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### Author Topic: Is joule thief circuit gets overunity?  (Read 584894 times)

#### TinselKoala

• Hero Member
• Posts: 13958
##### Re: Is joule thief circuit gets overunity?
« Reply #495 on: April 08, 2013, 01:07:49 PM »
Another way of looking at the issue, of course, is to view the battery CSR as part of the power supply, just an additional internal resistance of the battery. In that case, the uncorrected reading from the voltage probe, across the battery and the resistor, is the "correct" input voltage to the rest of the circuit. But.... then you don't get to include the power dissipation in the CSR itself as output. So both your input power and your total output power go down, because your battery  CSR is effectively "wasting" power that isn't getting to the rest of the circuit.

#### ltseung888

• Hero Member
• Posts: 4363
##### Re: Is joule thief circuit gets overunity?
« Reply #496 on: April 08, 2013, 01:12:15 PM »
No, Lawrence, your true input battery voltage is HIGHER than what your probe reads. Your probe is reading the battery and the resistor in series, not just the battery alone, and so is reading low, by the amount of the voltage drop across the resistor.

Hmm. Let me see if I can give my explanation without confusing things too badly.

Is the battery CSR to be considered part of the "power supply", or part of the circuit being powered? Since it's dissipating some power that the battery is supplying, I tend to think of it as part of the circuit. So the battery voltage that should be used for input power to the complete circuit is that which is read directly from the battery terminals without this resistor in series.

But the probe arrangement that Lawrence must use reads the battery voltage _with_ the resistor in series, and so must be reading _lower_ than the true battery voltage that we seek. Right?

What is the magnitude of this difference, and how can we correct for it? Since we know we need an answer that is Higher than what we are reading on the battery probe, we know that we have to _add_ something positive to our reading.

The difference is the voltage drop across the resistor. The true battery voltage is higher than what the probe reads, by the value of the voltage drop across the resistor, which is given directly by the "current" probe. The only problem is the negative sign of the reading from the current probe, which, as we recall, is an artefact of the way we need to position probes in this circuit.

So you take the reading from the battery probe, and ADD the _absolute value_ of the voltage drop across the resistor given by the current probe. The result gives the true battery voltage, as if the resistor wasn't there between the battery and the probe leads.

Note that this is NOT different from what .99 said. It just puts it in a different way. Subtracting a negative number is equivalent to adding its absolute value.

Of course if the resistor is considered part of the circuit, then the power dissipation in the resistor itself must be included in the circuit's total power dissipation as output.

ETA: I think your scope itself has the ability to do this "live" by selecting the Subtract function in the Math setup screens. Subtracting the voltage drop seen by the Ch2 probe from the battery-resistor voltage seen by the Ch 1 probe will yield the correct answer, because subtracting a negative is equivalent to adding a positive value. Again, this is the same thing that .99 has said and that I have explained above.
@TK or poynt99,

I am sorry to say that the explanation is adding more confusion.  The attached is an extract of the raw data from the Input CSV file.

It has three columns.  The first is the time which we can ignore in our analysis.  The second is Ch1 which represents the Input Voltage A1-A2.  The third is CH2 which represents the Input Current (A4-A3).  Please add the necessary columns and equations.

Hope that helps all to understand and do the correct analysis....

#### TinselKoala

• Hero Member
• Posts: 13958
##### Re: Is joule thief circuit gets overunity?
« Reply #497 on: April 08, 2013, 02:05:25 PM »
Lawrence, when you  hook a voltmeter to a battery, just by itself, you read the battery voltage, right?

Now put a light bulb in series with your battery and your voltmeter. Is the voltmeter reading higher, or lower? It is LOWER, by the value of the voltage drop across the light bulb due to its resistance. To find the true battery voltage in this case you must ADD the value of the voltage drop -- its unsigned, absolute value -- to the reading on the voltmeter.

The probe that you are using to read your INPUT BATTERY VOLTAGE is in series with a resistor. Thus it reads LOWER than the true, unresisted battery voltage. To find the true battery voltage, you must ADD the value of the voltage drop caused by the presence of the resistor. What could be plainer?

