Capacitive coupling of energy.

Hi Guys,

in the transverter or rotoverter alternator the energy is transferred capacitively via the capacitors. With the alternator, I can understand it because you want to maintain the reactive power but with the diode plug, why is it done there?

Or do the 2 capacitors act like a filter that only transmits one frequency and compensates and converts the others.

Unfortunately, I'm not a radio engineer, maybe someone could explain that.

Many Thanks

Best regards

Sven

Hi Sven
 I think you should or need to  use diode plug (or two-stage cap discharge isolation circuit to the load) so that the system does not "see" a resistive load, and system only fills up capacitors-first, and in this way capacitors (according to UF value chosen) will work as "resonate chambers" to fill, rather than a resistive load - this UF value (along with discharge rate and particular resistive loading, and pulse wid h of discharge too) are all factors if the output to load is going to reflect upon input and cause much more input draw upon loading of system (which is what you do not want)
 Anyways this my feeble opinion...
 ciao
 Kone

Hi Sven, Kone,

Capacitors only block DC. AC goes right through them when the frequency and/or capacitance is high enough. Pulsed DC also passes through capacitors. This is used in the Ćuk converter: https://wiki2.org/en/%C4%86uk_converter https://wiki2.org/en/%C4%86uk_converter. Capacitors can be used for keeping apart two different steady DC voltage levels in a signal chain while passing only the changing part of the signal. This is for example the AC filter in oscilloscope inputs.

If the energy goes primarily in one direction the power is primarily active. The active power reflects back to the generator as resistive loading. Just take a look at switched mode converters. They're taking active power from the source even though at some part it is switched on and off and the energy is momentarily stored in a capacitor or inductor or both. When the energy storing component (reactive component which are capacitors and inductors) discharge its energy to a resistive load the energy is dissipated as heat. Thus using capacitors or inductors doesn't mean that the power becomes reactive if the load is an active load i.e. the load dissipates the power instead of returning it to the source.

Regards
Ole
 

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

 Hi Sven
 I think you should or need to  use diode plug (or two-stage cap discharge isolation circuit to the load) so that the system does not "see" a resistive load, and system only fills up capacitors-first, and in this way capacitors (according to UF value chosen) will work as "resonate chambers" to fill, rather than a resistive load - this UF value (along with discharge rate and particular resistive loading, and pulse wid h of discharge too) are all factors if the output to load is going to reflect upon input and cause much more input draw upon loading of system (which is what you do not want)
 Anyways this my feeble opinion...
 ciao
 Kone

Hello Ole, Douglas,

I think soon that the repercussions of diode plug discharges are filtered so that it does not reflect or the energy remains completely reactive.

In my whole experiments, there was always a minimum reflection but then remained constant no matter what I have connected as a load on the diode plug. It is discharged when unloading a strong DC pulse that certainly can not be perfectly inserted back into the resonance wave.

I actually wanted to charge the batteries that I got via the transverter directly with VAR via an FWBR. The batteries are too bad for me to try this way. Hector has warned us that the batteries may not be good if you charge them reactive and at the same time get active power from them for the inverter.

That's why I prefer to further perfect or learn the diode plug.

I hope you can see the drawing of Hector I uploaded it via email unfortunately I can not upload anything at evgray apparently yahoo is overloaded?!?!

What I find particularly interesting are the reactors or inductors in each diode plug. With this I would have to delay the recharging of the capacitor so that I can completely discharge the capacitor so that the thyristor switches off safely. The inductor stores the energy and then releases it when the capacitor is fully charged. The capacitor acts like a snubber which then stores the energy.

Thoughts and knowledge are very welcome.

Best regards

Sven

Empty body

Hi Sven,

For the series connected capacitors a single capacitor can be used and it does the same job when having the value of the series connection. Cserial = 1/(1/C1 +1/C2).

The free energy is captured or liberated in the transformer and that free energy is real energy (active power). Reactive power only heats any resistance it passes through converting it into active power. Thus reactive power isn't useful in itself. Because a powerful magnetic field is wanted in the core of the transformer this is why reactive power can be used. But the field mustn't be removed as active power before capturing the free energy. Just as much energy from the magnetic field returns back to the capacitor forming the LC-tank. If using of that energy the power being used is active (real) power. The active part captured is directed to the load through the diode plug circuit.

