyristor. before I installed the diodes, the thyristors broke, but at the same time we installed the series resistor on the anode side of the auxiliary thyristor, maybe it is the reason that everything runs stable. Somehow you're right, but why did Hector draw the diodes between Variac and capacitor? Greetings Sven
 
 
Von Samsung Mobile gesendet


-------- Ursprüngliche Nachricht --------
Von: "onielsen@... [EVGRAY]"
Datum:09.02.2018 00:10 (GMT+01:00)
An: [email protected]
Betreff: Re: Aw: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug.

 
Hi Sven,

"There are many auxiliary thyristors gone without the diode a series resistor for current limiting I have installed for the auxiliary thyristor as in the drawing. The ignition of the power thyristors has worked very well so far, my idea is more likely to high sensitive thyristors that can be switched with μA thereby would be wasted little energy for the ignition and perhaps rather the synchrony of the system can be achieved."

Diodes don't limit the current very much as they're good conductors in the forward conduction direction. They have a forward conduction voltage drop of around 0.7V or a little more when at heavy load.

"The diode between reactor (Variac) and discharge capacitor should prevent a resonance that can arise between the reactive components"
The other two diodes at the other end of the variac already prevent the current from going in the opposite direction of the first diode at the top end. The top end diode is in series with the double split diodes and the current form either one of the split diodes also goes through the top end diode. Thus the top diode is only introducing another voltage drop (power dissipation) for current going through the variac. This is the brother in law effect (if one's brother in law sells diodes).

Regards
Ole
 



---In [email protected], <s.friedrich@...> wrote :
 
Hi Ole, first of all many thanks for your help, I will consider the change with positive diode plugs the ignition energy is removed after the transformer coil and the left diode directly from the capacitor. The charge curve and discharge curve of the two capacitors are almost identical now perfect, now you can work on the vote.



"The small diodes are protecting the small thyristor as the gates of the thyristor are very low ohmic in both directions."
Use resistors instead of diodes when they start conducting. A slight change in voltage gives a big change in current as they are very nonlinear devices "

There are many auxiliary thyristors gone without the diode a series resistor for current limiting I have installed for the auxiliary thyristor as in the drawing. The ignition of the power thyristors has worked very well so far, my idea is more likely to high sensitive thyristors that can be switched with μA thereby would be wasted little energy for the ignition and perhaps rather the synchrony of the system can be achieved.
I think that is still a long way, but a very interesting experiment.


"And why did you mark the Variacs and the protective diodes at the gate of the thyristor power?"
The variacs have the two bottom diodes prevent any current from changing direction. Thus the top diodes at each variac can be omitted as the current through the variacs never changes direction. The voltage across them will change depending on receiving energy or supplying energy."

The diode between reactor (Variac) and discharge capacitor should prevent a resonance that can arise between the reactive components. I will send you the original circuits.

regards

Sven
 ...

Hi Sven,

The voltage divider is at signal level. I don't think power resistors are necessary for triggering the power thyristor. The break down voltage is still quite high for the alternative small signal thyristor which makes the voltage divider only decrease the voltage a small part.

"While the circuit is not reflective to the source, a lot of extra power is applied to charge and discharge the capacitors"
 Charging the capacitor is real power when not given back the charge to the source. Only by giving back the charge makes the power reactive.

"Somewhere the power factor shifts again direction 1 but how do I find out?"
 The power factor is given by which way the phase shift is. A power factor of one means no phase shift between the voltage and current. If the voltage increases/decreases before the current does the power is inductive reactive. If the phase shift is opposite the power is capacitive reactive.

"The stray field energy is also a problem I can not get rid of them so easily because this converter works with alternating polarities for the supply."
 To keep stray fields low avoid current running through big loops. This can be done by twisting the wires in pairs. E.g. twist the forward current carrying wire with the return path to minimize the current taking a loop. The twisting makes the small loops left alternate in different directions thus canceling the fields from each other. Alternatively use coaxial cables which also cancel the fields.

Regards
Ole
 

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

 Hello Ole, you mean a voltage divider that ensures that the auxiliary thyristor is not so high voltage is applied. This resistance creates additional active power and the capacitor discharges. While the circuit is not reflective to the source, a lot of extra power is applied to charge and discharge the capacitors. I do not know how Hector did that without additional active power. The circuit works perfectly as you can see on the scope images. Somewhere the power factor shifts again direction 1 but how do I find out? The stray field energy is also a problem I can not get rid of them so easily because this converter works with alternating polarities for the supply. I have to keep the power reactive. Greetings Sven
 

 

 Von Samsung Mobile gesendet



-------- Ursprüngliche Nachricht --------
Von: "onielsen@... [EVGRAY]" 
Datum:10.02.2018 21:36 (GMT+01:00) 
An: [email protected] 
Betreff: Re: AW: Re: Aw: Re: AW: Re: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug. 

   Hi Sven,

Reconsideration

The small thyristor (SCR) can be used as a non-variable triggering device that trig at its breakdown voltage if the cathode is connected to the gate of the main SCR. The anode is then fed through the current limiting resistor to the opposite side of the capacitor for monitoring the voltage across the capacitor. Perhaps it even can be made variable by some resistor voltage dividing network like shown here if the pasting gets through:

Empty body

Hi Sven,

The gate trigger current of the main SCR http://ixapps.ixys.com/datasheet/mcd162-16io1.pdf can be up to 200mA. If the voltage across the capacitor is 400V at the point of triggering then a gate resistor at 1.1kOhm makes the gate current around 364mA. With 400V across and 364mA through the gate resistor some 146W of power is dissipated in the resistor during the brief moment of triggering.

"That's right, but you should only give back a certain part of it, part should be free energy that we store in the capacitors, otherwise the circuit makes no sense."
The part given back is the reactive part of the power. The part being consumed is the active (or real) power. As long as only making the free energy into active power the active power then is free energy.

"I meant only with the shift of the power factor, the additional active power is taken from the reactive circuit, if one finds the point where this happens explicitly in the circuit could you perhaps do something about it?"
The parts of the circuit having Ohmic resistances are the ones with the phase factor of one. Any reactive impedance which is caused by inductance or capacitance has the phase factor different from one. At resonance the inductive and capacitive reactances cancel each other as one is positive and the other is negative.

"I meant the stray field of the push pull trafos or the energy which is switched on the push pull back into the primary windings. Without secondary load you can not set the diode plug and this load must also be ohmic nature, a transformer with no load on the secondary of the push-pull transformer stops the switching in the diode plug."
Yes there could be a problem here. At no load the power is pure reactive being reflected back to source.

"You could work with freewheeling diodes on each winding and burn the energy through resistors, but that further increases the losses and only work with freewheeling diodes is too risky..."
If the energy is free it could be dissipated as heat or the device could be automatic shut off when there is no demand for power. When the demand for power returns the device then must also start automatic.

"I will try to work with smaller discharge capacitors so that the voltages on the diode plug side and the transverter are pretty much the same."
It's not the capacitors that determine the voltage when to be discharged. It's the triggering circuit parameters of the voltage divider that fires the SCR. Smaller capacitance only makes the capacitors charge/discharge faster and use less cycles for given voltage build up.

Regards
Ole
 

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

 Hi Ole,
 
 The voltage divider is at signal level. I do not think power resistors are necessary for triggering the power thyristor. The break down voltage is still quite high for the small signal thyristor which makes the voltage decrease.
 
 That's right, it worked very well only until I had set the diode plug started to heat the 2 watt resistors, so abe I replaced it with 2x 9Watt ceramic resistors. Before I had 2.2 kohm now 1.1 kohm. I can not imagine that the resistors are responsible for the positive phase shift, but in fact they consume active power.
 
 
 "While the circuit is not reflective of the source, a lot of extra power is applied to charge and discharge the capacitors"
 Charging the capacitor is real power when not given back the charge to the source. Only by giving back the charge makes the power reactive.
 
 That's right, but you should only give back a certain part of it, part should be free energy that we store in the capacitors, otherwise the circuit makes no sense.
 
 
 "Somewhere the power factor shifts again direction 1 but how do I find out?"
 
 The power factor is given by which way the phase shift is. A power factor of one means no phase shift between the voltage and current. If the voltage increases / decreases before the current does the power is inductive reactive. If the phase shift is opposite the power is capacitive reactive.
 
 I meant only with the shift of the power factor, the additional active power is taken from the reactive circuit, if one finds the point where this happens explicitly in the circuit could you perhaps do something about it?
 
 
 
 
 
 
 "The stray field energy is so a problem I can not get rid of them so easily because of this converter works with alternating polarities for the supply."
 
 To keep stray fields low avoid current running through big loops. This can be done by twisting the wires in pairs. E.g. twist the forward current carrying wire. The twisting makes the small loops left alternate in different directions thus canceling the fields from each other. Alternative use coaxial cables which therefore cancel the fields.
 
 I meant the stray field of the push pull trafos or the energy which is switched on the push pull back into the primary windings. Without secondary load you can not set the diode plug and this load must also be ohmic nature, a transformer with no load on the secondary of the push-pull transformer stops the switching in the diode plug.
 
 You could work with freewheeling diodes on each winding and burn the energy through resistors, but that further increases the losses and only work with freewheeling diodes is too risky, because I switched through directly with a defective diodes without load and the power thyristor was defective.
  
 I will try to work with smaller discharge capacitors so that the voltages on the diode plug side and the transverter are pretty much the same. So always creates a sink, that is 400V at the transverter and 250V at the diode plug more create the coupling capacitors not, that is principle that I try too much energy to extract.
 
 regards
 
 Sven
  ...

Empty body

Hi Sven,

"...sometimes I have the feeling that the diode plug is just a machine to generate free energy in the next stage ie the pushpull trafo by saturating it with sharp short pulses."
I thought the big ferroresonant transformer "transverter" had that purpose. Else get rid of it to make the device simpler. The ferroresonant transformer has a magnetic path that is supposed to go into saturation. This path acts as a very fast magnetic gate. The output push-pull transformer is then for impedance matching and putting the half phases together to be used in a single load.

"I have the feeling that the circuit is not yet complete as the diodes separate the inductor and capacitor, which are not necessary, but why are they drawn?"
Separation is only in the reverse conducting or blocking direction. The other half part has the diode in the opposite direction. Each half part is semi-resonant with one part for each direction of current flow.

"So the adjustment between inductor and capacitor must fit?!? Just how should that look on the scope that fits?"
They will oscillate at the resonant frequency and the curve is a sinusoidal wave or actually only a half oscillation for each half part of the diode plug. When the capacitance (or inductance) is decreased the oscillation becomes faster which is seen as a shorter distance in the repetition rate. I.e. the frequency is increased.

"But with smaller capacitors and the same trigger voltage less energy is taken, the frequency of 50 Hz remains, but I fear most of the energy is in the inductor Variac. When I tune to the best possible values, the lamps shine as brightly as with large capacitors."
But are the half sines (parabolic curves) not higher and of shorter duration even though the repetition rate is the same? The semi-resonant frequency should go up as well as the peak voltage if the discharge frequency stays at 50Hz and the load receives the same power.

"I do not have any ideas anymore. I wanted to connect the identical 230V / 24V transformers to each diode plug and shoot the pulses into the battery."
Transformers are for AC. If used at pulsed DC they may not demagnetize in the opposite direction and thus start to saturate. At least if used for pulsed DC (half phase) they must have an air gap to be able to proper demagnetize. Else the remanence of the magnetic core makes it a permanent magnet. Ed Leedskalnin calls the effect a perpetual motion hanger when the circuit is permanent magnetized.

Regards
Ole

 

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

 Hello Ole,
 
 SCR can be up to 200mA. If the voltage across the capacitor is 400V at the point of triggering then a gate resistor at 1.1kOhm makes the gate current around 364mA. With 400V across and 364mA through the gate resistor some 146W of power is dissipated in the resistor during the brief moment of triggering.
 
 The trigger is adjustable via pots, I would like to cover a large voltage range so even switch below 200 volts or earlier trigger. But I also do not think that they are responsible for losses.
 
 
 
 
 Otherwise, the circuit makes no sense. "That's right, but you should just give it a certain part of it.
 The part given back is the reactive part of the power. The part being consumed is the active (or real) power. As long as only the free energy into active power becomes the active power then is free energy.
 
 Well written, but how do you get it that way, sometimes I have the feeling that the diode plug is just a machine to generate free energy in the next stage ie the pushpull trafo by saturating it with sharp short pulses. Somehow I do not know how to continue. I have the feeling that the circuit is not yet complete as the diodes separate the inductor and capacitor, which are not necessary, but why are they drawn?
 
 
 "I meant only with the shift of the power factor, the additional active power is taken from the reactive circuit, if one finds the point."
 Ohmic resistances are the ones with the phase factor of one. Any reactive impedance or capacitance has the phase factor different from one. At one and the other is negative.
 
 So the adjustment between inductor and capacitor must fit?!? Just how should that look on the scope that fits?
 I always try to get a clean picture and adjust the distortions and tips are filtered out.
 
