Is joule thief circuit gets overunity?

[xee2]

ltseung888
Perhaps this is a better explaination of the problem:
When you are measuring volts times amps you are measuring both real and reactive power combined. A battery delivers real power. Therefore, to measure efficiency of a circuit, the output power measured must also be real power. Real power is measured using a resistive load. Thus the output power must be measured using a resistor.

[poynt99]

Real power is measured using a resistive load. Thus the output power must be measured using a resistor.

Real power measured in a load (PL) of any kind can be performed by taking the average of the instantaneous voltage across the load times the current through that load. It is critical however to ensure that the current measurement is across a pure resistance such as a high quality CSR resistor.

PL(AVG) = AVG[v(t) x i(t)], where i(t) might be obtained from the instantaneous voltage across a known pure current-measuring resistance; i(t)=v(t)/Rcsr.

[xee2]

Hi poynt99
Thanks for the explaination. To measure the real component of power, as I recall, the voltage vector and the current vector need to be in phase with each other. If the voltage vector is measured across a reactive load I do not see how averaging the values will produce the correct answer. I will have to think about that for a while. However, this is certainly true when both voltage and current are measured across the same resistor.

[poynt99]

To measure the real component of power, as I recall, the voltage vector and the current vector need to be in phase with each other.

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.

If the voltage vector is measured across a reactive load I do not see how averaging the values will produce the correct answer.

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.

[Pirate88179]

xee2 and .99:

We in the JT topic discussed long ago that if there was any more power in than out, a self-runner was the proof.  Especially using a supercap, which I believe Lawrence is, would this not be the best proof of even a tiny bit of more P in than out?  It would continue to run right?

Once that is achieved, then he, or someone, can attempt to place values on how much extra is there.

I do not think that his circuits are any where near 100% and certainly not above.  Looping one to a supercap would prove this very quickly.

Just my thoughts.

Bill