PhysicsProf Steven E. Jones circuit shows 8x overunity ?

[nightwind]

Hi Forest et All:

My appoligies, I was not aware of the earlier thread on this subject.  Not sure I understand your response,  we may be agreeing with each other.  It takes energy to create either reactive power or real power.  Converting from one to another doesn't create new energy when drawing from the same energy source.  So nothing is gained from the Canadian device except the money normally paid to the power company.  The power co. still has to generate the original energy.  They seem to be the looser here.  Hope I am wrong and some new energy is being created.   The worlds energy problems would be solved.   The Canadian idea is very creative, but not sure what it could lead to for overunity?

You absolutely correct about utility metering.  Their meters only record real power on residences, but they do record power factor (reactive power) on industrial and large commercial facilities.  Funny thing is that the more efficient a homeowners electric motors become, the more visible they are to the power company's meter.

[albertouno]

Greetings   JouleSeeker:

Per your request,  I completed several tests on your oscillator circuit.   First, I tested it with a resistive load of 1000 ohms.  The base resistor was 55k.  A 2.67 volt battery source was used.  The input power was primarily based on a 3.6 ma DC input current present.  Pin was 9.6 mw.   The resistor output was a more complex pulse waveform, which was calculated to yield about 1.4 mw.  The LED was bright during the test, and was estimated to consume  about 4 mw.  This results in a gain for the resistor load of 0.113 or 11.3%.  Counting the LED, the GAIN is  0.47 or 47%.
Going on to the next test, with a second  battery to be charged, the input power remained about the same.   I decided , because of time constraints, to do the “Alberto Capacitor Estimation test” for Pout.  This was described earlier where a depleted battery is charged, and then discharged into a fixed resistor to estimate Pout.  The discharge curve is compared to an ideal capacitor discharge, and energy stored on the battery estimated.  The charging time was 1.5 hours or 5400 seconds.  A copy of the discharge curves is attached, and a capacitor equivalent of 10.3 farads was used.  The curves are not as close as I have observed in the past, possibly due to over discharging the battery before the test,  but  should enable a “ballpark” estimate of Pout to be made.  The results for Pout for the battery was 8.41 joules, Pout for the LED of 21.6 joules, and total Pin of 51.7 joules.  This results in a GAIN of  0.163 for the battery, and 0.581 (58.1%) for the total circuit.
In summary, this circuit did not exhibit overunity, which is disappointing.  Hope this info is helpful, and I will be very interested in your battery test results. 

Alberto

[nul-points]

Nul-Points:
[...]
Is it better to do the test with depleted batteries for this reason?  Does just the voltage difference accurately measure the energy input to a more fully charged battery?

Alberto

Alberto:

apologies, i forgot you mentioned earlier that you've been using alkali cells - i don't have any experience trying to recharge these

i find it useful to use very depleted NiMH & NiCd cells when i'm looking for evidence of charging (or self-charging) because the cell (or battery) voltage slope is much larger in this condition and so it is easier to observe a change in charge (ie. the voltage changes more per unit charge than it does in the 'working range' - as seen in my earlier 'discharge' graph of the two)

however, i only use this method as a 'ballpark' indication of charging/discharging

i found that it is possible to approximate actual charge when operating within the working range (because the slope is very approximately linear), but it's necessary to have a reasonably good measurement of the equivalent load impedance (or power draw directly)

i used the actual cell discharge graph (with known load resistance) when i needed a more accurate indication of 'full-charge' to compare with 'actual charge' achieved by one of my circuits under test

i hope this helps!


i'm intrigued by your approximation of the cell charge using the capacitor equation - a neat approach for low-powered experiments!!


Steven:

yes, the cell terminal voltage graph on-load with a known resistor does give apparently conflicting information when reaching the fully discharged level - by 'bouncing back' when disconnected

but i think the explanation has to do with the equivalent 'internal resistance' of the loaded cell: i believe that unlike the capacitor (which reflects its level of charge directly in its terminal voltage), the cell voltage is mainly dictated by the chemical equation in the cell

i believe that when we discharge a cell, the re-distribution of the ions causes the internal resistance to increase - so we mostly see the drop in terminal voltage when the cell is on a reasonably significant load - and the ratio of the loaded to unloaded voltage decreases with increase in discharge

that would explain the effect we see of the terminal voltage rising back as we unload the cell after a good amount of discharge

co-incidentally, i happened to have recorded exactly this effect in the 2nd 'discharge' graph above - you can see that after discharging the 8.4V battery down to just below 6V on-load, i remove the load before stopping the datalogging and the unloaded terminal voltage returns to approx 8V within a few minutes!

at this level of charge, it's the loaded terminal voltage which gives a true indication of charge level - which is why for better accuracy we ought to measure a given cell with a known resistive load always to get a comparative value for the 'real' state of charge

however, within the working range (for light to medium load tests) it's probably acceptable to use the unloaded terminal voltage as a reasonable measure of charge - allowing for the cell to 'settle back' first, if necessary, following some power draw from it


i'm away on leave soon, so i'm going to post an update below of the latest data for my SJ1 variant circuit using 2x NiMHs in series to charge 2x NiMHs in parallel with occasional swapping of the two pairs

you can see that although i'm swapping the cell pairs when the supply pair discharge to approx 1.255V, the max voltage reached by the charging pair has been steadily decreasing

the experiment has been running for over 850 hours continuously now (just over a month)

so - no evidence so far that the generic 'Tesla Switch' approach maintains the total charge level of the batteries

if this behaviour continues, then i'll move on to the next test, which is to compare the length of time taken to discharge the supply with a similar 'work' load but without charging another battery pair

i'll have to stop this test run whilst i'm away, but i'll continue again when i return - it's almost at the point where i'll need to lower the discharge setpoint chosen for the swap


hope your experiments go well guys

all the best
np


http://docsfreelunch.blogspot.com

[JouleSeeker]