 
 
 
 
 
 "I mean, the transformer pulls into the primary windings, and the load does not pull back into the primary windings." load on the secondary of the push-pull transformer stops the switching in the diode plug. "
 Yes there could be a problem here. At no load the power is pure reactive being reflected back to source.
 
 
 
 "I want to work with smaller discharge capacitors so the voltage on the diode plug and the transverter are pretty much the same."
 It's not the capacitors that determine the voltage when to be discharged. It's the triggering circuit parameters of the voltage divider that fires the SCR. Smaller capacitance only makes the capacitors charge / discharge faster and use less cycles for given voltage build up.
 
 But with smaller capacitors and the same trigger voltage less energy is taken, the frequency of 50 Hz remains, but I fear most of the energy is in the inductor Variac. When I tune to the best possible values, the lamps shine as brightly as with large capacitors.
 
 I do not have any ideas anymore. I wanted to connect the identical 230V / 24V transformers to each diode plug and shoot the pulses into the battery. Unfortunately, the impedance of the transformers is so high that the diode plug does not start even with the battery as a load. I could try it again with freewheeling diodes and incandescent resistance.
 
 Regards
 Sven
...

Sven, OleAgain, it makes me think of harvesting the energy using permanent magnet motors. They seem to love pulsed DC input. The ac generator they drive must run at either1800 or 3600 rpm, depending on the type.Pulleys and belts?Cheers Warren

Sent from Yahoo Mail on Android 
 
  On Sun, Feb 11, 2018 at 2:06 PM, [email protected] [EVGRAY]<[email protected]> wrote:       
Hi Sven,

"...sometimes I have the feeling that the diode plug is just a machine to generate free energy in the next stage ie the pushpull trafo by saturating it with sharp short pulses."
I thought the big ferroresonant transformer "transverter" had that purpose. Else get rid of it to make the device simpler. The ferroresonant transformer has a magnetic path that is supposed to go into saturation. This path acts as a very fast magnetic gate. The output push-pull transformer is then for impedance matching and putting the half phases together to be used in a single load.

"I have the feeling that the circuit is not yet complete as the diodes separate the inductor and capacitor, which are not necessary, but why are they drawn?"
Separation is only in the reverse conducting or blocking direction. The other half part has the diode in the opposite direction. Each half part is semi-resonant with one part for each direction of current flow.

"So the adjustment between inductor and capacitor must fit?!? Just how should that look on the scope that fits?"
They will oscillate at the resonant frequency and the curve is a sinusoidal wave or actually only a half oscillation for each half part of the diode plug. When the capacitance (or inductance) is decreased the oscillation becomes faster which is seen as a shorter distance in the repetition rate. I.e. the frequency is increased.

"But with smaller capacitors and the same trigger voltage less energy is taken, the frequency of 50 Hz remains, but I fear most of the energy is in the inductor Variac. When I tune to the best possible values, the lamps shine as brightly as with large capacitors."
But are the half sines (parabolic curves) not higher and of shorter duration even though the repetition rate is the same? The semi-resonant frequency should go up as well as the peak voltage if the discharge frequency stays at 50Hz and the load receives the same power.

"I do not have any ideas anymore. I wanted to connect the identical 230V / 24V transformers to each diode plug and shoot the pulses into the battery."
Transformers are for AC. If used at pulsed DC they may not demagnetize in the opposite direction and thus start to saturate. At least if used for pulsed DC (half phase) they must have an air gap to be able to proper demagnetize. Else the remanence of the magnetic core makes it a permanent magnet. Ed Leedskalnin calls the effect a perpetual motion hanger when the circuit is permanent magnetized.

Regards
Ole




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

Hello Ole,

SCR can be up to 200mA. If the voltage across the capacitor is 400V at the point of triggering then a gate resistor at 1.1kOhm makes the gate current around 364mA. With 400V across and 364mA through the gate resistor some 146W of power is dissipated in the resistor during the brief moment of triggering.

The trigger is adjustable via pots, I would like to cover a large voltage range so even switch below 200 volts or earlier trigger. But I also do not think that they are responsible for losses.




Otherwise, the circuit makes no sense. "That's right, but you should just give it a certain part of it.
The part given back is the reactive part of the power. The part being consumed is the active (or real) power. As long as only the free energy into active power becomes the active power then is free energy.

Well written, but how do you get it that way, sometimes I have the feeling that the diode plug is just a machine to generate free energy in the next stage ie the pushpull trafo by saturating it with sharp short pulses. Somehow I do not know how to continue. I have the feeling that the circuit is not yet complete as the diodes separate the inductor and capacitor, which are not necessary, but why are they drawn?


