PhysicsProf Steven E. Jones circuit shows 8x overunity ?

[poynt99]

AS I SAID,  test requires that you replace the LED with a 1N4148 for the "final" test, to permit comparison.

This does not solve the problem of having a baseline for comparison. What criteria would there now be for the circuit operating?

With your current criteria, I am certain I can beat everyone. I can build a circuit that will drop from 2.5V to 2.4V in 10 seconds.....do I win?

.99

[JouleSeeker]

This does not solve the problem of having a baseline for comparison. What criteria would there now be for the circuit operating?

With your current criteria, I am certain I can beat everyone. I can build a circuit that will drop from 2.5V to 2.4V in 10 seconds.....do I win?

.99

Required is a JT-type circuit, with at least one transistor and one bifilar-wound toroid and it must light an LED to observable brightness in a lighted room (all as I said before IIRC) before the final test with a 1N4148 replacing the LED.

And the winner will check the FREQUENCY also... 
I'm going to add a provisional stipulation that the frequency of input-power pulses be above 200 Hz, to prevent pulsing "once in a blue moon" and also to avoid 60-Hz pick-up.  If you can persuade me this is not a "fair" stipulation, I'm listening.

The winning device cannot be poaching from the grid. 

How's that?  and I do appreciate your checking the criteria, .99.  Rigor without rancor.

[poynt99]

Required is a JT-type circuit, with at least one transistor and one bifilar-wound toroid and it must light an LED to observable brightness in a lighted room (all as I said before IIRC) before the final test with a 1N4148 replacing the LED.

And the winner will check the FREQUENCY also... 
The winning device cannot be poaching from the grid; that may require some additional tests.

How's that?  and I do appreciate your checking the criteria, .99.  Rigor without rancor.

No problem.

So this contest excludes anyone that does not have an oscilloscope.

Best to measure and use output power then. You need a baseline Pout and baseline Vbat.

Pout:

1) remove all output diodes/LEDs, and replace with only a fixed non-inductive resistor, of some chosen value that all will use.

2) measure Vout across Rout with scope probes (very close to body of resistor).

3) Have the scope compute rms of measured voltage.

4) Pout = Vrms²/Rout

5) Don't Panic! This works and has been proven.

.99

[JouleSeeker]

No problem.

So this contest excludes anyone that does not have an oscilloscope.

Best to measure and use output power then. You need a baseline Pout and baseline Vbat.

Pout:

1) remove all output diodes/LEDs, and replace with only a fixed non-inductive resistor, of some chosen value that all will use.

2) measure Vout across Rout with scope probes (very close to body of resistor).

3) Have the scope compute rms of measured voltage.

4) Pout = Vrms²/Rout

5) Don't Panic! This works and has been proven.

.99

It's true that some means of getting the frequency would be required at the end, if only to preclude the question of a prosaic 60-Hz source.   but you know, a guy can get a decent DSO for about $300 these days, or go to the local high school and check the frequency.  There are three guys that I know of within a few blocks of my house with oscilloscopes, Abe, Les and Bob.

Now, with regard to output power -- you're changing the system dramatically here and it's not the same system when you do this:

1) remove all output diodes/LEDs, and replace with only a fixed non-inductive resistor, of some chosen value that all will use.

Are you kidding?  You're drastically altering the dynamics of the device!

What I'm proposing is to replace each output resistor(s) with a resistor(s) whose heat-rise is calibrated for power measurement.  That way we actually leave the diodes and everything in place, and don't destroy the system in an attempt to measure it.  (Misses the power dissipated in the diode still.)

What you're suggesting is like throwing a sledge-hammer at a finely-tuned watch in a (vain) effort to see what's inside, how it works.

[poynt99]

Now, with regard to output power -- you're changing the system dramatically here and it's not the same system when you do this:
Are you kidding?  You're drastically altering the dynamics of the device!

What I'm proposing is to replace each output resistor(s) with a resistor(s) whose heat-rise is calibrated for power measurement.  That way we actually leave the diodes and everything in place, and don't destroy the system in an attempt to measure it.

What you're suggesting is like throwing a sledge-hammer at a finely-tuned watch in a (vain) effort to see what's inside, how it works.

Then I've misinterpreted this response of yours?

In any case, no I would not make a "specific requirement to have a diode or LED in the output" if you're going to measure Pout.  The LED in the output was for the cap/stopwatch test for Pinput, with the LED glowing... something any one can do without a scope.

That is why I asked, so that it could be simplified. If removing the diode changes the circuit too much and invalidates the proposed contest, then so be it. It was simply a suggestion and I was trying to help out.

.99