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Author Topic: Is joule thief circuit gets overunity?  (Read 600425 times)

TinselKoala

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Re: Is joule thief circuit gets overunity?
« Reply #525 on: April 10, 2013, 02:38:07 AM »
TK,

I still think Board 80's waveforms look a bit different than yours.  Not just that the spikes are missing, but the general slopes on the tops of bothe the V and I waveforms appear to have a different rate.

As with regards to your last sentence above, I agree that the LED turns on at the bottom of the input voltage valleys, at the point where the slope reverses, BUT, I would state: the power represented by the decreasing voltage up to that point is the power being drawn from the battery and stored in the toroid (and as well disipated in Q1, the input CSR and the battery's Rint).

The LED turns on when no current is being pulled from the battery (Q1 is off). 
Actually the light curve shows that there is still some current drawn during the time the LED is on. The current doesn't return to (near) zero until the LED turns off. Please see the trace below, which is the phototransistor output and the input battery current.
Quote
As well, if there is any battery recharging from the collapse of the toroid, it is occurring during the rising portion of the Vin waveform from just past the most negative peak when the LED is on.  During that period (whilst the waveform is rising) the LED is on and current must flow thru the LED, output CSR, Input CSR, input battery, and the toroid to complete he circuit.  The collasing toroid acts as an additional battery in series with, and with a voltage higher than, the input battery.  The polarity is such that the current thru the LED and toroid tends to raise the terminal voltage of the input battery slightly during the LED on time.  So, if anything, the battery is actually recovering a bit of its charge when the LED is on.

The internal R of the input battery will affect the amount of ripple seen on Vin, and the amount of battery depletion determines the battery internal R.

What happens to your circuit with a fresh alkaline?

PW

I don't know, I'll have to try to find one.
In the traces below, the top trace is the output from the phototransistor looking at the LED's light. The bottom trace is the Input Current. Both traces are using the exact center graticule marker as the zero baseline.
The crossing of the zero baseline by the input current might be a DC offset error in the scope itself, especially since this signal is displayed at such a high amplification. In the shot below my input current does not cross the zero baseline, whereas using the weaker button cell and a different vertical attenuation, it did.

TinselKoala

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Re: Is joule thief circuit gets overunity?
« Reply #526 on: April 10, 2013, 02:55:05 AM »
Using an absolutely fresh Duracell AA that measured 1.617 V unloaded, I got the PT light detector output and current input traces below, at the same vertical settings and same center baseline. I also tried a much weaker battery. (The horiz setting is different, the fresh battery allowed quite a bit higher frequency for a while.)

The residual current is greater for the fresher battery. As the battery voltage decreases, the current trace gets closer and closer to the zero current baseline. For my system, I need to use a pretty weak battery to get the input current to appear to cross the zero baseline.


TinselKoala

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Re: Is joule thief circuit gets overunity?
« Reply #527 on: April 10, 2013, 03:32:05 AM »
Yes, I suppose the current in Lawrence's shots is decaying faster than in mine. And the significance of this is.... ? I think in his recent shots he's using some kind of rechargeable battery, isn't he?
Now I can't even remember what battery I used for this shot, but the two comparisons of Lawrence are the board 80 and another one from earlier in the thread.

picowatt

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Re: Is joule thief circuit gets overunity?
« Reply #528 on: April 10, 2013, 03:35:01 AM »
Using an absolutely fresh Duracell AA that measured 1.617 V unloaded, I got the PT light detector output and current input traces below, at the same vertical settings and same center baseline. I also tried a much weaker battery. (The horiz setting is different, the fresh battery allowed quite a bit higher frequency for a while.)

The residual current is greater for the fresher battery. As the battery voltage decreases, the current trace gets closer and closer to the zero current baseline. For my system, I need to use a pretty weak battery to get the input current to appear to cross the zero baseline.

TK,

What are these traces again?  Is the lower trace really the input current?  If so, where is the zero ref?  If it is at the masking tape with "2", then I am confused.

Could the lower trace actually be input V?

