Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[cHeeseburger]

Hey Rose

Before we were cutting off the source when the receiver cap reached the source voltage. We are now cutting off at a much lower voltage, and the gain is apparent more than before.

We are going with woopys method here of using a cap as the source. Both the source cap and receiver are 10uf for this test.

The source is charged to 1000v, for purposes of Big Show.
And we have the recycle diode in place in the middle that continues the cap charging after we release the source cap at our desired voltage.

1 The 1000v cap is charged

2 The switch is closed

3 We wait till the receiver cap reaches 10v  yes 10v

4 we open the switch and the recycle diode takes over to continue charging the receiver with the flywheel.

The outcome

The 1000v source cap is now 990v

The receiver, being cut from the source at 10v, reached 132v!!!

And this is the best part. We only have to replace 10v worth into the source cap to fill it up again. Not replace a complete 1000v worth at 10uf. Is that a savings?
I think so. It may be easier than if empty. Dunno. 



Mags

Very interesting, Mags, but there is just one problem:

The energy held in a capacitor is 1/2 CV^2, as I'm sure you know.

To make the math easy, we also know that both caps are 10uF (C) and 1/2 is a constant when calculating the energy held in each cap.  So all we have to look at is V^2 to get a number that relates each capacitor's energy content which can be stated as Joules if we multiply the V^2 in each case by the constant 1/2 C (5uF).

Now, if you are following that okay, let's see what the total energy is before and after one cycle of your circuit's operation:

We start with 1000V on the one cap and zero on the other, so all the energy is in the one cap at first and it is 1,000^2  times our constant 5uF (1/2 C) or 1,000,000 * 5/1,000,000 = 5 Joules of total energy in the circuit.

After the cycle is complete, we have 990 Volts on the source cap and 132 Volts on the "collection" cap.  Seems like a nice energy gain.  But let's do the math before jumping to conclusions, okay?

990 ^2 is 980,100 times our constant 5uF (1/2 C) = 4.9005 Joules still remaining in the source capacitor.

How much is in the collection capacitor at 132 V?

Well...132^2 = 17,424 times our constant 5uF (1/2 C) = 0.08712 Joules...

Add those two up 4.9005 + 0.08712 = 4.98762 Joules total energy left in the sum total energy of both capacitors.  So on each cycle of the circuit, we have lost 0.01238 Joules of energy.  Do that about 400 times and you have lost all 5 Joules of the initial 1000V capacitor's energy, assuming you add ten volts before each cycle to start again at 1000 Volts. 

So, it is clear by using simple math and standard energy formulas for capacitors that your voltage numbers and simulations are probably correct but it is also clear that the circuit does not gain energy on each cycle but rather loses energy on each cycle.

On each cycle, you'll have to put a new 0.0995 Joules of energy from an outside source into the source cap to get it up from 990V back to 1000V and on each cycle, only 0.08712 Joules gets put into the collection capacitor.  COP = 0.8756

Humbugger

P.S.  Here's a good link where the secrets of using an inductor to transfer energy between capacitors is discussed:

http://www.smpstech.com/charge.htm

The approach and technique you are discovering and exploring has been used in switch-mode power supplies for almost fifty years, so your discovery is not new but it's very cool knowledge that lots of other smart circuit guys use all the time.  I really enjoyed the the last few sentences in the Personal Anecdote section.  Good advice! 

[Rosemary Ainslie]

Guys I thought I'd better answer this in depth or I'll be accused on 'closing my ears' to argument. 

Stefan is correct here.  Rosemary has explained that she sees very large spikes in the shunt (reported by her to be +10V and -30V in amplitude) at the moments the function generator switches.  These come from the signal generator trying to suddenly change the voltage on the huge gate capacitance and are further hugely exaggerrated in amplitude by the fast-changing nature of the resulting current spike and the inductance of the shunt.

Notice that it is stated as a fact that the spikes come from 'the gate capacitance - hugely exaggerated - by the fast changing nature of the resultant current spike'?  I would have thought he meant the resultant oscillation - or resonance.  Hardly a spike.  Anyway.  I keep needing to say this and clearly Humbugger is not getting the point.  The result of factoring in inductance and impedance would be to INCREASE the resistance on the resistor.  This would, correspondingly - reduce the value of the current flow measured from that voltage.  Which is valid.  Now here's the anomaly.  When this is done - when we factor in for this inductance and impedance - then the result is that there's an even greater benefit - an even bigger return TO the system - than before.  I've said this EVERYWHERE.  We've highlighted it in the report.  For some reason this is completely ignored.  I wonder if this is deliberate?  INDEED HUMBUGGER.  We're all interested in this fact.  Because it results in a greater gain when you do the required integrated power analysis.  Could it be that the phase relationship is the critical factor and NOT the actual measure of the current flow?  Or is the argument now that the phase relationship - which is required to sustain that current - else  it would simply dribble away at the zero crossing - is purely an aberration of the system?  Does the phase angle - shown by the scope meter, in fact NOT take place?  Strange.  Becasue it's seen through the battery, on the drain AND at the shunt.  Then - as mentioned - our delusions are also being shared by our oscilloscopes and our circuit components.  And you can't medicate those. 