In my experiments with Hector's bucking secondaries transformer the output is just AC. The load is critical as well as many other parameters.

For size of capacitance and power dissipation here is a document for calculating the ripple current: 'AC Ripple Current Calculations' https://www.vishay.com/docs/40031/apprippl.pdf https://www.vishay.com/docs/40031/apprippl.pdf. The impedance of a capacitor can be calculated and then the current at a given voltage can be calculated. The (parasitic) resistive part of the capacitor (because made of metal) dissipates energy as heat which has to be removed to prevent melt down. This is the power rating. The dielectric when not being a vacuum also dissipates heat when the frequency is high enough. Those values form the equivalent series resistance (ESR).

To prevent a high voltage drop the impedance of the capacitor must be much less than that of the load. Of course the voltage source also has to be of adequate low impedance. Capacitors limit the current through them without (ideally) dissipating power. For high power let through low impedance is required depending on the voltage. How much power goes through them then depends on the voltage and current. For sinusoidal waveforms complex math can be used together with Ohm's law. For other waveforms (as well as sinusoidal waves) Laplace transformation is used with Ohm's law.

Regards
Ole


 

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

 Hi Ole, Douglas,
 
 I know only capacitors for smoothing or charging, that they do not let through DC is new to me. Turning on one of the capacitors in series shines one more time, but I connect both poles with capacitors.
 
 I upload the circuit from Hector again.
 
 I would also be interested in how big they have to be to transmit a certain power at a certain frequency.
  
  
 
 regards
 
 Sven
   Gesendet: Samstag, 09. Dezember 2017 um 14:24 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: [EVGRAY] Re: Capacitive coupling of energy.
   Hi Sven, Kone,
 
 Capacitors only block DC. AC goes right through them when the frequency and/or capacitance is high enough. Pulsed DC also passes through capacitors. This is used in the Ćuk converter: https://wiki2.org/en/%C4%86uk_converter https://wiki2.org/en/%C4%86uk_converter. Capacitors can be used for keeping apart two different steady DC voltage levels in a signal chain while passing only the changing part of the signal. This is for example the AC filter in oscilloscope inputs.
 
 If the energy goes primarily in one direction the power is primarily active. The active power reflects back to the generator as resistive loading. Just take a look at switched mode converters. They're taking active power from the source even though at some part it is switched on and off and the energy is momentarily stored in a capacitor or inductor or both. When the energy storing component (reactive component which are capacitors and inductors) discharge its energy to a resistive load the energy is dissipated as heat. Thus using capacitors or inductors doesn't mean that the power becomes reactive if the load is an active load i.e. the load dissipates the power instead of returning it to the source.
 
 Regards
 Ole
 
 
 ---In [email protected], <konehead@...> wrote :
   Hi Sven
 I think you should or need to  use diode plug (or two-stage cap discharge isolation circuit to the load) so that the system does not "see" a resistive load, and system only fills up capacitors-first, and in this way capacitors (according to UF value chosen) will work as "resonate chambers" to fill, rather than a resistive load - this UF value (along with discharge rate and particular resistive loading, and pulse wid h of discharge too) are all factors if the output to load is going to reflect upon input and cause much more input draw upon loading of system (which is what you do not want)
 Anyways this my feeble opinion...
 ciao
 Kone

Hello Ole and interested,

I have now read that you can build a so-called capacitor bridge with which you can electrically isolate the circuits.
It has the same structure as a FWBR, but capacitors are used instead of diodes. With this circuit is achieved that the same voltage is applied to the other circuit. Maybe that would be the better choice.

Best regards

Sven

Hi Sven,

In a series circuit the same current passes through all the components in series. Thus removing one of the capacitors in a series circuit involves just putting a short instead of it and changing the other one to the value the two original capacitors had when in series. The voltage rating of the new capacitor must then be big enough for the full voltage across the two original capacitors and the power rating also must be high enough for any dissipated power. An ideal capacitor doesn't dissipate power though. Generally in a series circuit it doesn't matter in which order the components are placed. But it may matter where to have high and low voltages across the components when seen from a safety perspective. I prefer to have the high voltages close to the live line and those of lower voltage closer to the ground.

The C1 and C2 capacitors in the schematic reads 45 mF. This is probably 45 uF as 45 mF are quite big capacitors.