PW

poynt99

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Re: Is joule thief circuit gets overunity?
« Reply #529 on: April 10, 2013, 03:54:31 AM »
@PW, .99:
In Lawrence's recent scopeshots of Board 80, the one that I have edited to show the _true_ baseline for the CH2 current measurement.... note that the scope is reporting + 40.00 mV RMS for that signal...... which barely ever actually goes positive, never even reaching 40 mV positive that I can see, and extending deeply negative during its spikes.  RMS, of course, is mathematically always a positive value..... so how are we to interpret a _positive_ RMS value for a signal that is actually almost always, if not strictly always, negative? The RMS value gives the wrong direction for the current being measured in this channel, doesn't it?
Lawrence has explained that the RMS boxes are a "legacy" from his early work. I hope it's clear now that these values are not relevant and may even be confusing the issue, due to things like the positive RMS value for a clearly negative-average signal. They should not be displayed at all. If the scope has a simple "average" parameter available, that might provide some rough and useful information. An artificially positive RMS value does not.
Tk, dare I say you answered your own question. The scope is reporting a positive rms value because rms inherently is always a positive number.

TinselKoala

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Re: Is joule thief circuit gets overunity?
« Reply #530 on: April 10, 2013, 04:31:19 AM »
TK,

What are these traces again?  Is the lower trace really the input current?  If so, where is the zero ref?  If it is at the masking tape with "2", then I am confused.

Could the lower trace actually be input V?

PW
Yes, I think you are confused, but not about what you are thinking you are confused about. You are confused about your confusion, probably.

"At the same vertical settings and the same baseline" in my post you quoted was meant to refer to the immediately previous shot in post 525, where I explain that for these two screens I am using the exact center horizontal graticule marker for both displayed traces.  Sorry I wasn't too clear about that; taking shortcuts sometimes often seems to wind up taking longer in the long run. This is to bring them closer together so that it's easier to see the temporal relationships.
The lower trace is the input current, measured as usual at point A4 so it shows negative values, and it's displayed at 100 mV per division. ETA: and the baseline is at the center horizontal graticule line.

TinselKoala

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Re: Is joule thief circuit gets overunity?
« Reply #531 on: April 10, 2013, 04:43:14 AM »
If anyone's interested in the NTE3037 phototransistor measurement, I've made another short video where I drive a white LED with the function generator, using a triangle and a sine wave, at around 3kHz, and I show the response of the phototransistor to the changes in brightness of the LED. It seems quite satisfactory at this low frequency.

The video is uploading and should be available in a few minutes at:
 http://youtu.be/me3kPvrOLi0

ltseung888

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Re: Is joule thief circuit gets overunity?
« Reply #532 on: April 10, 2013, 04:44:14 AM »
One thing you must realize is that those traces were done with the tiny hearing aid battery that is mostly depleted. Here are some traces from this board for comparison to Lawrence's traces, that I just made with an alkaline AA battery that reads 1.304 volts unloaded... that is, it too is fairly well depleted.

I don't usually use this much input power, but this is more comparable to Lawrence's input.

In the first shot, Output, the top trace is the Voltage Output at B1, the bottom trace is the Current Output at B3. The top trace is set to 2 volts per division, the bottom trace at 100 mV per division. (The top trace is using the center horizontal graticule line as the baseline, not the line indicated by the number "1" to the right. The lower trace is using the baseline indicated by the "2".)  I am showing only a single pulse to make the point that the waveshapes are the same, except that my "60 MHz" analog scope isn't displaying the spike amplitude.... but I assure you that it is there. Timebase is at 10 microsec/div and the delay function is used to bring a pulse onto the screen window.

The second shot is the Input, the top trace is the Input Battery Voltage at A1 and the lower trace is the Input Current at A4. The top trace is at 500 mV per division and is using the center horizontal graticule marker as its baseline, not the number to the right. The lower trace is at 100 mV per division, is using the numbered graticule line "2" as its baseline, and clearly and repeatably shows values both above and below the baseline. No "invert" is used and the probe is positioned just like Lawrence's is wrt current direction. Timebase is at 50 microsec/div.

So you can presumably see clearly that, if Lawrence's board is OU.... then mine must be too, since it gives the same instrumental readings when powered and probed in the same way that his is..... excepting the appearance of the high-frequency spikes.

