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

Hi Sven,

"If I could connect the scope somehow so I can see where I have to flow back into the reactive power with the energy from the transformer ?"
I'm not sure that I understand this question. It is possible to see if the power is reactive or active by using the scope in X-Y mode and having the voltage across a component in the horizontal direction and the current in the vertical direction. A pure resistive load is then shown as a tilted line. The slope of the line is the resistance. If showing an ellipse there's a phase shift between the voltage and the current which is then reactive power. If the ellipse is tilted there is both real and reactive power in the component. I.e. the component both stores and dissipates energy at the same time like a resistive coil or a capacitor having big resistance.

An even better way is to use a scope with math functions to multiply the channel representing the voltage by the channel representing the current and having it show the curve representing the power. This way it can be seen if the component is consuming or delivering power. E.g. this can be used to see if a motor is actually working as motor or generator.

Regards
Ole
 

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

 Hello Ole, in the video I have 3x 400 Watt spotlight and the DC motor connected. I could slow down the engine with my hand a big difference did not make that. The total active power dropped by 30 watts at this load. Finally, I have a 2 kw fan heater connected now you can hear the scr switch hard. Soon, only the ohmic resistance of the transformer will work. If I could connect the scope somehow so I can see where I have to flow back into the reactive power with the energy from the transformer ?. Greetings Sven
 

 

 Von Samsung Mobile gesendet



-------- Ursprüngliche Nachricht --------
Von: "onielsen@... [EVGRAY]" 
Datum:13.02.2018 00:04 (GMT+01:00) 
An: [email protected] 
Betreff: Re: Aw: Re: AW: Re: AW: Re: AW: Re: Re: Re: AW: Re: AW: Re: Re: AW: Re: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug. 

    Hi Sven,

The impedance is reduced by mechanically loading the motor or if it is a transformer the secondary is loaded which cancels some of the field in the primary coil.

Regards
Ole

 

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

   Hi Ole,
 
 Here is my video what I did back then. At high load, the lamp also went out and at idle it shone brightly. So you could extract some of the energy from the primary coil to reduce the impedance. Or am I wrong.
 
 regards
 
 Sven
  
 https://www.youtube.com/watch?v=yTORuC57G0g
  
  
  
 Gesendet: Montag, 12. Februar 2018 um 13:10 Uhr
 Von: "'Sven Friedrich' s.friedrich@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Aw: Re: AW: Re: AW: Re: AW: Re: Re: Re: AW: Re: AW: Re: Re: AW: Re: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug.
    
 Hi Ole,
 
 I had rather thought of a snubber, about a freewheeling diode, I could unload the primary coil immediately partly.
 
 I tried that with a DC motor and a self-built IGBT push pull inverter. At idle, the coils were disposed of energy via the freewheeling diodes and the capacitor in series and a light bulb. The lighter the secondary, or the lighter the DC motor was, the brighter the bulb was. Under load, the impedance dropped and the energy thrown back, I make a drawing or cut.
 
 regards
 
 Sven   Gesendet: Montag, 12. Februar 2018 um 11:59 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: AW: Re: AW: Re: AW: Re: Aw: Re: Re: AW: Re: AW: Re: Re: AW: Re: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug.
   Hi Sven,
 
 The boost and buck converters change the transformation ratio by changing the duty cycle or ratio between input and output part of the cycle. When using thyristors this gives the problem of shutting them off while current are going through them. Shut off has to be done externally by bypassing the current through them. Transistors are usually used instead of thyristors as well as much higher switching frequencies for being able to use smaller transformers for the same power.
 
 Regards
 Ole
 
 
 ---In [email protected], <s.friedrich@...> wrote :
   Hi Ole, I have already connected a 400w spotlight and a 116w bulb. Is not it possible to directly discharge the primary winding like a boost or buck converter to lower the impedance? Greetings Sven
  
  
 Von Samsung Mobile gesendet

 .....

Hello Ole, in the video I have 3x 400 Watt spotlight and the DC motor connected. I could slow down the engine with my hand a big difference did not make that. The total active power dropped by 30 watts at this load. Finally, I have a 2 kw fan heater connected now you can hear the scr switch hard. Soon, only the ohmic resistance of the transformer will work. If I could connect the scope somehow so I can see where I have to flow back into the reactive power with the energy from the transformer ?. Greetings Sven


Von Samsung Mobile gesendet

-------- Ursprüngliche Nachricht --------
Von: "[email protected] [EVGRAY]" <[email protected]> 
Datum:13.02.2018  00:04  (GMT+01:00) 
An: [email protected] 
Betreff: Re: Aw: Re: AW: Re: AW: Re: AW: Re:  Re:  Re: AW: Re: AW: Re:  Re: AW: Re:  Re: AW: Re:  Re:  Re: AW: [EVGRAY] Re: All information about Diode Plug. 

