Bedini solid state oscillator and Tesla switch combination

[quantumuppercut]

I've been doing experiments about this setup. It seems like an efficient energy transfer process. What I see so far is depends on how you charge your inductor, you can get more or less heat for the same input and output.

Let's say you have an inductor L and wants to charge to current i . The amount of energy is 1/2 L i^2 . The question is how much energy do you put in to charge the inductor and does the amount of heat generated while charging related to input? If it does not, then the so call copper loss on power line needs investigation.

Below is the analysis I made. Inductor L is being charged to current i and 1/2i in separate cases. The charging time is taken near zero to get linear simplicity. The question we want to ask is input/output and heat loss.

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Energy when charge to i is 1/2Li^2 and energy charge to 1/2i is 1/2L1/4i^2, which is about 1/4 of the other case. Now what about the input. We can calculate the charges draw from the source by integrate the area under the line. The area for current i are 4 triangles and the area of 1/2 i is 1 triangle. This is no surprise. We can charge to 1/2i four times and it'll be the same as charge it to i . In both cases, energy input = output. However, when looking at the heat produced they are different. We can calculate the heat with I^2Rt . One case is I^2Rt and the smaller one is 1/4I^2R(1/2t), which is about 1/8 times.

So by pulsing it with shorter time we can alter the heating while there is no effect on input and output. This heating seems to have no origin. You can also charge an inductor faster (to the same energy) with higher voltage to minimize heating. I think it's call copper loss.

[Guruji]

Best charging

I've been doing experiments about this setup. It seems like an efficient energy transfer process. What I see so far is depends on how you charge your inductor, you can get more or less heat for the same input and output.

Let's say you have an inductor L and wants to charge to current i . The amount of energy is 1/2 L i^2 . The question is how much energy do you put in to charge the inductor and does the amount of heat generated while charging related to input? If it does not, then the so call copper loss on power line needs investigation.

Below is the analysis I made. Inductor L is being charged to current i and 1/2i in separate cases. The charging time is taken near zero to get linear simplicity. The question we want to ask is input/output and heat loss.

ImageShack® - Online Photo and Video Hosting

Energy when charge to i is 1/2Li^2 and energy charge to 1/2i is 1/2L1/4i^2, which is about 1/4 of the other case. Now what about the input. We can calculate the charges draw from the source by integrate the area under the line. The area for current i are 4 triangles and the area of 1/2 i is 1 triangle. This is no surprise. We can charge to 1/2i four times and it'll be the same as charge it to i . In both cases, energy input = output. However, when looking at the heat produced they are different. We can calculate the heat with I^2Rt . One case is I^2Rt and the smaller one is 1/4I^2R(1/2t), which is about 1/8 times.

So by pulsing it with shorter time we can alter the heating while there is no effect on input and output. This heating seems to have no origin. You can also charge an inductor faster (to the same energy) with higher voltage to minimize heating. I think it's call copper loss.

Hi Quantum so which is the best spot for effieciency of charging when inductor is coldest you're saying in this setup?
Thanks for the analysis.

[quantumuppercut]

Hi Quantum so which is the best spot for effieciency of charging when inductor is coldest you're saying in this setup?
Thanks for the analysis.

Hi Guruji,

Before I think charges are conserved and can be transfer completely via transformer. My recent experiment shows that it's energy that is conserved and not charges. I have put capacitors in parallel and series to recover inductor energy and it shows the more in series, the more efficient the process. So I think this set up recover at most 1:1 but not limit to the possibility of RF energy entering the system. It seems like the faster you pulse it, the less heat loss you get but where did the heat go... I think it takes on a form of RF energy but not sure.

[quantumuppercut]

Today I've done calorimetry to see if the heating match with the energy in the capacitor. It shows that the energy in the capacitor equal the energy release in the resistor.

1/2CV^2 = I^2R

This means the energy loss in the capacitor is proportional the the charges loss. But I've shown that we can have less heat with shorter pulse, so the energy within the capacitor does not necessary be in the form of heat when discharge. Energy balance equation would be:

1/2CV^2 = energy loss + inductor energy

The energy loss could either be heat or radiation or both. If we can recover with the ratio of 1:1, then we must have a gain in energy since neither heat or radiation would likely return to the circuit.

[mbrownn]

Nice work.

If you want to push the envelope and get high COP's, increase frequency (shorter pulse) and voltage. Your power can be kept the same by shortening the pulse preventing burn out. It has something to do with the ratio between voltage and impedance so low resistance coils are good but the inductance sets the frequency (or length) of the pulse, higher inductance means lower frequency. Once your frequency exceeds a given ratio on the inductance, the output goes up. At 60v, COP's>1 are there, at 220v there is much more. A problem with semi conductors is the switching is not so abrupt and this abruptness is needed to get best performance.

[Guruji]

Nice work.

If you want to push the envelope and get high COP's, increase frequency (shorter pulse) and voltage. Your power can be kept the same by shortening the pulse preventing burn out. It has something to do with the ratio between voltage and impedance so low resistance coils are good but the inductance sets the frequency (or length) of the pulse, higher inductance means lower frequency. Once your frequency exceeds a given ratio on the inductance, the output goes up. At 60v, COP's>1 are there, at 220v there is much more. A problem with semi conductors is the switching is not so abrupt and this abruptness is needed to get best performance.

Hi Mbrown do you mean that air core coils are less efficient if I understood you rightly? How can we short the pulse to a bedini circuit?
Thanks

[mbrownn]

Hi Mbrown do you mean that air core coils are less efficient if I understood you rightly? How can we short the pulse to a bedini circuit?
Thanks

No air coils may be more efficient because there are no eddy currents in them.

If we are using an SSG with small coils we are having to push a little more energy through to power the wheel and this may consume a little more current from the source. With a solid state Bedini oscillator the current can switch off a little sooner as we don't need to power that wheel, also the magnets may induce a BEMF in the coil increasing impedance and lowering effective input voltage and so the radiant spike.

When we pulse a coil the current ramps up charging the inductor, as it approaches saturation the current starts to level off and energy begins to be wasted by just maintaining the magnetism in the coil. We want to switch off before that happens. The inductance of a coil sets the speed that happens, so a large inductance is slower than a small one. As we switch off the inductor gives us the transient radiant spike and then releases the current effectively stored by the inductor.

If we shorten the pulse we consume less power from the source and get less current in the field collapse but we still get the radiant spike usually at about 10 times our input voltage. We do not need current to flow to get the radiant, all we need is the voltage but of course we have to flow a little to allow the voltage to enter the coil. If we increase the voltage input this happens faster so we can switch off faster. My tests seam to show that once we pulse faster than a given speed for an inductor the radiant remains the same but the current is less. It makes sense that to maximize the radiant, we put in as high voltage as our semiconductors can stand and switch it as fast as possible while still getting the radiant spike.

Using a Bedini at 12v the charging effect of the radiant was around 97% of the input power but increasing voltage and shortening the pulse we put the same power into the coil but the spike is much bigger and the spikes effect on charging seems to be proportional to voltage and not the power.

Of course there is a limit on batteries, we don't want to create an arc in the battery. Tesla has told us this is the way to increase the radiant and my results seam to confirm it but my equipment is not that accurate so more testing is needed.

[EMCSQ]

Why is the left of ossie's circuits not suitable for swapping batteries?

[Guruji]

No air coils may be more efficient because there are no eddy currents in them.

If we are using an SSG with small coils we are having to push a little more energy through to power the wheel and this may consume a little more current from the source. With a solid state Bedini oscillator the current can switch off a little sooner as we don't need to power that wheel, also the magnets may induce a BEMF in the coil increasing impedance and lowering effective input voltage and so the radiant spike.

