Is joule thief circuit gets overunity?

[ltseung888]

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

@PW,
Yes.  I have done more measurements since pressing the "default setup" button.  I also deliberately press the "default setup" button first after warming up the scope for 1 hour and then do the usual experiment.  The crossing 0 reference behavior is still there.  I shall meet Mr. Zhou this afternoon.  I shall leave the Atten issue with the Atten Experts.

[poynt99]

Lawrence,

As you described your experiment, of course it will be easy to extend the life of the battery's charge that way, but you are not measuring output power and energy and comparing.

With the battery, the LED load will be at a relatively high output power and will slowly decline as the battery voltage declines. This is normal. But by only charging the cap for 10 seconds every 2 minutes, you are allowing the output power to drop quite a bit in that 2 minute span of time, so in effect you are simply stretching out the time the battery will last and still have the LED put out a decent amount of light.

There is no OU in this scheme, but you may indeed extend the life of the battery on its single charge.

If you looked at this from the Energy perspective, both cases will be very close, with possibly less total energy going to the load in the case with the capacitor. If you could somehow turn down the voltage on the battery and leave it connected at all times, it would deliver less average power to the LED, and this would be accomplishing the very same thing you are doing with the cap.

What you have with your capacitor scheme, is a very crude step-down convertor.

[jbignes5]

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.

Really half the energy is lost due to heating? Really? What kind of wire you using resistive heating wire?


"If there's no resistance (even the smallest parasitic) in the circuit, with an inductor and capacitor, the oscillations will go on forever, and in that case the resistor-capacitor equation doesn't apply. But then how can we say that the capacitor reaches a final state of charge?

I've attached an analysis of the problem for the R-C and the R-L-C cases.

But, the fact that the energy that ends up in the capacitor in these two cases is only half the energy provided by the voltage source doesn't mean that every method of charging the capacitor loses half the energy from the source. I described two methods that don't suffer from that loss in a previous post."

So I would have to believe that not all the ways to charge a capacitor end up with 50% loss. As real technicians know not every circuit conforms to the ideal computations we present sometimes.

The previous post:

"As near as I can tell, the link http://hyperphysics.phy-astr.gsu.edu...capeng.html#c1

merely gives the classical formula for the energy stored in a capacitor, c*v^2/2, or q*v/2 which is the same thing. This formula is correct no matter how the capacitor gets charged.

However, it isn't correct to say that whenever a capacitor is charged from some kind of energy source, without exception, only half the energy provided by the source ends up in the capacitor.

Sometimes it is true, as in the case where a source of constant voltage is suddenly connected to a series combination of a resistor and capacitor.

If the circuit consists of an inductor in series with a capacitor, with negligible parasitic resistance, then suddenly connecting this circuit to a constant voltage source will transfer almost all of the energy provided by the source to the capacitor, IF the voltage source is disconnected at the right time, namely when the current goes to zero at the end of one half cycle of oscillation. A simple diode can provide this disconnect.

Another way to ensure that all the energy provided by the source ends up on the capacitor is to connect the capacitor to a variable voltage source initially set to zero, and then gradually turn up the voltage to some final value.

Even if there is a resistor of non-negligible size in series, if the voltage is turned up slowly so that the charging current remains small at all times, most of the energy ends up in the capacitor, with very little dissipated in the resistor.

This is what is described in your original third reference, in the vicinity of equations 13 and 14. By making the voltage steps smaller and more numerous, the effect is the same as turning up a variable voltage source gradually."

Post reference  http://www.physicsforums.com/showthread.php?t=292838

[TinselKoala]

So we are agreed, apparently, that the energy on a capacitor, in Joules, is (CV²)/2, no matter how lossilly or ideally the capacitor is charged. (C in Farads, V in volts).

Why, then, is everyone avoiding doing the correct test of the JT's performance using a charged capacitor?

The circuit's output power can be monitored, and the total electrical energy output to the load can be easily calculated from the instantaneous power curve and the duration of the test run. The input energy over the test run can be known precisely, by knowing the voltage on the capacitor at the beginning and at the end of the timed test run interval. The input energy can then be directly compared to the output energy. What is so hard about this test? What does it matter how the capacitor is charged, as long as the output energy is greater than the input energy? Or is it......? We will never know, as long as those with the proper equipment (and an alleged OU device to test)  refuse to perform it.

[ltseung888]

Lawrence,

As you described your experiment, of course it will be easy to extend the life of the battery's charge that way, but you are not measuring output power and energy and comparing.

With the battery, the LED load will be at a relatively high output power and will slowly decline as the battery voltage declines. This is normal. But by only charging the cap for 10 seconds every 2 minutes, you are allowing the output power to drop quite a bit in that 2 minute span of time, so in effect you are simply stretching out the time the battery will last and still have the LED put out a decent amount of light.

There is no OU in this scheme, but you may indeed extend the life of the battery on its single charge.

If you looked at this from the Energy perspective, both cases will be very close, with possibly less total energy going to the load in the case with the capacitor. If you could somehow turn down the voltage on the battery and leave it connected at all times, it would deliver less average power to the LED, and this would be accomplishing the very same thing you are doing with the cap.

What you have with your capacitor scheme, is a very crude step-down convertor.  *** But it can save electricity costs!

@poynt99,

In this case, I am not talking about OU.  I am talking about savings in electricity bills. 

It is possible that the same energy is "stretched out".  The energy is pulsed to light the LED.  So some of the time is idle (no electricity) time.  The pulsing is so fast that the naked eye cannot see the difference.  The LED appears to be of equal brightness.  That is good enough for the end user.  I also believe less energy is converted to heat. 

*** The oscilloscope showed that the voltage at the capacitor dropped to 1.1V approximately at the end of the 2 minutes.  So you are right in saying that we are actually supplying less energy to the LED as the NO capacitor case is almost a constant at 1.5V.