I think you can probably add your OWN column to your OWN spreadsheet to do the computation of ADDING the ABSOLUTE VALUE of the VOLTAGE DROP indicated on CH2, to the raw reading of your CH1. It does no good to have other people do work for you that you do not understand. You must strive to understand these very basic facts about voltage, current, and power measurement if you presume to claim that your measurements indicate anything special.

#### TinselKoala

• Hero Member
• Posts: 13958
##### Re: Is joule thief circuit gets overunity?
« Reply #498 on: April 08, 2013, 02:14:43 PM »
For every time instant:

V
Batttrue = VCH1measured + ABS(CH2measured)

#### ltseung888

• Hero Member
• Posts: 4363
##### Re: Is joule thief circuit gets overunity?
« Reply #499 on: April 08, 2013, 07:19:42 PM »
For every time instant:

V
Batttrue = VCH1measured + ABS(CH2measured)

@TK,

Your explanation is correct if the circuit is pure conventional DC.  The current can only be in one direction even if there were fluctuations.  The CH2measured  can only be in one direction.

In the actual experiment, no matter how you measure or view it, the current (CH2measured ) always has a positive and a negative component.  There is something not quite right in using the ABSOLUTE value.

Since both Poynt99 and PhysicsProf have used oscilloscopes to measure Input and Output Power, I would like to hear their comments before redoing all experiments.

My point of view is to use the attached spreadsheet sample but with A4-A3 connected as in the latest Board 80 arrangement in reply 485.

In any case, as far as Board 80 is concerned, the resulting COP is greater than 1 using any of the above analysis!

#### picowatt

• Hero Member
• Posts: 2039
##### Re: Is joule thief circuit gets overunity?
« Reply #500 on: April 08, 2013, 11:34:16 PM »

@TK,

Your explanation is correct if the circuit is pure conventional DC.  The current can only be in one direction even if there were fluctuations.  The CH2measured  can only be in one direction.

In the actual experiment, no matter how you measure or view it, the current (CH2measured ) always has a positive and a negative component.  There is something not quite right in using the ABSOLUTE value.

Since both Poynt99 and PhysicsProf have used oscilloscopes to measure Input and Output Power, I would like to hear their comments before redoing all experiments.

My point of view is to use the attached spreadsheet sample but with A4-A3 connected as in the latest Board 80 arrangement in reply 485.

In any case, as far as Board 80 is concerned, the resulting COP is greater than 1 using any of the above analysis!

Lawrence, TK, .99,

Would someone please explain to me why the data for CH2 Lawrence presents in his recently posted Excel only has a range of -.008 to +.016?

Visually, the CH2 capture appears to have a range of approximtely -.150 to +.020 (or there about, might be +.016).

The spreadshees seems to have a very small number of sample points, are the negative peaks just being missed/omitted?

Thanks,
PW

Actually, now that I see that the seconds column is always at -.0006, I don't quite understand what this data is supposed to represent!

#### ltseung888

• Hero Member
• Posts: 4363
##### Re: Is joule thief circuit gets overunity?
« Reply #501 on: April 09, 2013, 01:06:20 AM »

Lawrence, TK, .99,

Would someone please explain to me why the data for CH2 Lawrence presents in his recently posted Excel only has a range of -.008 to +.016?  Just an extract

Visually, the CH2 capture appears to have a range of approximtely -.150 to +.020 (or there about, might be +.016).

The spreadshees seems to have a very small number of sample points, are the negative peaks just being missed/omitted? The full analysis will be shown later

Thanks,
PW

Actually, now that I see that the seconds column is always at -.0006, I don't quite understand what this data is supposed to represent! That column was the sample time from -0.0006 to 0.0006 seconds.  The actual values are a few more places after the decimal point.  You need to see the full raw data of 10,000 samples to fully understand.

