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New theories about free energy systems => Understanding OverUnity => Topic started by: nul-points on February 02, 2016, 07:23:16 AM

Title: To be deleted
Post by: nul-points on February 02, 2016, 07:23:16 AM
....Temporary home for the flyback flashlight thread data, while we wait to see if Stefan has any ca jones...
Title: Re: To be deleted
Post by: nul-points on February 02, 2016, 07:24:23 AM
...single 5mm LED light level...
Title: Re: To be deleted
Post by: nul-points on February 02, 2016, 07:25:54 AM
...runtime extended by swapping o/p store back as i/p...
Title: Re: To be deleted
Post by: nul-points on February 02, 2016, 07:27:15 AM
...Itsu replication - LED waveform & power...
Title: Re: To be deleted
Post by: nul-points on February 02, 2016, 07:29:59 AM
...Illumination intensity comparison - a) standard 'grounded' LED drive - b) flyback flashlight drive
Title: Re: To be deleted
Post by: nul-points on February 02, 2016, 07:31:19 AM
...flyback flash-lite configuration...
Title: Re: To be deleted
Post by: nul-points on February 02, 2016, 07:33:09 AM
...Itsu replication - I/p, o/p & LED power traces...
Title: Re: To be deleted
Post by: nul-points on February 02, 2016, 07:36:33 AM
...comparison of Illumination intensity vs runtime; both using 3x 750mAh NiMH & commercial 24 white LED head- a) commercial flashlight (straight DC drive) - b) np flyback flash-lite drive - runtime of flyback flash lite is 10 hours vs commercial flashlight at 5 hours...
Title: Re: To be deleted
Post by: nul-points on February 04, 2016, 02:38:44 PM
Flyback 'Vampyre' circuit variant; sucks out the last drops of energy from 'Used' primary cells - and transfers the energy into secondary (rechargeable) batteries

Can also be used with solar cell i/p to trickle charge storage devices

Photo below shows circuit being used to drain a discarded AA 1.5V alkaline cell, and transfer the energy to an 8.4V NiMH battery

C1 - 100uF 16VTantalum;
VR1 - multi-turn 10K Varpot;
Q1 - low power, hi-gain NPN transistor, eg. BC547, BC337, etc;
D1 -  low-leakage Schottky diode, eg. BAT 42;
T1 - approx 40:120 turns magnet wire on ferrite toroid;
device shown uses EMI filter split-ferrite core, approx 25mm OD, 10mm high with 9-filar x 0.09mm litz
(mono-filament wire ok, just keep pri:sec ratio approx 1:3)


Title: Re: To be deleted
Post by: nul-points on February 08, 2016, 10:26:31 PM
Start of comparative tests between straight DC drive of a compound resistive load
and pulse drive of the same load, using the same fully-charged NiMH batteries and the
Flyback Flash-lite circuit

These results will give an indication of the efficiency of this pulse circuit compared to
a benchmark of the equivalent amount of DC energy converted

The compound load is a multiple-watt, ceramic-encased, 50ohm resistor, in series with
a BAT42 Schottky diode. The diode & resistor are pressed together in close proximity
with a temperature probe, completely encased inside approx 18mm (0.75") thick
compressed paper insulation

The datalog of the DC test run shows that the fully charged NiMH battery (3x 750 mAh cells)
was able to provide power to the load for approx 13.75 hours over a voltage range of 4.3 to 3.3V
(ie. fully-charged to fully-discharged)

The DC test was able to achieve a maximum of approximately 13.5 degF of temperature differential
over ambient (the temperature-differential profile for the complete test run, of course,
varied slightly as the battery voltage decreased with discharge)

The next test will be to operate the Flyback Flash-lite cicuit with the same insulated load,
powered by the same fully-charged NiMH cells (having adjusted the power level of the circuit
to give a similar initial load-temperature rise over ambient) and compare the run duration
with that of the DC test

IF (and this is a big 'if'!) this pulse circuit could deliver the same level of heat at the load for
the same duration as the straight DC drive achieved, then the circuit would be 100% efficient.
This in itself would be a remarkable achievement

If the circuit can match the same level of heat as the DC drive, but for less time,
then the pulse circit will be underunity (ie. some level of efficiency less than 100% - situation normal)

Moving into the realm of flying pigs, if the circuit can match the same level of heat as
the DC drive, but for a longer duration, then the pulse circit will have provided
evidence of overunity operation

Mesdames et messieurs, faites vos jeux...

(results of comparative test should be ready towards the end of today, GMT, Tuesday 9th Feb)

np
Title: Re: To be deleted
Post by: nul-points on February 09, 2016, 12:09:11 PM
Cerdos Voladores!

Hah  - that's a close call !

