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Overunity Machines Forum



Testing the TK Tar Baby

Started by TinselKoala, March 25, 2012, 05:11:53 PM

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0 Members and 19 Guests are viewing this topic.

TinselKoala

Quote from: poynt99 on April 22, 2012, 02:35:02 PM
(snip)
Actually in my sims, I found that it was indeed the battery voltage trace that caused the negative power computation. As you measure closer and closer to the battery, the product gets less negative until finally you get a positive product when looking directly across the battery(s), assuming the superimposed oscillation drops to negligible levels.
(snip)

How can that be? How can a negative current value multiplied by a positive battery value be anything but negative?  If the battery trace is always positive then the product will have the same sign as the current, no matter the magnitude of the oscillations, I think.  Can you explain further? Are you seeing battery oscillations that go below the zero reference line? I haven't been able to make those in Tar Baby, no matter how much I drive it.

Ah... are you referring to the sign of the overall "average" power?
Certainly the _average_ power would change sign  as the positive-going areas outweigh the negative going areas in the average... but a negative current sample times a positive voltage sample will still yield a negative power for that sample, yes? This is what I mean when I say that the sign of the power is determined by the sign of the current.... I'm talking about the instantaneous sample power values, with battery oscs that don't go below the zero line.


TinselKoala

Quote from: picowatt on April 22, 2012, 02:23:33 PM
TK,

If you are going to do .99's scope shots at the batteries, I'd like to see, as .99 stated, with all other probes disconnected, one probe ground at the Batt- and then a shot of the scope with the 'scope probe tip placed at each battery terminal starting with the positive terminal of the first battery terminal above Batt-.

This would provide some evidence as to the battery impedance at Fosc, and the effects of the interconnect lead inductance.

You have already done so much, I am reluctant to ask more from you.

PW
Don't worry, as long as I get to eat once in a while to displace the coffee, I'll be fine. A bit "techy" maybe but fine.

I'll do the tests that you and .99 want to see on the batteries in a little while.


No, I'm not going to try to run any more heavy currents through my probe shield leads! That's the kind of thing that causes batteries to catch fire!
;)

Sometimes isolated ground references do come in handy; that's the major advantage of the Fluke-o-Scopes (ScopeMeter 123 and 199). And I sure could use a good highvoltage differential voltage probe.



hartiberlin

Quote from: TinselKoala on April 21, 2012, 09:35:30 PM
I think I  must still not be understanding you.

The "conventional" current, the one that works with calculations and right-hand-rules and all of that, is (thanks to Benjamin Franklin) assigned to be "out" of the positive pole of the source, around the circuit, and "into" the negative pole of the source.
Of course now that we understand things a bit better than Franklin or even Faraday, we know that what really happens is that _electrons_ carrying their negative charges bump into other electrons which bump into other electrons and so the _charge_ is transferred along the conductor from the Negative polarity to the Positive polarity. The electrons themselves bump along rather leisurely, but the _charge_  and whatever signal or power carried by it transfers at the speed of light in the conductor.
So in this video when I am describing the "conventional" and "anti-conventional" current directions, I am referring to the convention that Ben Franklin left us with, so that all the calculations make sense. 
It doesn't really matter, it's just a matter of sign, and so the convention remains with us and continues to confuse freshman EE students every September.

So I'm not sure what you are meaning about the current direction, and I'm still not clear on your use of "AC". Where is the AC in the circuit from the 9v battery (or power supply) to the gate input? It's all DC with a slight ripple on top. If I put a cap in there, nothing will get through, will it? The only place I see true AC is across the CVR... that is, in the main circuit itself when it is oscillating strongly.

So I suppose I'm still not following your meaning.

Sorry, maybe I was just a bit sloppy in my explanation.

Yes, I meant the Benjamin Franklin current direction.

But I did expect the current to flow OUT of the 9 Volts battery .
But as your scope shots show, the current through the 10.3 Ohm resistor comes
from the Q2 transistor, when it conducts.

So just put an ampmeter or a shunt in series with the plus pole of the 9 Volts battery before the 10 K pot in your circuit
and you will probably see, that there is probably not the 155 mA flowing but surely a much lower DC current.

The 155 mA you measured probably only flow from the main battery plus pole over the heater and Q2 via the 10.3 Ohm
resistor and the almost shorted pot to the ground of the main battery.
Thus it showed a POSITIVE voltage on the scope and not a negative voltage, what I suspected...

Hope it gets now clearer..

Regards, Stefan.
Stefan Hartmann, Moderator of the overunity.com forum

hartiberlin

Sorry, I was confused...
My last statement is not right.

I now see, that the current ala Benjamin Franklin direction
flows from the Plus pole of the main battery through heater -> Q2 -> 10.3 Ohm
shunt-> wiper of pot->negative pole of 9 Volts battery.

As the 9 Volts battery is in series via the Rshunt with the main battery and the minus pole
of the 9 Volts battery is the lowest voltage potential, due to Kirchoff´s law also the same 155 mA current
flows in the plus pole of the 9 Volts battery back via Rshunt to the ground of the main battery´s minus pole...

A bit hard to see with this not optimal drawn circuit diagramm...

Regards, Stefan.
Stefan Hartmann, Moderator of the overunity.com forum

TinselKoala

I've been doing inductance measurements of the commercial marked inductors using the resonance method with known capacitors, poly, tantalum and good ceramic ones. I can get within 10 percent of the marked value.

Here's how I do it: I hook the cap and the inductor in parallel with the scope and the function generator set to sine wave output, then I sweep the FG until the resonance is found by the trace amplitude peaking. Then I very carefully tune the FG so that I am as sure as I can be that I am at the peak amplitude oscillation. Then I look at the FG's frequency using its synch output going to the Philips 6676 counter... that has a crystal timebase oscillator in it and always passes its selfcheck to eight sig digs. Then I "do the math" (tm RA).
1/L=C*((2*pi*f)^2) with L in H, C in F, and f in Hz.  Sounds like some strange crew of freaks on parade... but it works out if you don't forget where the decimal points go.

So anyway, there is something about those resistors that the ProsKit meter just doesn't like. The resonance method agrees with the meter for the commercial inductances, but not for the power resistors used in the CVR. Since I confirmed the resonance reading with a couple of different capacitors and resonant frequencies, I guess I have no choice but to believe it, and call the inductance of the 1 Ohm, 10 Watt ordinary power resistors as 0.5 microHenry, which means the stack of 4 parallel would be 125 nanoHenry (although I have not measured this by resonance directly....yet.)

So the inductive reactance at 2 MHz will be XL = 2pi*f*L, or 2*3.142*0.000000125*2,000,000  == about 1.6 Ohms. Since the DC resistance is 0.25 Ohms, the total AC impedance at 2 MHz will be Z= sqrt(R^2 + XL^2), or sqrt(1.6^2 + 0.25^2) == also about 1.6 Ohms, closer to 1.62 Ohms.

I've reflected this new value in the updated spreadsheet I posted a bit earlier, and completed the power calculations using it. The mean power is _NEGATIVE_ about 13 Watts or so.

I'd still like to see another measurement of the Ainslie resistors made though, by the resonance method if possible... which I doubt.