And your detailed explanation seems to concur with what I said: the LED turns on at the bottom of the valleys, at the point where the slope reverses, and the power represented by the decreasing voltage slope to that point is the power that is pulsed into the LED.
@TK,
 
From my experience of looking at the many JTs scope shots, your JT is likely to be OU.  If you have a Digital Scope with saving  CSV data capability, you can easily confirm it with Excel.  Your finding that a weaker battery shows more "crossing" of the zero line for Input Current is also correct - same as my results.
 
I have scheduled to go to Shenzhou tomorrow to work with Mr. Zhou and take more photos - probably with pretty models too.  That will be fun.....
« Last Edit: April 10, 2013, 07:07:23 AM by ltseung888 »

picowatt

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Re: Is joule thief circuit gets overunity?
« Reply #533 on: April 10, 2013, 04:54:00 AM »
Yes, I think you are confused, but not about what you are thinking you are confused about. You are confused about your confusion, probably.

"At the same vertical settings and the same baseline" in my post you quoted was meant to refer to the immediately previous shot in post 525, where I explain that for these two screens I am using the exact center horizontal graticule marker for both displayed traces.  Sorry I wasn't too clear about that; taking shortcuts sometimes often seems to wind up taking longer in the long run. This is to bring them closer together so that it's easier to see the temporal relationships.
The lower trace is the input current, measured as usual at point A4 so it shows negative values, and it's displayed at 100 mV per division. ETA: and the baseline is at the center horizontal graticule line.

TK,

Me?  Confused?  Yeah sure, that would be a first...

Did you mean to say at "A2" point s opposed to "A4"?  A2 is the input CSR/battery junction.  A3 and A4 should be the ground ref at the opposing end of the input CSR to which probe ground leds are attached.

So, if the center of the graticule is the zero ref, it is strange that you never get to zero current (or even go slightly above the zero ref) like Lawrence does.  Could your Q1 base be leakey?  Just a thought, as that could be a sneak path for current.  Otherwise I can't figure why you're showing current draw ALL the time.  Am I correct to assume that connecting the probe tip used at A2 to the ref ground at A3-A4 puts the trace at the zero ref line?


PW

picowatt

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Re: Is joule thief circuit gets overunity?
« Reply #534 on: April 10, 2013, 05:19:50 AM »
TK,

Forgive me, I see that I am indeed confusing the A1, etc points.  Back to Vin, Iin, Vout, Iout for me!

Can I use the "must be gettin' old" card?

In you reply #522, The input current trace looks as it should, going to zero and possibly just above zero.

What is different now?

PW

picowatt

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Re: Is joule thief circuit gets overunity?
« Reply #535 on: April 10, 2013, 05:47:54 AM »
TK,

When you get the time, see if you can duplicate the reply 522 scope shots.

Particularly with regard to the Iin trace.

Could the Iin channel not be correctly set at your zero ref line "2" in that post?

PW

TinselKoala

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Re: Is joule thief circuit gets overunity?
« Reply #536 on: April 10, 2013, 06:53:58 AM »
TK,

Me?  Confused?  Yeah sure, that would be a first...

Did you mean to say at "A2" point s opposed to "A4"?  A2 is the input CSR/battery junction.  A3 and A4 should be the ground ref at the opposing end of the input CSR to which probe ground leds are attached.

So, if the center of the graticule is the zero ref, it is strange that you never get to zero current (or even go slightly above the zero ref) like Lawrence does.  Could your Q1 base be leakey?  Just a thought, as that could be a sneak path for current.  Otherwise I can't figure why you're showing current draw ALL the time.  Am I correct to assume that connecting the probe tip used at A2 to the ref ground at A3-A4 puts the trace at the zero ref line?


PW

Now you are really confusing me. I am using the letters and numbers as shown in this schematic attached below. I have always used these numbers, and I thought that everyone else was doing so as well. Is there a more recent schematic showing the test point locations? The only difference is that there is no capacitor on my board, nor on the ones Lawrence is testing now, right?