They are only present in the gate-current loop and are not present in the battery loop, as I explained.

Also a gross misrepresentation of the fact - or, in other words, a humdinger of that 'L' word.  It is evident across the battery in a HUGE positive spike and it is simultaneous with the spike at the shunt.  Just that they're in antiphase, showing that its 'RECHARGING' that battery.

I believe it is these spikes that she agrees are far more negative than positive which cause her scope average on the current trace to often show a small negative value (millivolts) which is then misinterpreted as battery charging current. This only shows up in the scope-averaged current at the low-power operating level because the operating currents are very small, so these spikes throw off the average  In the higher power mode of operation, these spikes still subtract from the measured average current, but that current is much larger in this mode so the overall average still always shows as a positive number..

I've highlighted it.  Another HUMdinger from our Humbugger.  He HOPES that it's these spikes that throw off the average mean voltage across the shunt.  What then would he then do with the mean average voltage in consistently NEGATIVE mode when we're dissipating frantically high wattage from the resistor element?  I KEEP referring to this.  Clearly I need to show more of these waveforms.  I have some on the system.  I'll look for them.  And I will certinly concentrate on showing this in future.  AGAIN, Humbugger, this is an emphasis that you need to lose if you're depending on it to win this argument. 

To get a true measure of the actual battery current, either the function generator must be eliminated (as Stefan suggests) or the shunt must be moved out of the gate circuit loop and placed where it sees only battery currents.

This is where?  There is nowhere on the circuit that is not connected to every other part of the circuit.  And it makes not a blind bit of difference if we put the shunt on the source or the drain.  And it amuses me that when we used a 555 their call was for a functions generator.  Now that we are complying to this the call is to get back the 555.  Makes one think that there's nothing will satisfy these extensive reaches into the bottom of that barrel.  The downside is that it - unfortunately - does seem to be bottomless.

In either case, the shunt must have an inductive reactance  that is far below the shunt resistance at 1.5MHz or the scope sampling and multiplying technique cannot be used due to large phase shifts in the apparent current versus the true current at any given sampling instant.

There it is again.We depend on the DISTORTIONS of those phase shifts.  Mainly because they seem to be resulting in a wave form that is continually reinforced.  There is NO QUESTION that if we eliminated them then we would lose that advantage.  The advantage is NOT IMAGINED.  Or is it that you'd simply prefer that we eliminate that advantage?

This is extremely difficult if not impossible to do, since even the length of a one inch straight wire will add 20nH to the shunt, causing many degrees of phase shift.  Even the very best "non-inductive" shunt resistors will exhibit several nanohenries of inductance and skew the phase at 1.5MHz significantly.[/b].

Well there you go.  That's exactly what's needed.

But all is not lost!  Even with a highly inductive shunt, as Rosemary is using, the true average current in the shunt is easy to obtain...without even using a scope that features averaging!

And here, guys we get to the nub of the issue.  The eternal requirement to AVERAGE.  Humbugger - hold your horses.  I intend showing you the AVERAGE on HIGH WATTAGE OUTPUTS so that I can finally silence this argument.  Then you'll need to retract this argument.  RIGHT NOW you're assuming that we ONLY GET HIGH WATTAGE ON POSITVE VOLTAGE AVERAGES ACROSS THE SHUNT.  THIS IS WRONG.  I will say this as often as is required.  We've shown it.  You are quite simply WRONG.  If your final supporting argument is based on the assumption that we cannot get a continual negative mean average at higher voltages then - again - you are wrong. 

Recall my demonstration from an earlier post in this thread where I showed a 0-2A trapezoid wave and the effects on the voltage displayed that a large inductance would have.  The true average current in that setup was +1A and ramped evenly back and forth between zero and +2A, never going negative at all.  Yet the inductive voltage was well below zero half the time, whenever the true shunt current was down-ramping toward zero.  See the first picture, reproduced here for your convenience.

Now we take that same circuit, doing the same thing, with the same values (printed out for you on the second picture) and add two little simple  RC filters to average both traces.  Guess what!  Both traces now show the exact true average current of 1 Ampere positive (50mV on a 50m Ohm shunt resistance).  They are exactly superimposed.  So now we know the magnitude and direction of the actual DC equivalent average current flowing even though we have measured it across a highly (grossly in this case) inductive shunt resistor!

This is hardly relevant.  Just another red-herring from that bottomless barrel of fishy facts.  We have entirely different results.  Probably because we're using live test apparatus and because we all have open minds.  It reminds me of those endless videos that you post of TK's where you show variations to waveforms from the same wire.  Unfortunately you ALWAYS show the reference at a junction and not on that wire.  I don't think we can accuse you of impartiality in the way you marshall your facts.