Well, the datalog results below, for the Flyback Flash-lite test run, show that it operated for 14.5 hours (for the same supply drop, using the same load, and set to the same initial temperature differential as the DC drive run)

So this pulse circuit operated for an extra 45minutes, heating the same output load to the same level as the DC circuit (as monitored for at least the first half of its run duration)

Further testing will be required to confirm its performance, but the results are very promising

More to follow
np
Title: Re: To be deleted
Post by: nul-points on February 13, 2016, 05:58:09 AM
Flyback Flash-lite test, 9th Feb '16:-

Use Flyback Flash-lite circuit supplied by 3x NiMH 750mAh cells (battery 'A') to charge 3x emptied NiMH 750mAh cells (battery 'B') in series with 50 ohm resistor; output from circuit via a BAT42 Schottky diode (see schematic above, Reply #5)

The 50 ohm resistor/ series diode arrangement is 'standing-in' for an LED head, to help quantify some test values; it presents a comparable load level in its own right, energy-wise, and it demonstrates the underlying principle of this Flyback Flashlight operation that, since work is not conserved,  the same energy can be used more than once, en route to the Entropy trashbin


Datalog results attached below, show discharge voltage profile for battery A (red trace), and charging voltage profile for battery B (blue trace)

Previous results posted above (Reply #9) show discharge profile for fully-charged battery A:-
(Straight DC drive of 50 ohm resistor in series with BAT42 diode)

Approx. avg. supply volts:  ((4.25-3.3)/2)+3.3  = 3.77V

Approx. volts drop across series BAT42: 0.3V

Approx discharge current: 3.47/50 = 69.5 mA

Load power:  (0.3 + 3.47) x  69.5 = 262 mW

Runtime duration: 13.75 hrs

Supply energy (NiMH): 262 x 13.75 = 3600 mWh

Back calc. of NiMH mAh: 
  energy per cell = 3600 / 3 = 1200 mWh
  av volts per cell = 3.77 / 3 = 1.26V
cell capacity = 1200 / 1.26  = 950 mAh


The same DC discharge test procedure was then used to determine the amount of energy transferred to battery B, in this latest test below (using relative discharge time into same load as for fully-charged battery A, above)

DC discharge duration test of Battery B (via 50 ohm / BAT42 in series):  5.5 hours  (from 4.2V to 3.3V) approx

Approx. Energy in Batt B:  (5.5 / 13.75) x 3600 = 1440 mWh

Coulometric charging efficiency of NiMHs:  70% approx
(source: Powerstream.com)

Approx charge input energy:  1440 / 0.70 = 2060 mWh

Approx energy transferred via 50 ohm/ BAT42 : 2060 mWh
(because this series load transferred all the energy used to charge battery B)

approx COP of Flyback Flash-lite when charging battery B, from empty, and heating 50 ohm resistor, from fully-charged battery A
(2 x 2060)  / 3600 = 1.14

(ie. an Efficiency of 114%)


NB.  i shouldn't have to spell this out, but we *do* have some er, 'excitable' members (on both sides of the Unity fence) so, E&OE, YMMV, Caveat Emptor, etc, etc - these are 'ballpark' figures - the clue is in the hefty use of the term 'approx'

np
Title: Re: To be deleted
Post by: nul-points on February 20, 2016, 06:31:49 PM
Hello reader

...an interesting development here at Flyback Fortress

First, some context...

One of my early projects, back in 2007, when i started to get interested in FE,  was the Jensen Unidirectional Transformer (UDT)

There wasn't much info about it, and since i didn't have any transformer core laminations at the time, i decided to try and create a toroid version using EMI ferrites

(NB. This was about a year before Mr Heins hit the scene with his BiToroidTransformer)

I'm not proud of my build quality then, but the 'Triformer' was born - see photo below for that first version (3x thin toroids)

[Edit 1] added photo of more recent Triformer, 2x thin primary toroids, 1x thick secondary toroid (T Heins recommended this improvement to my original 2007 version); similar to current Flyback flash-lite test run, but without conventional secondary mod


Although Mr Jensen's notes seemed to suggest that the UDT be used with AC waveforms, i've only ever used the Triformer with pulsed DC waveforms

The only time, before now, that i've seen any benefit from the Triformer configuration, was when i used it with a flyback oscillator to charge batteries from a solar panel - keeping all other settings the same, the charge current could be increased just by shorting an unused winding

Ok - you've guessed it - i've now decided to try one of my Triformers with the Flyback flash-lite circuit

On this version of the Triformer i've added a conventional secondary over the top of the primary, this leaves two secondaries available at the junctions with the primary toroids and the 3rd toroid

2nd test still in progress (because its runtime has been extended by at least a day!) - 1st test runtime lasted 24 hours illuminating a 1W LED using a 2x 750mAh cell NiMH battery with the Triformer connected as a conventional transformer

On this 2nd run i repeated the test but shorted both secondaries - so far, the circuit has been illuminating the same LED for nearly 40 hours and looks like it could double the runtime of the 1st test!

When i started the 2nd run, i wasn't expecting anything significant, so i'll need to rerun these two tests and include the photometer output in the datalog results, to check if the extended runtime has been traded off against illumination level

More later
np

[Edit 2]  Test run 2 has just completed - 48 hours (double the runtime than with secondaries unshorted) - datalog traces of battery discharge below -  Unshorted Test :   0-24hrs;  Shorted Test starts @ 24hrs, finishes @ 72hrs

i'm rerunning those tests now with photometer readings added - i'll try & adjust the shorted test power to give equivalent illumination to compare runtimes directly

np
Title: Re: To be deleted
Post by: PARAV on February 26, 2016, 02:43:51 PM
Hi nul-points,


I have been eagerly watching and waiting for your latest test which I think was showing some real promise with the " triformer".


Please let us know how it's going and give us an update.


Thanx in advance--Paul
Title: Re: To be deleted
Post by: nul-points on February 26, 2016, 07:22:46 PM
Quote from: PARAV on February 26, 2016, 02:43:51 PM

I have been eagerly watching and waiting for your latest test which I think was showing some real promise with the " triformer".

Please let us know how it's going and give us an update.

