Testing the TK Tar Baby

[evolvingape]

TK:

Thanks for the pictures, really appreciated to see the setup.

The AC superimposed on the battery voltage is interesting.  Interesting because I think the NERDs always took it literally and used it in their DSO calculations.

Poynt believed that Rosemary had her battery probe in the wrong place and the AC would not be manifested if your probe was right on the battery terminal itself.  Now I am not so convinced of that in looking at your scope traces because I am assuming that you are measuring the battery voltage properly.  Have you tried a direct connection with your scope probe to the negative terminal of bat 1 and a direct connection to the positive terminal of bat 3?

I would not be surprised if you would see the oscillations like that.  What is likely happening is that the Q2 MOSFET array, acting like an RF waveform generator, is simply superimposing AC on everything.   The batteries probably appear like just another load (not sure of the impedance) to the AC source coming from the Q2 array.  I am pretty sure that the RAT team was completely faked out by that superimposed AC and interpreted it as the literal battery voltage for the DSO crunching.

It's pretty obvious that a big fat battery, kind of like a slow electronic sloth, is not going to react to the buzzing fly of the high frequency AC that is superimposed on it's output.

So the moral of the story is the suggestion that if indeed there is high frequency AC literally on the terminal posts of the batteries, that is a fake-out, and you would have to divine the true battery voltage for purposes of DSO number crunching by filtering it out.  It's garbage data that needs to be filtered out.

MileHigh

It appears that the AC oscillation you have discovered over the battery is the entire argument for the "second part" of the 2 part paper.

Here is an "interesting quote" post #232 from the blog:

http://newlightondarkenergy.blogspot.co.uk/

"This may be a better way to explain the anomalies and it may also get to the heart of Bubba's objection.  The oscilloscope probes are placed directly across the batteries that ground is at the source rail and the probe is at the drain.  Which is standard convention.  Then. During the period when the oscillation is greater than zero - in other words - when the battery is DISCHARGING - then it's voltage it falls. And it SERIOUSLY falls.  It goes from + 12 volts to + 0.5.  Given a  6 battery bank, for example, then it goes from + 72 volts to + 3 volts. At which point the oscillation reaches its peak positive voltage.  And this voltage increase is during the period when the applied signal at Q1, is negative.  WE KNOW that this FAR EXCEEDS THE BATTERY RATING.  In order for that battery to drop its voltage from + 12V to + 0.5V then it must have discharged A SERIOUS AMOUNT OF CURRENT.  Effectively it would have had to discharge virtually it's ENTIRE potential as this relates to its watt hour rating.  We EXPECT the battery voltage to fall during the discharge cycle.  But we CERTAINLY DO NOT expect it to fall to such a ridiculous level in such a small fraction of a moment AND SO REPEATEDLY - WITH EACH OSCILLATION.

Now.  If we take in the amount of energy that it has discharged during this moment - bearing in mind that it has virtually discharged ALL its potential - in a single fraction of a second.  And then let's assume that we have your average - say 20 watt hour battery.  For it to discharge it's entire potential then that means that in that small fraction of second -  during this 'discharge' phase of the oscillation it would have to deliver a current measured at 20 amps x 60 seconds x 60 minutes giving a total potential energy delivery capacity - given in AMPS - of 72 000 AMPS.  IN A MOMENT?  That's hardly likely.  And what then must that battery discharge if it's rating is even more than 60 watt hours?  As are ours?  And we use banks of them - up to and including 6 - at any one time.  DO THE MATH.  It beggars belief.  In fact it's positively ABSURD to even try and argue this.

NOW.  You'll recall that Poynty went to some considerable lengths to explain that the battery voltage DID NOT discharge that much voltage. Effectively he was saying 'IGNORE THE FACT THAT THE BATTERY VOLTAGE ALSO MEASURES THAT RATHER EXTREME VOLTAGE COLLAPSE'. JUST ASSUME THAT IT STAYS AT ITS AVERAGE 12 VOLTS.  Well.  It's CRITICAL - that he asks you all to co-operate on this.  And in a way he's right.  There is NO WAY that the battery can discharge that much energy. SO?  What gives?  Our oscilloscope measures that battery voltage collapse.  His own simulation software measures it.  Yet the actual amount of current that is being DISCHARGED at that moment is PATENTLY - NOT IN SYNCH.

