Testing the TK Tar Baby

[MileHigh]

Part 2.....

So, let's look at a simplified model for the MOSFET oscillator itself.  The power input is the average current + voltage drop between the drain and the source.   So in simplified terms the drain-source is the pair of terminals that represent the power input of the MOSFET oscillator.

The power output of the MOSFET oscillator is the four parallel and connected gate terminals.  These are capacitively coupled to the actual MOSFET oscillator engine.  So, the output from the MOSFET oscillator is a 2 MHz AC output voltage via the gate terminal.  How much AC power is output through this terminal is dependent on the AC source impedance, the AC voltage, and the AC load impedance.

I hope that you are following here, I am kind of describing the circuit with a bit of a level of abstraction in the hope that it connects with yourself and some of the readers.

Finally, let's look at most of the AC power output paths that form the AC load on the MOSFET oscillator gate output:

1.  The Q2 gate output to the 0.25 ohm CSR to ground.  This where you have already observed pure AC with your scope.

2.  The Q2 gate output going across the Q1 source-gate capacitance, through the 0.5 ohm resistor, through the 50 ohm resistor in the function generator, then to the negative potential supplied by the function generator.

3.  The Q2 gate output going across the Q1 source-drain capacitance, through the inductive resistor, and then to the positive terminal of the battery set.  (This may be the path that explains the AC voltage that we see superimposed on the battery voltage)

4.  The Q2 gate output going across the Q1 source-drain capacitance, and then to the Q2 drain.  (This might be part of the feedback mechanism to sustain the oscillation)

So, that's a "simplified" model for the Q2 MOSFET oscillator.  The fact is that we know that there is AC on the Q2 source also, and that also will be disseminating AC power through more signal paths.

The skinny from this too-long posting is that I showed a mechanism for superimposing an AC voltage signal on the battery positive terminal.  There is a clear path between the Q2 MOSFET oscillator and the battery positive terminal.

I said before that this whole circuit was buzzing with high-frequency AC, and hopefully this posting helped explain that.

MileHigh

[TinselKoala]

I think that's clear enough.

Meanwhile... how efficient is the Tar Baby as a heater?  In other words, how much of the power that the device draws, is actually dissipated in the load, where it can do useful work, and how much is "wasted" by dissipation in heating the mosfets and resistors and RF radiation?

Once the batteries are all recharged, I'll set Tar Baby up with strict negative-going pulses on the gate, and maximum attainable Q2 drain current as measured by the inline DMM, which for me seems to be something under 200 mA. I'll record a time-temperature curve if I can stay awake. My load cell is pretty well insulated so it might not reach equilibrium temperature in a reasonable time, or it might. We'll see. Regardless, I will then take a straight DC power supply and provide the load with the same average current as read on the DMM, and record the time-temperature data. Then I'll plot the curves for comparison.
Since I only have one load and it will have to cool back to ambient between runs, this will take a day or two to complete.

Meanwhile, I encourage the NERD team... if, that is, all the RATs haven't yet abandoned ship .... to hurry up and get on with their own testing, to refute my results and show that they actually do have what they have claimed so loudly in glowing bold CAPS to have.

(My goodness... am I about to lose my membership in the Rabid Debunkers and PseudoSkeptics Society (RDPSS)? What kind of a thing is that for a Rabid Debunker to say... encouraging testing and publication of results.... ? Everybody knows that a RDPSS member in good standing must SUPPRESS all discussion, DELETE all possible evidence of OU, COVER UP and DISCOURAGE testing that shows OU, and DIVERT INTEREST onto other topics, like Nitinol motors and LENR. I guess I had better watch out, or the MIBs will take away my black Buick.....)

[hoptoad]

snip..
I guess I had better watch out, or the MIBs will take away my black Buick.
..snip

You've got a black Buick ? You lucky B...... LOL

Seriously though TK, I like your methodology.

One thing that still mystifies me a bit, is the the idea of true AC manifesting in the circuit. Varying DC superimposed onto the supply rail is easy to picture in this circuit, but the dual LED experiment you carried out seems to indicate true AC.

This intrigues me a little, because the polarity of any Collapsing EMF during the off time of the pulsing of the inductive resistors is such that current would not normally flow though the load (even with mosfet body diodes) as a result of the current/magnetic field collapse.

The polarity of the resultant EMP is such that a diode would need to be across the load in flyback arrangement, but then the current from the collapse event would still be in the same (uni) direction through the resistors, as the supply.

It's easy to picture actual BEMF from a motor/rotor inducing a current opposite to the supply via the body diodes of the mosfets, and therefore superimpose true AC, but there is no such motor/rotor in this circuit. I find that true AC seems to be present as opposed to just varying DC quite interesting.

Thanks for your presentations thus far !

Cheers

[TinselKoala]

Thanks, hoptoad, I'm glad you find it all entertaining.   

The NERD RAT team have been using the scope data dumps of the voltage drop across the CVR (equivalent to the current through the CVR by applying Ohm's law to the resistance value and vdrop) and the voltage on the mosfet drains to calculate their power data, have they not?