Thanx in advance--Paul

Hi Paul, apologies for the delay - interesting things seen with the Triformer, so i'm now backtracking and trying it with some of the flyback flashlite circuit variants which i've explored & parked

I'll post updates again asap

thanks for your interest ...and Hi to all the other folk quietly watching this thread in the background - appreciated!
np
Title: Re: To be deleted
Post by: seaad on February 27, 2016, 07:05:47 AM
.
Title: Re: To be deleted
Post by: nul-points on February 27, 2016, 08:25:47 AM
Great!  Got it - thanks, Arne

np
Title: Re: To be deleted
Post by: ramset on February 27, 2016, 08:28:00 AM

NP
I did just want to make you aware of this thread

http://www.energeticforum.com/renewable-energy/20370-thats-not-knife-knife-er-ou-flashlight.html#post286432

respectfully

Chet
Title: Re: To be deleted
Post by: nul-points on February 27, 2016, 02:52:04 PM
Appreciate what you're doing there - many thanks Chet

np
Title: Re: To be deleted
Post by: nul-points on December 02, 2018, 06:14:21 PM

firstly, apologies for the long gap in posting results here - i've had health issues


i've greatly appreciated the positive and interesting exchange of ideas and results with some good friends who i met here on OU.com - you know who you are!


this is likely to be my last posting on OU (for a number of reasons, not just health) and i wanted to share some results which have followed on from my previous experiments with charge-switching and feedback


i have attempted to 'distil' all that i've observed and learnt from my previous solid-state experiments

as in my orevious experiments, the switching circuit shown below operates at higher than audio frequencies (approx 130 kHz here), in 2 half-cycles a) & b):

a) the transistor switches on, its collector current illuminates the output LED(s) and also stores energy in the coil field and electrolytic cap; when the coil saturates then the transistor turns off;  enough energy remains in the electrolytic cap to supply the output LED(s) until the next cycle (the cap voltage is DC with minimal ripple)

b) current supplied from the energy stored in the collapsing coil-field and the electrolytic cap is fed back to the battery (while the transistor is off) via the feedback LED - IFF the feedback LED is illuminated then the battery is receiving charge in this half-cycle (in this case, equivalent to approximately one-fifth of the output current)

(over the last few years i've tested a LOT of exotic transformer designs, solenoidal/'partnered'/(multi)/toroidal - i don't think that there is anything unusual about the transformer here - approx 1cm^3 volume of ferrite, solenoidal wound, almost completely enclosed by the ferrite;  the wire is 2x45x0.45cm diam. insulated copper, 5:5 turns)

(Fig.1 below, Schematic)


for these tests, the battery is an LIR providing 4.15V;  supply current is 5.4mA

pk voltage of feedback pulse across feedback LED + battery is approx 7V;  pulse width approx 1.5 us

(Fig 2. below, Voltage pulse across Feedback LED)


so-called 'conventional' current flows OUT of the positive terminal of the battery in half-cycle a) and INTO that terminal in half-cycle b)

voltage across 1ohm Current Sensing Resistor in +ve supply line shows approx same peak magnitude current flows, into and out of the battery, matching the 20% duty cycle for the respective average current flows

(Fig 3. below,  bidirectional Voltage across 1ohm CSR in +ve supply line)


(NB.  all values measured using a True RMS meter and confirmed in order of magnitude using a 'scope)

feedback current:  1.18mA (True RMS)
feedback LED load:  approx 2.6V * 1.18mA = 3mW approx

main LED branch current:  6.5mA (True RMS)
main LED branch load:  4.15V * 6.5mA = 26.98mW

voltage across electrolytic//main LEDs is 2.7V DC
main LED load:  2.7V * 6.5mA = 17.55 mW

switching cct load:  26.98 - 17.55 = 9.43mW

total LED load:  17.55 + 3 = 20.55mW

total power load:  20.55 + 9.43 = 29.98mW

supply current:  5.4mA (True RMS)
total power supply:  4.15 * 5.4mA = 22.41mW

Efficiency  n  =  (total load / supply)  =  (29.98 / 22.41) = 134%

results shown are instanteous power, mW (these are proportional to the energy being converted, mWh)


*** Edited to add...***


Feeding back the flyback energy, plus some stored in the electrolytic cap, has taken the whole device OU - more work is being achieved for less supply because energy is being 'recycled' back into the battery.  energy is conserved - but work isn't.


if the feedback is removed, the supply current increases; it becomes the only current path, however it is lower than the load current with feedback - less work is achieved for a higher supply level

Title: Re: To be deleted
Post by: tinman on December 03, 2018, 09:34:37 AM
Quote from: nul-points on December 02, 2018, 06:14:21 PM
firstly, apologies for the long gap in posting results here - i've had health issues


i've greatly appreciated the positive and interesting exchange of ideas and results with some good friends who i met here on OU.com - you know who you are!


this is likely to be my last posting on OU (for a number of reasons, not just health) and i wanted to share some results which have followed on from my previous experiments with charge-switching and feedback


i have attempted to 'distil' all that i've observed and learnt from my previous solid-state experiments

as in my orevious experiments, the switching circuit shown below operates at higher than audio frequencies (approx 130 kHz here), in 2 half-cycles a) & b):

a) the transistor switches on, its collector current illuminates the output LED(s) and also stores energy in the coil field and electrolytic cap; when the coil saturates then the transistor turns off;  enough energy remains in the electrolytic cap to supply the output LED(s) until the next cycle (the cap voltage is DC with minimal ripple)

b) current supplied from the energy stored in the collapsing coil-field and the electrolytic cap is fed back to the battery (while the transistor is off) via the feedback LED - IFF the feedback LED is illuminated then the battery is receiving charge in this half-cycle (in this case, equivalent to approximately one-fifth of the output current)

