George Egely's Magic Wand

    I have a question. I am putting together a high-voltage PSU, which will provide DC at 2-3 kV and a fe mA to the primary circuit of the Egely device. Like this:-



    I have two 7.5 KV 0.1mF capacitors. Are they (one or both, series or parallel) going to provide a reasonable amount of smoothing, or should I try something else.?

    your voltage and current specs imply an effective load of about 1megohm. With C= 0.1 microfarad, time constant is RC = 0.1 seconds, so it should provide fair smoothing at 50 or 60 Hz. Using both capacitors in parallel would help, or you could just save one for future use.

    Quote from Alan Smith

    I have two 7.5 KV 0.1mF capacitors. Are they (one or both, series or parallel) going to provide a reasonable amount of smoothing, or should I try something else.?

    It also depends on the output impedance of the HV AC generator and the level of "smoothness" that is required. But in general something around 0.1uF should be fine.

    I know George has been using some sort of series inductor for smoothing (I'm now much less worried about that, since discovering the feed to the "device" is DC and not pulsed).


    If the post capacitor ripple still looks a bit much, on the oscilloscope, maybe running a few turns of the output wire around a lump of ferrite (or soft iron) might help. Whether you'd need an extra series diode, like George also uses, I don't know.

    "The most misleading assumptions are the ones you don't even know you're making" - Douglas Adams

    Thanks 1
    Quote from Frogfall

    If the post capacitor ripple still looks a bit much, on the oscilloscope, maybe running a few turns of the output wire around a lump of ferrite (or soft iron) might help. Whether you'd need an extra series diode, like George also uses, I don't know.

    As a decades ago radio ham I'm familiar with inductors, but it's a good idea so thanks for the reminder. There are some classy iron-cored toroids in the 'maybe useful' box, also dig out a radio to look for RF emissions, which I suspect will be plentiful and broadband


    I have a mid-way review of another project this week, so that's currently keeping me busy re-writing data into civil service speak, also waiting for a bunch of 470k resistors to build a resistor chain to feed the input-side scope and multimeter with something less lethal than 3kV. All good fun of course.

    Quote from Alan Smith

    There are some classy iron-cored toroids in the 'maybe useful' box

    I guess the supply ripples will only be at 100Hz, so it would take a fair lump of iron to slug down at that frequency. But yes, the spark device is probably going to create a lot of RF noise.

    "The most misleading assumptions are the ones you don't even know you're making" - Douglas Adams

    Thanks 1 Like 1
    Quote from Frogfall

    I know George has been using some sort of series inductor for smoothing (I'm now much less worried about that, since discovering the feed to the "device" is DC and not pulsed).


    If the post capacitor ripple still looks a bit much, on the oscilloscope, maybe running a few turns of the output wire around a lump of ferrite (or soft iron) might help. Whether you'd need an extra series diode, like George also uses, I don't know.

    There is from me no concern about the 50Hz ripple on the supply. Nor for any HF ripple.


    The issue is that any measuring device connected to this system anywhere is going to experience HF interference from the spark - as seen on the scope trace. This will be enough to induce an ac voltage on any unscreened lead or circuit.


    Now, as I understand it, input power can safely be measured via two DC quantities (the voltage is stable enough for Vdc * Idc ~ P to work). However DC measurements from a DMM should not be assumed accurate when its leads have large amounts of HF interference. They may be accurate - or there may be a DC offset introduces in the internal amplifier circuits if these pick up the HF.

    Which is why I view DC values measured from scope - even though they will be less accurate then the DMM, as more certain in this case. You can do both, and they should be the same. But if not you cannot assume the DMM is more accurate than the scope even though its spec says it should be so.


    if there is too much HF stuff on the scope you can connect the probe via a 1cm wire threaded through a ferrite bead which with the scope input capacitance X1 will knock out the HF stuff.


    Technically, a large HF amplitude ~ 5kV on the PSU output could pull more power from the PSU than the DC measurements suggest. However I don't see this as a worry because such a large HF amplitude is not feasible from RFI.