"Maybe the free energy is in the VARs."
The definition of reactive power is that the energy going into a component is given back again instead of being dissipated. This is clearly seen when plotting the curves of voltage and current 90 degrees apart and then multiplying those curves to get the curve of power. The curve representing the power has equal area above and below the zero line. This indicates equal amounts of energy entering and leaving the component during a full cycle. Thus no free energy here per see. The curve of power should be asymmetric in the direction of energy going out of the component to indicate an energy source.

Your picture needs the neutral wire unless using the ambient as the return path (radiant energy) which requires high frequency or high voltage to transfer the power through the very small capacitance formed between the two circuits.


Regards
Ole
 

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

 Hi Ole,
 
 and how would the single capacitor then be installed only on one side or in parallel to separate the circuits as two adjacent wires are exchanging electrical fields such as noise. Maybe the free energy is in the VARs. I see the circuits being posted only converting the power into low voltages and high currents as do the rectifying for battery operation. As with diode plug, capturing some of the pendulum energy VARs to make them real and useful in capacitors.
 
 I found an impedance calculator on the internet for capacitive reactances in my 50 Hz technique I would have very large capacitors to get a low resistance. at 600 μF I still lie at 50Hz at 5.3 ohms.
 
 I wonder if you can connect the circuits as well as in the picture I uploaded. It would simplify the construction because one would only have to increase the capacity at one point.
 
 For me that will be the point to try it out to achieve the best values.
 
 I send everything by mail at yahoo I can not upload any pictures.
 
 regards
 
 Sven  ...

Empty body

Hello Ole,

I think Hector was thinking of inserting 2 capacitors into the circuit, if one would have thought I had only one recorded.

The coupling via the two capacitors also ensures galvanic isolation of the two circuits. Perhaps this is important so that no repercussions are generated by the diode plug when discharging the capacitors.

I think the mF in the US is the same as the μF in Europe.

The capacitor bridge has different capacitors.

1 capacitor with high capacity and 1 capacitor with low capacity in series and that 2 times to generate a high potential difference I think. Whether it is still possible to retain the filter properties I doubt it. Since the impedances of the capacitors is indeed very different by their capacity.

I think these coupling capacitors have their justification.
Maybe Hector will contact you about this topic.

Best regards

Sven

Hi Sven,

Using capacitors as galvanic isolation may be a bad idea as they transfers deadly current if having big enough capacitance. There are safety capacitors for decoupling touchable parts to the neutral wire. This is the safety class Y capacitors (https://www.allaboutcircuits.com/technical-articles/safety-capacitor-class-x-and-class-y-capacitors/ https://www.allaboutcircuits.com/technical-articles/safety-capacitor-class-x-and-class-y-capacitors/)  that have double insulation.

Capacitors are good at transferring transients or high frequency while inductors are good at transferring DC or low frequency.

Regards
Ole 

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

 Hello Ole,

I think Hector was thinking of inserting 2 capacitors into the circuit, if one would have thought I had only one recorded.

The coupling via the two capacitors also ensures galvanic isolation of the two circuits. Perhaps this is important so that no repercussions are generated by the diode plug when discharging the capacitors.

I think the mF in the US is the same as the μF in Europe.

The capacitor bridge has different capacitors.

1 capacitor with high capacity and 1 capacitor with low capacity in series and that 2 times to generate a high potential difference I think. Whether it is still possible to retain the filter properties I doubt it. Since the impedances of the capacitors is indeed very different by their capacity.

I think these coupling capacitors have their justification.
Maybe Hector will contact you about this topic.

Best regards

Sven

Sven,

Your correct in USA we use mF or uF interchangeably.    Besides what Ole 
said one has to make sure of low ESR and some caps react differently 
depending on if a signal is symmetric or asymmetric, one can see this 
with a curve tracer.  I think the attribute is called absorption but I 
forget.

For sound we prefer oil or poly or teflon film over mylar film and 
ceramics or tants.   Also some manufacturers use steel for the leads of 
caps and resistors which has a magnetic property we don't like to hear. 
Foil resistors with copper leads sound best but are prohibitive with cost.

Teflon caps are really nice sounding but although and amazing insulator 
is a poor carrier so it takes some larger sizes in teflon than polyfilm 
materials but they can usually handle a higher voltage.  I don't know if 
any of this is relevant to your work but have a gut feeling sound has 
similar properties to some of the aether magic we chase.