(over the last few years i've tested a LOT of exotic transformer designs, solenoidal/'partnered'/(multi)/toroidal - i don't think that there is anything unusual about the transformer here - approx 1cm^3 volume of ferrite, solenoidal wound, almost completely enclosed by the ferrite;  the wire is 2x45x0.45cm diam. insulated copper, 5:5 turns)

(Fig.1 below, Schematic)


for these tests, the battery is an LIR providing 4.15V;  supply current is 5.4mA

pk voltage of feedback pulse across feedback LED + battery is approx 7V;  pulse width approx 1.5 us

(Fig 2. below, Voltage pulse across Feedback LED)


so-called 'conventional' current flows OUT of the positive terminal of the battery in half-cycle a) and INTO that terminal in half-cycle b)

voltage across 1ohm Current Sensing Resistor in +ve supply line shows approx same peak magnitude current flows, into and out of the battery, matching the 20% duty cycle for the respective average current flows

(Fig 3. below,  bidirectional Voltage across 1ohm CSR in +ve supply line)


(NB.  all values measured using a True RMS meter and confirmed in order of magnitude using a 'scope)

feedback current:  1.18mA (True RMS)
feedback LED load:  approx 2.6V * 1.18mA = 3mW approx

main LED branch current:  6.5mA (True RMS)
main LED branch load:  4.15V * 6.5mA = 26.98mW

voltage across electrolytic//main LEDs is 2.7V DC
main LED load:  2.7V * 6.5mA = 17.55 mW

switching cct load:  26.98 - 17.55 = 9.43mW

total LED load:  17.55 + 3 = 20.55mW

total power load:  20.55 + 9.43 = 29.98mW

supply current:  5.4mA (True RMS)
total power supply:  4.15 * 5.4mA = 22.41mW

Efficiency  n  =  (total load / supply)  =  (29.98 / 22.41) = 134%

results shown are instanteous power, mW (these are proportional to the energy being converted, mWh)


*** Edited to add...***


Feeding back the flyback energy, plus some stored in the electrolytic cap, has taken the whole device OU - more work is being achieved for less supply because energy is being 'recycled' back into the battery.  energy is conserved - but work isn't.


if the feedback is removed, the supply current increases; it becomes the only current path, however it is lower than the load current with feedback - less work is achieved for a higher supply level

Where are your CVRs in the schematic?
Surly you are not using a current probe at these low power levels?


Brad
Title: Re: To be deleted
Post by: nul-points on December 03, 2018, 10:30:51 AM
hi Brad, how are ye?   nice to meet you too


"voltage across 1ohm Current Sensing Resistor in +ve supply line shows approx same peak magnitude current flows, into and out of the battery, matching the 20% duty cycle for the respective average current flows

(Fig 3. below,  bidirectional Voltage across 1ohm CSR in +ve supply line)

(NB.  all values measured using a True RMS meter and confirmed in order of magnitude using a 'scope)"


[edited to show context]





Title: Re: To be deleted
Post by: itsu on December 03, 2018, 02:47:01 PM

Hi NP,

sorry to hear about your problems, i hope they change for the better soon.

Your "Efficiency  n  =  (total load / supply)  =  (29.98 / 22.41) = 134%"  warrants a further investigation.

Could you be more specific about the transformer though?

You mention:
approx 1cm^3 volume of ferrite, solenoidal wound, almost completely enclosed by the ferrite;  the wire is 2x45x0.45cm diam.
insulated copper, 5:5 turns).


I cannot picture this in my mind, so could you perhaps show a picture?
The 5:5 and 2x45x0.45cm  (i guess you mean mm) is not clear to me (2x 5 turns, or 2x 45 turns, ontop of each other
or next to each other, insulated copper = magnet wire?).

Thanks,   regards Itsu
Title: Re: To be deleted
Post by: nul-points on December 03, 2018, 05:46:29 PM
hi itsu, good to 'see' you again, i hope you are well

thanks, my motivation and attention to detail are just 2 of the things affected, as you may sense from my lack of posting, and also thinking that 0.45cm could possibly be correct as a wire diameter fitting inside 1cm^3 volume of ferrite - hah!

...i tried so hard to make sure that i reported all the details correctly, LOL


yes, the excess power value took me by surprise, too - but the circuit is so simple, with only 2 main current paths between supply, so that it was more obvious that there was something unusual happening

previous circuits which had LED loads and also stored some energy in a 2nd battery showed energy could be stored and reused but this is an interesting development to achieve with just the 1 supply battery


ok, the transformer...

yes, i should have described 2 equal lengths of 0.45mm magnet wire, cut from 1 piece 45cm long, wound as 5 turns each - i guess they could be wound on as bifilar, but i wound 1 turn of say primary, then 1 turn of secondary, then 2nd turn of primary, then 2nd turn of secondary, etc

i didnt have a fully enclosed ferrite 'pot' of 2 halves, so i improvised, i'll try to attach a photo of something equivalent** - the total volume of ferrite is approx 1cm^3  just enough to enclose the windings inside a thin shell of ferrite

(**my xfr is small, dark ferrite taped up with black tape - so i don't think it will photo well)

[Edit: corrected winding details;  added photo to show comparative size of transformer 'pot core', black rectangular object, centre of proto-board, approx 20x10x5mm, some wire is exposed at each end, topologically similar to a hollow torus containing 2 interleaved solenoidal windings, similar to bifilar wound.  poor quality photo using phone - sorry!]


i hope this helps!

thanks for reading, good luck with your experiments
Title: Re: To be deleted
Post by: doktorsvet on December 04, 2018, 01:54:34 AM

Found a picture of the OS on the Internet.
I wound 5 sections of 10 + 5 turns in each
Title: Re: To be deleted
Post by: itsu on December 04, 2018, 04:07:29 AM

Hi NP,

i am fine thanks.