Regards,
Rosemary

[Rosemary Ainslie]

looks like that link was correct about possible damage to the MOSFETS

wonder if it's correct about the other 2 things i picked out from it, too?

Indeed you were on the money here.  Regading the rest I've done my best to comment.  Sorry it took so long.  I see now we need to pay attention better here nul-points.

<<EDIT #1>>
interesting, also, that rensseak's link to PETT describes the parasitic oscillation as a negative resistance event!  (although i don't recall it clarifying whether it's negative differential resistance, or the real thing)

Also not sure.  The thing is that the voltage seems to cross zero which introduces a negative component.  But there's still negative spiking at the transitions to the 'on' switch.  Then there's not, typically any further negative crossing until the next cycle.  Just don't know.

<<EDIT #2>>i've been a bit concerned about the suggestions to generate the parasitic oscillations by just connecting a negative voltage across the gate - the driving waveform from the SigGen is after all a dynamic waveform, not just a collection of two DC levels - ie., it also contains transients  so we shouldn't overlook the possibility that the parasitic oscillation is 'triggered' by a transient, before being able to sustain during a suitable state of the input (ie. the negative level)

Not sure if I understand this.  But I think that the actual trigger is the negative inductance established on the load during the on period.  But really not qualified to comment.  Better left to you guys.  Perhaps Neptune or Paul.  But I absolutely agree with you.  It needs to be better explored.  The first anomaly that we demonstrated was that this negative oscillation can occur with absolutely ZERO current being measured through the shunt.  That was strange. 

Kindest regards,
Rosemary

Not edited.  Just took forever to find the referenced post. 
R

[nul-points]

thanks for the comments, Rosemary

i think there are a couple of things we can learn from those points:

- we can expect to have more MOSFET failures if we explicitly try to reproduce these parasitic oscillations (because the necessary gate drive is 'off-label')

  and this would be one valid reason that the manufacturer recommends avoiding parasitic oscillations
(another reason being that such an output would be a 'distortion' of the input - and for most digital or analogue signal applications that would be an undesirable feature of a product!)


- 'positive' resistance is the characteristic of a component to directly convert electrical energy into heat energy

  'negative resistance' is considered to be the inverse  - ie., the ability of a component to directly convert heat energy into electrical energy

NR has been the subject of much discussion and debate - basically it would mean that electrical energy could be generated without having to expend energy - ie. no user-provided work would be required - because ambient heat which exists all around us, indoors and out  (and waste heat from other machines) could be used as input to a negative resistance system to provide 'free' electrical energy as output

one of the few genuine 'negative resistance' components (based on carbon fibre material) was recently discovered in the US (by another lady researcher, Deborah Chung)

NR is not to be confused with negative differential resistance (NDR) where the component does exhibit the inverse Volts/Amps relation to usual Ohms Law, but there is still a positive DC offset, so in this case we are still expending energy as work to drive the system

so - the links quoted previously describe the parasitic oscillation area of the MOSFET characteristic graphs as exhibiting 'negative resistance' - but i didn't see them clarify if it is NDR - or if it is true NR

if the MOSFET data refers to true NR, then this would explain any anomalous electrical energy gained by systems such as Rosemary's

if the MOSFET data refers to NDR, then the parasitic oscillations are likely to be converting electrical energy to heat, as normal, not converting heat to electrical energy

- i think we've all made note of the third point now, that some voltage transitions might still be necessary to get the oscillations going, even if the gate input is a fixed negative DC voltage (eg. battery)


i hope this provides a reasonable summary
np


http://docsfreelunch.blogspot.com

[nul-points]

So figure, we took the source away at 10v on the receiver, and it went to 132v after?  Is that an increase of 1320% of from our flywheel?   

Magzzzz

hi Magzzzz

it would be great to think 'contrariwise', but sadly the only thing here that is Overunity is your enthusiasm!! 

at the start, the input 10uF cap charged to 1000V holds 5000mJoules

at the end, the input 10uF cap at 990V holds 4901mJoules

so the input 10uF cap has supplied 99 mJoules to the 'Believe' circuit

the output 10uF cap at 132V holds 87mJoules
(ignoring the 0.5mJoule of 10V on 10uF)


Efficiency = 87/99 = 88%


if you want to even approach 100% then you need to transfer the charge from input to output in many steps of smaller energy transfer

it took me a year, using the same circuit arrangement in 2008, to confirm that the only things which are incorrect in the EE text books are that the value of charge-separation in an isolated circuit like this is NOT constant - and the energy dissipated : energy stored ratio is not always 1 : 1 as claimed

i've already given you & woopy a link to the thread which contains my results so i won't include it again here

i'm not trying to be negative about your work - i admire your zeal - i would like to save you guys from wasting your valuable time

regards
np


http://docsfreelunch.blogspot.com