Author Topic: Is joule thief circuit gets overunity? (Read 605275 times)

poynt99 · « **Reply #300 on:** March 03, 2013, 07:05:11 PM »

Hi Bill.

Yes, a self-looped arrangement should prove if the device is OU or not.

I've not seen any proposals on how to do it exactly though, have you?

xee2 · « **Reply #301 on:** March 03, 2013, 08:14:52 PM »

Quote from: poynt99 on March 03, 2013, 04:29:07 PM

To be precise, allow me to rephrase the above;
In order for real power to manifest and be measured as real power, the current and voltage vectors must be in phase.
Keeping in mind that the stated method is to acquire instantaneous samples of current and voltage, what happens to the product of v(t) x i(t) if they are 90º out of phase? Is the p(t) for these samples not 0W?

This method inherently compensates for the samples where the phase is skewed. In other words, it takes care of the fact that the current and voltage may not always be in-phase. Only those samples that are in phase, or partially in phase will result in a real power product, and if the wave forms are periodic (in 99% of the cases they are), then taking the average of this product will yield an accurate measurement of the over-all average power.

poynt99
If 90 degrees out of phase the real power will be 0 watts. No problem. However when not in phase and not 90 degrees out of phase, the power will be part real and part reactive. Would not the correct method be to compute the dot product between the two complex number values for voltage and current? That way only the in-phase parts of the vectors would be used. I am not real good at math, so I may have this wrong, but I think that would be correct.

In any case, I am not trying to re-design the experiment. I am just trying to understand how the output power is being measured. Thanks for your help.

Pirate88179 · « **Reply #302 on:** March 03, 2013, 08:21:28 PM »

Quote from: poynt99 on March 03, 2013, 07:05:11 PM

Hi Bill.

Yes, a self-looped arrangement should prove if the device is OU or not.

I've not seen any proposals on how to do it exactly though, have you?

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

poynt99 · « **Reply #303 on:** March 03, 2013, 08:44:29 PM »

xee2,

Try to read what I wrote several more times until it makes sense. I think you are missing some salient points.

The beauty of the method is that it computes only the real power, no matter what the wave forms are nor what the phase relationships are.

Picture two wave forms, one above the other, synchronized in time. One is the voltage, and one is the current. Pick a number of points along the horizontal time axis and multiply the voltage and current values together at each of those points.

Now, with our method, the oscilloscope is doing precisely this but at a fixed sample rate. With the math function in the scope you multiply these two values together, then apply a running average to the result. What you end up with is a real time display of the average power.

NickZ · « **Reply #304 on:** March 03, 2013, 09:06:53 PM »

Bill and All:
I've been asking for this one simple proof of concept, using an AA, or cap, to test the discharge rate, so that it can also be further verified by us here as well as on other forums. Even by those of us that don't have any test instruments. Just a cap or battery on the circuit. I've never ever heard of any reply back, only scope shots, after scope shots, after even more scope shots. With no self runner (still presently running), to show for it all though, as yet. Hopefully this will all be verified, some time soon.

conradelektro · « **Reply #305 on:** March 03, 2013, 09:11:43 PM »

Quote from: Pirate88179 on March 03, 2013, 08:21:28 PM

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 · « **Reply #306 on:** March 03, 2013, 10:18:40 PM »

Quote from: poynt99 on March 03, 2013, 08:44:29 PM

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 · « **Reply #307 on:** March 03, 2013, 10:49:49 PM »

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 · « **Reply #308 on:** March 03, 2013, 11:27:19 PM »

Quote from: xee2 on March 03, 2013, 10:18:40 PM

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 · « **Reply #309 on:** March 03, 2013, 11:43:37 PM »

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?

poynt99 · « **Reply #310 on:** March 04, 2013, 12:01:57 AM »

Quote from: xee2 on March 03, 2013, 11:43:37 PM

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?

Yes you are doing this wrong.

Look at the scope traces in my above post. How can you have more than one value for voltage or current at any particular instant of time?

You can't. There is only one voltage and one current. Their product is the power at that instant of time.

xee2 · « **Reply #311 on:** March 04, 2013, 12:37:55 AM »

Quote from: poynt99 on March 04, 2013, 12:01:57 AM

How can you have more than one value for voltage or current at any particular instant of time?

poynt99
In the example there is only one value of current (1 amp) and one value of voltage (11 volts) on the scope. Those values are the total current and total voltage the scope sees across the non-resistive load. For AC circuits, when the load is not a pure resistance the reactive components can add voltage without adding real power. Like I said, I have not thought this all through. But this is one of the things that I think needs more thought.

poynt99 · « **Reply #312 on:** March 04, 2013, 02:05:16 AM »

Quote from: xee2 on March 04, 2013, 12:37:55 AM

poynt99
In the example there is only one value of current (1 amp) and one value of voltage (11 volts) on the scope. Those values are the total current and total voltage the scope sees across the non-resistive load. For AC circuits, when the load is not a pure resistance the reactive components can add voltage without adding real power. Like I said, I have not thought this all through. But this is one of the things that I think needs more thought.

In your example your total voltage and current consisted of two values added together. I am trying to tell you that there is no such thing. The values are simply what they are, nothing more nothing less.

Regardless if the voltage across a load is increased due to its reactance, when that voltage is multiplied against the current at that instant, the p(t) might be quite small, even though the voltage is higher.

Anyway, I have provided all the information you need, in a clear manner. It is up to you to study and try to understand it.

xee2 · « **Reply #313 on:** March 04, 2013, 03:39:04 AM »

poynt99
Thanks for your help. You seem to be on top of this. I was always taught to measure real power using a resistor, so I am always suspicious when that is not done. I am sure that if there are any problems you will spot them.

ltseung888 · « **Reply #314 on:** March 04, 2013, 04:02:41 AM »

Some pleasaant surprise first. This new batch from Shenzhen used toroids that are machine wire-wounded. The COP on board 70 jumped to -27.32. I shall check on other boards. With more boards at my disposal, I can try the two or more stage JT. (Output of one becomes Input to another) Preliminary tests were encouraging. More on that later.

Now see the new scope shots - not all JTs are equal even though they all light up the LED!!!

Not all JTs can demonstrate OU - once they can, I call them FLEET... That is why JT research is interesting...

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