Ok, i think i get the picture now  :)

I have some pot cores, but not that small, so have to look for one in my junkbox or improvise.
Rest of the parts should be no problem.

I will try to follow your line of measurements and calculations and hopefully reach similar results.


Be well,   regards Itsu

Title: Re: To be deleted
Post by: Void on December 04, 2018, 05:50:38 AM
Quote from: nul-points on December 02, 2018, 06:14:21 PM

(NB.  all values measured using a True RMS meter and confirmed in order of magnitude using a 'scope)

feedback current:  1.18mA (True RMS)
feedback LED load:  approx 2.6V * 1.18mA = 3mW approx

main LED branch current:  6.5mA (True RMS)
main LED branch load:  4.15V * 6.5mA = 26.98mW

voltage across electrolytic//main LEDs is 2.7V DC
main LED load:  2.7V * 6.5mA = 17.55 mW

switching cct load:  26.98 - 17.55 = 9.43mW

total LED load:  17.55 + 3 = 20.55mW

total power load:  20.55 + 9.43 = 29.98mW

supply current:  5.4mA (True RMS)
total power supply:  4.15 * 5.4mA = 22.41mW

Efficiency  n  =  (total load / supply)  =  (29.98 / 22.41) = 134%

results shown are instanteous power, mW (these are proportional to the energy being converted, mWh)


Hi nul-points. I tried to fairly evaluate what you have described above, but right away see the following issues:
Most true RMS meters or regular DVM's are not spec'd for use to anywhere near 130 kHz, although a few may
possibly be spec'd for use at that high of frequency. What true RMS meter were you using (make and model #)?
Do the specs for that meter indicate it can be used to measure AC currents up to at least 130 kHz?

Your terms and calculations such as 'main LED branch load', 'main LED load', 'switching cct load', 'total LED load' just aren't
making much sense to me. You would have to clarify those terms and explain or show exactly how and where you are making
those measurements if you want people to be clear about the measurements you are describing. When measuring
power in AC circuits, the phase angle between the voltage and current is very important. When you are talking about
AC circuits with pulse or spikey waveforms, power measurement can be a lot a trickier. At any rate, you can't just throw
a DVM or true RMS DVM in at different points in AC circuits to measure current and ignore phase angles.

Also, above you said "results shown are instanteous power, mW", but you can't calculate efficiency based on instantaneous power,
and you can't measure instantaneous power if you are using a true RMS meter to measure current. To calculate efficiency,
you must measure/calculate average output power and average input power. So what you said about 'instantaneous power,' really doesn't
appear to make any sense.

Not trying to be difficult at all. Just some honest feedback on what was presented. I am trying to be as nice about it as I can, but I
know from experience that honest feedback will more often either be ignored or attacked rather than being fairly considered. :)
My comments are directed to anyone in general who cares about really trying to understand how things are really performing
in circuits such as this. IMO, an experimenter must be willing to examine and question all assumptions being made and check all
measurements over many times with the frame of mind of what might I being doing wrong here or overlooking here, and even then
chances are that a person could still be overlooking one or more things that could be throwing their measurements/calculations off,
if they think they are measuring OU. I have seen it many times over. :) Unless your circuit is a self-runner, it is highly recommended
to ask for feedback from other people with experience in this area to point out any issues they might see. You must also be willing to
clearly explain or show exactly how you are making your measurements. If a person doesn't approach this sort of experimentation with
this type of mind set, then in my opinion they have most likely lost before they have even started. That might possibly sound harsh to newbies,
but long experience does prove this out. AC circuits with complex waveforms such as pulse waveforms, etc. really can be quite tricky to evaluate
(unless you can make it into a self-runner that is). :)

Title: Re: To be deleted
Post by: nul-points on December 04, 2018, 06:34:50 AM
hi Void


yes, i see what you're saying, point taken - i'm obviously much more confused these days than i thought i was  :(
Title: Re: To be deleted
Post by: tinman on December 04, 2018, 06:35:13 AM
 author=Void link=topic=16384.msg527698#msg527698 date=1543920638]

QuoteNot trying to be difficult at all. Just some honest feedback on what was presented. I am trying to be as nice about it as I can, but I
know from experience that honest feedback will more often either be ignored or attacked rather than being fairly considered. :)
My comments are directed to anyone in general who cares about really trying to understand how things are really performing
in circuits such as this. IMO, an experimenter must be willing to examine and question all assumptions being made and check all
measurements over many times with the frame of mind of what might I being doing wrong here or overlooking here, and even then
chances are that a person could still be overlooking one or more things that could be throwing their measurements/calculations off,
if they think they are measuring OU. I have seen it many times over. :) Unless your circuit is a self-runner, it is highly recommended
to ask for feedback from other people with experience in this area to point out any issues they might see. You must also be willing to
clearly explain or show exactly how you are making your measurements. If a person doesn't approach this sort of experimentation with
this type of mind set, then in my opinion they have most likely lost before they have even started. That might possibly sound harsh to newbies,
but long experience does prove this out. AC circuits with complex waveforms such as pulse waveforms, etc. really can be quite tricky to evaluate
(unless you can make it into a self-runner that is). :)