But science is science.  And if we're going to ignore measurements - then we're on a hiding to nowhere.  So.  How to explain it?  How does that voltage at the battery DROP to +0.5V from +12.0V?  Very obviously the only way that we can COMPUTE a voltage that corresponds to that voltage measured across the battery - is by ASSUMING that there is some voltage at the probe of that oscilloscope -  that OPPOSES the voltage measured across the battery supply.  Therefore, for example, IF that probe at the drain - was reading a voltage of +12 V from the battery and SIMULTANEOUSLY it was reading a negative or -11.5 volts from a voltage potential measured on the 'other side' of that probe - STILL ON THE DRAIN - then it would compute the available potential difference on that rail +0.5V.  Therefore, the only REASONABLE explanation is to assume that while the battery was discharging its energy, then simultaneously it was transposing an opposite potential difference over the circuit material.  WHICH IS REASONABLE.  Because, essentially, this conforms to the measured waveforms. And it most certainly conforms to the laws of induction.

OR DOES IT?  If, under standard applications, I apply a load in series with a battery supply - then I can safely predict that the battery voltage will still apply that opposing potential difference - that opposite voltage across the load.  Over time.  In fact over the duration.  It most certainly will NOT reduce its own measured voltage other than in line with its capacity related to its watt hour rating. It will NOT drop to that 0.5V level EVER.  Not even under fully discharged conditions.  So?  Again.  WHAT GIVES?  Clearly something else is coming into the equation.  Because here, during this phase of the oscillation, during the period when the current is apparently flowing from the battery - then the battery voltage LITERALLY drops to something that FAR exceeds it's limit to discharge anything at all.  And we can discount measurement errors because we're ASSURED - actually WE'RE GUARANTEED - that those oscilloscopes are MEASURING CORRECTLY.  Well within their capabilities.

SO.  BACK TO THE QUESTION?  WHAT GIVES?  We know that the probe from the oscilloscope is placed ACROSS the battery supply.  BUT.  By the same token it is ALSO placed across the LOAD and across the switches.  It's at the Drain rail.  And its ground is on the negative or Source rail. And we've got all those complicated switches and inductive load resistors between IT and its ground.   Could it be that the probe is NOT ABLE to read the battery voltage UNLESS IT'S DISCHARGING?  UNLESS it's CONNECTED to the circuit?  Unless the switch is CLOSED.  IF there's a NEGATIVE signal applied to the GATE then it effectively becomes DISCONNECTED?  In which case?  Would it not then pick up the reading of that potential difference that IS available and connected in series - in that circuit?  IF so.  Then it would be giving the value of the voltage potential that is still applicable to that circuit.  It may not be able to read the voltage potential at the battery because the battery is DISCONNECTED.  It would, however, be able to read the DYNAMIC voltage that is available across those circuit components that are STILL CONNECTED to the circuit?  In which case?  We now have a COMPLETE explanation for that voltage reading during that period of the cycle when the voltage apparently RAMPS UP.  What it is actually recording is the measure of a voltage in the process of DISCHARGING its potential difference from those circuit components.  Which ONLY makes sense IF that material has now become an energy supply source.

It is this that is argued in the second part of that 2 part paper - as I keep reminding you.  Sorry this took so long.  It needs all those words to explain this.  The worst of it is that there's more to come.

Kindest regards as ever,
Rosemary"

[TinselKoala]

Facepalm.

Evolve... you do realise what a word salad that post from RA is, I hope.

(But first, I didn't "discover" this RF oscillation, I merely have demonstrated it unambiguously for what it is. It is feedback caused by a little bit of energy ringing back and forth between all those crazy wires and the capacitances of the mosfet and wiring. Circuit designers know this oscillation well and take pains to avoid and eliminate it. In fact if I just put those mosfets up close on TarBaby's board they would probably go away.)

And of course her whole argument about the battery "discharging its whole potential in an instant" in bold and caps is bogusity of the first magnitude.

Come, let us reason together.

The drain trace drops to or near zero in the oscillations because the mosfet is turning on briefly then.
I have demonstrated what happens to the drain trace when a mosfet turns on.

The current is limited by the circuit's resistance.

The Drain voltage drops to zero when the battery is seeing the minimum resistance in the circuit, which is the mosfet's Rdss of 2 ohms plus the CVR of 0.25 ohms plus the 11.11 ohms of the load. Call it 14 ohms, or 15 to be conservative and allow for cliplead crimps and such.
(Actually since it is the Q2 stack that is oscillating the resultant Rdss for the four mosfets in parallel is only 0.5 Ohm. Like that makes a big difference. Well, so instead of 15 ohms use 13.5 ohms in the following calculation.)

Ohm's Law tells us that V=IR, so I = V/R and so a 60 volt battery will push 4 amps of current through a 15 ohm load, dissipating power at the load of P=I^2R, or 240 Watts continuous power if it was on 100 percent of the time. (Here the "load" of course means the entire circuit including the warm mosfets.)

But it's not.

In the single oscillation period, modelling it as a square wave at 1.2 MHz, the pulse is On for less than a half of a microsecond, but call it 500 nanoseconds per pulse.

