Author Topic: Is joule thief circuit gets overunity? (Read 602723 times)

picowatt · « **Reply #465 on:** April 05, 2013, 01:31:54 AM »

Quote from: ltseung888 on April 05, 2013, 12:22:56 AM

@picowatt

The pictures on reply 450 were wrong because I did not press the "default setup" button first before doing the calibration. The Atten Oscilloscopes have many options and settings. For normal operations, I had to use specific settings depending on what I wanted to display. For example, I displayed Ch1 Vpp, Vrms and Ch2 Vpp, Vrms, etc. To do the calibration of the probes correctly, very specific setting was required. The Atten Scope has the "default setup" or "factory setup" to help to achieve that.

I did not press that button first and thus the "calibration pictures" were wrong. A simple error.

*** This is actually a good experience. When I bring my Atten Scopes (or use new ones) at one of the Top Universities, I shall do the full calibration procedure in front of the Academics. The results will be more convincing. It also points out the difficulty if I rely on "their oscilloscopes". There is no possibility that I will be able to master the operations of all different models of oscilloscopes. Bringing one or two Atten Scopes for demonstration is a MUST. (Bringing a laptop computer is NOT.)

The signal at the calibration post is a 1 K square wave of approximately 3Vpp. It doesn't matter what the scope's timebase or vertical sensitivity is set to, your scope should always indicate approximately 3Vpp when the calibrator is probed (assuming the probe and channel probe factor settings are correct).

If at anytime you when you are using your scope, you probe the calibration terminal with any probe connected to any channel at any vertical sensitivity or timebase setting, and the display is anything other than approximately 3Vpp, something is wrong. You should not have to be in "default setup" to make the scope read correctly.

Besides incorrect probe compensation, your post 450 scope 1 channel 2 shows the 3Vpp cal signal as being 15Vpp. Does this mean that all of your input current measurements made with scope 1 channel 2 were being indicated as 5 times higher than they actually were? How did you have the scope set so that channel 2 reads higher than it should by a factor of 5?

I suggest you recheck all probes and channels using the calibrator post to ensure that all channels are indicating approximately 3Vpp, adjust your probe compensation for minimal under/overshoot and then repeat a few of your tests.

To lend validity to any further scope shots of your JT circuit, I suggest you also provide scope shots of the probes connected to the cal terminals as you have the scope set to make your JT measurements (adjust only the vertical sesitivity if needed to prevent display clipping).

The very fact that you would post the scope shots of 450 in response to TK's request to see your probe compensation without realizing the errors contained in those captures is in itself quite telling.

PW

TinselKoala · « **Reply #466 on:** April 05, 2013, 06:37:26 AM »

Quote from: ltseung888 on April 04, 2013, 07:36:28 AM

The correct Oscilloscope Probe Calibration diagrams.

@TK,

I read the Manual and found the correct Oscilloscope Probe Calibration Procedure. The correct photos are shown here. Thank you for all your comments - they refer to incorrect photos. All my posted data are correct as the calibration was done right for me when I bought the Scopes.

@testers and All researchers.

False Alarm from TK..... I did not give him the correct calibration diagrams..... The pictures on reply 450 are definitely not calibration pictures. Should not blame TK for not recognizing them as oscilloscope manufacturers have different displays.

Lawrence, your protest here is simply wrong. You are grasping at straws in an attempt to get yourself out of this major error. False alarm from me? No, dear Sir..... the story is somewhat different.

You provided what I asked you for, which was the traces from the scope's calibrator output without making any adjustments to your probes. This means that the traces you showed in post 450 ARE INDEED traces which reflect the conditions of your probe/channel combinations as you have been using them. Is this, or is this not, the case?

The fact that you have NOW calibrated your probes according to the first chapter of your owner's manual... or at least one of them... after all this time.... is rather "telling", as Picowatt might say. How long have you had your manual?

Your owner's manual instruction is garbled somewhat though: Your probe/channel combo must be calibrated together. Not just "channel 1 factory settings". You manually set your channel vertical atten to 1 volt/div. You assure that your probe switch is at 10x and that the channel setting is also at 10x internally. You manually set your timebase to read conveniently for a 1 kHz signal. You assure that you are triggering on the channel of interest and at the correct voltage level and rising or falling slope. Then you probe the scope's calibrator output with the probe/channel combo you are calibrating. You carefully adjust the probe's compensating capacitor until you get the nicest looking square wave, and on analog scopes you adjust the channel attenuator knob's "cal" portion to make the displayed waveform show exactly three divisions (three volts), or whatever its calibrator output is supposed to be. You repeat this process for each channel/probe combination. Generally, the channels themselves will be "equal" in input capacitance, and once the probes are set properly they should not require resetting much and it shouldn't matter which probe you have in which channel, on a given scope/probe set. But just in case, for my own two in-use scopes, I use only one probe set for one scope and the other for the other scope, and I have them all marked 1,2, and A,B, so they always go in the same channels in the same scopes.