 Hi Sven,

The impedance is reduced by mechanically loading the motor or if it is a transformer the secondary is loaded which cancels some of the field in the primary coil.

Regards
Ole



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

 
Hi Ole,

Here is my video what I did back then. At high load, the lamp also went out and at idle it shone brightly. So you could extract some of the energy from the primary coil to reduce the impedance. Or am I wrong.

regards

Sven
 
https://www.youtube.com/watch?v=yTORuC57G0g
 
 
 
Gesendet: Montag, 12. Februar 2018 um 13:10 Uhr
Von: "'Sven Friedrich' s.friedrich@... [EVGRAY]" <[email protected]>
An: [email protected]
Betreff: Aw: Re: AW: Re: AW: Re: AW: Re: Re: Re: AW: Re: AW: Re: Re: AW: Re: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug.
 
 

Hi Ole,

I had rather thought of a snubber, about a freewheeling diode, I could unload the primary coil immediately partly.

I tried that with a DC motor and a self-built IGBT push pull inverter. At idle, the coils were disposed of energy via the freewheeling diodes and the capacitor in series and a light bulb. The lighter the secondary, or the lighter the DC motor was, the brighter the bulb was. Under load, the impedance dropped and the energy thrown back, I make a drawing or cut.

regards

Sven
 
Gesendet: Montag, 12. Februar 2018 um 11:59 Uhr
Von: "onielsen@... [EVGRAY]" <[email protected]>
An: [email protected]
Betreff: Re: AW: Re: AW: Re: AW: Re: Aw: Re: Re: AW: Re: AW: Re: Re: AW: Re: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug.
 
Hi Sven,

The boost and buck converters change the transformation ratio by changing the duty cycle or ratio between input and output part of the cycle. When using thyristors this gives the problem of shutting them off while current are going through them. Shut off has to be done externally by bypassing the current through them. Transistors are usually used instead of thyristors as well as much higher switching frequencies for being able to use smaller transformers for the same power.

Regards
Ole



---In [email protected], <s.friedrich@...> wrote :
 
Hi Ole, I have already connected a 400w spotlight and a 116w bulb. Is not it possible to directly discharge the primary winding like a boost or buck converter to lower the impedance? Greetings Sven
 
 
Von Samsung Mobile gesendet
.....

Hi Ole, I assume that I have a clean sine wave at the reactive power generated in the transverter in the transmission over the coupling capacitors changes nothing except that everything is shifted by 180 degrees. If I now turn on the diode connector, this sine wave will change, I hope, at least by switching the thyristors. If you now set the switch so that is within the voltage time, the half LCs should be united again. It all works perfectly well thanks to your help, but that you can just connect any load as in a conventional way is a misconception of me or I have probably misinterpreted in the descriptions.

regards

Sven

Hi Ole an dSven
 A group of 4 electrical engineers at a science meeting I was at many years ago (a Bill Beatty weird science meeting) was approached and asked by me how to do accurate real-power measurements for a DC pulsed "konehead" motor that I brought to the meeting.....they thought about it and talked about it for awhile, and then said this to me, and I would assume it would work in AC too:
 On graph paper, make an exact drawing of how the voltage looks on scope....
 On the same graph paper, but translucent type, make another exact drawing of how the current looks on scope....
 Lay the translucent graph paper (current) on top of the other piece of graph paper...line up the small graph squares so it sits on top perfectly.
 With pencil or pen, shade-in the areas where current ovelaps the voltage, then after that, calculate how much power the shaded-in area represents...
 ciaoKone

Hi Sven,

The Edison way of distributing power with big impedance mismatch by having the voltage staying almost constant during different loads doesn't work too good for a resonating circuit as any change in impedance will affect the resonant frequency. Without any compensation the load must stay constant. I believe that is what the diode plug is for. It keeps the resonant frequency at a constant frequency by only allowing the load to see the capacitors instead of loading the LC-tank of the Ferro resonant transformer. This probably requires that each capacitor is connected for a full charge transfer like reaching the peak voltage before switched to the load.

