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RF Power Generator Over Unity

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The Flavored Coffee Guy

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overunity thermoelectric

Hello,

The Circuit Examples, Part Values are included with the zip file. If you would like rules of thumb, you can find them here at this link at my website. Tutorial

Simulator model files that are provided in the attached file were produced by Switcher CAD III Bode plots of the successive stages of the circuit establish how the circuit is constructed and reveals the importance of fine tuning. If you are familiar with capacitive current doublers, they don't have a ring value and resonant circuits do. It does work as a product of Q, rate and at times, bad impedance matching and precision tuning.

The model files include a circuit that if constructed will convert 20W to 2.4KW. Another model file which may require very serious parts searches and research to acquire or construct the required parts would convert 30W to 13KW, if constructed. The supply is intended for plasma generation for the sake of heating. After which the heat may be used to heat a building cheaply, or to provide heat to a pile of thermoelectric generator cells and convert it to a usable form of electricity. Plasma at high enough temperatures acts like a piece of wire and the resistance drops to a few places after the decimal point if the vacuum pressure is right. Which is similar to Ionization (gas-filled) Tubes, which can be used to switch on to conduct millions of amperes of current. Very similar to a xenon camera flash. One the plasma is excited by a 2 to 3kv pulse, 200 to 450 volt dump from a capacitor bank through bulb. At higher frequencies the domain and temperature of the plasma doesn't have time to loose conductivity and produces allot of heat.
 

rf power generator circuit

Hello,

rather surprizing material. I didn't see at first look, where the energy is coming from, can you tell in brief?

Regards,
Frank
 

microwave overunity

You are making experiments without knowing the basics of electricity.
 

overunity coil

FvM said:
Hello,

rather surprizing material. I didn't see at first look, where the energy is coming from, can you tell in brief?

Regards,
Frank

In brief, DC analysis of a simple capacitive current doubler can show you. Let's say I have a power supply with an output impedance of 100 ohms. I place a 10µF capacitor across the output and a 25 ohm load in parallel. What happens with the DC circuit is that if the output voltage of the supply is 10 volts, I wind up measuring 5 volts across the 25 ohm load and a current of 200 mA. Where the output with an impedance matched load of 100 ohms would result in 10 Volts at 100 mA.

With a parallel resonant circuit the holds true, except at resonance each half cycle of the capacitor can build up a higher voltage. So, if you look at the turn ratio of the transformer it should be putting out 100 volts at 500 mA, now if you drop that into a 1:2 turn ratio you should get 200 Volts at 250 mA, but you are more likely to see 180 volts at 1 ampere if Xc and XL are low enough at resonance and the Q of the resonant circuit high enough.

A Q of 5,000, tells you that the circuit will only loose 1/5,000th of the energy stored in it per cycle.

Building the real circuits is not like the simulators all of the time. Switcher CAD III does not treat a coil as a coil when it's modeling a transformer. Take a wall transformer and a Inductance Meter. Measure the inductance of the primary with the secondary open and write that value of inductance down. Then measure the inductance with the secondary shorted and write it down. For any resistive load, the last capacitor across the primary of a transformer's actual value will depend upon the resulting value of inductance. There cannot be any short cuts taken in the di/dt*L, and mutal inductance at work in a transformer for the math to exactly model reality. Any circuit that is resonating for the most part appears to be an open circuit in most cases. You can calculate how close that is when you use this formula Ω=1/√LC.

Resistive loads allow you to backwards tune the circuit from output to input to place the power on any given or chosen load. If you produce heat, you can then use the heat to generate electricity using thermoelectric generator cells or thermocouples. Thermoelectric Generators produces better results in the search engines. HI-Z has stock cells rated at 20 Watts per Cell.

A changing load farts with resonance and any effective energy available.

For the most part it's a bad impedance match until you get to the product of Ω=1/√LC, were you'll find a voltage drop and an increase in current and power of the input is lower than the power in the following stage, or output. Keeping Xc and XL as low as possible and Q of the inductor as high as possible provides the power out of an impedance mismatch by most standards.

I'm not good at brief. I went over some details. Basically, a step up transformer can become a step down transformer under the conditions of a bad impedance match. XC and XL can each equal 1Ω and the total impedance of the two in parallel as DC resistances in parallel represent the circuit's start up impedance. Which is 0.5Ω. The if the secondary has 10 volts at 1 ampere available, the voltage will drop until there is a volt to the ohm power ratio. 10 Watt across 1 ohm from a transformer with that output would give you 3.16 Volts at 3.16 Amperes. Or the square root of power. Typically, that's a bad thing because the secondary windings of a step up transformer are not designed to handle that current. That effect in event also depends upon the winding resistance of the secondary and the current will only rise to that limit.

The resulting impedance of resonance is Z = 1/(1/Q + (LC)), and Q for the most part is calculated based upon the inductor Q = (2pifL)/rw rw = DC resistance of the coil's windings.

When the impedance of power circuit equals the start up impedance of a parallel tank circuit, it will resonate at 1.414 times the source voltage. So, 20 Volts pp converts to 28.28 volts pp at that point in the circuit. So, converting RMS, you take you peak times 0.707 and wind up with the RMS and right back where a push pull amplifier would have you reversing current through a coil with a 10 Volt source, near Unity. It looks like more energy.

If XL and XC in parallel = 0.5 ohms, and Z > the output impedance of a transformer, there can be more amps at the same voltage, more volts and amps at a higher voltage. XL and XC do not need to be at 0.5 ohms, typically you would use XL as a current limiter, and step up the voltage even though XL and XC in parallel are lower or much lower than the output impedance of the previous tranformer stage. Impedance matching from the transformer stage to the next is strictly based upon Z = 1/(1/Q + LC)

I have a microwave oven transformer the primary's inductance measures 68.2mH when the secondary is open. When the secondary is shorted the primary measures 13.02mH. This is where some simulators fail to model transformers corretly. Now, when the primary is open the secondary measures 15.5H, and when the primary is shorted it measures 3.375H. This means that if you want power to show up in your final transformer stage, the capacitor on the primary winding of the transformer would have to be based upon matching the reflected impedance's resulting inductance. If the primary winding were resonant Xc = XL at resonance. XL = 2*pi*f*L, at 1KHz that equals 428.51Ω with the secondary open and only 81.8Ω, and XC will not change at 1KHz for any one given value of capacitor. That gives you two different values of capacitor to make the circuit resonant at 1KHz depending upon whether the circuit is open or shorted, and it will actually land somewhere inbetween.

It gets all the way to the point that you almost need to dig for a good old fashion tuning capacitor and mount it in mineral oil to handle the voltages without arcing.
 

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