Since the traces you showed in post 450 seem to reflect the conditions under which your data were gathered.... you need to calibrate, fix the voltage problem, and redo your experiments and calculations. You may still find your OU results.... I don't think this is the only error.... but this much, at least, must be done.

I pointed out the discrepancy in voltage on one of your traces some time ago, or rather I "asked" if anyone noticed anything peculiar, and .99 came back with the abnormally high voltage, but nobody made any further comment about it.

Quote

All my posted data are correct as the calibration was done right for me when I bought the Scopes.

And that is probably why ONE of your probes has correct compensation. Whoever did it, used the instructions in the manual, which only cover one probe in one channel. SO all your posted data FROM THAT ONE SINGLE PROBE/CHANNEL COMBO may indeed be correct. Or it may not. Only redoing the experiment will allow you, or us, to tell.

Quote

Should not blame TK for not recognizing them as oscilloscope manufacturers have different displays.

Further.... you, Lawrence, should not presume to attempt to teach me about oscilloscope displays or the usage of oscilloscopes in general. I asked you for something specific and you supplied it. Now, you are weaseling, because your ignorance and your misuse of the scopes is made plain for all to see. Your only possible "out" is to prove that the scope shots in 450 were not taken recently but were taken before your experiments, and the problems shown were corrected before you began taking data. But that's not the case, is it.

ltseung888 · « **Reply #467 on:** April 06, 2013, 03:10:37 AM »

Board 80 results after all probes were recalibrated. Same characteristics. can wait for poynt99 and Physicsprof to post their results with their oscilloscopes.

TinselKoala · « **Reply #468 on:** April 06, 2013, 07:18:25 AM »

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.

And it's clear from your "Board 80 output" that you will no longer have any negatives in your output result, as all your measured values in both traces are now positive.

I think it's pretty clear to anyone who can read a scope trace that your results are now going to be considerably different.... now that you have properly calibrated ALL your probes. Yes, you do need to retract and redo your previous results, because THEY WERE DONE INCORRECTLY. This is an issue of scientific integrity here, Lawrence. You took data with uncalibrated probes with substantial artefactual errors, and you recorded and reported that data as correct, when it was not. A retraction is indeed absolutely in order. If a redo still supports your claims, that's fine, at least you did the right thing. If the redo doesn't still support your claims, then we have saved a lot of people a lot of trouble trying to replicate your errors that you published as correct.

Now.... the input current trace in your recent example Board 80 shots. You are here measuring a very small signal at the scope's high gain setting. I asked you before: Do your scope traces change AT ALL as you move cabling around to different locations? I don't mean moving the probe points, just the way the cables themselves are routed from the board under test, over to the oscilloscope. Often, at high gain settings reading weak signals, the system can be very sensitive to how wires are routed, and even the length of the probe's ground cliplead can affect readings greatly, especially for signals like yours with substantial high-frequency components (the spikes).
So please, set up as normal, look at the current trace as in the post above, and move the probe wires around while measuring. Does the waveform change at all?

TinselKoala · « **Reply #469 on:** April 06, 2013, 08:15:51 AM »

Lawrence, do the two slides below show your input current and voltage measurement positions that you have been using for your recent data? Why do they not agree with the circuit diagram?

Just to be clear: The circuit diagram has the common ground for all four channels on one side of the input current viewing resistor, away from the battery, as is correct. It shows the probe tip (incorrectly called "+ve" in the caption) of the current channel connected to the battery-resistor connection, as is correct to make this measurement, and the reference or ground lead (incorrectly called "-ve" in the caption) connected to the common ground.

But the photograph shows this probe reversed, with its TIP connected to the circuit common ground and the reference lead connected to the battery-resistor junction. This of course inverts the signal -- shows it going backwards -- as well as introducing a magnitude error into the current reading, especially if you have another probe hooked up _correctly_ to read the input voltage.

Please explain this discrepancy, which to my mind would fully account for your anomalous "negative" input power computations that you have recently been reporting.

TinselKoala · « **Reply #470 on:** April 06, 2013, 08:58:58 AM »

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 · « **Reply #471 on:** April 06, 2013, 09:48:27 AM »

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 · « **Reply #472 on:** April 06, 2013, 10:58:19 AM »

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 · « **Reply #473 on:** April 06, 2013, 01:39:10 PM »

Quote from: TinselKoala on April 06, 2013, 07:18:25 AM

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 · « **Reply #474 on:** April 06, 2013, 06:25:41 PM »

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.

poynt99 · « **Reply #475 on:** April 06, 2013, 07:15:39 PM »

One other important issue to be keenly aware of Lawrence, is that your CH1 (A1-A2) probe is NOT giving you an accurate measurement of the true battery voltage for making the input power computation. You are in fact measuring across both the battery and the battery CSR resistor.

In order to obtain the true battery voltage measurement from the A1-A2 difference, you must subtract the voltage drop across the battery CSR (A4-A3) from the A1-A2 measurement.

Since A1-A2 is positive, and A4-A3 is negative, subtracting the two is equivalent to adding the two. So in fact your battery voltage is actually higher than what your A1-A2 probe is capturing, and as such, your input power result will be higher (in the negative direction) as well.