On 12/9/2017 1:19 PM, [email protected] [EVGRAY] wrote:
>
> Hello Ole,
>
> I think Hector was thinking of inserting 2 capacitors into the 
> circuit, if one would have thought I had only one recorded.
>
> The coupling via the two capacitors also ensures galvanic isolation of 
> the two circuits. Perhaps this is important so that no repercussions 
> are generated by the diode plug when discharging the capacitors.
>
> I think the mF in the US is the same as the μF in Europe.
>
> The capacitor bridge has different capacitors.
>
> 1 capacitor with high capacity and 1 capacitor with low capacity in 
> series and that 2 times to generate a high potential difference I 
> think. Whether it is still possible to retain the filter properties I 
> doubt it. Since the impedances of the capacitors is indeed very 
> different by their capacity.
>
> I think these coupling capacitors have their justification.
> Maybe Hector will contact you about this topic.
>
> Best regards
>
> Sven
>
>

Sven If you just happened to have made a Rene charger, it seems to do a good job charging lead acid batteries when run off the capacitor circuit of an RV by way of a step down transformer, so that it runs the primary winding on 20 volts AC.I found no measurable draw on the little 5 amp battery running the 125 watt inverter which powered the RV whether the Rene was connected, or not.For that experiment, I was also driving a 12 volt electric outboard motor, being used as a test load.The load at the 5 amp battery measured 1.5 amps (@ 12 volts) and the load at the brushes of the outboard motor measured 2.5 amp ( @ 12 volts dc)The Rene output could not be measured with amps going in, yet one could see the voltage rapidly rising on a multimeter, and the batteries so charged able to run the RV.If you have a working RV, and some multitap transformers, a load motor, a few good batteries, rectifiers, smallish inverter, and lot, and lots, of switches, you can replicate the experiment, as it seems it works all the time, so far that I have fired it up.I have to start on grid AC, then switch over to the battery/inverter power to the running Rotoverter. Tune the Rotoverter roughly. Open the switch shunted across the capacitor circuit in series transformer primaries, now including it into the capacitor circuit. Retune your capacitor box. Close the secondary winding switch to the load which is an electric DC motor, that draws more than input, at least. Tune the capacitor box.Now close the Rene charger input circuit while watching the battery amp meter.Other than the rising battery voltage, there is nothing to report except the specific gravity of the ekectrolyte also improves, increasing the specific gravity reading, as it charges.Cheers Warren
Sent from Yahoo Mail on Android 
 
  On Sat, Dec 9, 2017 at 9:44 AM, [email protected] [EVGRAY]<[email protected]> wrote:       
Hello Ole, Douglas,

I think soon that the repercussions of diode plug discharges are filtered so that it does not reflect or the energy remains completely reactive.

In my whole experiments, there was always a minimum reflection but then remained constant no matter what I have connected as a load on the diode plug. It is discharged when unloading a strong DC pulse that certainly can not be perfectly inserted back into the resonance wave.

I actually wanted to charge the batteries that I got via the transverter directly with VAR via an FWBR. The batteries are too bad for me to try this way. Hector has warned us that the batteries may not be good if you charge them reactive and at the same time get active power from them for the inverter.

That's why I prefer to further perfect or learn the diode plug.

I hope you can see the drawing of Hector I uploaded it via email unfortunately I can not upload anything at evgray apparently yahoo is overloaded?!?!

What I find particularly interesting are the reactors or inductors in each diode plug. With this I would have to delay the recharging of the capacitor so that I can completely discharge the capacitor so that the thyristor switches off safely. The inductor stores the energy and then releases it when the capacitor is fully charged. The capacitor acts like a snubber which then stores the energy.

Thoughts and knowledge are very welcome.

Best regards

Sven
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Sven,

Often we try not to use capacitors in direct coupled circuits but they 
have many uses.  In most cases they are used to isolate the DC power 
supply/ bias from the AC musical output.  In cases where sound needs to 
flow through capacitors such as separating two stages that have 
different power requirements from interaction with one another or 
preventing the power supply for a microphone leaking into the recording 
chain such as the blocking capacitor for phantom power, in these cases 
we look for devices with pleasing sound quality or rather most 
transparency.   In other cases where they are used purposefully in the 
signal path is for equalizers and other analog filters like crossovers.  
Like when one puts a small capacitor with a tweeter in order to protect 
it with a 6db single pole high pass filter.