There is No probe tip at A2. A2 is NOT the junction of the input battery and the input CVR. CSR, whatever. A3 and A4 are NOT connected together, there is the input CVR between them. There is a probe tip, the input current probe, at A4 and its reference ground lead is at the common circuit ground at A2A3B2B4. As shown in the schematic below, and as noted in the legend of my photo of the PCB board above.

A2, A3, B2 and B4 are all connected together as shown in the diagram as the common "ground" and as I have noted in the legend to my photo of the PCB above. A1 is the battery positive terminal. A4 is the battery side of the input CVR, and this gets the current probe TIP, and so winds up being inverted.

Yes, grounding the probe tip to its reference lead produces the same baseline as switching the channel coupling switch to "ground". Yes, with weak input battery I get the input current trace "apparently" crossing the zero line and with stronger batteries it does not do so.

As Lawrence has confirmed, the stronger the battery used, the greater the "residual" or leakage current and the further negative the input current trace goes. What I find remarkable is that the wave shape itself is the same, even though it can move up or down wrt the reference depending on the battery's state of charge.

In the photos that show the PT output, there is NO connection between the PT circuit and the JT circuit, except thru the instrument ground leads, and there is no way that power can get from the PT supply to the JT, or vice versa.

Can't handle the letters and numbers, eh...?






TinselKoala

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Re: Is joule thief circuit gets overunity?
« Reply #537 on: April 10, 2013, 07:00:01 AM »
TK,

Forgive me, I see that I am indeed confusing the A1, etc points.  Back to Vin, Iin, Vout, Iout for me!

Can I use the "must be gettin' old" card?

In you reply #522, The input current trace looks as it should, going to zero and possibly just above zero.

What is different now?

PW
Ah, OK, I replied to your last post before I saw this one. So we are clear on the letter-number designations now?  ;D

Lawrence says that he too has noticed that stronger input batteries lower the entire input current trace and that weaker input batteries are more likely to show the zero-crossing behaviour in the input current trace. The difference you ask about is that the 522 trace was made with a depleted button cell, either a LR44 or a 392, I've used both, and these cells cannot provide the current output that a AA can, even if the AA is relatively depleted. The later traces were made with a little stronger AA, and then the last ones by your request with a new AA, and it showed the most negative current trace (most residual or "leakage" current). Since Lawrence's systems behave the same way, this is more "evidence" that our circuits have similar performance when tested similarly.

EDIT: Sorry, I see now that my notes say I used a AA battery that was at 1.3 v unloaded for those traces, not a button cell. Apologies, things are moving fast and my lab is in chaos, as you can probably tell.

TinselKoala

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Re: Is joule thief circuit gets overunity?
« Reply #538 on: April 10, 2013, 07:14:38 AM »
TK,

When you get the time, see if you can duplicate the reply 522 scope shots.

Particularly with regard to the Iin trace.

Could the Iin channel not be correctly set at your zero ref line "2" in that post?

PW
Ah, I see now that I used a AA battery that was at about 1.3 volts unloaded for those traces. Sorry, I thought I had used a button cell.
I'm sure I can duplicate it, I've got plenty of dead and half-dead AA cells lying about. But the scopes are cold for the night, I may get to it tomorrow or later in the morning.
Yes, I checked the baseline position several times both by the coupling switch set to Ground and also by manually shorting the probe tip to its ground clip.  The zero-crossings are as real here as they are in Lawrence's boards.
Check back in the morning, I'll see about reproducing it again.

ltseung888

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Re: Is joule thief circuit gets overunity?
« Reply #539 on: April 10, 2013, 12:52:16 PM »
@TK,
 
From my experience of looking at the many JTs scope shots, your JT is likely to be OU.  If you have a Digital Scope with saving  CSV data capability, you can easily confirm it with Excel.  Your finding that a weaker battery shows more "crossing" of the zero line for Input Current is also correct - same as my results.
 
I have scheduled to go to Shenzhou tomorrow to work with Mr. Zhou and take more photos - probably with pretty models too.  That will be fun.....

I am puzzled that wih only an analog oscilloscope, how can one do the spreadsheet analysis?  Without the spreadsheet, how can one do the multipliction and averaging of thousands of sample points?