So very very true Void
Title: Re: To be deleted
Post by: Void on December 04, 2018, 06:48:34 AM
Quote from: nul-points on December 04, 2018, 06:34:50 AM
hi Void
yes, i see what you're saying, point taken - i'm obviously much more confused these days than i thought i was  :(

Well, not saying it is all necessarily incorrect, but was just pointing out
some likely problem areas and areas that would need clarification which I noticed.
Thanks for sharing here however. :)

Quote from: tinman on December 04, 2018, 06:35:13 AM
author=Void link=topic=16384.msg527698#msg527698 date=1543920638]
So very very true Void

Yes, I have tripped myself up on various occasions in the past in my own experimenting.
It is easy enough to overlook things when trying to evaluate circuit performance in these types of circuits.
Can happen to anyone for sure. :)

Title: Re: To be deleted
Post by: itsu on December 05, 2018, 01:11:17 PM

While waiting for the correct BC327 transistor i setup the circuit on a breadboard using a 2N5401 transistor instead.

Smallest pot core i had using 0.45mm dual bonded magnet wire (green/red) 5 turns measuring 184uH each and when shorting
the other windings i have 0.84uH left (for leakage inductance calculations).
The capacitance inbetween the 2 windings is 96pF.

Using a 3.7V lion battery pack.

Using the 50K pot i set the frequency to 130Khz.

Input current was measured in 3 ways,
1 using a normal DMM in mA DC (10mA),
a 10 Ohm 1% inductionfree resistor in the return line measured with the yellow probe 97.mV / 10 = 9.7mA and
finally my current probe showing 9.6mA (green trace).

With a 1 Ohm csr it was hard to correctly measure the current.

So i am fairly confident that the DC input current is around 10mA and i will use the DMM for continue monitoring.

Video here: https://www.youtube.com/watch?v=22IEuVtAIIk&t=24s (https://www.youtube.com/watch?v=22IEuVtAIIk&t=24s)


Itsu
Title: Re: To be deleted
Post by: itsu on December 05, 2018, 04:02:10 PM

From the above video we see that in this setup the input power is 10mA x 3.88V = 38.8mW

The green led is pulling 5.7mA rms at 2.52V rms for a power of 3.985mW see screenshot 1

The red (load) led is pulling 11.22mA rms at 2.056V rms for a power of 23.07mW see screenshot 2

The transistor path (main led branch??) is pulling 16.08mA rms at 3.912V rms for a power of 43.38mW see screenshot 3


Not sure yet where to take additional measurements to get to the same conclusion as NP did, but will figure that out.

What i notice is that the lion battery pack voltage is raised somewhat (3.88 v 3.91), not sure where that comes from, probably some chemical reaction.

Yellow is the voltage across the led / transistor
green is the current (current probe) through that led / transistor
red is the math trace yellow x green (voltage x current)
blue is the voltage across the lion battery pack.
Itsu 
Title: Re: To be deleted
Post by: nul-points on December 05, 2018, 04:34:05 PM

thanks for the info, itsu, a very thorough inspection as always! - it looks like you have the whole circuit covered for measurements!

i agree about the 1 ohm CSR - i only used the CSR with the scope, to make a ballpark check on the supply current duty cycle and relative magnitude & current direction (for which it performed acceptably and supported the DMM current readings); so most written readings taken using DMM,  no CSRs, no supply capacitor (so far)


i also checked the supply to the drive branch (ie. the non-feedback branch) and confirmed that the current there is unidirectional from the supply to the transistor/primary/main LEDs, and also that the average of that current is higher than the average supply current by approx 20%

i see that i forgot to mention that the 2 main LEDs are 'cold white' type (hence 2.7VDC across the electrolytic cap //main LEDs); and just for info the bias pre-set resistor value is approx 20k Ohm

i hope this additional helps confirm the circuit i'm using

thanks
Title: Re: To be deleted
Post by: itsu on December 05, 2018, 04:45:02 PM

NP,

thanks for this additional info,  good to know, i will make some led changes then.
I measured my potmeter to be at 8K for the 133Khz

My measurements are just some praktice runs untill i get the BC327.



During the few hours of testing, the DMM monitoring the input current went from 10mA to 10.7mA and
the voltage across the battery pack (not monitored from the beginning  :-[ ) slowly drops using a Fluke 179 meter
from 3.791V to half an hour later 3.788V.

Itsu

Title: Re: To be deleted
Post by: nul-points on December 05, 2018, 05:04:12 PM
that all sounds pretty much nominal  - your cct is drawing approx 2x current as mine using a bias resistor value about half of mine

my battery is 60 or 70 mAh and a full charge takes approx 10-11hours to deplete, and i believe your battery is approx 840 mAh, so a full charge on your battery should deplete in approx 84 hours


thanks
Title: Re: To be deleted
Post by: nul-points on December 06, 2018, 11:17:40 AM

some observations about this circuit SO FAR:-
(plugging in example measurements displayed for itsu's preliminary practice run)

a) the current between the supply and the circuit is bi-directional
     (confirmed using scope probe on CSR in supply lead)

b) the CSR can be considered to be resistive
     test example:  10 Ohm non-reactive, negligible phase-shift component

c) the 'supply' current is the average of the RMS current values for the 2 half-cycles
     test example: 10mA

d) Input voltage is a DC voltage, essentially constant, negligible phase shift
     test example: 3.88V DC

e) the Power In (per cycle) is the product of the static supply voltage and the RMS 'supply' current
    = 3.88 × 10 = 38.8mW

f) the positive half-cycle RMS current enters the circuit into the transistor branch
    test example: 16.08 mA

g) the total dissipative load across supply rails (per cycle) is the product of the static supply voltage, as above, and the positve half-cycle RMS 'supply' current
     = 3.88 × 16.08 = 62.4 mW

h)  the Efficiency n, for driving this load is    dissipative load / power in
      = 62.4 / 38.8  =  1.6   (or 160%)