That is, they are taking the peaks of the oscillations on the mosfet drains, times the current at those times, and taking the time periods and multiplying these together to get a quasi-time-integration of a power signal, I think. Am I right about this?

So they are looking the area under the curve, formed by the peaks of the oscillations on the drain trace wrt the baseline of that trace. Am I right about this?


If so..... there is a minor problem.   

[TinselKoala]

As I've noted before, in the second part of the NERD demo, they are using a POSITIVE OFFSET of their FG signal which as we know will turn on Q1 during the non-oscillating phase of the signal from the FG.

Here's the proof: an explanation and analysis of the LeCroy toy oscilloscope screen shot. Please download and display the figure so you can refer to it as you read the captions.

1. This is the trace of the Voltage Drop across the CVR. The value of the CVR is nominally 0.25 Ohm, so the Current will be, by Ohm's Law, I=V/R, or I = V/0.25. So the Current, in Amps, will have the value of 4 x the indicated voltage displayed by this trace. The part of the trace indicated by the red line is about 1 1/2 minor divisions ABOVE the baseline level, indicating current flow during the non-oscillating portion of the trace. Flow in the normal direction, giving a positive voltage drop across the CVR.

2. The baseline zero voltage level of the CVR trace.

16. The channel setting for the CVR trace. It is displayed at 2.00 volts per division. Therefore each minor division is 2/5 of a volt. Therefore, the voltage level indicated in (1) above is about 3/5 of a volt, and this corresponds to 12/5 Amp or 2.4 Amps. THIS IS WHERE THE HEAT IN THE ELEMENT IS COMING FROM, not from the oscillation portion of the trace at all.

3. The level of the Battery Trace during the NON-oscillating phase.

4. The zero baseline marker below the number 2. Since this trace is set to 100 V / div, we can see that this trace is sitting at battery voltage, and the "means" in the parameters box agree. Therefore, during the NON-OSCILLATING portions of the trace again applying Ohm's law, we find V=IR, or R=V/I, or R(total) = 50/2.4 = 20.8 Ohms. The resistance of their load is 11.1 Ohms, the Q1 mosfet's on-state resistance is 2 Ohms, the CVR is 0.25 Ohms, the cliplead connections and wires add some resistance.... and the initial figure of the 2.4 amps current read off the scope trace by me here a year later isn't all that precise. THEREFORE, the figures given by Ohm's Law and the actual circuit measurements are in nearly exact agreement. The circuit is dissipating (2.4)x(50)= 120 Watts during the NON-oscillating phase of the signal, and (2.4)(2.4)(11.1)= 64 Watts of that is getting to the load itself and the rest is wasted heating the Q1 mosfet and the CVR and the other resistances in the circuit during the NON oscillating portion of the gate signal.

5. The level of the GREEN trace, the common mosfet drains, during the NON oscillating portion of the signal.
6. The zero baseline, the little green line below the number "4".
8. The channel setting, 100 Volts per division.
10. The zero baseline level again.
11. Where the "50 volt" level... the battery voltage level.... would be on that trace at 100 v/div.
12. One full division above the baseline, or 100 V, for comparison.
Note that this is the common mosfet drain signal, or alternatively the load itself. We see that during the NON_oscillating part of the signal, the voltage sags well below the battery voltage and even sags below horizontal during the individual periods themselves. We know that when the mosfet is OFF the voltage here should be HIGH, at or just below the battery voltage. But it's not. Therefore the mosfet is at least partially ON here, as corroborated by the CVR trace.

13. The channel setting for the Gate Drive Signal trace, 20.0 V/div.
14. The baseline or zero volts level of the Gate Signal trace, just below the number "3".
15. The top level of the Gate Signal trace, sitting at least 1 minor division ABOVE the zero volt baseline indicated at (14). This is indicating a POSITIVE level of at least 4 Volts during this non-oscillating portion of the cycle. This is enough to at least partially turn on the Q1 mosfet.

9. The "math trace" channel setting: 500 VxV (not W) per division, and the function being performed: the simple multiplication of the CVR _raw_ trace and the battery voltage trace. Since these are both voltages, the scope knows that the display then will be V V, or the product of two voltages. Had a current probe been used here, as I have illustrated in other videos using a similar LeCroy scope, the scope WILL display V A or even Watts here. But the simple math being used by the presenters does not account for Ohm's Law on the CVR.... therefore the math is wrong. Where the presenter uses the displayed "-5.29 V V" figure  (item 7) as "Watts" he is simply talking garbage.

In short, the circuit is NOT giving anything like the bogus figures they have calculated from the scope using the oscillating portions of the waveform.

It is dissipating 120 Watts during the NON-oscillating portions, of which less than half is heating the resistor load. Very little or no power is actually passing during the oscillations.

In other words, this ENTIRE DEMONSTRATION is a combination of ignorant error, arrogance, mendacity, and outright DECEPTION (the way they chose to display the Gate drive signal on the TEK, discussed earlier).

I hope somebody will check my math and interpretation of the scope screen. Somebody OTHER than the NERD RATs, I mean.