The So-Called Don Smith Generator

[MenofFather]

So first things first: Is this worth a try?

Skysabre

Maybe worth, maybe not, that is another model?

[mscoffman]

Skysabre,

If you find a system like this Don Smith system that you build and find it is OU
and you are willing to produce the input/output power table for that is a least
1500Watts in excess OU or more maximum. I would be willing to trade a balance
of system design for that design information and link it to a either household
inverter front-end or a portable generator type front end. For example an 8KVDC
at 20amp (measured) continuous output power supply rating linked to an
inverter would be no problem.

You will never find the complete balance of system design for a system that
requires any kind of complex controls. The best you could find in that case
would be to build a start/stop control with constant power production capability
or stop and a large and expensive battery bank capacity.

Keep this offer (by agreement only) in mind and you will see.

:S:MarkSCoffman

[Jeg]

Hi Skysabre,

I went through the pdf file you altruistically uploaded.  I am strongly "amazed" by seeing the wiring diagram shown in pdf Page 27: they use voltage divider resistors to match the alleged 8kV output voltage (stored in the 2uF capacitor) to the input of an off-the-shelf 12V or 24V DC to 110V AC power inverter. Using resistors for dividing voltages makes sense when reasonably small currents are involved because of the unavoidable heat loss in the resistors.

Let's do some simple calculations backwards,  let's choose a power inverter which is able to output just 2kW at 110V AC, ok? Then let's have its own efficiency, say, 90%, so the input DC power to it should be 1.1 x 2kW=2.2kW i.e. 10% higher.

Now if this inverter receives 24V DC input (let's say), then the input DC current demand is about I=2.2kW/24V=91.6A.

Now lets calculate the resistor values for dividing the 8kV DC to 24V. Here is a link I used, it also calculates the power loss in the two resistors:  http://www.bowdenshobbycircuits.info/r2.htm 

Using the example described in the text for finding the series resistance needed for a LED as a sample, I entered for battery voltage 8000, for current 91.6 and for Vout 24, I got 87 Ohm for R1 (and R2 would be represented by the 24V input of the inverter) but with this value the dissipated power in R1 would amount to 730.6kW....  the idea to use resistors to drop the 8000V to 24V is ABSURD, INSANE to say the least. And there is no any other suggestion to reduce the 8kV (or whatever kV coming from the LC tank via the diodes to the 2uF storage capacitor) to 24V or 12V.

The other thing is that while it may be possible to store a few kW power in a resonant LC tank circuit (it would need a coil and a capacitor with extremely low loss i.e. very high Q) but when directly loading such high Q resonant circuit, the circulating power in it would get reduced just because the external load reduces the high Q (the resonant AC impedance of such a high Q LC tank may range from several ten to several hundred kOhm and shunting it say with the 87 Ohm resistor via the inverter input, the high AC impedance gets killed, so the kV voltage reduces accordingly, there is no 'juice' left in the tank to feed the inverter. AND there is no any high wattage resistor shown in the pictures, which would imply they really used them to reduce the kV output...

Now no wonder why this pdf file contains NO any measurement results done by those 'compiling' it. It is a "pathwork", a "botch", taken text and pictures from some sources, and sorry to say I consider it as a hoax.
At other forums and also here, several members attempted to replicate Don Smith setups but nobody succeeded so far (or did not report, that is) and even Don Smith never showed any working device, only talked about the kW output and showed the assembled boards with those components like in the pictures in the pdf file.

Regardless of Don Smith or from this pdf file, the problem to be solved would be to find a circuit topology which does not reduce an LC tank resonant impedance too much and dissipates only a reasonable amount of power while dividing the kVs to a lower more workable value.  This would be challange circuit to design and build for sure, still strongly assuming that the input power to the LC tank could be kept at a lower value than that of the output power received from THE circuit.

Gyula

Gyula and others

I already made a lot of experiments and i have learn a lot of things around this stuff. But even with the most well designed secondary, i still cant see any OU results. I fooled my self many times as i wanted to believe it, but things are not so good like in the way it presented to be. Perhaps there are other secrets here like NMR as elementsix told us. Or perhaps there is something else which i cant recognize it.

Can someone help to find the NMR freq of copper in combination with the local magnetic field? As i remember gold's nmr frequency under earths magnetic field is very low... 30-36Hz!!! Perhaps element six can define his statement to help us a little more. I dont know how low freqs like that can help.

[gyulasun]

Hi Jeg,

I am not an expert on NMR,  unfortunately and I recommend to turn to member verpies in a Personal Message if you like, he is an expert on it.  of course element six may also help.

Gyula

[Jeg]

Hi Jeg,

I am not an expert on NMR,  unfortunately and I recommend to turn to member verpies in a Personal Message if you like, he is an expert on it.  of course element six may also help.

Gyula

Thanks a lot Gyula
Even it looks unlikely to me, i should try it.