Is joule thief circuit gets overunity?

[ltseung888]

 Step back to the "energy saving" part of the demonstration.

1.    Put the 2.3V 1 Farad Capacitor in parallel with the Battery or DC Power Supply.  Let the initial charging be about 1 minute.
2.    Disconnect the Battery for 2 minutes.  The LED should be brightly ON at the end of the 2 minutes.
3.    Connect the Battery back for about 10 seconds.  The LED appeared to be brightly ON with no dimming.
4.    Repeat Step 2 and Step 3 a couple of times.  Convince yourself that the LED appeared to be equally bright from the naked eye.
5.    Hook up the DSO (may need the Tektronic 4-CH to be accurate).  Get the Average Input Power during the 10 Second Interval.  Get the Average Output Power during the 2 Minute Interval.
6.    The Input Energy can be estimated as the Average Input Power x 10 seconds.  The Output Energy can be estimated as the Average Output Power x 120 seconds.
The Atten gives a much higher Output Energy value but since its accuracy is in doubt, I shall delay posting the figures.  I hope Poynt99 can do this above experiment with his Scope.

An end user used an automatic timer to implement the above and claimed considerable energy savings compared with supplying energy all the time.  Is this a better energy saving lighting system?

The setup may not be OU but is it energy saving???

[TinselKoala]

No. The process of charging a capacitor from a battery is a lossy process.

However, using a capacitor as an input power source is a good way to check your claim of overunity or excess efficiency, because the energy in a capacitor can be known precisely, unlike the energy in a battery. The energy in a cap in Joules is simply (CV²)/2  where C is capacitance in Farads and V is the charged voltage in volts.

So you can charge a cap to some voltage. Then you know the total input energy E_in available from the formula. Then you can let your JT run, monitoring the output as usual. The average power output in Watts times the time in seconds is the energy output E_out. Does E_out > E_in? (More properly the energy is equal to the time integral of the instantaneous power curve.)

More properly, you would charge your capacitor to a known precise voltage V₁, then run the system until the capacitor voltage decreased to V₂ and stop. The total energy input to the system is then (CV₁²)/2 - (CV₂²)/2, and this can be compared to the total E_out measured in your usual manner.

[poynt99]

Lawrence,

It is highly unlikely that this is an energy saving setup. In fact it may be more wasteful than just leaving the battery connected all the time.

Every time the battery is connected to the capacitor, about half the energy is lost due to heating in the connection.

[picowatt]

Lawrence,

Have you attempted to measure one of your boards using your scopes, and as you normally do, since pressing the "default settings" button?  If not, you might consider doing so.

Before doing any tests, press "default settings" and then adjust settings as required (sweep, vertical sensitivity, trigger level and source, etc).

I am curious as to whether or not pressing "default settings" may have corrected some measurement errors (based on recent scope shots).

I continue to suggest that you perform the "3 step test" I recently proposed and post the results.

PW

[ltseung888]

Lawrence,

It is highly unlikely that this is an energy saving setup. In fact it may be more wasteful than just leaving the battery connected all the time.

Every time the battery is connected to the capacitor, about half the energy is lost due to heating in the connection.

@TK and poynt99,

I shall repeat the following experiment.

(1) Use a rechargeable AA battery with known voltage as source to the Board with NO capacitor.  The LED will be ON for a long time.  The Battery will be connected all the time.

(2) Use a rechargeable AA battery with same known voltage as source to the Board with the capacitor.  A twin timer will be connected.  The AA battey will be connected 10 seconds and disconnected for 2 minutes.  We can then compare the drop in voltage of the two batteries from time to time.

As I recall, case (2) Battery showed a much lower voltage drop - indicating an energy saving?  But I should not rely on my memory.  A new experiment is in order.  I shall do it without the DSO first.  I shall swap the boards to ensure validity.  Will this be a conclusive experiment?