    If you look on the LHS of the Egely lab scope screen here, before the oscillator kicks in you can see the ripple is not huge.


    @THH - I will do both scope and DMM measurement of the input continuously as Egely does -see above.

    I'd like to highlight just how important Frogfall 's thoughts are about heat dissipation on two resistors, one before, one after the reactor cell being irrelevant information about reactor cell efficiency/COP. I wouldn't go as far as stating "it is impossible to use these two values to infer any kind of cell efficiency", because there is one (and one only) scenario where it is possible in this exact circuit, even Dr. Egely was very careful to make a note on this during ICCF24 presentation:


    So Dr. Egely's calorimetry and COP statements are valid only if the capacitor have been fully charged and discharged in my opinion.


    Unfortunately I did not see data which suggests this criteria was honored. In the presentation capacitor was shown to be only partially charged and discharged:


    Expanding further on this topic, I don't believe it's even possible to apply t≥5τ criteria with the electrical setup presented, because:

    >>1. there is no timing mechanism which guarantees t≥5τ neither at charge, nor discharge, only charging current and ark discharge breakdown voltage seems to provide a timing;

    >>2. charging current (rate of voltage rise) and capacitor discharge threshold level cannot be set to match t≥5τ criteria, arguments:

    >>>>2.a. at t≥5τ charge current should be near zero, voltage on capacitor should be near exact the input voltage;
    >>>>2.b. discharge trigger level might not be constant, because it's governed by an arc discharge device with gas pressure and composition that can drift and with electrode characteristics that can drift;

    >>>>2.c. in order to make a suitable timing, input power supply voltage level should exactly match the breakdown voltage of gas discharge device at every discharge event (which is practically not likely), then one might need a pure luck at every event to force the discharge to happen, and not to wait perhaps minutes, hours, or even to infinity.

    >>3. capacitor discharge by voltage breakdown in gas (ark) usually stops before reaching 0V, in other words to keep the ionized gas in ionized state, so it can carry current down to near 0V on the capacitor, there is need to keep the excitation up of the gas, which usually is not there anymore at low voltages. But I have hopes in this regard, maybe the nuclear events provide this ionization energy.


    As arguments on why the presented calorimetry does not work if the capacitor is not fully charged and fully discharged, in other words t≥5τ criteria is not met:

    - consider the charge resistor being similar to a current monitoring shunt resistor, and the output resistor to be a load resistor. Similarities can be drawn with any other power converter using a shunt to monitor a current and a final load resistor to dump all useful power: small power is dissipated on the shunt resistor by design (why waste energy at some monitoring component) and most of the power (for ex. >95% in a converter having good efficiency) gets dissipated on the load resistor.

    Ex.: a converter of 100W drawn at input can have 1W dissipated on a shunt resistor and 95W dissipated at output load resistor (other 4W are dissipated on other internal components such as MOSFETs, tranformer/inductors, capacitors). Does this mean the converter COP is 95W/1W = 95 by looking at heat on two resistors? Over-unity of 94? Wow! Absolutely not! The converter is COP = 0.95, 95% efficient. :thumbup:

    - I've made a quick spice simulation to get some crude estimate on powers for 2 resistors in a circuit similar to Dr. Egely's. I intended to replicate parcial charge and discharge of capacitor based on the limited info I had from ICCF video.

    Charge a capacitor to almost 3kV, discharge to almost 1.5kV, with repetition rate of aprox. 2ms.

    The beauty of a spice simulation that it accurately simulates and stores many datapoints (voltages, currents) of which we can easily apply some simple numerical maths: multiply voltage (V) and current (A) on desired components obtain power (W), average instantaneous power or integrate it over time to obtain energy (J). For example on first resitor:


    LTspice simulation file attached ("CapacitorDischarge.zip/CapacitorDischarge.asc").