(additional energy is dissipated in the feedback LED which, when included in the results, increases the efficiency of the circuit
Title: Re: To be deleted
Post by: Void on December 06, 2018, 11:21:53 AM
Hi Itsu. Nice test. I would guess your LED power consumption measurements/calculations
are reasonably close to actual since your Tectronix scope RMS readings are probably
reasonably close, and since the current and voltage spikes across the LEDs
appear to be reasonably in phase.

However, your transistor power calculation looks quite suspect to me.
Ignoring the base current, the power consumption for the transistor should roughly be the
RMS collector to emitter voltage times the RMS collector current (assuming they are
reasonably in phase), when the transistor is conducting. The collector to emitter voltage should be quite low
when the transistor is conducting. When the transistor is not conducting, power consumption should be near 0.
Maybe you already explained it in your video, I can't watch it right now, but where are you getting that
3.912 V from? That doesn't look right to me at all.
All the best.
Title: Re: To be deleted
Post by: itsu on December 06, 2018, 01:34:42 PM

Thanks NP, Void,

for Void, yes i was wondering the same about that measurement, but its how NP measured it, so i did the same, but i too don't think it is the correct way.
The 3.912V i mention at the transistor is the voltage across the whole 'main led branch', so including the load led voltage etc.


In my below new setup i have changed that to as what i think is correct (voltage across E / C).



New setup using 2x 10mm white leds.
Also the 10 Ohm csr is removed.

The frequency had to be adjusted again to 130Khz by adjusting the 50K pot, its now set at 3.8K.

All measurements are done by the scope to be able to compare (backed up by the DMM's).

Current probe and voltage probes are deskewed as much as possible to rule out unwanted phase shift (important in AC like signals), but i am left with a 8ns delay (phase shift) between current and voltage probes.
This has no significant influence as shown by comparing with the 10 Ohm csr.

Below the circuit with some measurements.

The combined power users (leds and transistor) make up for a total of 21 + 1.4 + 2.3 = 24.7mW .
The input power is 28.4mW, so i think we have accounted for most of all the power into the circuit.

There are severall current measurements done, and with some i have a problem, like in the 'main led branch'.
The 9.2mA at the transistor comes short for supplying the follow on current through the load led (7.2mA) and to the feedback link to the elco (5.4mA)

Anyway, these are the results up till now:



Regards Itsu
Title: Re: To be deleted
Post by: Void on December 06, 2018, 02:09:00 PM
Quote from: itsu on December 06, 2018, 01:34:42 PM
for Void, yes i was wondering the same about that measurement, but its how NP measured it, so i did the same, but i too don't think it is the correct way.
The 3.912V i mention at the transistor is the voltage across the whole 'main led branch', so including the load led voltage etc.

Hi Itsu. I see. Thanks for the clarification.
Yes, the transistor power dissipation should be relatively small, but I think it will be tricky
to measure it accurately with those pulse waveforms. 
How did you arrive at the 1.5V and 2.3 mW for the transistor?

Title: Re: To be deleted
Post by: itsu on December 06, 2018, 02:57:27 PM

Hi Void,

nice to have you take a look at this.

I let my scope do the hard work, by calculating the power from the voltage across the C/E and the current through this transistor.

In the above diagram, i have my yellow probe ground lead on the Emitter and the probe tip on the Collector (channel inverted on).

The current probe is at the arrow above the Emitter (also channel inverted on as my current probe is pointing the wrong way (up)).
 
The values and the math trace (yellow x green) is in the screenshot below.

Itsu
Title: Re: To be deleted
Post by: Void on December 06, 2018, 03:13:34 PM
Hi Itsu. Thanks for the details.
Yes, looks good to me for your transistor measurements.
Around 2mW transistor power dissipation looks reasonable.
Having that scope current probe and that math function is very handy. :)

Title: Re: To be deleted
Post by: itsu on December 06, 2018, 03:30:15 PM
Quote from: nul-points on December 06, 2018, 11:17:40 AM
some observations about this circuit SO FAR:-
(plugging in example measurements displayed for itsu's preliminary practice run)

a) the current between the supply and the circuit is bi-directional
     (confirmed using scope probe on CSR in supply lead)

b) the CSR can be considered to be resistive
     test example:  10 Ohm non-reactive, negligible phase-shift component

c) the 'supply' current is the average of the RMS current values for the 2 half-cycles
     test example: 10mA

d) Input voltage is a DC voltage, essentially constant, negligible phase shift
     test example: 3.88V DC

e) the Power In (per cycle) is the product of the static supply voltage and the RMS 'supply' current
    = 3.88 × 10 = 38.8mW

f) the positive half-cycle RMS current enters the circuit into the transistor branch
    test example: 16.08 mA

g) the total dissipative load across supply rails (per cycle) is the product of the static supply voltage, as above, and the positve half-cycle RMS 'supply' current
     = 3.88 × 16.08 = 62.4 mW

h)  the Efficiency n, for driving this load is    dissipative load / power in
      = 62.4 / 38.8  =  1.6   (or 160%)