"I meant only with the shift of the power factor, the additional active power is taken from the reactive circuit, if one finds the point."
Ohmic resistances are the ones with the phase factor of one. Any reactive impedance or capacitance has the phase factor different from one. At one and the other is negative.

So the adjustment between inductor and capacitor must fit?!? Just how should that look on the scope that fits?
I always try to get a clean picture and adjust the distortions and tips are filtered out.






"I mean, the transformer pulls into the primary windings, and the load does not pull back into the primary windings." load on the secondary of the push-pull transformer stops the switching in the diode plug. "
Yes there could be a problem here. At no load the power is pure reactive being reflected back to source.



"I want to work with smaller discharge capacitors so the voltage on the diode plug and the transverter are pretty much the same."
It's not the capacitors that determine the voltage when to be discharged. It's the triggering circuit parameters of the voltage divider that fires the SCR. Smaller capacitance only makes the capacitors charge / discharge faster and use less cycles for given voltage build up.

But with smaller capacitors and the same trigger voltage less energy is taken, the frequency of 50 Hz remains, but I fear most of the energy is in the inductor Variac. When I tune to the best possible values, the lamps shine as brightly as with large capacitors.

I do not have any ideas anymore. I wanted to connect the identical 230V / 24V transformers to each diode plug and shoot the pulses into the battery. Unfortunately, the impedance of the transformers is so high that the diode plug does not start even with the battery as a load. I could try it again with freewheeling diodes and incandescent resistance.

Regards
Sven
...
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Empty body

Hi Sven,

This text from Hector tells that the impedance of each part (input and output side) of the circuit has to match the frequency. The charging of the capacitors must be done in no longer time than a half cycle which is 10ms at 50Hz. The output as well must be done in less time than a half cycle.

To do this the LC constant of the input part formed by the output of the ferroresonant transformer and one of the capacitors must be shorter than the time of half a period. The same applies for the output part of the circuit. As the capacitor is a constant for both input and output only the transformers can be changed. Thus if the capacitors are chosen to match the ferroresonant transformer the output transformer has to be chosen for the proper output impedance.

The load has to discharge each capacitor in less than 10ms. The full period time is given as 2πf X square root(LC). Thus solve for L or C when choosing the other one of them. f is the frequency in Hz L is the inductance in Henry and C is the capacitance in Coulomb. The same formula is used for the charging part of the circuit where L is the output inductance ot the ferroresonant transformer. For the output transformer the inductance is decreased by increasing the load as this cancels more of the primary winding's inductance. The fine thing about transformers is that they actually change the impedance between the primary and secondary windings. Thus choose the transformation ratio to fit the wanted impedance which can go up or down or be one to one. This can be chosen for 24V at the output and the voltage of the capacitors at the input. The voltage at the capacitors is the peak voltage that must be transformed down to the peak of 24V. That peak is different if the period time is different from a full wave when combining the two semi-waves. A shorter pulse width at 50Hz requires a higher peak value for delivering the same RMS voltage.

Regards
Ole 

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

 Hello Ole, I found something here on evgray.
 
 I quote: Transverter Secrets Revealed:
Data from 1999 series of Experiments

480/220VAC Universal transformer design : 

The circuit here consists of 2 half LC tanks, AS know the RADIANT Energy 
flashover occurs at TURN ON, one capacitor charges to maximal resonant value
then the cycle goes in opposed sinewave as the other charges in Resonant
condition the first one is DISCHARGED in the second transformer with a 
nominal gain of 1.618 if properly tunned, this second circuit deplete C 
Voltage value as near to 0 Volts as SCR switches off at a minimal remanent 
voltage (minimal)before primary tank reverses sinewave to Reload half LC and 
discharge the other loaded capacitor.....


The discharge does not affect input tunning as its at opposed non-coupled 
relation this permits the full vector of the power components 
to be decopled from the source (totaly non-reflective) as to permit a PERFECT 
resonant tune of LC as to charge capacitor in RF radiant operation mode.


Now you have the BASIC and simple desing of the looped transverter circuit , 
just be Verry carefull with it ,use at minimal power and follow safety common 
sense I dont want you to have any incident with this device ,as you validate 
feel free to post (its quite demanding in its tuning) post this letter with
your results as EXTRA information for others to follow.


