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Overunity Machines Forum



Confirming the Delayed Lenz Effect

Started by Overunityguide, August 30, 2011, 04:59:41 PM

Previous topic - Next topic

0 Members and 14 Guests are viewing this topic.

SchubertReijiMaigo

@ Gotuluc: don't worry, I'm not on the drug or anything illicit...

When you posted the first curve page 31, sorry but it was obvious that your MATH curve was totally active (PF=1), you have reedited your post and display a curve which is reactive...

The first curve in your post above is reactive while the second is slightly active...

So the cap have the ability to correct both L and the reflect R in the trafo ?

One problem in this setup: the output power is very low, 0.42 Volts through 10 Ohms is not very much, here a huge risk of measurement error must be take into account before any conclusion...

gyulasun

Hi Luc,

I accept the 300W toroid transformator cannot show signs of saturation, albeit the scope shots on the current (i.e. voltage drop across the shunt) shows, this is why I suspected starting saturation somewhere in the setup. 
To explore where those spikes on the current peaks may come from it would be good to test the inverter output terminated with a resistor that draws about 1 Amper current from the 118V AC output and see on the 1 Ohm (or 10 Ohm) current shunt whether the spikes are there or not.  I say this test because perhaps the sinewave inverter includes a low pass filter at its output make the sinewave from the 'uggly' switched waveforms and the low pass filter may include a choke coil with a core that may start saturating...  a guess from me, one would not expect such behaviour from a 1000W inverter but those spikes must be caused by something...

Gyula

EDIT  It occured to me a better way would be to load the inverter output with a 22 uF capacitor and see on the 1 Ohm series shunt how the current waveform looks like for a capacitive load (because your resulting load from the series LC was capacitive nature). The 22uF (or near to this value) means a 1 Amper capacitive load current for the inverter output.


Quote from: gotoluc on October 28, 2011, 08:09:50 PM
Hi Gyula,

you may want to edit your post (if you still can) Please look at the Replacent scope shot above.

Regarding the peaks, it's not Saturation of the core. This is a 300VA Toroid, it takes much more current then that to Saturate it. Also, please note that these peaks only happen when connected to the sine wave Inverter. When connected to Grid there are no Peaks. See post with Grid scope shot on the next page

The Inverter is rated at 1000 Watts Continuous. Paid $300. for it

Thanks for your time

Luc

CRANKYpants

Quote from: gotoluc on October 29, 2011, 02:38:20 AM
Hey Thane,

have a look at these Shots... Let me know what you think

Luc

LUC,

WE NEED TO KNOW HOW HIGH THE TRANSFORMER EFFICIENCY HAS TO BE TO COMPENSATE FOR THE INVERTER EFFICIENCY...  :P

CAN YOU PUT A LOAD IN THE INVERTER AND FILL IN THE BLANKS BELOW SO WE CAN CALCULATE THE EFFICIENCY OF THE INVERTER?

EVERY TEST YOU DO OUGHT TO INCLUDE THE DATA BELOW BECAUSE THAT INVERTER EFFICIENCY CHANGES AS IT HEATS UP RIGHT?

DC INPUT SIDE

DC VOLTAGE =
DC CURRENT =
DC INPUT POWER =

AC OUTPUT SIDE

LOAD RESISTANCE =
LOAD VOLTAGE =
LOAD CURRENT =
LOAD PF =
(DON'T ASSUME YOUR RESISTORS ARE PURELY RESISTIVE BECAUSE THEY ARE NOT)

AC OUTPUT POWER =
DC INPUT POWER =
INVERTER EFFICIENCY = (AC OUTPUT / DC INPUT) x 100 = ? %

CHEERS
T


CRANKYpants

Quote from: gotoluc on October 28, 2011, 12:34:19 PM
THIS POST WAS EDITED

Amp meter on the 12.2VDC side of the inverter shows the inverter consumes 400mA idle (nothing connected on output) Under load it goes up to 1.130 A
Luc

IF WE WANT TO KNOW WHAT'S REALLY GOING ON LET'S LOOK AT THE (SOURCE) INPUT REACTION TO LOADING...

DC INPUT CURRENT ON IDLE (NO-LOAD) = 0.4 A
INPUT POWER ON NO-LOAD = 4.88 WATTS

DC INPUT CURRENT ON LOAD = 1.13 A
INPUT POWER ON ON-LOAD = 13.8 WATTS

INPUT POWER INCREASE % NO-LOAD TO ON-LOAD = 182%

THIS MEANS THAT THE INVERTER HAS TO WORK HARDER TO DELIVER POWER TO THE LOAD REGARDLESS OF WHAT THE SCOPE SAYS...

MY SUGGESTION IF I WERE YOU WOULD BE TO DELIVER POWER ON THE AC SIDE WITHOUT ANY CURRENT INCREASE (REACTION) ON THE DC SIDE  ;)

CHEERS
T

PS

IT WOULD ALSO BE GOOD TO KNOW WHAT THE CURRENT AND VOLTAGE IS TO THE INVERTER WITH NO COIL ATTACHED AND THEN WITH COIL ATTACHED?

kEhYo77

@gotoluc

Maybe you should try this driving circuit from Romero at your secondary, it looks promissing...

Generator Coil Speedup Circuit Simulation

In some of the 'acceleration under load' videos Romero showed this attached circuit to be responsible for the speedup effect.
I did a simulation of that replacing Hall sensor with a zener diode.
The result is blinking led with power disipation shown on the right graph and on the left we've got power taken from the rotor...

This can be done in 1:1 transformer too, I think so... ;)

This sim is just an example. You would have to play with parameters (right mouse button on a componetnt to edit) to make it right...