@picowatt and All,

Now that I have every last doubt cleared, I shall do a complete description and possibly a video starting from
(1) Building the oscilloscope test-ready board using the standard Joule Thief.
(2) How to select the standard Joule Thief that show possible OU behavior - FLEET.
(3) Calibration of the Atten Oscilloscope and the Probes.
(4) Full explanation of the oscilloscope test-ready board circuit diagram.  (especially the need to have common ground for the 4 probes and the poynt99 explanation of negative power as measured on the oscilloscope.)
(5) The test procedure in every minute detail. (especially the capturing of average input and output power and why simultaneous capture of Input and output may or may not be significant in various situations.)
(6) Capturing the raw data into CSV files.  (analysis of the full data - not just extracts.  The raw data will be developed to full pictures and how such pictures corresond to the screen BMP files)
(7) The full EXCEL analysis. (Every column and formula will be explained)
(8) How this ties with a theoretical model assuming a "hidden lead-out energy source".  (This is actually the high-light of the whole exercise.  It will clearly prove OU beyond any shadow of doubt.)
(9) Suggest the full set-up for a "battery recharger" system and why a simple loopback cannot achieve a self-runner.
(10) I shall let the "farmers" do much more research to achieve the commercial OU products.

Thank you to the Almighty for sending the many helpers to clarify the process.

#### picowatt

• Hero Member
• Posts: 2039
##### Re: Is joule thief circuit gets overunity?
« Reply #502 on: April 09, 2013, 02:09:57 AM »
Thanks Lawrence.

I realized it was just a short "snippet" of data after the fact...

PW

#### ltseung888

• Hero Member
• Posts: 4363
##### Re: Is joule thief circuit gets overunity?
« Reply #503 on: April 09, 2013, 06:18:35 AM »
The full paper will take weeks if not months.  Meanwhile, I shall post some raw data for those who would like to do the analysis themselves.

The data is from Board 80 meant for the Hong Kong Government.

#### TinselKoala

• Hero Member
• Posts: 13958
##### Re: Is joule thief circuit gets overunity?
« Reply #504 on: April 09, 2013, 09:34:10 AM »

@TK,

Your explanation is correct if the circuit is pure conventional DC.  The current can only be in one direction even if there were fluctuations.  The CH2measured  can only be in one direction.

In the actual experiment, no matter how you measure or view it, the current (CH2measured ) always has a positive and a negative component.  There is something not quite right in using the ABSOLUTE value.
Yes, strictly speaking you are correct. I was trying to use the strict DC case as the example so that you could understand that your probe is measuring across the battery AND resistor and therefore must always read a lower value than simply measuring the voltage alone. Sorry I wasn't too clear about that. The only way the current could reverse across this resistor is for the voltage at the battery negative terminal to be HIGHER than the voltage at the circuit end of the resistor. But you seem to think that your measurements are indicating that this is the case, at least part of the time. I still don't think that it is, but nevertheless... let's amend my attempt at a simple explanation, and remind you of this: the probe position you are required to use produces a NEGATIVE current reading when the current is flowing CONVENTIONALLY around the circuit, from the battery to the circuit, dissipating power in the circuit.
So following .99's advice, since the CH2 probe is reversed and giving a negative value most of the time, you must subtract its reading from the combined reading that the CH1 probe gives instead of adding it. You are subtracting a negative number _most of the time_ from a positive number, which is equivalent to adding it.  1.5 minus (-0.5) == 2.0 .

