Is joule thief circuit gets overunity?

[TinselKoala]

Furthermore.... it is important to realize that a bench power supply very often has its NEG output lead connected to the chassis ground, which is connected to the line cord ground, which is connected to the line cord grounds of other instruments like the scopes, which is connected to their chassis grounds, which are connected to the scope probe "grounds" or reference leads. This means that if you are using a bench power supply, your scope probe grounds may be altering the power input to the circuit if they are connected anywhere else other than the exact same place the power supply negative lead is connected.

This means that in the circuit as diagrammed above you CANNOT use a ground-connected power supply as a substitute for the battery, because its negative is connected back to the grounds of the scope probes which are connected in a different place--- on the other side of the input CVR, bypassing it.

[Rodelu]

Hey Lawrence, why are you SO stubborn?

Why you keep pushing your 'guaranteed OU' results base only on oscilloscope readings, when in post 450 it was clear that you don't even know how to calibrate the probes/scope nor how a square wave should look like on the scope's screen?

[ingyenenergiagep]

Welcome everybody!

I am Robert from Hungary.

Mr. Ltseung888!

Why dont you try this effect on toroid with higher voltage(48V or higher, 200V)? Higher voltage, lower amp-> better efficiency.
2V 10 winding, 48V 240winding.

[ltseung888]

 

Now your "Board 80 input" looks like you have the current sensing probe hooked up backwards. The voltage drops indicated on the top trace reflect the times when the oscillator is drawing the most current from the battery, so these instants should be reflected by +peaks+ not drops in the current trace. Your trace doesn't represent "negative input power", more likely it represents a probe hooked up backwards or a channel not properly inverted for the measurement hookup your board requires.

@TK

You raised too many points.  I shall try to answer them one at a time.

If you look at all the Input Waveforms starting from reply 240 on page 17, you will see mostly voltage drops and not peaks.
For example, check:
Reply 240, 241, 247, 250, 262, 264, 314, 323, 326, 332, 351, 378, 382, 386, 387, 390, 423, 434, 443, 467.


In reply 393, I deliberately set CH1 to AC and compared that to Ch2.  It is clear that a Voltage drop below the 0 axis corresponds to a current drop.  I attribute this to the "back emf"  feedback by the circuit.  Thus there were NO experimental measurement errors.

When I added the capacitor (reply 424, 430), the drop was no longer that pronounced.  This is expected as the capacitor will smooth out the voltage.

In reply 260, 261, the DC voltage was wet to 1.56V.  The drop changed to peak.  The same applies to reply 425 when I used the 2n3055 with the voltage at 2.04V.  Hope that will help in your thinking.....

The Almighty will help us all to learn tegether and overcome the personal attackes or insults...

[poynt99]

TK,

Actually, according to the latest scope shot and the circuit diagram of the probe locations, the power polarities seem correct.

The output power will be a positive result, and the input power a negative result. This is exactly what one would expect from a circuit where the battery is supplying a net power, IF the scope probes are NOT reversed polarity with reference to the battery and battery CSR.

In such a case, the supply always results in a negative power, and any load results in a positive power. I've gone over this issue several times in the past.

The voltage drop across the battery CSR will be inverted wrt the polarity of the voltage drop across the battery, and in Lawrence's case, since the scope probes across these two points are NOT reversed, this will result in a negative power product, as it should.

However, this does NOT indicate the battery is being charged, quite the contrary; it clearly indicates that it is supplying a net power to the circuit, and it is depleting normally.