Pulsed DC Transformer with Embedded Magnets

[The Observer]

Rave,

The Embedded Magnets are Atomic in Size and Rotate on a Dime to Magnify the Magnetic Field of a Coil by 1000s of Times !

See below...

Best Regards,
                         The Observer

[Paul-R]

tsueng,

until you can definitvely show watts IN....against watts OUT..... then you can light as many led's as you want....... you can have 1.2V IN...and show 60V OUT

It needs to be watts in versus watts out. (Not volts).

When it comes to the equation, don't forget that spikes are hard to handle, watts-wise.
It would be better to start with a charged battery, run the equipment which will be
outputting to accurately depleted batteries.

After a number of hours,
1. measure the amount of volts/amps needed to bring the first battery up to fully charged.
2. Measure the number of volts/amps needed to bring the output collecting batteries
up to fully charged.

COP is the second figure over the first.

[rave154]

paul_r.... i agrree with you, that was my point of pointing out the need the show watts IN against watts OUT.and not just 1.2V IN against 60V OUT..or even 600V OUT....V against V shows nothing..

also, thanks for the image of the permiability of iron etc.... but what i was referring to was that earlier in this thread Tseung was propounding the values of a core with actual neo magnets embedded in them to show that the flux from the magnets when positioned in the correct direction wouls somehow Add to the overall power OUT.....

now he seems to have hijacked the joule theif scheme and immediately claims it is his "Lead Out Tech"

so which is it?......Joule Thief?..embedded magnets?...flux manipulation transformer? flynn transformer?

[XS-NRG]

now he seems to have hijacked the joule theif scheme and immediately claims it is his "Lead Out Tech"

so which is it?......Joule Thief?..embedded magnets?...flux manipulation transformer? flynn transformer?

Actually Mr.Tseung now wants to label the TPU as a his "Lead Out Tech"

Reply #493 on: July 20

The chance of our team producing a 1KW lead-out device for the poor countries within three months is excellent. 

So that will be soon  , we just need to wait some more day's.

However:

  I must listen to the Voice that directs me to benefit the World.

Lawrence Tseung

But the voice that directs Mr.Tseung did not see this:

http://www.youtube.com/watch?v=ppAH5ztirOg&feature=related

And This:

http://www.youtube.com/watch?v=tRVF0rG-m0A&feature=related

And so he does not realize the World isn't going to benefit...
But i will just wait to see what he comes up with on the 20th of October 

[mscoffman]

It needs to be watts in versus watts out. (Not volts).

When it comes to the equation, don't forget that spikes are hard to handle, watts-wise.
It would be better to start with a charged battery, run the equipment which will be
outputting to accurately depleted batteries.

After a number of hours,
1. measure the amount of volts/amps needed to bring the first battery up to fully charged.
2. Measure the number of volts/amps needed to bring the output collecting batteries
up to fully charged.

COP is the second figure over the first.

@All

For my money the only way to measure wattage is to convert the
voltage to filtered 12VoltsDC. This will give you a nice high current
for the DVM to measure accurately. Then reconvert the 12VDC back
up to 120VAC at 60Hz. or 240Vac at 50Hz. to reconverted again
by standard DC supplies or battery charger. The o'scopes should
be reserved for showing that there are *no* pulses on this 12VDC
line and that the voltage or current does not change significantly
during the *one second* DVM instrument gating period. Voltage
and current measured with a separate DVM then multiplied.


The downconversion to 12VDC (using Schottky bridge rectifiers and
transformers can allways be considered 100% efficient.) I can
handle the fact that the upconverter is not going to be 100%
efficient but that inefficiency can be measured as xx% and presented
and I will accept that.

If someone thinks it is easy to correctly calculate the wattage of a
pulsed signal I wish they would present it because it is not very easy.
It involves Fourier decomposition of both the current and voltage
waveforms because the phase of each component relative to the
other is important. Then the sum needs to be built up from parts.
Inexpensive instruments that do this are designed for 50 or 60Hz *sine wave*
waveforms. If you really want true pulse power you will need 40K-50K$
USD high bandwidth calculating equipment.

I think one should run through this 12VDC method at least once,
because I think you will find the numbers that result come in very
different from what the cheap AC instruments are showing. You
are going to need to explain those differences.
I trust the instrumentation setup that LT is showing - not.
The stuff that he is doing is too important to have instrumentation
errors.

:S:MarkSCoffman