Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[Rosemary Ainslie]

Stefan - here again is the point.

Energy is vi dt.  Therefore we multiply the current determined by voltage across the shunt with the voltage at the battery. 

We addressed this at the demo.  We used a TYPICAL waveform - that SHOWS NO NEGATIVE MEAN AVERAGE - and showed that the antiphase condition of those voltages - across the battery and across the shunt - INEVITABLY results in COP INFINITY.  This is because the battery voltage is at its LOWEST when energy is delivered and at its HIGHEST when energy is being returned.  That way - REGARDLESS - the gain is ALWAYS to the battery.

The anomaly is this.  If we apply CLASSICAL PROTOCOLS then the result is Infinite COP.  The question - as you rightly point out - is does the energy go through the battery?  I do not know.  But what I do know is that it is in line with the voltage measured through the drain.  If it is argued that this energy on the drain is from the FET - then the FET would need to be discharging something in the region of 60 amps.  I think this is unlikely.  But I'll try and find a condition that can prove this more conclusively.

Rosemary

[hartiberlin]

I would do this with PLEASURE.  I cannot use the oscilloscope probes.  I've written this in an email reply to you.  I've mentioned this to Poynty here on this thread.  THE SCOPE PROBES DO NOT SPAN THOSE TERMINALS.  I can CERTAINLY do it if I add wire.  But then we're back to where we started. 

Yes, add some thick wires there and measure with the scope
head directly at the positive terminal
and with a thick diameter wire connected directly
connect the ground line of the scope to the neative pole of the battery.

Good point Harti.  I would say that there's more above than below which definitely CONFLICTS with the displayed values.

Hmm, but it could also be, that the green area is bigger all in all.
As the scope says it is a negative nanoVolts ,
the negative area seems to be only very minuscule bigger...



Regards, Stefan.

[Rosemary Ainslie]

When the function generator has the low signal,
then the oscillation is running.

Then the oscillation amplitude of the 1.5 Mhz overlayed on the low
signal of the function generator is around 5 Volts
at maximum.
If we calculate the gate to drain-source resistance as a short
at this high frequency, there is only the limiting
resistor of 50 Ohms at the output of the function
generator.

Thus the maximum power can only be 5 Volts ^2 /50 Ohm / 2
cause the maximum power can be put out, when the
gate to drain-source capacitive resistance would be equal
to the internal output resistance of the function generator.

So at 5 Volts oscillation amplitude it could be a maximum
of 0.25 Watts.
( the function generator output resistance and the
gate to drain-source capacitive resistance are voltage
dividers and thus at the gate to drain-source capacitive resistance
only 1/2 of the output voltage of the function generator can occur at maximum)

At 10 Volts oscillation amplitude of the overlayed 1.5 Mhz signal it would be about 1 Watts max, what the
function generator could provide into the circuit.

Regards, Stefan.

Ok.  This is more comforting.  We put a .5 Ohm resistor at the gate to measure the energy.  We established that there was something in excess of 5 watts - but that this was being returned to the functions generator.  One of the guys there said that there was enough capacitance associated with the generator to absorb this energy.  I am not qualified to comment.

Regards.
Rosemary

[hartiberlin]

Stefan - here again is the point.

Energy is vi dt.  Therefore we multiply the current determined by voltage across the shunt with the voltage at the battery. 

The problem is, you did NOT measure the voltage at the battery !
Only inside the circuit, where you have a long cable to the batteries,
which has too much inductance at 1.5 Mhz !

Please measure the voltage directly at the batteries with very big cables !

[Rosemary Ainslie]

The problem is, you did NOT measure the voltage at the battery !
Only inside the circuit, where you have a long cable to the batteries,
which has too much inductance at 1.5 Mhz !

Please measure the voltage directly at the batteries with very big cables !

Still not sure what you mean.  Do I add wires across the batteries?  Or do I leave the cables there and simply apply the probe directly to them?  I'll gladly do whichever - or both, as required.

Rosemary