Holcomb Energy Systems:Breakthrough technology to the world

[rakarskiy]

An interesting picture emerges. Analyzed the result of the experiment from the video: 

https://www.youtube.com/shorts/0LmQzSx06uE

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Compare with Ohm's law. It is very interesting how the wattmeter counts. In the secondary circuit, the voltage drop is exactly three times (117/39=3). At the generator terminals, the phase under load is 39 volts, the idle EMF is 117 volts. The wattmeter cannot measure EMF, only calculate. Therefore, its value is based on the EMF.

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Now think about it, it's enough to add 2 volts to the network a couple of times a day, and your account will be increased, you won't even notice.
And in the experiment, the power delta value is always greater than 1.

In red on the highlighted yellow field, I believe it is correct, in terms of Ohm's law.

[rakarskiy]

Expanded addition: An interesting experiment was carried out with a set of a single-phase generator with electromagnetic excitation ROTOR + STATOR:

https://youtu.be/0LmQzSx06uE

In the experiment, the rotor electromagnet winding, connected in series, is connected through LATR and WATTMETER to a 220 V/50 Hz network. At the output, an excitation winding is used, with a resistance of 8 ohms. Further, through the WATTMETER, a 220V / 700W boiler is connected to it as an active load. Switching scheme and indicators on a visual slide

http://rakarskiy.narod.ru/_fr/0/8488233.jpg

Judging by the readings of the WATTMETER, we get a power equal to 45 W from the network, and 140 W enters the load from the excitation winding (efficiency = P2 / P1 = 140/45 = 3.0). Let's try to figure out the CVC in the circuit sections of the primary and secondary circuits of our converter without moving parts. The circuit from the network to the LATR, which includes the FIRST WATTMETER. Mains voltage U1 = 220V, Power P = 45 W. The current strength for this indicator will be: I = P / U = 45/220 = 0.2A. The current in the primary circuit will be 0.2A. Perhaps the current strength indicator will have an increased value, but LATR is essentially a voltage divider. In our case, the armature electromagnet winding and the lower part of the LATR winding have a resistance calculated according to the parallel connection rule. The overall resistance will decrease. In the video, the author shows that the idle power of LATRA is 12 W, which means that the current in the circuit will be 0.05A. The real current in the rotor winding circuit will be 0.2 - 0.05 = 0.15A. Let's leave it for dessert, we will also consider the current in the rotor winding to be 0.21A, taking into account the decrease in voltage at the LATR output, to 210 volts.

Further, an interesting approach to determining the power in the circuit on the WATTMETER. I mean, consider the situation. The wattmeter at the input measures the voltage up to LATR, there are no complaints about it. Data on the voltage at the rotor terminals (in fact, at the output of the LATR: idle U2=210V; under load U3=111. We see that there is a voltage drop that is associated with the connection of the resistance of the rotor winding. According to the power calculation rule, it is necessary to take into account the real voltage and effective current in the circuit passing through the load P = I * U: 1) Option P= 0.2A * 111V = 22.2 W; 2) Option P = 0.15A * 111V = 16.6 W; 3) Option P = 0.21 A * 111 V = 23.3 W. As you can see, far from 45 W, but at idle LATRA has 0.12A * 220V = 26.6 W. How are you? The diagram of this process is below:

http://rakarskiy.narod.ru/_fr/0/4768894.jpg

Now the most interesting thing is the measurement by the WATTMETER of the power of the secondary circuit - the winding of the generator and the active load. The WATTMETER shows 140 W of effective power, with a voltage at the phase terminals of 39V. You can roughly, very roughly calculate the current strength I = P / U = 140W / 39V = 3.59A. But I want to note that the total current for our load of 0.7 kW at 220V will be 3.18A. Our boiler was supposed to be hot, but it barely heats up. Here the reason is different, this current is a variant of compensation by the electric field of its potential difference. Our indicators: open circuit voltage U2 = 117V; under load U3 = 39V; winding resistance R1 = 8 ohms, you can calculate the load resistance by the formula R2 = U2 / P = 2202/700 = 69 ohms. We calculate the current strength of the secondary circuit I = (U2-U3) / (R1 + R2) = (117-39) / (8 + 69) = 1.2A. We can calculate the power in the secondary circuit P = 1.2A * 39V = 46.8 W. How do you like the difference 140W / 46.8W = 2.99. This is how to understand the algorithm for calculating the WATTMER, the indicator turned up, 117V * 1.2A = 140 watts. The formula of the so-called total power P = E*I, which is absurd in its essence, is used. Why you can read in my material "The resultant force of Ampere". How does this threaten the consumer and benefit the seller? If during the day the voltage in the consumer's network is raised by 2 volts several times, then your final result of power consumption will be more. You won't even notice it, you won't be able to check it. For that, on a regional scale, earnings from scratch from the seller.

Back to actually evaluating our transformation: let's do this in every aspect of our data:

1) According to input/output wattmeters: efficiency = 140 W / 45 W = 3.1;

2) According to the results of the calculation of the CVC of primary and secondary circuits: efficiency = 46.8 W / 16.6 W = 2.8 or 46.8 W / 22.2 W = 2.1;

3) For real power from the network and at the load: efficiency = 46.8 W / 45 W = 1.04.

I believe the red highlighted net conversion is in the ROTOR-STATOR assembly and is greater than one.

This data is within the conversion limits of Holcomb Energy Systems solid state generators. His reports and marketing posts show about twice the result in relation to the power of entry.

[alan]

No, Holcomb didn't invent magnetic field gain. If he's for real then what he has invented is a way to utilize the magnetic gain without reflecting the load back to the primary windings.

I've spent this last winter experimenting with gain from magnetic flux. I can tell you this much, gain can be had by avoiding counter emf (Lenz) produced by secondary windings during the magnetization phase of the core primary windings. Ordinary transformers and ordinary generators do not do this.

Cadman

Agree, I also thought it is doing this.

Who is saying that transformers or generators are doing this?

Flyback converter is doing this.
Boost converter is doing this.
Joule Tchief is doing this .
Cheers,
Pix

Flyback has 2 self-inductance phases, continuous forward convertors & xformers reflect self-inductance as mutual inductance and bemf in the other coils, 'Holcomb' is like a continuous mode forward convertor or regular transformer but without self-inductance reflected as bemf towards the input, which normally is used as a natural gain feedback because it pumps up input current as a result output has more power, but output=input.  DZ gen has variable output vs steady input, more load, less resistance, more current according to ohm, more self-inductance, but no mutual inductance and their bemf in primaries.

[alan]

Spinning magnetic field across AC motor stator makes it an ordinary generator, the only difference is there is no mechanical rotation  to create rotation of magnetic field. Lenz law apply.

The across part doesn't happen, the axis of rotation is in the same position as the center axis of the coil (this becomes obvious with the DZ gen). Otherwise you were right.

Vronk
I rewatched the DZgen video, my previous statements regarding 'axis' of rotation are wrong. The N-S-N-S poles are rotating across the faces of the coil, flux is changing.

[Jimboot]

Tonight I put a 100v 68uf cap across my 240v incandescent lamp. My 24vac supply rated at 1000ma hooked up to the 8ohm rotor coils in parallel. Output from stator coils connected to the lamp with the cap across it.
Measured with my DMM so take it with a grain of salt. However there was a significant difference visually in brightness.
Input from 24vac supply with rotor and load connected.
23vac 519ma 11.93 watts
Output from the stator connected to the / lamp cap 40vac 970ma. 38.8watts
So that was interesting.

I have also attached the model of the generator I have recycled from.