Think about the voltages at the ends of the resistor when current is flowing. "Conventional" current is taken to flow from positive to negative, and a meter hooked up this way will read positive current. What does this mean for the resistor? When current is flowing normally, the voltage at the negative terminal of the battery is the very lowest voltage in the system, so you can call it "zero". The voltage on the other side of the resistor is higher than this. So your probe, instead of measuring the value of the voltage difference between the battery positive, and ZERO, is measuring between the battery positive and something HIGHER than zero. This means it reads LESS than the true battery voltage, and to get to the true battery voltage you need to ADD the _positive_ value of the voltage drop across the resistor. Since your CH2 probe is reversed and gives a negative value, you need to Subtract this Negative value (minus a minus is the same as adding a plus) to get to the right answer: a number greater than what your CH1 probe indicates.
In the case of _reversed_ current, the negative pole of the battery must be at a HIGHER voltage than the other end of the resistor towards the circuit. Current can only flow from higher to lower voltages. This means that your CH1 probe is measuring the battery voltage, PLUS the voltage "drop" across the CVR-- it is reading high this time instead of low. So to get to the correct voltage, you need to SUBTRACT the value of the voltage drop in CH2 from the total voltage. However, in this case the numbers reported by your CH2 probe will be POSITIVE, due to the reversed orientation of the probe. So now you will be subtracting a positive value, which leads to the required lower true battery voltage.
So... your spreadsheet reports what the probe reads, regardless of any trace invert display selection. When the current is normal, conventional, ordinary, discharging the battery current, the probe and the spreadsheet will call this "negative" and this negative value must be subtracted... which means adding its absolute value. When the current is "reversed", according to you charging the battery, supplying energy from the circuit back to the battery, the probe and the spreadsheet will call this "positive" and this positive value must be subtracted, in the ordinary way, from the CH1 reading.
So the spreadsheet formula becomes simply
VBattTRUE = VCH1 - VCH2;     the spreadsheet knows that subtracting a negative is the same as adding the positive value, and this handles both directions of current whenever they occur.

Quote

Since both Poynt99 and PhysicsProf have used oscilloscopes to measure Input and Output Power, I would like to hear their comments before redoing all experiments.

Yes, I concur. By all means let us hear from .99, who has the proper equipment for making these determinations, and the skill to use it.
By the way, I note that the Atten scopes can measure the Humerous content of a Power cable. That is indeed a valuable feature... around here.

Quote

My point of view is to use the attached spreadsheet sample but with A4-A3 connected as in the latest Board 80 arrangement in reply 485.

In any case, as far as Board 80 is concerned, the resulting COP is greater than 1 using any of the above analysis!

Ah.... no it's not. It is COP >1 using conceptual and operational errors and one particular measurement system that is not concurrently valid with others.

#### TinselKoala

• Hero Member
• Posts: 13958
##### Re: Is joule thief circuit gets overunity?
« Reply #505 on: April 09, 2013, 09:44:27 AM »

In the first place your RMS voltage reading is irrelevant. Your scope trace in CH1 indicates fully 1.50 volts or more except in the "valleys" which are not "back emf", they are simply the dips in voltage caused by the LED turning on for a brief instant. Only the instantaneous voltage values are relevant to your power measurements, the RMS values only obfuscate the true issues.

In the second, most important place: Your WHITE LINE in your CH2 trace is a misrepresentation if it is supposed to be "zero" volts. Look at where the little bluegreen "2" symbol is to the left of the display area. THAT is where your "zero" value is, slightly above the major graticule line, not where you have drawn your white line which is almost a full minor division BELOW the graticule line. And the only part of your trace that is in any sense crossing this true zero line is .... NOISE, and perhaps a tiny ground bounce. And of course we recall that this probe is inverted, so negative values mean.... entirely conventional, battery discharging, power dissipating in the circuit, current flow.

#### TinselKoala

• Hero Member
• Posts: 13958
##### Re: Is joule thief circuit gets overunity?
« Reply #506 on: April 09, 2013, 09:58:02 AM »

Once again, you are citing RMS voltage values, which are entirely irrelevant here. Only the instantaneous voltage and current (voltage drop) values have any relevance to power and energy calculations done in this manner.

But more importantly..... you are STILL not putting your channel baselines directly and exactly on a horizontal graticule line. This is IMPORTANT for observers to be able to interpret easily your scope images. Those wiggly lines have meaning! Lots and lots of it. You are making it harder than necessary to interpret those meanings, and indeed actively obfuscating some details, by your sloppy positioning of your baselines. Positioning the baselines correctly allows one to see at a glance the voltage levels concerned.

Also, you are still putting your Trigger right on top of your baseline. Fortunately the Atten scope has a fairly intelligent trigger and isn't too upset by your attempts to fool it. Please place your trigger decently above your baseline level, so that you know that you are not going to be triggering on noise, and so that the T indicator doesn't obscure the 2 of the baseline indicator, displayed to the left of the trace area.

Your parameters panels do not provide the information necessary to determine energy performance at all, so please do not suggest that they "indicate possible OU".