Of course you would need to do this computation in the spread sheet.

ltseung888 · « **Reply #476 on:** April 07, 2013, 06:54:33 AM »

Quote from: poynt99 on April 06, 2013, 07:15:39 PM

One other important issue to be keenly aware of Lawrence, is that your CH1 (A1-A2) probe is NOT giving you an accurate measurement of the true battery voltage for making the input power computation. You are in fact measuring across both the battery and the battery CSR resistor.

In order to obtain the true battery voltage measurement from the A1-A2 difference, you must subtract the voltage drop across the battery CSR (A4-A3) from the A1-A2 measurement.

Since A1-A2 is positive, and A4-A3 is negative, subtracting the two is equivalent to adding the two. So in fact your battery voltage is actually higher than what your A1-A2 probe is capturing, and as such, your input power result will be higher (in the negative direction) as well.

Of course you would need to do this computation in the spread sheet.

@poynt99:

Thank you for the reminder. All my spreadsheets have the above mentioned computation. You have Board 33 and can easily do the actual experiments. I am showing the details of Board 80 for those who want to replicate and improve. Please show the results of Board 33 on your 4 channel Scope. There will be more to discussions as far as the theory is concerned. For now, we just need to ensure that there are NO experimental or equipment errors.

TinselKoala · « **Reply #477 on:** April 07, 2013, 08:59:04 AM »

Quote from: poynt99 on April 06, 2013, 06:25:41 PM

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.

I think you are missing my main "poynt". If you look carefully at the circuit diagram, and look carefully at the photo that goes with it, you will see that the probe in the example photo is reversed from the hookup given in the circuit diagram. The photo shows the probe _TIP_ connected to the common circuit ground. How is it possible to make simultaneous input voltage and current measurements if the probe is hooked up as the _photo_ shows?

TinselKoala · « **Reply #478 on:** April 07, 2013, 09:14:02 AM »

Replicate and improve? How about this. You've seen this before but you obviously didn't get the message. Ease of manufacture, elimination of circuit errors, consistency between individual units, cost reduction, complexity reduction.....

Of course, since my test points are directly at the respective current monitoring resistors and my current paths are in general shorter and more direct than yours... I am likely to see less artefactual waveform distortions than you are.

Proudly displaying my construction "improvements" in my "replication" of a circuit that I actually built well before you started using it:

(It's hard for me to believe that after all this time, someone hasn't spent ten minutes with a circuit CAD program and generated a nice template for Lawrence, that he could send off and get made into boards professionally for minimal cost. I put this together completely manually in an afternoon as a sort of a joke.)

TinselKoala · « **Reply #479 on:** April 07, 2013, 09:56:49 AM »

Quote from: poynt99 on April 06, 2013, 06:25:41 PM

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.

Perhaps the following attached pix will help clear up the issue.
In the first simple diagram below I have shown a battery, a voltmeter (separate and isolated), a current-viewing resistor , the connecting wire and two test points A and B. A would usually be considered the "common circuit ground".... except when also monitoring output in Lawrence's circuit too, "B" must be used as the common reference point for all 4 scope probes.
"Conventional" electric current flows from positive to negative and this would indicate normal battery discharging as it dissipates energy in the circuit. The isolated voltmeter reads the battery voltage and at the correct polarity. Now hook a scope to points A and B. Clearly, point B will be at a higher positive voltage than point A when current is flowing conventionally. So if you hook your scope TIP to point B and your reference "ground" lead to point A, you will see a _positive_ voltage indicated on the scope when current is flowing conventionally and the battery is discharging. This positive voltage is the "drop" across the resistor and gives the current thru the resistor by Ohm's law.
But the circuit Lawrence uses requires that the common ground actually be at point B. So the current must be monitored with probe TIP at point A-- where the voltage is _lower_ than at the probe's reference lead at point "B". This means the scope will indicate a NEGATIVE voltage when current is flowing conventionally and the battery is discharging. This negative voltage _reading_ when multiplied by the voltmeter's positive reading will yield a "negative" power value. To avoid confusion, many people (like me) would simply press the trace invert button to make the reading agree with the _conventional_ definition of current flow.
Now, look at the schematic of Lawrence's circuit. He correctly has the common circuit ground at the point corresponding to my point "B", and the input current probe TIP at the battery negative, my point "A". This means that, _{without using the trace invert function}, the computed power _should_ be "negative" when the current is flowing normally, just as Poynt99 has explained many times.
However.... the PHOTOGRAPH that Lawrence has recently provided shows the probe TIP at the common circuit ground ( my point "B") and the probe ground at the battery negative (my point "A"). This will invert the current reading, and also create a ground path through any other instrument grounds that might be connected to the correct common circuit ground.
So I have asked about this. What's the explanation for the discrepancy between the PHOTO showing how Lawrence has recently been measuring, apparently incorrectly, and the SCHEMATIC which shows what I would consider to be correct probe hookups, but which will give an _apparent_ negative power unless the channel trace invert function is used?

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Author Topic: Is joule thief circuit gets overunity? (Read 602723 times)

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