So the energy in each pulse of the oscillation is actually (less than, because it's not a rectangular pulse) 0.0000005 second x 240 Joules per second, or about 0.00012 Joule, or a bit less than the battery's "whole potential" of over 10 MegaJoules or so. Ten orders of magnitude ! And she thinks her math might sometimes be a "tad out".

And, as we may have determined earlier in the thread, much of this energy is "recycled" into the next period. Only the radiation and Joule heating losses need to be made up by the battery's power to sustain this oscillation.

[TinselKoala]

Here, Rosemary and the NERD RATs: These ladies might be just what you need, to come up to speed on your circuit math skills.
http://www.youtube.com/watch?v=lUwdvAYEb3Q
http://www.youtube.com/watch?v=Ees3b0R4rJ8

[Magluvin]

Tk

Back in Roses thread, the vid of removing the battery from the circuit. When you removed the battery, there was still oscillation?  What Im getting at is, if you remove everything from the circuit other than the mosfets(the mosfets in Roses schematic sown in the vid, just imagine the transistors as a block with the sig gen and the 2 circuit leads, 1 going up to the inductor and the other lead going down to the shunt), when you power up the sig gen, you still get the oscillations?

If so, then the source battery is not even necessary to get the transistor block to oscillate. So, is it possible that the transistor block is only allowing the source to conduct through the circuit during half of the block oscillation, like the block is just a fast on and off switch? 
If so, then are we only getting pulsed DC through the heater from the source?

Or, from your experience, is there actually any current flowing back to the battery, or just out flow period?



MH

We have talked on this cap transfer issue, heat losses, inductors, before. ;]

I can fully understand that if we have a 12v batt connected to a 12v motor with a resistor/resistance(conductors) in series with the motor, that not all the energy being released from the battery is being dissipated by the motor.

Yes, I agree. To have resistance in the conductors or adding a resistance in series in the circuit but wishing that all the source energy were being supplied to the motor, then these resistances are unwanted losses. But Im not sure that heat is the loss in general. You know I have been thinking on this for a while now. ;] Well, still thinking.

If we take a case of where heat is the desired output, are we still losing anything from input to output? Even if we collect the heat from the heater, the battery and the circuits connections?

Here is where Im stuck on losses...

If we have a super conductive battery(all theoretical), and a super conductive buss bar, and we short the battery with the buss bar, we are losing energy even though we ended up with no heat. We could say that we lost it all in the incredible EMP. But I see it as we discharged the super battery by leveling out the charge/potential difference at the 2 terminals of the battery, and thats all. And, that heat and magnetic fluctuations are byproducts.  I know there are other things to consider, but Im just giving my example.

Like TK said about the caps esr.  What if we had(theoretical) super conducting resistor(current limiter). Its purpose is just for loss less current limiting.

If we insert this loss less current limiter in series with the motor and battery, this is where I have my issue.

If the motor and battery connected alone pulls say 10A, and if we add the SC current limiter inline, we now only have 5A going through the motor, did we lose anything? Did any energy taken from the battery not get to the motor, even if the motor had a desired output at 5A?

Now, we have the batt and motor alone at 10A running. We add a resistor(real) in the loop, one with a value that would bring the current down to 5A.  Well, the motor is only going to put out 5A of work. AND, the battery is draining only half as fast as the battery/motor alone at 10A. 

Sooo.  Why would we say we are losing anything?  By adding the resistor in series, to limit the current through the motor, we are not still pulling 10A from the battery and only getting 5A through the motor. Can ya feel where Im coming from?

So, if we have 2 superconducting caps, 1 charged to 10v and the other empty, I can see that if we were to discharge the charged cap into the empty, that they might oscillate forever.  Well what if we rectified the discharge? Would all of the charge from the charged cap end up in the empty one?
Or if we added our superconducting current limiter in the discharge loop. Would we end up with still 5V in each cap? The same as regular caps? Without losses?  Thats my drift.

Ok, Ill let that cook for a bit. then I will expand on it a bit if these posts are welcome. Been thinkin on this for a while.

Mags

[TinselKoala]

@Mags:
With these mosfets, no, I see no oscillations at any gate drive settings with the main battery disconnected.

I don't see any trace of current flowing "back to the battery" either. The power that can light up the LEDs of Doom is bouncing off the inductor and would bounce off the battery too if the inductor wasn't there.

(I am running without these LEDs for the present test, though.)

ETA: My oil is at 250 F at least (the thermometer has gone all the way around and is past zero again....) and batt voltage is 32.3 at 2007 hours. I'm going to switch back to the low-current osc mode using strict negative gate drive, only the Q2 mosfets now, providing 100 mA drain current.

(Scope parameters as before.)