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Author Topic: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011  (Read 741428 times)

Sprocket

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #300 on: March 25, 2011, 12:27:57 AM »
@Rosemary - Thanks for the info. Yes, I am definitely thinking of specific applications as well, hence my reason for wanting a 'real-world' heater element.  I presumed that the one you were using was somewhat specialised as it seemed fairly low-power - most of the heater elements I've googled were in the 2-4KW range.  Since any garden-variety should do, I'll probably source one locally.

@neptune - that certainly describes my pulse-generator, simple - just pulses, nothing else, and definitely no offset option.  I've been threatening to get a decent function-generator for years but inconveniences like eating/beer etc. always seem to win out!  And recently one of the channels went in my scope.  I finally managed to trace the fault, but the needed i.c. (UB1202AM) has been relegated to the status of "obsolete stock" so it's gonna take mucho deneros if I opt to fix it - most US firms will only sell wholesale and don't even respond to emails, but I managed to find a Chinese supplier who will sell me a minimum of 5 for $10 each, so we're talking at least 100 bucks. The function-generator will have to wait for another while.  As for the 555 circuit, I'm sure I will be able to scare up something to generate the required waveform without too much difficulty.

Magluvin

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #301 on: March 25, 2011, 01:21:48 AM »
Hey Woops

I stuck us in the Tesla Igniter thread as it is appropriate and this is Roses deal here and the cap shorting is getting away from mine and her discussions. Shorting the cap is just putting direct source across the inductor, thus more voltage seen, and more current seen. ;]   The idea though is on the same lines with the flywheel effect.

You can check us out over there Rose. =]  Always welcome. And I have another treat for ya. Its a way to use 1 energy 2 times, well almost 2 times. Very simple.  ;]

Mags

cHeeseburger

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #302 on: March 25, 2011, 03:41:45 AM »

On a different subject, but related, is another observation that I think is worth considering.  Seems we have all in the past (self included) assumed that the current in the shunt represents the current in the battery.  To and from, as it were.  This would certainly be the case under a DC analysis, where the MOSFET gate is correctly considered an open circuit without current flow.

At a frequency of 1.5MHz and each MOSFET having (from the data sheet) 2800pF of capacitance from Gate to Source, with five in parallel that is 14nF or 0.014 microfarads  which is substantial.  We see the 1.5MHz oscillations appear on the gate voltage traces in Rosemary's scope shots. 


Here I forgot to mention that the very large narrow spikes we see in all the shunt current traces all correspond to the switching pulses from the pulse generator charging and discharging the gate capacitance through the shunt.

The shunt inductance enlarges these spikes.  They are confined to the gate-source current loop and thus do not appear in the actual battery current.

To obtain an accurate waveform and measure of the battery current ONLY, I would recommend highly that a Kelvin-sensing shunt of extremely low inductance be used, placed on the battery minus side of the common ground point (rather than on the MOSFET source side).

Such a shunt can be easily fabricated in a few minutes for about $5.00 US.  See the picture below for the details.

Kindest regards,

Bryan

CHeeseburger
« Last Edit: March 25, 2011, 05:38:06 AM by cHeeseburger »

hartiberlin

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #303 on: March 25, 2011, 04:43:13 AM »
Hi  Hartiberlin

in response to your inquiry on youtube

hope i have undesrtood your mind

http://www.youtube.com/watch?v=RQFD1cDlEUU

good luck at all

Laurent


Okay, thanks,
but I meant , what would happen,
if you pulse the power supply
to the circuit, not shorting the capacitor.

Anyway, as you did it looks good and you
could use a incandescent lamp like
a 12 Volts 10 Watts Halogen lamp
in series with your switch when you short out the cap,
then this lamp will light up.

P.S: I will move this Forward EMF discussion postings to a different
thread, as it does not apply to the Rosemary circuit..
Regards, STefan.

hartiberlin

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #304 on: March 25, 2011, 04:46:38 AM »
Here I forgot to mention that the very large narrow spikes we see in all the shunt current traces all correspond to the switching pulses from the pulse generator charging and discharging the gate capacitance through the shunt.

The shunt inductance enlarges these spikes.  They are confined to the gate-source current loop and thus do not appear in the actual battery current.

CHeeseburger

Yes, through the switching of the function generator
additional energy can be flown into the circuit via the
Gate Source and Gate Drain capacitances.

So it will be wise to just use a negative DC power
supply on the Gates to start and keep the oscillations.

Magluvin

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #305 on: March 25, 2011, 04:59:38 AM »
Hey Stefan
If it is ok, you could put the posts in the igniter thread. If you insert them just before my first post today, it will fit the time period as there wasnt any posts for a long time. In it fits.  =]  Just a suggestion.


Thanks
Mags

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #306 on: March 25, 2011, 05:00:09 AM »
Yes, through the switching of the function generator
additional energy can be flown into the circuit via the
Gate Source and Gate Drain capacitances.
Really?  Where is this energy coming from?  The plug?  Or from ground?  I can prove that it does not come from the plug and I intend proving that it does not come from ground through the simple expediency of applying a groundless connection.  And that capacitance would need to generate in the order of 60 amps in BOTH directions.  The ONLY supply that is capable of that much current is from the battery.

Rosemary

hartiberlin

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #307 on: March 25, 2011, 05:00:21 AM »
Rosemary,
please measure the battery voltage directly across the battery terminals
with the scope, not inside the circuit.


Here are the 2 scopeshots.
You decide which areas are bigger, the
red ones (positive ones) above the black
ground line or the green areas below the ground
line (negative current recharging the battery).

The math function of the scope says, the green area, is bigger.
You decide !

The question is, when the low potential of the
function generator also oscillates with 1.5 Mhz,
does this supply much power from the function generator
into the circuit via the capacity of gate to drain and source ?

As the funtion generator has about 50 Ohms output resistance,
I just calculated that it could add about 0.5 Watts max into the
circuit at these oscillation amplitudes via the capacitive coupling.

Regards, Stefan.
@ Poynt, please post your simulation file.
Thanks.

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #308 on: March 25, 2011, 05:15:27 AM »
Rosemary,
please measure the battery voltage directly across the battery terminals
with the scope, not inside the circuit.
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. 
 
Here are the 2 scopeshots.
You decide which areas are bigger, the
red ones (positive ones) above the black
ground line or the green areas below the ground
line (negative current recharging the battery).
Good point Harti.  I would say that there's more above than below which definitely CONFLICTS with the displayed values.  I suspect that this was taken from a stored shot of a full cycle.  Else the numbers would have adjusted accordingly.  I intend looking into this tomorrow.

The question is, when the low potential of the
function generator also oscillates with 1.5 Mhz,
does this supply much power from the function generator
into the circuit via the capacity of gate to drain and source ?
This can be easily proved or disproved.  And this will CERTAINLY be tested tomorrow.  I'll let you know.

As the funtion generator has about 50 Ohms output resistance,
I just calculated that it could add about 0.5 Watts max into the
circuit at these oscillation amplitudes via the capacitive coupling.
That EXACTLY matches the energy that we measured.

Stefan, will you PLEASE carefully read the email that I sent you.  We're missing each other by a mile.

Kindest regards,
Rosemary

hartiberlin

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #309 on: March 25, 2011, 05:23:02 AM »
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.

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #310 on: March 25, 2011, 05:28:40 AM »
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

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #311 on: March 25, 2011, 05:32:41 AM »
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.

Quote
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

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #312 on: March 25, 2011, 05:34:29 AM »
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

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #313 on: March 25, 2011, 05:39:31 AM »
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

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #314 on: March 25, 2011, 05:43:40 AM »
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