Testing the TK Tar Baby

[TinselKoala]

I should add, for the spectators, that this system, suggested way back in the old days by Stefan Hartmann our good host, and by many other people since then, is specifically designed to emulate the CONTINUOUS OSCILLATION, LONG DUTY CYCLE mode that Ainslie used to claim that she preferred, before she realized that high heat in the load required significant Q1 on time as well.
SO this negative bias supply circuit does the SAME THING as the FG does when it's making the oscillations: it is supplying a NEGATIVE VOLTAGE to the SOURCE of Q2 and doing it with the ability to "source" the required up to 200 mA of current, just as the FG was doing. It is achieving the oscillations by exactly the same method and application of polarities that the FG does.

(posted schematic again to keep everything on the same page)

[poynt99]

Part 1.

A closer look at what makes up a FG, with some particular insight into how its output stage works.

R_OUT is the 50 Ohm resistance usually included as a standard in most FGs.

R_INT is the resistance associated with the output transistors and their ability to control the load they are driving. This resistance is represented by the R_INT resistors shown going back to the associated power supply for each transistor.

The +15.5V and -15.5V power supplies are typical voltages for a FG output stage, and demonstrates how and why the FG can generate both a positive-going and negative-going voltage swing at its positive terminal (with respect to its negative terminal).

Carefully note that the power supplies and negative output terminal are commoned together.

[fuzzytomcat]

Howdy all,

If your not confused yet on Rosemary's understanding of her COP>INFINITY device(s) operation on mosfets Q1 and Q2 ..... well 

http://newlightondarkenergy.blogspot.com/2012/01/232-another-summation-based-on.html#links

Rosemary's BlogSpot Quote #232 - "another summation based on countering the counter arguments"

..... snip rubbish .....

The red pencilled lines are intended to show the commonality of the switches. Therefore, in effect, one has those transistors arranged that the battery can access either Q1 or Q2 - REGARDLESS. Which argument then claims that there is ALWAYS a path for the positive flow of current from the battery supply. And therefore too - the battery is NEVER disconnected. Which, by default - means that when we measure energy being delivered from the supply - which is that voltage measured above zero - then correctly it IS INDEED being discharged by the battery. This conclusion makes not the slightest difference to our claim. We still measure that negative wattage number. There is still, evidentially, more energy being delivered back to the battery than was first supply by the battery. But there's a nicety that needs to be factored in which goes to the real anomaly and not to standard assumptions about anything at all. Lest we lose the significance of this data - I'm taking the trouble to show this here - as it's been argued on the forum.

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.

..... snip rubbish .....

FTC

[TinselKoala]

See.... Since 2009 we have been trying to explain to these people that when an N-channel mosfet is wired to switch a highside low impedance load, the DRAIN VOLTAGE is HIGH, at battery voltage, when the mosfet is OFF. And when the mosfet is ON, the DRAIN VOLTAGE measurement drops to near the negative rail voltage. This is completely normal and I have demonstrated it several times in such simple manner as to be bordering on the mindless. And it doesn't indicate any huge current flows or complete battery discharges.

YET THEY STILL MAKE THE SAME ERROR, as evidenced in the quote Fuzzy posted.

This error is also at the heart of the inverted duty cycle issue from the COP>17 circuit. They measure the duty cycle by monitoring the mosfet DRAIN VOLTAGE with respect to the source. They see a duty cycle here of 4 percent HIGH, that is, four percent of the time the voltage at the drain is at battery voltage. And they think that means the mosfet is ON during that time and OFF when the drain voltage falls.

Do the Math. She has the audacity to say that and then commit the most egregious computational lunacies one after the other in the same breath. I am starting to think that this is some kind of parody of a parody, and the reality show will premier in September, and Ainslie will turn out to be Red Green, or somebody like that as host.

[MileHigh]

Rosemary:

If Chess won't remove it, then I am asking you to remove it yourself: