Is joule thief circuit gets overunity?

[conradelektro]

I have been using the germanium diode method in my most recent efforts.  It is not easy to get the energy flowing in the direction that you want, and to keep it going that way.  In other words, it is being powered initially by the supercap but when you feed back to the cap, you need to keep the power from flowing in the reverse direction.  I came up with this method not knowing any other way to do this without using some electronic switching mechanism which uses more energy.  So far so good but, I need more time to play with it.

Bill

@Bill: Why do you want to use a supercap, it has much more leakage current than a Tantalum capacitor (Tantalum 0.5 µA, supercap 0.32 mA)?

If the circuit is OU, a large capacitor is not needed, 0.1 µF should be enough to provide a reserve for little fluctuations.

I do not know how fast a supercap accepts a charge, but the Tantalum capacitor would react in the GigaHz range.

In a Joule Thief circuit which is a little bit OU the LED would shine a bit brighter than expected when it is looped back to a capacitor. The Voltage would self adjust to about 3.2 Volt (in case of a white LED). Of course this is only speculation, because nobody ever showed a OU Joule Thief.

0.1 µF Tantalum Capacitor 50 V, Leakage  Current 0.5 µA
http://uk.mouser.com/ProductDetail/AVX/TAP104K050SCS/?qs=sGAEpiMZZMtZ1n0r9vR22X84dCiTW0Oj4kJEQkHEXDQ%3d
Supercapacitors / Ultracapacitors 350F 2.7V , Leakage Current 0.32 mA
http://uk.mouser.com/ProductDetail/Cornell-Dubilier/CDLC351E2R7T11/?qs=sGAEpiMZZMtKtLvoHD9hC99dCDgohi7U

Greetings, Conrad

[xee2]

What you end up with is a real time display of the average power.

poynt99
Yes. But unless the measurements are across a pure resistive load I think the average power will be of both real and reactive power combined since the scope measures total voltage and total current from what every power is there (real and reactive). I may be missing something, but I do not see how to remove the reactive power except by using a resistive load or computing the dot product. I am not saying you are wrong, I have not spent a lot of time studying this. So this is just a first impression opinion.

[MileHigh]

Xee2:

The scope will measure positive power when the supply source is supplying power to the load.  If the load is reactive, it will store some of that power and then push it back to the supply source.  That will be measured as negative power.

Assuming the DSO that can do mathematical operations then the net measured average power will simply be the positive power minus the negative power averaged over multiple cycles.

So the measuring DSO does not care at all what the load looks like.  The load is just a black box that can do anything.

MileHigh

[poynt99]

poynt99
Yes. But unless the measurements are across a pure resistive load I think the average power will be of both real and reactive power combined since the scope measures total voltage and total current from what every power is there (real and reactive). I may be missing something, [snip]

The scope does not measure TOTAL voltage or current, it measures on a sample by sample basis.

The scope measures only what is there, agreed? Forget about phase and real vs. reactive power. It does not matter what kind of load is being measured, by default the scope only computes REAL power when we use this method.

The picture is a wave form (not 1 full cycle) in which we will look at only 6 samples.

Sample1 = -100V x +10.5A = -1050W
Sample2 = -323V x -0.73A = +235.8W
Sample3 = +20.9V x +0.35A = +7.3W
Sample4 = +15V x +0.21A = +3.15W
Sample5 = -12.6V x -0.121A = +1.52W
Sample6 = +17.9V x -0.050A = -0.9W

Next we would take the average of all 6 measurements: -803.13W/6 = -133.85W

Of course the scope is going to sample at a  much higher rate than this, so it will get a more accurate measurement of the average power over all.

[xee2]

poynt99
Supose at a given instant of time the reacitve power is 10 volts and 0 amps while the real power is 1 volt and 1 amp. I think the scope will read 11 volts and 1 amp (11 watts) while the real power is actually only 1 watt. Am I doing this wrong?