The triggering circuit fires at a preset voltage level no matter if this is actually the peak voltage. Perhaps this is wrong. Discharge should happen after or earliest at the peak voltage.

Regards
Ole
 

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

 Hi Ole, I assume that I have a clean sine wave at the reactive power generated in the transverter in the transmission over the coupling capacitors changes nothing except that everything is shifted by 180 degrees. If I now turn on the diode connector, this sine wave will change, I hope, at least by switching the thyristors. If you now set the switch so that is within the voltage time, the half LCs should be united again. It all works perfectly well thanks to your help, but that you can just connect any load as in a conventional way is a misconception of me or I have probably misinterpreted in the descriptions.

regards

Sven

Hi Kone,

Modern digital oscilloscopes can do those calculations on the fly. If having the multiplication function the curve of power can be shown. If also able to integrate the power curve a curve representing the energy transfer can be shown.

I've heard of the paper method where the areas are cut out and weighed instead of counting the area when knowing the mass per area of the paper.

Regards
Ole
 

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

 Hi Ole an dSven
 A group of 4 electrical engineers at a science meeting I was at many years ago (a Bill Beatty weird science meeting) was approached and asked by me how to do accurate real-power measurements for a DC pulsed "konehead" motor that I brought to the meeting.....they thought about it and talked about it for awhile, and then said this to me, and I would assume it would work in AC too:
 On graph paper, make an exact drawing of how the voltage looks on scope....
 On the same graph paper, but translucent type, make another exact drawing of how the current looks on scope....
 Lay the translucent graph paper (current) on top of the other piece of graph paper...line up the small graph squares so it sits on top perfectly.
 With pencil or pen, shade-in the areas where current ovelaps the voltage, then after that, calculate how much power the shaded-in area represents...
 ciaoKone

OleNice description of the phenomena.
How do we cause the discharge of the capacitors at a preset voltage, into another circuit, where it can go to a load?Cheers Warren
Sent from Yahoo Mail on Android 
 
  On Tue, Feb 13, 2018 at 6:37 AM, [email protected] [EVGRAY]<[email protected]> wrote:       
Hi Sven,

The Edison way of distributing power with big impedance mismatch by having the voltage staying almost constant during different loads doesn't work too good for a resonating circuit as any change in impedance will affect the resonant frequency. Without any compensation the load must stay constant. I believe that is what the diode plug is for. It keeps the resonant frequency at a constant frequency by only allowing the load to see the capacitors instead of loading the LC-tank of the Ferro resonant transformer. This probably requires that each capacitor is connected for a full charge transfer like reaching the peak voltage before switched to the load.

The triggering circuit fires at a preset voltage level no matter if this is actually the peak voltage. Perhaps this is wrong. Discharge should happen after or earliest at the peak voltage.

Regards
Ole



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

HiOle, I assume that I have a clean sine wave at the reactive powergenerated in the transverter in the transmission over the couplingcapacitors changes nothing except that everything is shifted by 180degrees. If I now turn on the diode connector, this sine wave will change, I hope, at least by switching the thyristors. If you now set the switch so that is within the voltage time, the half LCs should be united again. It all works perfectly well thanks to your help, but that you can justconnect any load as in a conventional way is a misconception of me or Ihave probably misinterpreted in the descriptions.

regards

Sven  #yiv5670004108 #yiv5670004108 -- #yiv5670004108ygrp-mkp {border:1px solid #d8d8d8;font-family:Arial;margin:10px 0;padding:0 10px;}#yiv5670004108 #yiv5670004108ygrp-mkp hr {border:1px solid #d8d8d8;}#yiv5670004108 #yiv5670004108ygrp-mkp #yiv5670004108hd {color:#628c2a;font-size:85%;font-weight:700;line-height:122%;margin:10px 0;}#yiv5670004108 #yiv5670004108ygrp-mkp #yiv5670004108ads {margin-bottom:10px;}#yiv5670004108 #yiv5670004108ygrp-mkp .yiv5670004108ad {padding:0 0;}#yiv5670004108 #yiv5670004108ygrp-mkp .yiv5670004108ad p {margin:0;}#yiv5670004108 #yiv5670004108ygrp-mkp .yiv5670004108ad a {color:#0000ff;text-decoration:none;}#yiv5670004108 #yiv5670004108ygrp-sponsor #yiv5670004108ygrp-lc {font-family:Arial;}#yiv5670004108 #yiv5670004108ygrp-sponsor #yiv5670004108ygrp-lc #yiv5670004108hd {margin:10px 0px;font-weight:700;font-size:78%;line-height:122%;}#yiv5670004108 #yiv5670004108ygrp-sponsor #yiv5670004108ygrp-lc .yiv5670004108ad {margin-bottom:10px;padding:0 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Hi Ole,

So I always understood that the resonance is not influenced by optimally adjusted system and takes no additional active power.

