PhysicsProf Steven E. Jones circuit shows 8x overunity ?

[nul-points]

ok, i find this interesting - i just replaced the LED with an OA93 (germanium diode)

the discharge time decreased to 359 seconds (5min 59sec)

Pav = 2.126/359 = 5.9uW


i wasn't ready for that!   maybe the 1N914 discharge time was correct after all?


perhaps there's a good reason why someone mentioned earlier about using several LEDs!!! 

[jmmac]

You are correct, i didn't recheck my calculations. I get the same values you mentioned. Sorry about that.

For the setup where i charge the capacitor, the diode anode is connected to the ground, the cathode is connected to the capacitor plus and the capacitor minus is connected to the transistor emitter.

I disagree with you regarding the use of the 1N4148 in order to compare results. The only thing it achieves is to make sure everyone has the same diode (voltage drop). It does nothing to solve the problem of comparing led brightness (power consumption).

My led is so sensitive that i can light it up connecting one end to the ground of my scope and touching the other end with my finger!

Thank you for doing the replication AND the cap/time method test, Jaime!

1.  "I'm not sure we can conclude much from these results and it's very difficult to compare results from different experimenters since there is no way to measure the led brightness. If i use the diode i get similar discharge times (a little longer)." 
  The use of a common diode like the 1N4148 allows us to make direct comparisons between different experimenters; we don't have to measure the LED brightness!
   @Xee2-  your suggestion of another "standard" diode Rk44 is noted; let me try it out experimentally and see how it differs.  The 1N4148 is so common, that provides some advantage for a "standard input-power test" .

2.  I need to check your numbers; pls spell out the algebra in the future -- that would help to see where there may be a discrepancy.  I get:

Ecap = 1/2 C V**2
so
Pinput = 1/2 C (Vstart**2 - Vstop**2)/time

With your numbers,
Pinput = 1/2 C (2.48**2 - 1.5**2)/time
          = 1/2 * 10mF * (6.15-2.25)/time
          = 19.5mJ/104s
          = 0.188mW = 188 uW  (not 37uW -- pls re-check)

For your longer discharge time, 296 s, I get
Pinput = 19.5mJ/296 s =66uW

Please re-check the numbers, would you?  we should agree on the calculated average input power.

Also, I note that with a charging cap on the output leg I typically found n~0.5-0.7 so we're not far different.  Of course, this method for Pout neglects the power dissipated in the LED (and other components), but is a first estimate (as I noted earlier).  Thanks for doing this test.

Important:  what was the POLARITY on this output cap (compared with the direction of the LED)?

[xee2]

Of course, this method for Pout neglects the power dissipated in the LED (and other components)...

input  = 0.5*(10e-6)*(2.48^2 - 1.5^2)/104 = 1.875e-7 = 188 uW

I can not calculate output power since end voltage was not given.

LED should not be used when measuring output power. The LED should be replaced with rectifier diode. The power lost in the rectifier diode is I^2 x R where R is the on resistance of the diode (very small for RK44).

I do not think the waveform matters, since all of the energy going to output is from energy stored in magnetic fields of coils. How this energy dissipates is not critical, only the "total amount" of energy matters.

If all of the energy dissipated in all of the components is added up it should theoretically always be exactly equal to the total input energy (n=1.00). Only the load power needs to be measured, the rest is irrelevant. When charging a capacitor, the capacitor is the load.

[xee2]

My led is so sensitive that i can light it up connecting one end to the ground of my scope and touching the other end with my finger!

This should not happen. That means the scope ground is sourcing high frequency current and voltage.

[NickZ]

  Possibly the adding of more than one led may help to obtain the magic numbers.  Kooler (my Hero) used three leds on one of his 5 month long test units.  Worth a try. 
Resonance has everything to do with this, but a steady voltage also needs to be maintained, as the battery is part of the draw in these circuits. The oscillator is also charging the battery or the cap to a degree. In most cases, without the use of the resonance factor the feed-back charge is not of a high enough degree to keep the voltage from dropping. So, it is a flow balancing act to keep it going strong. One volt input,  8 volts back to the battery.  Similar to the Joule Ringer.  I don't belief that anyone has really hit the nail on the head yet.  But, I do hope to see it done.
   Good luck with your tests... we may all learn something from them.