Greetings   JouleSeeker:

Per your request,  I completed several tests on your oscillator circuit.   First, I tested it with a resistive load of 1000 ohms.  The base resistor was 55k.  A 2.67 volt battery source was used.  The input power was primarily based on a 3.6 ma DC input current present.  Pin was 9.6 mw.   The resistor output was a more complex pulse waveform, which was calculated to yield about 1.4 mw.  The LED was bright during the test, and was estimated to consume  about 4 mw. This results in a gain for the resistor load of 0.113 or 11.3%.  Counting the LED, the GAIN is  0.47 or 47%.

How do you estimate the Poutput?  sorry, what you have said is just insufficient to tell how accurate your estimates are.

Going on to the next test, with a second  battery to be charged, the input power remained about the same.   I decided , because of time constraints, to do the “Alberto Capacitor Estimation test” for Pout.  This was described earlier where a depleted battery is charged, and then discharged into a fixed resistor to estimate Pout.  The discharge curve is compared to an ideal capacitor discharge, and energy stored on the battery estimated.  The charging time was 1.5 hours or 5400 seconds.  A copy of the discharge curves is attached, and a capacitor equivalent of 10.3 farads was used.  The curves are not as close as I have observed in the past, possibly due to over discharging the battery before the test,  but  should enable a “ballpark” estimate of Pout to be made. The results for Pout for the battery was 8.41 joules, Pout for the LED of 21.6 joules, and total Pin of 51.7 joules.  This results in a GAIN of  0.163 for the battery, and 0.581 (58.1%) for the total circuit.
In summary, this circuit did not exhibit overunity, which is disappointing.  Hope this info is helpful, and I will be very interested in your battery test results. 

Alberto

A "ballpark estimate" is not much to go on.  Again, you see as we have discussed at some length above, the difficulty in measuring Pout -- also the importance of optimizing the circuit.

Thanks for your comments also, NP.

Now for my latest results, where I actually charge 2 batteries on the output circuit as I describe above, and measure the voltage Vin and Vout.  Also, I use a much higher base R (Rb) about 0.9 mega-ohms as I described above, and I've measured it at 0.91Mohms.

The results are interesting, I suppose.  Takes a lot of patience.  I seem to be running into the problems you mentioned, NP, with battery voltages relaxing etc.  But I let the batteries "rest" for hours before beginning the run.  Because of the "relaxation" problem with rechargeable batteries, I'm thinking about going back to capacitors as I will describe in a subsequent post.  But first the data using batteries in the circuit -- schematic above.

10 Oct 2011:
@20h00, Vinput = 1.934V (two AA's in series),  Voutput = 1.031V (on the AA's in parallel, charging).


@22h048, Vinput = 1.943V,  Voutput = 1.0340V.
So both voltages actually went UP a little, per the Keithley DMM.

The next morning, 11/11/11
@8h00, Vinput = 1.9447V,  Voutput = 1.03460V

@10h44, Vinput = 1.9438V,  Voutput = 1.03464V
Vinput is down, Voutput continues up.

@0:15 (this morning 12 Oct), Vinput = 1.9466V,  Voutput = 1.0356V

@6h21 (this morning), Vinput = 1.9441V,  Voutput = 1.0362V

At this point, with not enough sleep (yawn), I accidentally touched the DMM probe wires in such a way as to discharge the input batteries for about a second; this ends the run.


Observations:  Vinput seems to go up and down, but does not vary much.  Clearly the input power is very small...  The output power is also small with Rb = 0.91Mohms, but the voltage on the two rechargeable AA's on the output climbs rather steadily... and slowly.   

The efficiency?  Voutput rose from 1.031V to 1.0362V so the 2 AA's on the output leg did charge as expected given the direction of the diode in the LED, while the Vinput was roughly the same... but I don't think this is particularly conclusive due to battery-relaxation questions.

Conclusion:  this run was about 46 hours.  This is too long for rapid progress -- to change variables such as Rb, or the transistor or the capacitor, and to see results.   I will go back to capacitors and see what I can learn (in the absence of a calorimeter... sigh...)

[JouleSeeker]

   I have obtained some fairly high-quality, low-leakage caps and have done some encouraging tests today using these.

By using a charged 60,000 uF (from now on, 60mF) cap and connecting this to a 10mF cap, I find:

Vinitial = 1.97 V, V final = 1.78V and
Vinitial = 2.54 V, V final = 2.28V in another quick test -- both consistent with theory.  Using E = 1/2 CV**2, initially and finally, we find that the efficiency is about 93% in both cases.

  We EXPECT to lose some energy in such a process, but note that with 60mF initially and 70mF finally, we lose only about 7% of the initial energy.  If I had used 10mF initially and 20mF finally (that is, connecting a charged 10mF cap to another (uncharged) 10mF cap), we would have lost half the initial energy -- this follows from charge conservation and E = 1/2 CV**2.

The caps also "leak", losing energy by leakage through the dielectric (no doubt).   So far, I have found  two caps that together leak at almost exactly the same rate as my 60mF cap, so I can keep track of the leakage that way.  I will input into the 10mF cap from my circuit, in place of the AA's on the output leg of the circuit, so as to lessen "theory" losses as described above. Tthe leakage on that cap is remarkably small.

Test results later, looks like tomorrow given duties today.