When we pulse a coil the current ramps up charging the inductor, as it approaches saturation the current starts to level off and energy begins to be wasted by just maintaining the magnetism in the coil. We want to switch off before that happens. The inductance of a coil sets the speed that happens, so a large inductance is slower than a small one. As we switch off the inductor gives us the transient radiant spike and then releases the current effectively stored by the inductor.

If we shorten the pulse we consume less power from the source and get less current in the field collapse but we still get the radiant spike usually at about 10 times our input voltage. We do not need current to flow to get the radiant, all we need is the voltage but of course we have to flow a little to allow the voltage to enter the coil. If we increase the voltage input this happens faster so we can switch off faster. My tests seam to show that once we pulse faster than a given speed for an inductor the radiant remains the same but the current is less. It makes sense that to maximize the radiant, we put in as high voltage as our semiconductors can stand and switch it as fast as possible while still getting the radiant spike.

Using a Bedini at 12v the charging effect of the radiant was around 97% of the input power but increasing voltage and shortening the pulse we put the same power into the coil but the spike is much bigger and the spikes effect on charging seems to be proportional to voltage and not the power.

Of course there is a limit on batteries, we don't want to create an arc in the battery. Tesla has told us this is the way to increase the radiant and my results seam to confirm it but my equipment is not that accurate so more testing is needed.

Ok I am using a solid state now but what circuit should I add now to shorten the pulse too for bigger spikes?
Thanks for the info.

[wings]

Ok I am using a solid state now but what circuit should I add now to shorten the pulse too for bigger spikes?
Thanks for the info.

US 5514918 _Pulse_generator ?

[quantumuppercut]

Oscillator gives out much radiant energy from spikes but until now I do not see a good method to capture those energy. I don't think it's practical to make a large Tesla tower and send huge spikes around. Those radiation diminish with the squared of the distance and quite disturbing. Right now I'm thinking could transformer mechanics works on radiant energy except near distance. If so, we just need to manipulate the transformer to adjust power factor, I think it's more practical.

[quantumuppercut]

I've been looking up on electromagnetic waves and how it travels and transmitted. If we have an antenna that send out EM waves, it is desire to have the receiver antenna position in certain orientation to get maximum induction. Below is a drawing of transmitting and receiving antennas. The left antennas transmit EM waves. The top picture shows how receiving antenna correctly positioned. I believe this is discovered and demonstrated by Heinrich Hertz. My opinion, however, this orientation of induction would subject to Lenz.

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I've been experimenting with setup like the bottom figure. I believe the E field can also establish with orientation parallel to the Poynting vector. Such setup, I believed, does not subjects to Lenz. I will borrow a video which I think best fits this description. Starting at 1:00, one can see the diode orientation effects induction. If this is correct, then we have another type of induction that does not effect the source.

StifflerDr's Channel - YouTube

[quantumuppercut]

According to parallel resonance circuit, when at resonance the only loss in the circuit would be resistance. It is stated that this energy is the input energy. However, when examine closely from this video, the input doesn't seems to account for this dissipation energy.

Coil Resonance Tutorial 3 - YouTube

At resonance, the equivalent circuit for this would look like two light bulbs connected in series with the source, which dictates that two bulbs will light up equally. I think this is it.

The mechanism for OU may lies in resonant rise.

[mbrownn]

According to parallel resonance circuit, when at resonance the only loss in the circuit would be resistance. It is stated that this energy is the input energy. However, when examine closely from this video, the input doesn't seems to account for this dissipation energy.

Coil Resonance Tutorial 3 - YouTube

At resonance, the equivalent circuit for this would look like two light bulbs connected in series with the source, which dictates that two bulbs will light up equally. I think this is it.

The mechanism for OU may lies in resonant rise.

When tuned correctly the loss is only the ohmic resistance of the circuit. In this condition you can have say 100w oscillating in the LC circuit with a supply being only a tiny fraction of that.

[Guruji]

Resonance

When tuned correctly the loss is only the ohmic resistance of the circuit. In this condition you can have say 100w oscillating in the LC circuit with a supply being only a tiny fraction of that.

Well said mbrown the thing is how to get that perfect tuning to batteries