Is joule thief circuit gets overunity?

[ltseung888]

Much Thanks to TinselKoala

TinselKoala pointed me to the correct use of Instantaneous Voltage on the Oscilloscope - Use DC Coupling.  I used this technique to greatly improve the resonance tuning process.

At BSI, we used to have a demonstration to show how to cut electricity bills.  The circuit turns 30 LEDs ON for 15 seconds and OFF for 2 minutes and repeats.  The values selected were more or less random with the human eye judging the brightness.

With the new tuning technique, we can detect the range that the circuit is in overunity mode.  It turned out that the frequency range from about 3 KHz to 24 KHz fits such description.  If we want the circuit to remain in this range and still show acceptable brightness, the best time is 3 Seconds ON and 10 minutes OFF.  We can use the oscilloscope instead of the human eye.

Resonance tuning and obtaining overunity is now a piece of cake.  Thanks to the Almighty and TinselKoala.

[TinselKoala]

Lawrence: the positive lead of your power supply is connected. Disconnect that and your "forever light" will fade normally just as mine does. Maybe slower, I haven't spent any time optimising my circuit variant... I just built it last night !! But I've been able to discover a few things and explain them to myself to my _preliminary_ satisfaction.

As I have emphasized in my recent videos, you _cannot_ have any  line connected device at all connected to this circuit or its variants or you will fool yourself. I mean by the ground lead, the positive lead or any lead at all.

Even a scope probe or reference will screw up your circuit _unless_ the scope is fully isolated. I don't know if this is the case for your scope. Are the probe reference leads (cable shields, outer conductors) connected together inside the scope, and is the internal chassis grounded to the mains ground back thru the line cord? Does it use a 2-prong "wallwart" style power supply, or is the power supply internal?


And please don't mention me and the Almighty in the same breath. He and I don't get along too well. I don't believe in him..... and I don't think he believes in me, either.

[MileHigh]

Fausto:

Let go slow: you said: energy goes from the battery to the inductor. Good. Inductor discharges that stored energy back to us (or another system, whatever). Good, all the same, what goes in, goes out.


Is that correct? Good, now, think again, battery give 10 watts of power in a second, so 10 joules. Great. Inductor discharges and gives back (minus losses) 10 joules or less back.


Now, this process heats up the resistive wire TWICE, not once like a regular resistor. Can you explain that? How come a resistor for the same 10 joules of input power ONLY heat once ? in others words a regular resistor will give you only an equivalent amount of 10 joules of heat, while the SAME resistance in a inductor will give you more than that in heat.


I would love to hear how you going to explain that out!

You are not correct here.  So we know any real-world inductor is made of wire and has resistance.  So what we can easily do is model the real-world inductor as an ideal inductor with zero resistance in the wire in series with a small resistor.

So what happens when we energize this inductor?   Let's say it takes 5 seconds to energize the inductor.  So after 5 seconds some of the supplied battery energy was used to create the magnetic field to energize the ideal inductor.  At the same time during the 5 seconds some of the battery energy was burned off in the resistor.

After five seconds the only thing that is happening is that battery power is being burned off in the resistor.  Also after five seconds there is energy stored in the inductor.  That energy came from the battery.  Important:  Note that this energy stored in the inductor has not gone anywhere else, i.e.; it has not passed through a resistor.

Then after some time the inductor discharges its stored energy.  This stored energy will discharge through the resistance of the inductor itself, and some sort of a load resistance.   Therefore this energy is only discharged ONCE, not twice.

Here is the energy path:  [energy in battery] -> [energy stored in inductor] -> [energy dissipated in inductor internal resistance and load resistance]

There is no 'magic' in the inductor that allows the same energy to get used twice.

MileHigh

[TinselKoala]

@Lawrence: I take it that you are charging the cap for the short interval , then running on the stored energy for the longer interval. Here's a tip: the capacitor will charge up to very near the battery or supply voltage quickly. Once it gets to that voltage (actually it approaches it asymptotically) you can't put any more charge on the cap!! So leaving it connected to the power supply beyond that point is just wasting power from the supply. You should be able to rig a system that shuts off charging just at the time the cap reaches the PS voltage setting and no longer; this will improve (lower) your input power figure, I think. You could use a simple voltage comparator made from , eg, a 741 op amp, comparing the voltage from the supply, dropped by a quarter of a volt or so by a resistor divider network, to the voltage on the cap. When the comparator flips, have it trigger another switching stage to cut off the cap from the power supply.

[MileHigh]

Fausto:

I want to see how any one can explain this to not be overunity. 2000 pounds doing work, pumping water is NOT fake and it is not running on current, since ALL the batteries are in series.


If this is not the power of induction at work, I don't know what is power than.

Here is where you have to analyze things without any emotion.  First of all, the 2000 pounds is not doing any work at all.  The 2000 pounds is the dead weight of the rotor of Joe's big motor (or it is the weight of the entire motor assembly, I can't remember.)

What do you mean the motor is not running on current?  How can you say that after all these years?  You know electrical power is voltage times current.  There is no such thing as a motor running on voltage only and there is no such thing as a motor running on current only.  Those are nonsensical concepts.

The "power of induction" is not doing any work and there is no power associated with induction.  Inductance can only store and release energy provided by an external power source.

So, we know that Joe is using what?  I think it's about 170 batteries in series?

I challenge you Fausto to explain how what Joe Newman demos is perfectly explainable with his 2000 pound rotor and the 170 AA batteries.   You are claiming that is overunity and I am challenging you to explain how it is in fact under unity.

MileHigh