    Here are the results:


    The "shunt" resistor before the capacitor dissipates p_r1 ~= 0.52W

    The "load" resistor after the capacitor dissipates p_r2 ~= 1.37W

    Apparent efficiency by comparing the two powers/energies is COP ~= 2.6, which is of course falsely assumed in this case, because t≥5τ criteria is not met

    Real efficiency is by comparing input power to sum of power at both resistors: COP ~= 1 --> this is that we should strive to achieve: measure input power in DC domain reliably and integrate to obtain input energy, then place the whole unit (all converters, resistors, ark discharge devices, everything) in a heat calorimeter and measure output energy. What do you think?

    Thank you - a lot to think about there. The problems of measuring input are much discussed at the Egely lab. Attached is a picture I took of the blackboard in the lab during a discussion about this, which was (of course) in Hungarian.


    Its good that some people are running simulations ( Tibi.fusion  SM6FIE ) - but we do need to ensure that we don't introduce even more confusion.


    If I had the skills to use the graphical software, I would have a go too (I'm not an electronics engineer)


    As I mentioned further up this thread, I was more concerned about the twin resistor calorimetry technique when I thought the device was being driven by a primary "pumping" oscillator (**). But as R1, C1 & the spark gap are acting as their own oscillator, I am less concerned. After all, even if C1 only partially discharges (and videos show that the discharge level does indeed vary when the input voltage is only marginally above the spark initiation voltage) - that means that an equally partial energy pulse drives the spark.



    I'm afraid I don't know how this circuit detail is meant to model the spark gap Tibi.fusion - maybe you can explain the logic in more detail, thanks.


    (n.b. (**) It appears that some of the Moray machines used some kind of pumping oscillator to drive or trigger the "tubes" - quite possibly using crude forerunners of the Tunnel / Gunn diode (initially his Swedish Stone, and then later a semiconductor "detector" of his own devising). He once claimed that his machine would still run without the peculiar detector, but its oscillations would be very uneven.)

    "The most misleading assumptions are the ones you don't even know you're making" - Douglas Adams

    Edited 2 times, last by Frogfall ().

    Thanks 1

    Egely also mentions the importance of an inductor and a diode in series on the input-side because of "explosions" (= energy generation) in the cell.

    Maybe not that important for the simulation, but to protect the PSU.

    By the way a nice Spice.


    If we think about the COP-calculation and assume that the PSU is perfoming well (efficiency e.g. ~ 90%), then the input average energy

    to the cell could be calculated by Ec_in = Ein - ER1. Here:

    Ec_in = the actual energy send to the cell

    Ein = the measured energy to the PSU

    ER1 = the measured resistor energy with a calorimeter


    Here it doesn't matter how the capacitor C1 voltage actually varies. The only interesting thing that matters is how much energy is transferred from

    the capacitor to the cell, but this way it can be measured without a direct measurement.


    The output energy ERL is also measured with a calorimeter.

    The final COP would then be simply ERL / Ec_in
    This requires of course the input power to the PSU measurement, but that is standard measrument.

    At least a this way it should be possible to get a rough estimate of the COP.

    If the real COP would be e.g > 2, then even this method should give some interesting results.

    To look (as ever) on the bright side, even if the COP of this device is only 2, the low cost and simplicity suggests that it would be economically feasible to daisy-chain them, with the first single system feeding say 3 others, and those in turn feeding power to another 9. A bright smiley thought for Sunday :) .

    Quote from Alan Smith

    This is a very nice series of screen captures from George Egely's ICCF-24 presentaion. Plenty of detail for those with sharp eyes.


    Egely Screen Captures from ICCF-24.pdf

    My prediction: COP = 1. But then I am an LENR skeptic.


    From the detailed discussion - COP measurement relies on:

    1. Calorimetry for input resistor power
    2. Assumption that average capacitor discharge power = average input resistor power
    3. Assumption that power transferred to cell is ONLY that from discharging the capacitor
    • I agree with 1.
    • 2 is in wrong - but it is not clear what that error does to the COP calculation.
    • 3 is wrong, and underestimates input power by a (pencil & paper) possibly large amount.


