Partnered Output Coils - Free Energy

[picowatt]

Now,here is a kicker. The DMM read's 0.000A on DC when the diodes are emmited from the output circuit. Do you know what that means PW?.

Tinman,

Without a better understanding of the measurement point (schematic?) and whether you are measuring AC or DC current, no, not really, I do not know what that means. 

Are you saying that the DC ammeter inline with the positive terminal of the battery that was normally showing 400ma or so during previous tests now reads zero when the scope is disconnected from the circuit and the diodes are omitted from being in series with the globe?

OR,

That a DC ammeter placed in series with the secondary, and the lamp connected to that secondary, indicates zero amps when neither diode is in series with the lamp? 

If the ammeter "reading zero" is in the secondary leg, is it set to measure AC or DC current?  Is it averaging the on and off time current flow and telling us that the pos and neg current flow averages to zero?  (which, if you look at the V waveform, it does appear to be close to that case, i.e., the "area" above and below the zero V line appears similar, and is why I initially considered an inadvertent AC coupled path or AC coupled leakage)

PW   

[TheCell]

@Tinman
Normally the average of the current and the voltage at a secondary of a transformer is zero . 
We need the normal average value of the volts across the shunts, not the rms-value.
If one of these current average values is not null (I guess evt. the inner secondary current av. val.)
then you have somethin special.
Normally the flyback current is fed by the strength of the magnetic field which decreases until the next on pulse.
I think the decrease is logarithmically not linear, which is not the case in the inner secondary.

[poynt99]

Now,here is a kicker. The DMM read's 0.000A on DC when the diodes are emmited from the output circuit. Do you know what that means PW?.

It most likely means (and I agree with PW) that the average DC current through the load is 0A.

Remember Brad that current alone is not power. This was clearly illustrated when you isolated the forward and reverse currents using the diodes. What's really pertinent for this analysis is energy (i.e. power and time).

You should buy an inductance meter next time you have a few bucks. It's quite clear that the secondary is vastly more inductive than your primary. As such, the secondary can be "energized" quite rapidly by the primary, but when the primary goes open (i.e. the MOSFET switches OFF), the secondary is left to its own inductance and kickback "rate". It should not be surprising that the IK is soooo looong, should it? Remember the lessons that have been taught here time and time again? The length of the IK is largely determined by the load seen by the inductor, but the ENERGY released during IK is the same, regardless how long it takes. Remove that bulb (globe) and you will see the IK you are familiar with, i.e. a huge magnitude, short duration spike. Short out the output, and you will see a much longer but lower magnitude "spike", just like you are seeing in your results.

[tinman]

You should buy an inductance meter next time you have a few bucks. It's quite clear that the secondary is vastly more inductive than your primary. As such, the secondary can be "energized" quite rapidly by the primary, but when the primary goes open (i.e. the MOSFET switches OFF), the secondary is left to its own inductance and kickback "rate". It should not be surprising that the IK is soooo looong, should it? Remember the lessons that have been taught here time and time again? The length of the IK is largely determined by the load seen by the inductor, but the ENERGY released during IK is the same, regardless how long it takes. Remove that bulb (globe) and you will see the IK you are familiar with, i.e. a huge magnitude, short duration spike. Short out the output, and you will see a much longer but lower magnitude "spike", just like you are seeing in your results.

It most likely means (and I agree with PW) that the average DC current through the load is 0A.

Thats exactly what it means.

Remember Brad that current alone is not power. This was clearly illustrated when you isolated the forward and reverse currents using the diodes. What's really pertinent for this analysis is energy (i.e. power and time).

Yes-agreed.
But lets think about this a bit. When the diodes are emmited,and the current value is 0,and the globes filament is the same temperature,then the resistance value is the same throughout a full cycle. So we now have an over all DC current flow value of 0 across a resistance that is constant,as the filament simply will not have time to cool between cycles at these frequencies-->this i have checked with a solar pannel on top of the globe,and i can see no ripple across the solar pannel with the scope-the voltage line is smooth.

Also note that this is not inductive kickback,as the current flow is in the opposite direction. We also have a case where the primary is loaded in both directions,unlike inductive kickback ,where the primary becomes open,and current continues to flow in the same direction. This is a conventional generating/transformer effect,not an inductive kickback effect-->the current flow is in the wrong direction.

I do remember us going through current and voltage averages-if we average one,we dont average the other-correct? So in my last test,the scope showed 30v peak during the 5% on time,and the current on the DMM was 80mA. So as the current is an average value,then the voltage dosnt have to be averaged-correct? So we have 30 x .080,which = 2.4 watt's. Now the rated value of the globe is 2 watts from a 12 volt supply. So although the globe looks bright,we are only supplying it with .4 watts of extra power over it's rated value,and probably why it hasnt blown.

So for arguments sake,i swap the bulb out for a 22 ohm resistor,and the DC current value on the meter still reads 0. This means that we have the same amount of power being disipated across that resistor during the off time of 1 cycle,as we do during the 5% on time over 1 cycle. So although the voltage and current is lower during the off time,the time is far greater than the 5% on time at a higher current and voltage. If there was more current flowing in one direction than there was in the other,the DMM would show this offset of current,and read a diferential current value.

Remember Brad that current alone is not power.

That is correct. But current through a resistance is power. EG-100mA through say 22 ohms is 220miliwatts of power. Here we have a current through a resistance,and when we know the resistance value,we can calculate power.

[tinman]

@Tinman
Normally the average of the current and the voltage at a secondary of a transformer is zero . 
We need the normal average value of the volts across the shunts, not the rms-value.
If one of these current average values is not null (I guess evt. the inner secondary current av. val.)
then you have somethin special.
Normally the flyback current is fed by the strength of the magnetic field which decreases until the next on pulse.I think the decrease is logarithmically not linear, which is not the case in the inner secondary.

This is not flyback current,as it is in the opposite direction. This is standard transformer current(AC).