I again measured the main LC with switching and without switching the thyristors. I think he should voltage with the switch s.besten look exactly as without switching.

I tried to stop everything like that, but there is a point where the system collapses, maybe you have an idea why that is, I think there is an overlap here, but I'm not sure.

regards

Sven

https://youtu.be/BWObu6ZEDuc https://youtu.be/BWObu6ZEDuc

Hi Ole, Douglas,

Hmm, not bad. A good oscilloscope can determine that I have only a simple one and I must admit that I have not dealt with the mathematical calculations yet, but you once wrote that my device is very simple. Is not just Tektronix.

Really good oscilloscopes are also very expensive, I'm already on the verge to buy a used Tektronix, unfortunately, the price and age of the devices have always deterred me.

I have the possibility to measure the input power at 50Hz via a multifunction computer, where I can look at VAR, ampere, power factor, voltage, VA, frequency etc.

Of course, the comparison would be optimal from input to output power, but I'm still far from that to think because more energy could come out than I put in.

Unless, of course, there is an effect that these HV tips charge a battery super fast with little power and give off a lot. But then the battery is actually overunity and not the circuit, just then we just did not know how to properly charge it. laughing.

regards

Sven

SvenI am not any expert on oscilloscopes, but my electrical engineer friend Steve, got me one from China. Although, I didn't know much, I trusted him in his choice.As it turned out, it is a nice instrument, one that I can grow with as my skills improve. The exact same scope is branded by quite a number of Chinese based companies. They use the same cases and circuit boards calling them by whatever name they feel like.They are good scopes, cost very little, do two chanels, are easy to program.
Check out alibaba, or Toptom.If you feel uncomfortable ordering from China, try the Austrailian on line source for the same thing.They often have weekly special discount prices.They are very light weight, so shipping does not cost a fortune Cheers Warren
Sent from Yahoo Mail on Android 
 
  On Tue, Feb 13, 2018 at 9:16 AM, [email protected] [EVGRAY]<[email protected]> wrote:       
Hi Ole, Douglas,

Hmm, not bad. A good oscilloscope can determine that I have only a simple one and I must admit that I have not dealt with the mathematical calculations yet, but you once wrote that my device is very simple. Is not just Tektronix.

Really good oscilloscopes are also very expensive, I'm already on the verge to buy a used Tektronix, unfortunately, the price and age of the devices have always deterred me.

I have the possibility to measure the input power at 50Hz via a multifunction computer, where I can look at VAR, ampere, power factor, voltage, VA, frequency etc.

Of course, the comparison would be optimal from input to output power, but I'm still far from that to think because more energy could come out than I put in.

Unless, of course, there is an effect that these HV tips charge a battery super fast with little power and give off a lot. But then the battery is actually overunity and not the circuit, just then we just did not know how to properly charge it. laughing.

regards

Sven
  #yiv6876850758 #yiv6876850758 -- #yiv6876850758ygrp-mkp {border:1px solid #d8d8d8;font-family:Arial;margin:10px 0;padding:0 10px;}#yiv6876850758 #yiv6876850758ygrp-mkp hr {border:1px solid #d8d8d8;}#yiv6876850758 #yiv6876850758ygrp-mkp #yiv6876850758hd {color:#628c2a;font-size:85%;font-weight:700;line-height:122%;margin:10px 0;}#yiv6876850758 #yiv6876850758ygrp-mkp #yiv6876850758ads {margin-bottom:10px;}#yiv6876850758 #yiv6876850758ygrp-mkp .yiv6876850758ad {padding:0 0;}#yiv6876850758 #yiv6876850758ygrp-mkp .yiv6876850758ad p {margin:0;}#yiv6876850758 #yiv6876850758ygrp-mkp .yiv6876850758ad a {color:#0000ff;text-decoration:none;}#yiv6876850758 #yiv6876850758ygrp-sponsor #yiv6876850758ygrp-lc {font-family:Arial;}#yiv6876850758 #yiv6876850758ygrp-sponsor #yiv6876850758ygrp-lc #yiv6876850758hd {margin:10px 0px;font-weight:700;font-size:78%;line-height:122%;}#yiv6876850758 #yiv6876850758ygrp-sponsor #yiv6876850758ygrp-lc .yiv6876850758ad {margin-bottom:10px;padding:0 0;}#yiv6876850758 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Hi Warren,