I think you are correct concerning the oil filled capacitor.  As the 
devils advocate I would add that all all oil caps are not the same, 
plate material is a big consideration such as aluminum being a current 
mirror vs a material with low to no magnetic properties.  In one energy 
design, I forget which, lead plates were used for the capacitor.

Tesla said aetheric energy did not conduct with copper very well and he 
preferred graphite for this energy.


On 12/10/2017 6:37 AM, [email protected] [EVGRAY] wrote:
>
> Hello Mick,
>
> do you know yourself well with this matter, why do you do that with 
> the capacitors? As I've read it, it ensures that no direct current is 
> transmitted, for me it sounds like a filter to switch the disturbances 
> of diode plug scr not to stop reactions to the generator. In the audio 
> domain to avoid acoustic interference, I think?!?
>
> I think in energy technology, the oil-filled capacitors from 
> compensation systems are optimal.
>
> regards
>
> Sven
>
>

Hello Mick,

do you know yourself well with this matter, why do you do that with the capacitors? As I've read it, it ensures that no direct current is transmitted, for me it sounds like a filter to switch the disturbances of diode plug scr not to stop reactions to the generator. In the audio domain to avoid acoustic interference, I think?!?

I think in energy technology, the oil-filled capacitors from compensation systems are optimal.

regards

Sven

Make sure you build the very best capacitor tuning box that German engineering can make, as it will be used many years from now. Use only oil filled capacitors of at least 600 volts , if you can, as you never know what voltage you might use. Sturdy switches are good.Equip it with excellent connectors.Now you are ready to play with RV, as the tuning capacitance is central to the whole magic trick.By making this instrument easy to use, and absolutely reliable, it will make you happy with the results.Jumper wires all over the work bench just suck, and shock too.Cheers Warren

Sent from Yahoo Mail on Android 
 
  On Sat, Dec 9, 2017 at 10:53 AM, [email protected] [EVGRAY]<[email protected]> wrote:       
Hi Sven,

For the series connected capacitors a single capacitor can be used and it does the same job when having the value of the series connection. Cserial = 1/(1/C1 +1/C2).

The free energy is captured or liberated in the transformer and that free energy is real energy (active power). Reactive power only heats any resistance it passes through converting it into active power. Thus reactive power isn't useful in itself. Because a powerful magnetic field is wanted in the core of the transformer this is why reactive power can be used. But the field mustn't be removed as active power before capturing the free energy. Just as much energy from the magnetic field returns back to the capacitor forming the LC-tank. If using of that energy the power being used is active (real) power. The active part captured is directed to the load through the diode plug circuit.

In my experiments with Hector's bucking secondaries transformer the output is just AC. The load is critical as well as many other parameters.

For size of capacitance and power dissipation here is a document for calculating the ripple current: 'AC Ripple Current Calculations' https://www.vishay.com/docs/40031/apprippl.pdf. The impedance of a capacitor can be calculated and then the current at a given voltage can be calculated. The (parasitic) resistive part of the capacitor (because made of metal) dissipates energy as heat which has to be removed to prevent melt down. This is the power rating. The dielectric when not being a vacuum also dissipates heat when the frequency is high enough. Those values form the equivalent series resistance (ESR).

To prevent a high voltage drop the impedance of the capacitor must be much less than that of the load. Of course the voltage source also has to be of adequate low impedance. Capacitors limit the current through them without (ideally) dissipating power. For high power let through low impedance is required depending on the voltage. How much power goes through them then depends on the voltage and current. For sinusoidal waveforms complex math can be used together with Ohm's law. For other waveforms (as well as sinusoidal waves) Laplace transformation is used with Ohm's law.

Regards
Ole





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

Hi Ole, Douglas,

I know only capacitors for smoothing or charging, that they do not let through DC is new to me. Turning on one of the capacitors in series shines one more time, but I connect both poles with capacitors.

I upload the circuit from Hector again.

