Author Topic: Is joule thief circuit gets overunity? (Read 578500 times)

ltseung888 · « **Reply #750 on:** May 06, 2013, 04:56:06 PM »

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 · « **Reply #751 on:** May 06, 2013, 05:27:58 PM »

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 · « **Reply #752 on:** May 06, 2013, 05:31:21 PM »

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 · « **Reply #753 on:** May 06, 2013, 07:04:34 PM »

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 · « **Reply #754 on:** May 07, 2013, 12:19:08 AM »

Quote from: poynt99 on May 06, 2013, 05:31:21 PM

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?

ltseung888 · « **Reply #755 on:** May 07, 2013, 12:28:21 AM »

Quote from: picowatt on May 06, 2013, 07:04:34 PM

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 · « **Reply #756 on:** May 07, 2013, 01:35:10 AM »

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 · « **Reply #757 on:** May 07, 2013, 05:38:03 PM »

Quote from: poynt99 on May 06, 2013, 05:31:21 PM

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 · « **Reply #758 on:** May 07, 2013, 07:19:08 PM »

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 · « **Reply #759 on:** May 08, 2013, 12:51:05 AM »

Quote from: poynt99 on May 07, 2013, 01:35:10 AM

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.

ltseung888 · « **Reply #760 on:** May 08, 2013, 01:18:51 AM »

Trying to resolve the DC Offset Problem at Zhou's Shop.
(1) A UNI-T brand DSO was used as suggested by Poynt99. The display line was very thick and some crossing of the 0 ref line was seen.
(2) Mr. Zhou does not have the Tektronics in stock but he can order it at a relatively low price but the condition is that money has to be paid in full up front and there is no refund.
(3) Another new Atten was calibrated using the normal procedure recommended by Atten (pressing Default button and then the Auto buttorn etc.) The crossing of the 0 reference line was still seen but the actual value appeared less.
(4) Mr. Zhou believes that a 5-6mV variance or DC offset error is within the specification or limitation of the Atten. Using the 4-Ch tektronics may be the solution. (There were many requests for quotes with such high end equipment in the last few days. One Indian group asked for quotes while I was there.)
(5) Mr. Zhou suggested that I use different Atten Scopes on the same Board and check the variation. I have three in my bedroom. I have access to another two in Hong Kong. Mr. Zhou has dozens in stock. That may show clearly why the Tektronics is worth its price.
(6) Supporters' first reaction is - use the right equipment. The cost is small compared with the potential benefits. The "demonstrated" savings in electricity bills using the capacitor may already justify such expense.

More work but we are learning more. God Bless.

poynt99 · « **Reply #761 on:** May 08, 2013, 02:32:01 AM »

Lawrence,

I believe Instek is a fairly decent scope and I see Mr. Zhou sells those too. Could Mr. Zhou try one and see if there is an improvement?

ltseung888 · « **Reply #762 on:** May 08, 2013, 03:16:43 AM »

Quote from: poynt99 on May 08, 2013, 02:32:01 AM

Lawrence,

I believe Instek is a fairly decent scope and I see Mr. Zhou sells those too. Could Mr. Zhou try one and see if there is an improvement?

@poynt99.

I am sure he will try if I ask. He is already making money on this "Lead-out energy" project.

ltseung888 · « **Reply #763 on:** May 08, 2013, 03:42:44 AM »

First result from "Demonstrating longer battery life" with Demo Boards.
The results are as follows:
(1) Board 116 barebone JT with no capacitor.
Initial Battery Voltage = 1.318V
Battery Voltage after 24 hours = 1.252V

(2) Board 124 JT with capacitor and twin timer set to 10 seconds ON 2 minutes OFF
Initial Battery Voltage = 1.317V
Battery Voltage after 24 hours = 1.261V

This experiment will be continued for a few days/weeks until the LED is off?
The initial results indicate longer battery life. Does that translate to saving in electricity bills when properly implemented?

TinselKoala · « **Reply #764 on:** May 08, 2013, 08:26:02 AM »

1. Where do those voltage numbers come from? What is the accuracy of your voltage measurement? Have you calibrated your voltage measurements against a known standard? This is absolutely necessary if you are reporting a claim that depends on a few milliVolts difference, as you are here.

2. "Apparent" brightness here is no good as a criterion for several reasons, not the least being observer bias. You absolutely need to have a reliable and accurate _instrumental_ measure of light output, especially if you are making a claim based on "similar" light outputs and voltage levels that differ by a few milliVolts.

I've already given you links for local experts with good test equipment, links for affordable and accurate voltage standards, and so on. Should I now find you a suitable method of measuring light intensity, or can you handle that search yourself?

3. We need to see the schematic for your timer in the black box. Many people have made the error of allowing power from their timer/clock/control electronics to make it through and add power to the system under test. Please show the exact schematic and hookup for your timer so that we can evaluate whether or not it might be also acting as a power supply to the circuit under test.

9 mv / 1250 mv == less than one percent. A result that is not different from noise, being smaller than the voltage measurement error of your oscilloscope, anyway.

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Author Topic: Is joule thief circuit gets overunity? (Read 578500 times)

ltseung888

Re: Is joule thief circuit gets overunity?

TinselKoala

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poynt99

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picowatt

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ltseung888

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ltseung888

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poynt99

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jbignes5

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TinselKoala

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ltseung888

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ltseung888

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poynt99

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ltseung888

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ltseung888

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TinselKoala

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