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Overunity Machines Forum



Mysterious Resonant Circuit

Started by EMdevices, July 24, 2008, 10:04:51 PM

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0 Members and 4 Guests are viewing this topic.

innovation_station

sure you can have excess engery that is not the problem but to use it correctly apears to be my problem ....

just like em said  it is much harder than you think .....  not as easy as it looks i know this first hand....  for some time now....

but i had no training peroid in any of this ...  that is half my problem....

ist
To understand the action of the local condenser E in fig.2 let a single discharge be first considered. the discharge has 2 paths offered~~ one to the condenser E the other through the part L of the working circuit C. The part L  however  by virtue of its self induction  offers a strong opposition to such a sudden discharge  wile the condenser on the other hand offers no such opposition ......TESLA..

THE !STORE IS UP AND RUNNING ...  WE ARE TAKEING ORDERS ..  NOW ..   ISTEAM.CA   AND WE CAN AND WILL BUILD CUSTOM COILS ...  OF   LARGER  OUTPUT ...

CAN YOU SAY GOOD BYE TO YESTERDAY?!?!?!?!

Hoppy

@ EM

I suggest you measure your VBE and you will probably find it very high and almost certainly above the typical SOA limit of 5V. This will degrade the junction and bring the transistor gain / hfe down a peg or two until the transistor eventually destroys. Strange things happen during this process and I have experimented with a few configurations without a base emiter diode as per your circuit.

Hoppy

Hoppy

@ EM

I should have said measure the VBE on a scope not with a meter.

Hoppy

gyulasun

Quote from: EMdevices on July 26, 2008, 09:15:19 AM
....

@willitwork,  yes I've been thinking and planning to do this ultimate test to prove OU without a shadow of doubt.  However, it's a bit trick, and the reason is that the oscillator and the output power depends on the load resistor.  I tried different resistors and I get different performance.    This is not uncommon, most stages in an electronic system need matching between stages.  So, I am researching and thinking how I can do this effectively.  OU engineering is not as easy as it looks.

EM

Hi EM,

First I would suggest trying a simple linear regulator (I think of the adjustable LM317) set to  7 or 8V DC output just to match your present battery voltage  and load the regulator output with 100mA  to see how your oscillator behaves with that type of load (of course your output coil is lifted from gnd and drives a full wave bridge (4 x 1N4148 or 1N914 or similar HF diodes) with a puffer electrolytic capacitor as earlier was already suggested). 

And if your oscillator still consumes around 100mA from the battery while your regulator reproduces this power input then you could loop back its output to replace the battery...

I know the linear regulator surely wastes power which may amount to even loosing the higher than one COP margin but this would be the simplest first try to investigate and get a further insight on your circuit.
Next step would be to look for a simple DC-DC converter which can have over 90% efficiency that surely maintains the higher than one COP of your oscillator circuit.  Here is some switching type ICs for this job (of course there are many other types too): http://www.siongboon.com/projects/2005-08-07_lm2576_dc-dc_converter/   and maybe the LM2575-ADJ type (its output voltage is adjustable, not fixed) can be a simple candidate for this test.

Wish you good luck and experimenting.

Couple of questions:  is the capacitor in the order of several nF of value between the gnd and the transistor base? Also the value of the capacitor for the tank circuit that affects the running frequency?

rgds,  Gyula

Grumpy

Quote from: EMdevices on July 26, 2008, 09:15:19 AM

@grumpy,  the battery does not heat up.  I understand those battery charts, its the mAh, a measure of energy, so if you drain more you depleate faster, that's all.  But you can draw two amps if you really wanted.

EM

I am aware of that.  Shorting a 9v is a popular way to warm your pockets on a cold day.

I brought it up because I don't think that circuit should draw 0.1 amps.  Do you have another means to measure the current? 

Now you say the battery does not heat up - do a sanity check by placing an equivalent resistor across the battery to get your 0.1 amps and see if the battery heats up.
It is the men of insight and the men of unobstructed vision of every generation who are able to lead us through the quagmire of a in-a-rut thinking. It is the men of imagination who are able to see relationships which escape the casual observer. It remains for the men of intuition to seek answers while others avoid even the question.
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