I find it remarkable that this board is producing "exactly" 3 kHz signal.

#### TinselKoala

• Hero Member
• Posts: 13958
##### Re: Is joule thief circuit gets overunity?
« Reply #507 on: April 09, 2013, 10:19:13 AM »
And finally, for this round.... your input and output measurements are not synchronous, being taken on two different scopes. Would you please show a set of traces obtained in the following manner: Use Scope 1 to monitor the Input Voltage and the Output Voltage. Use Scope 2 to monitor the Input Current and the Output Current. This will allow us to see the time and magnitude relationships between Input and Output directly, something that we have not yet seen from your arrangement.

When a nice 4-ch scope is used we'll have all of this data on one screen which will make matters much easier to interpret.

#### ltseung888

• Hero Member
• Posts: 4363
##### Re: Is joule thief circuit gets overunity?
« Reply #508 on: April 09, 2013, 11:22:08 AM »
And finally, for this round.... your input and output measurements are not synchronous, being taken on two different scopes. Would you please show a set of traces obtained in the following manner: Use Scope 1 to monitor the Input Voltage and the Output Voltage. Use Scope 2 to monitor the Input Current and the Output Current. This will allow us to see the time and magnitude relationships between Input and Output directly, something that we have not yet seen from your arrangement.

When a nice 4-ch scope is used we'll have all of this data on one screen which will make matters much easier to interpret.

@poynt99:

We are now all earger to see your 4-CH scope results.  Sorry to put the pressure on you.

@TK:

The Vrms display is left from history.  Five Years ago, I did not use oscilloscopes and relied on voltmeters and ampmeters with AA batteries as source.  There was discussion that the correct way was to see the entire waveform and use the equation:
Intanstaneous Power = Instantaneous Voltage x Instantaneous Current.
If there were pulse elements, the closest value is the Vrms - not Vmax, Vavg etc.  At that time I called such displays "Tseung index" as a comparison for my many FLEETs.  They are still useful when I compare different JTs.

I enclose the scope analysis for Output and for Input as you suggested.  The numerical or absolute value for Average Output Power is greater than the Average Input Power.  If we agree that the COP is the ratio of Average Output Power over Average Input Power, we get COP > 1.

Let us let poynt99 comment on that - with the results from his 4-Ch scope.

#### picowatt

• Hero Member
• Posts: 2039
##### Re: Is joule thief circuit gets overunity?
« Reply #509 on: April 09, 2013, 11:33:43 AM »
TK,

Referring to the schematic in your post 469 a few pages back, if you draw a short circuit between the A3-A4 point (scope grounds) and the A1 point, you will have maximum current flow thru the input side CSR.  Assuming for the moment that the battery (and the short circuit) has an infinitely low internal resistance, the measured voltage at A1 would be zero volts, and at A2, the voltage would be -1.5V (assuming a 1.5 volt battery).

Current flow would threfore be 1.5 amps and the actual battery voltage would be (A1-A2), which is 0-(-1.5), or 1.5volts.

In the recent captures, the negative going ripple on A1 is due to Vdrop across both Rint and the CSR when current is being drawn from the battery.  Any negative voltage observed at A2 represents current drawn from the battery.

Therefore, the negative dips/ripples on the A2 voltage are due to Q1 turning on and loading the battery with the toroid, which  produces current flow thru the CSR and causes A2 to be a negaive voltage (base curren is also drawn to a lesser degree).  During the Q1 on time, the voltage across the LED is effectively Vce(sat), so the LED is turned off because the voltage at B1 is below Vled(on) during this time. (this assumes the schematic as drawn in your 469 without a second battery in series with the LED).

When Q1 turns off, the voltage at B1 rises above Vbatt as the energy stored in the toroid discharges in series with Vbatt.  The voltage at B1 is clamped at the LED voltage as the LED turns on briefly (the B1 LED voltage must also be corrected by subtracting the output CSR Vdrop similar to the A1 Vbatt correction).

As the waveforms contain some fairly high frequency components/harmonics, all of the caveats regarding wiring inductance relating to accurate voltage/current measurements must also be considered.

That's my take on it...

PW