For basic triggering you can use multy-vibrator optoinsolated mode a simple 
switching diode set up to triger the opossed capacitor when charged, Input ac 
regulates the timing, LOAD regulates the discharge time so overal time must be 
shorter (Lower impedance)to increase frequency (shorter pulse)(shorter 
discharge time as to reduce capacitor Voltage value to near 0 .


Read ZPEV2.pdf  ARK Files 

Hector D Perez Torres 
 I understand this so that the load may only have a low impedance so that the energy remains reactive best discharged against a short circuit, so the power thyristors and diodes that can make these strong discharges at all possible, well a short circuit even hold the circuit not from. But where does this lead us back to reactive power?
 
 I can not send such a potential directly through a 24Volt battery, it has only an internal resistance of 11 mOhm.
 
 Or do you understand this text differently?
 
 regards
 
 Sven
  ...

Correction C is the capacitance in Farad of course.

Regards
Ole

 

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

 Hi Sven,

This text from Hector tells that the impedance of each part (input and output side) of the circuit has to match the frequency. The charging of the capacitors must be done in no longer time than a half cycle which is 10ms at 50Hz. The output as well must be done in less time than a half cycle.

To do this the LC constant of the input part formed by the output of the ferroresonant transformer and one of the capacitors must be shorter than the time of half a period. The same applies for the output part of the circuit. As the capacitor is a constant for both input and output only the transformers can be changed. Thus if the capacitors are chosen to match the ferroresonant transformer the output transformer has to be chosen for the proper output impedance.

The load has to discharge each capacitor in less than 10ms. The full period time is given as 2πf X square root(LC). Thus solve for L or C when choosing the other one of them. f is the frequency in Hz L is the inductance in Henry and C is the capacitance in Coloumb Farad. The same formula is used for the charging part of the circuit where L is the output inductance ot the ferroresonant transformer. For the output transformer the inductance is decreased by increasing the load as this cancels more of the primary winding's inductance. The fine thing about transformers is that they actually change the impedance between the primary and secondary windings. Thus choose the transformation ratio to fit the wanted impedance which can go up or down or be one to one. This can be chosen for 24V at the output and the voltage of the capacitors at the input. The voltage at the capacitors is the peak voltage that must be transformed down to the peak of 24V. That peak is different if the period time is different from a full wave when combining the two semi-waves. A shorter pulse width at 50Hz requires a higher peak value for delivering the same RMS voltage.

Regards
Ole 

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

 Hello Ole, I found something here on evgray.
 
 I quote: Transverter Secrets Revealed:
Data from 1999 series of Experiments

480/220VAC Universal transformer design : 

The circuit here consists of 2 half LC tanks, AS know the RADIANT Energy 
flashover occurs at TURN ON, one capacitor charges to maximal resonant value
then the cycle goes in opposed sinewave as the other charges in Resonant
condition the first one is DISCHARGED in the second transformer with a 
nominal gain of 1.618 if properly tunned, this second circuit deplete C 
Voltage value as near to 0 Volts as SCR switches off at a minimal remanent 
voltage (minimal)before primary tank reverses sinewave to Reload half LC and 
discharge the other loaded capacitor.....


The discharge does not affect input tunning as its at opposed non-coupled 
relation this permits the full vector of the power components 
to be decopled from the source (totaly non-reflective) as to permit a PERFECT 
resonant tune of LC as to charge capacitor in RF radiant operation mode.


Now you have the BASIC and simple desing of the looped transverter circuit , 
just be Verry carefull with it ,use at minimal power and follow safety common 
sense I dont want you to have any incident with this device ,as you validate 
feel free to post (its quite demanding in its tuning) post this letter with
your results as EXTRA information for others to follow.


For basic triggering you can use multy-vibrator optoinsolated mode a simple 
switching diode set up to triger the opossed capacitor when charged, Input ac 
regulates the timing, LOAD regulates the discharge time so overal time must be 
shorter (Lower impedance)to increase frequency (shorter pulse)(shorter 
discharge time as to reduce capacitor Voltage value to near 0 .


Read ZPEV2.pdf  ARK Files 

Hector D Perez Torres 
 I understand this so that the load may only have a low impedance so that the energy remains reactive best discharged against a short circuit, so the power thyristors and diodes that can make these strong discharges at all possible, well a short circuit even hold the circuit not from. But where does this lead us back to reactive power?
 
 I can not send such a potential directly through a 24Volt battery, it has only an internal resistance of 11 mOhm.
 
 Or do you understand this text differently?
 
 regards
 
 Sven
  ...