I would also be interested in how big they have to be to transmit a certain power at a certain frequency.  
regards

Sven Gesendet: Samstag, 09. Dezember 2017 um 14:24 Uhr
Von: "onielsen@... [EVGRAY]" <[email protected]>
An: [email protected]
Betreff: [EVGRAY] Re: Capacitive coupling of energy. 
Hi Sven, Kone,

Capacitors only block DC. AC goes right through them when the frequency and/or capacitance is high enough. Pulsed DC also passes through capacitors. This is used in the Ćuk converter: https://wiki2.org/en/%C4%86uk_converter. Capacitors can be used for keeping apart two different steady DC voltage levels in a signal chain while passing only the changing part of the signal. This is for example the AC filter in oscilloscope inputs.

If the energy goes primarily in one direction the power is primarily active. The active power reflects back to the generator as resistive loading. Just take a look at switched mode converters. They're taking active power from the source even though at some part it is switched on and off and the energy is momentarily stored in a capacitor or inductor or both. When the energy storing component (reactive component which are capacitors and inductors) discharge its energy to a resistive load the energy is dissipated as heat. Thus using capacitors or inductors doesn't mean that the power becomes reactive if the load is an active load i.e. the load dissipates the power instead of returning it to the source.

Regards
Ole


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

I think you should or need to  use diode plug (or two-stage cap discharge isolation circuit to the load) so that the system does not "see" a resistive load, and system only fills up capacitors-first, and in this way capacitors (according to UF value chosen) will work as "resonate chambers" to fill, rather than a resistive load - this UF value (along with discharge rate and particular resistive loading, and pulse wid h of discharge too) are all factors if the output to load is going to reflect upon input and cause much more input draw upon loading of system (which is what you do not want)

Anyways this my feeble opinion...

ciao

Kone

 
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Hi Guys,

I have now learned a lot about coupling capacitors, they mainly ensure that no DC components are transmitted. Unfortunately, the designed for how high the load resistance after the capacitor is the ohmic load. The smaller the resistance, the higher the capacitor must be in order to pass low frequencies at 50Hz already very low frequencies. Since the thyristors switch in the diode plug on very low resistance, the capacitors would be huge to be able to transfer a lot of energy, at least for AC applications. I think that all coupling effects of the diode connector on the energy source are completely blocked by the coupling capacitors. I will test it if I have found suitable capacitors.

Best regards

Sven

Sven,

You are correct about this point and one reason why high impedance loads 
are easier to deal with as the tuning or filter capacitors can be 
smaller as you said.  Another reason Hectors designs use the highest 
voltage windings for a low voltage use so we don't burn up the energy, 
prior to storing/extracting in the capacitive load.

Sometimes with a current source on a MOSFET source follower for instance 
the drive impedance becomes very high.  Below is an article about MU 
followers which may provide some food for thought.

Also someone on here mentioned about regenerative and super regenerative 
radio circuits.  These are very interesting and important ways to use 
positive feedback as a multiplier as with such designs at the very high 
gain settings the super-regen frequencies show up and begin to quench 
the oscillation.  It is a fine line between quenching and adding some 
useful harmonics to pull more energy out of the matrix.

Keep up the solid work!

http://www.fetaudio.com/wp-content/uploads/2003/09/Mu-Stage.pdf


On 12/14/2017 5:52 AM, [email protected] [EVGRAY] wrote:
>
> Hi Guys,
>
> I have now learned a lot about coupling capacitors, they mainly ensure 
> that no DC components are transmitted. Unfortunately, the designed for 
> how high the load resistance after the capacitor is the ohmic load. 
> The smaller the resistance, the higher the capacitor must be in order 
> to pass low frequencies at 50Hz already very low frequencies. Since 
> the thyristors switch in the diode plug on very low resistance, the 
> capacitors would be huge to be able to transfer a lot of energy, at 
> least for AC applications. I think that all coupling effects of the 
> diode connector on the energy source are completely blocked by the 
> coupling capacitors. I will test it if I have found suitable capacitors.
>
> Best regards
>
> Sven
>
>

Hi Mick,

I try to stay tuned, unfortunately I do not have the time to give myself up completely.

I am trying to build this modularly and make it separately adjustable.

I have now cheap Electronicon oil condensers for 1000 VAC can buy. So I want to build a new platform for the sheath current filter which I can set optimally.

regards

Sven

Hi Mick,

I try to stay tuned, unfortunately I do not have the time to give myself up completely.

I am trying to build this modularly and make it separately adjustable.

I have now cheap Electronicon oil condensers for 1000 VAC can buy. So I want to build a new platform for the sheath current filter which I can set optimally.

regards

Sven