    Both 2 & 3 errors come from the same assumption that the plasma tube operates as a perfect switch - open circuit until some trigger voltage after which it is close circuit. In that case no power is transferred directly through the input resistor to the circuit (either V or I = 0 at all times). However we do not know this assumption is correct.


    Anyway, detailed measurements should be quite easy. To be clear how much of the input power is transferred to the plasma (which the COP=2 etc relies upon) it will be necessary to record EHT (Vs) and capacitor voltage (Vc) over discharge cycles and integrate:


    1a. integral (Vs*(Vs-Vc)/R1) dt / T = input power from EHT supply

    1b. alternately - add a shunt resistor Rs and integrate Vs*Vsh/Rs - this is easier if Vs is stable because an average of Vsh will be enough and is easy to find.

    2a . integral (Vs-Vc)^2/R1 dt / T = power going into R1

    2b. Work out power into R1 from calorimetry (as has been done)


    Those two measurements (1 and 2) avoid any assumptions. They are both low frequency measurements (1kHz waveform) - correct procedures to filter out higher frequencies should make it possible to do them accurately. It may require scope data download and external numeric integration from samples.


    Note that just measuring the EHT input power is not enough, since it is possible (if the plasma tube acts as a perfect switch as assumed in the calculations others have made) that only 1/2 of the input power goes into the plasma tube. The real amount could be more or less than this.

    Quote from Alan Smith

    Thank you - a lot to think about there. The problems of measuring input are much discussed at the Egely lab. Attached is a picture I took of the blackboard in the lab during a discussion about this, which was (of course) in Hungarian.


    I can definitely see two important things considered on the blackboard:
    1. RC time constant:

    τ = R * C,

    where:

    τ (pronounced "tau") is the time constant is seconds

    R is resistance in Ohms

    C is capacitance in Farads


    reference [1]: https://electricalacademia.com…onstant-rc-time-constant/


    2. Energy stored in capacitor

    E = 1/2 * C * U^2,

    where:

    E is the energy stored in the capacitor in Joules (or Watt-seconds)

    C is capacitance in Farads

    U^2 is voltage of capacitor square


    reference [2]: https://pressbooks.online.ucf.…gy-stored-in-a-capacitor/


    Allow me to highlight why these are important.

    Consider R = 1MegOhm, C = 1nF. Their time constant is τ = R * C = 1 ms. According to [1] if you apply a voltage on this RC group, then in 1τ = 1ms the capacitor will charge up to ~63% of input voltage.

    Simulation confirms this: voltage across capacitor is ~1.9kV, which is ~63% of 3kV input.

    Energy stored in 1nF capacitor at 1.9kV is according to [2]: E = 1/2 * C * U^2 = 0.5*1E-9*(1.9E+3)^2 = ~1.8mJ

    Let's check the simulation: I'm multiplying voltage across capacitor with current into capacitor, then integrate over all time duration:

    This gives the same value 1.798mJ = ~1.8mJ


    Now comes the interesting part: let's check the energy on the resistor with the same approach: voltage across resistor times the current, integrated over time:

    Well at time duration of t = 1τ = 1ms the energy on resistor is not equal to energy on capacitor.

    Let's try at t = 5τ = 5ms:

    Energies are almost equal (not exactly equal because 5τ gives ~99.3% voltage, for 100% you'd need to wait even more, theoretically to infinity; plus numerical integration errors):


    I'm sure the discussion in the lab also touched this point: I believe were trying to solve energy equations considering partial charge and discharge of the capacitor.


    I want to be very clear: this does not mean the reactor cell / ark discharge device does not produce excess! This only means the proposed calorimetry method is only valid if respecting t>=5τ criteria. We have no evidence this was not respected during COP measurements, we only have some demonstration waveforms which show some repetitive operation there criteria is not applied.

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