That's exactly what Sven's circuit does. It's adjustable to a wide range of voltages. When the preset voltage is reached the SCR(s) trig.

Regards
Ole
 

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

 Ole Nice description of the phenomena.
How do we cause the discharge of the capacitors at a preset voltage, into another circuit, where it can go to a load?
 Cheers Warren
 Sent from Yahoo Mail on Android https://overview.mail.yahoo.com/mobile/?.src=Android
 
 On Tue, Feb 13, 2018 at 6:37 AM, onielsen@... [EVGRAY]
 <[email protected]> wrote:

   Hi Sven,

The Edison way of distributing power with big impedance mismatch by having the voltage staying almost constant during different loads doesn't work too good for a resonating circuit as any change in impedance will affect the resonant frequency. Without any compensation the load must stay constant. I believe that is what the diode plug is for. It keeps the resonant frequency at a constant frequency by only allowing the load to see the capacitors instead of loading the LC-tank of the Ferro resonant transformer. This probably requires that each capacitor is connected for a full charge transfer like reaching the peak voltage before switched to the load.

The triggering circuit fires at a preset voltage level no matter if this is actually the peak voltage. Perhaps this is wrong. Discharge should happen after or earliest at the peak voltage.

Regards
Ole

 

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

 Hi Ole, I assume that I have a clean sine wave at the reactive power generated in the transverter in the transmission over the coupling capacitors changes nothing except that everything is shifted by 180 degrees. If I now turn on the diode connector, this sine wave will change, I hope, at least by switching the thyristors. If you now set the switch so that is within the voltage time, the half LCs should be united again. It all works perfectly well thanks to your help, but that you can just connect any load as in a conventional way is a misconception of me or I have probably misinterpreted in the descriptions.

regards

Sven

Hi Sven,

"So I always understood that the resonance is not influenced by optimally adjusted system and takes no additional active power."
Correct. It's like a mechanical pendulum. It keeps on swinging when energy has been applied to it. If the energy isn't dissipated it doesn't stop swinging.

It looks like the SCRs trig after the peak voltage. This is because of the two RC filters between the big capacitor and the DIAC and gate of the SCRs. It also looks like sometimes the triggering doesn't happen but then is done at the next half cycle.

Without knowing what is being measured it's hard to tell exactly what happens. If measuring the gate-cathode signal and the voltage across the anode-cathode of the SCR this would make it be possible to see when the SCR fires. This would require differential probes or having the ground of the scope floating. If the top bus line is used as reference the left hand side SCR could be measured that way as well as the voltages across each capacitor to see if they both are discharged in each cycle. 


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Regards
Ole

 

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

 Hi Ole,

So I always understood that the resonance is not influenced by optimally adjusted system and takes no additional active power.

I again measured the main LC with switching and without switching the thyristors. I think he should voltage with the switch s.besten look exactly as without switching.

I tried to stop everything like that, but there is a point where the system collapses, maybe you have an idea why that is, I think there is an overlap here, but I'm not sure.

regards

Sven

https://youtu.be/BWObu6ZEDuc https://youtu.be/BWObu6ZEDuc

Hi Ole,

did you watch my video with the attempted setting, if I go beyond a certain point loses the signal. But you clearly see that not much is missing to have the sinus again as it was before.

Yesterday I once again measured the impedances with 150 volts, unfortunately my Variac shaft no longer because he is not designed for such a performance.

About the inverter, I have connected the Variac gives off a much more constant performance as our power grid.

The individual push-pull windings are at 90 ohms impedance and with full load maybe 1-2 ohms. Since I can only give minimal tension on the winding and the power consumption increases extremely. The coupling of this transformer is extremely good, I think.

Now I wanted to test the secondary winding as a push-pull, although it has a little more impedance but the difference between the two windings is only 1 milliampere current flow at the same voltage. I think these windings are most similar.

regards

Sven