Is joule thief circuit gets overunity?

[ltseung888]

Ah, I just saw your post about the CSV file not changing in response to the channel invert setting. That seems strange to me. Is there some way we can confirm this?  Reviewing Table 2-47 and the associated text in the manual, I see nothing that is very helpful on this question.

But OK.... if the CSV storage is the raw, non-inverted data regardless of how the invert display button is set, then a negative value calculated here, with your probes positioned as in the schematic and apparently as in your latest picture above, indicates conventional +positive+ current flow in the normal direction, just as .99 has been telling you. See my rough sketch a few posts ago to understand which side of the current viewing resistor has the higher voltage in each case. The "negative" voltage reading comes from the fact that the probe TIP, at the negative pole of the battery, must be at a LOWER voltage than the ground clip on the circuit side of the current viewing resistor, when the current is flowing normally. So if the voltage at the ground clip is assigned "zero", then the probe tip will be negative.

@TK and poynt99,

Can you point out where the highlighted information is posted.  Thank you.

[Pirate88179]

http://security-world.blogspot.com/2011/01/undp-scheme-2011.html
http://naijaspleen.wordpress.com/2012/11/12/spleen-scam-mails-alert-undp-grant-programme/
http://419.bittenus.com/12/2/undpgrantprogram.html
http://emailscamscammersscamming.blogspot.com/2012/08/undp-grant-donation-programme.html

Seems legit???











NOT.

Does this mean that I will not be receiving all of that money?  I was really looking forward to building a new research facility.
Crap!

Bill

[TinselKoala]

@TK and poynt99,

Can you point out where the highlighted information is posted.  Thank you.

Most recently, .99 explained it here, just a few posts ago:
http://www.overunity.com/12686/is-joule-thief-circuit-gets-overunity/msg356491/#msg356491
My sketch, posted a few comments ago and attached again below, should help make it clear. When the current is flowing normally _out_ of the battery, supplying power TO the circuit, then the two ends of the current sensing resistor are at different voltages.... with the junction of the resistor and the negative battery pole ("A) at a LOWER voltage than the end of the resistor that is towards the rest of the circuit ("B"). So a conventional current flowing out of the positive pole, thru the circuit dissipating power, will, when measured with your schematic diagram's probe orientation and a NON inverted channel -- as in your CSV data, apparently -- then you will read a negative value for this voltage, and this negative will carry through the Ohm's law calculation and result in a _negative_ power value. This is because of the "reversed" orientation of the probe, which is required in the special case of your circuit where four channels must be used simultaneously and with a common "reference" point, what you are calling the common circuit ground. Claro?

[ltseung888]

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.

@TK and poynt99,

Please help me to get this confusion out from the spreadsheet analysis.  Let us assume that NO Invert function was applied in both Ch1 and Ch2.  In the spreadsheet, should I add the A1-A2 value (positive) and the A4-A3 (negative) values together to get the TRUE input voltage across the battery? 

The addition will effectively lower the value of the "Input Voltage" (A1-A2 value).  I should then use this lowered value in the calculation of the Input Power.  Multiple the lowered (A1-A2) value with the captured (A4-A3) value.  For the calculation of the COP or the Power Comparison, I should then add a negative sign to the resulting Input Power value.

Is that correct?

Thank you for clarifying this point.   I shall redo the experiments with the "correct" analysis if needed.

[TinselKoala]

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.