(additional energy is dissipated in the feedback LED which, when included in the results, increases the efficiency of the circuit

My reaction below each statement

a) the current between the supply and the circuit is bi-directional
     (confirmed using scope probe on CSR in supply lead)

   Not what i see, the supply current is a DC current of 10mA out of the battery pack.
   I do have a smoothing cap of 2200uF and a decoupling cap of 0.1uF on the supply rail.
   Without them, i see an all positive pulsing current.


b) the CSR can be considered to be resistive
     test example:  10 Ohm non-reactive, negligible phase-shift component

   agreed.


c) the 'supply' current is the average of the RMS current values for the 2 half-cycles
     test example: 10mA

   Not sure what you mean here.


d) Input voltage is a DC voltage, essentially constant, negligible phase shift
     test example: 3.88V DC

   Agreed.


e) the Power In (per cycle) is the product of the static supply voltage and the RMS 'supply' current
    = 3.88 × 10 = 38.8mW

   Both voltage and current here are DC, so no cycle, further agreed.


f) the positive half-cycle RMS current enters the circuit into the transistor branch
    test example: 16.08 mA
   
   Not sure what you mean here, the supply current (10mA) together with the feedback current (5.7mA) enters
   into the transistor branch (16.08mA).

   
g) the total dissipative load across supply rails (per cycle) is the product of the static supply voltage, as above, and the positve half-cycle RMS 'supply' current
     = 3.88 × 16.08 = 62.4 mW

   This is where i have problems with as the transistor only consumes little power. See also Voids remark
   
h)  the Efficiency n, for driving this load is    dissipative load / power in
      = 62.4 / 38.8  =  1.6   (or 160%)

    Also here i have a problem as the former statement g) is not correct in my mind.


Itsu
Title: Re: To be deleted
Post by: itsu on December 06, 2018, 04:20:48 PM

I did some tests on statement a) above and have to adjust my reaction there already  :D

When removing both the 2200uF elco and 0.1uF decoupling cap i have on the rail, i get the below shown current in green.

The inline DMM in mA setting still shows 6.85mA, but the scope now shows 9.164mA rms and 6.871mA mean current.

So it seems the DMM is averaging the current.

On the total input it makes little difference as the input power still calculates around 27.7mW

Itsu
Title: Re: To be deleted
Post by: Void on December 06, 2018, 04:33:53 PM
Quote from: itsu on December 06, 2018, 04:20:48 PM
So it seems the DMM is averaging the current.

Hi Itsu. Yes, that makes sense as ordinary DMMs will measure roughly the average AC voltage and current,
whereas true RMS meters will indicate the RMS values. I say 'roughly' because many cheaper DMM's are not
so accurate and most DMM's and true RMS meters are not spec'd to measure AC current and AC voltage at
higher frequencies such as 130 kHz, so their accuracy at such frequencies can't be counted on.

Title: Re: To be deleted
Post by: nul-points on December 06, 2018, 09:23:55 PM

thanks for the update Itsu, i think we're making progress - apologies for the delayed reply, i was out at band practice until late this evening (Thurs)

changing the Red LED to White has made the circuits more similar, that's good - my feedback LED is Green, as in my schematic, but if your circuit uses a White that should be ok

(if your feedback LED is White and it's illuminated by a 2mA current  however, then i question that the voltage across it is only 1.54V, so that measurement might need checking (did you use DMM or scope?)

yes, as i mentioned in post #32 the circuit doesn't have a supply capacitor, and CSRs are not included either (only added to take certain scope shots), so that was important that you removed the 2200uF elcap (& 0.1uF too?) to also make the circuits more alike

ok, so removing the supply caps changes your circuit's supply behaviour, but you're still only seeing positive current flow** into the circuit - so this is a major difference between our circuits (and the reason why steps (c) & (f) don't mean anything to you) - when i observe the supply current using a scope & 1 Ohm CSR (inserted inline with the positive supply) the current is bidirectional: the feedback current, Ifb, flows into the supply, and the main drive current, Iin, flows into the circuit as you'd expect

these 2 different current flows show as opposite polarity voltages across the CSR (scope trace shown below hopefully).  the duty cycle of these 2 approx. triangular pulses within each cycle is around 1.5:8 and the average current indicated by their relative areas and proportion of the cycle time gives reasonable confirmation of the values measured by DMM

the different current flow direction, or polarity, (Iin, current into main drive path; Ifb current out of feedback LED path) is confirmed (on my circuit, at least) with a DMM

hopefully this info will help you see what steps (c) and (f) achieve (assuming your circuit can achieve bidirectional current flow in the two parts of each cycle)

[Edit: ** actually, i believe your circuit does have bidirectional current flow in the supply lines: your latest scope shots in post #42, from the supply line probe (after removing supply caps) appears to show a narrow elongated negative going triangle pulse followed by a larger flattened somewhat triangular pulse of opposite polarity
   i also note the same approximate relationship between Isupply, Iin & Ifb in your circuit and in mine:
   Iin = Isupply + Ifb ]

the steps (a) to (h) of my result overview posted above in post #35 only use 3 current measurements - these are the 'combined' supply current and the 2 separate current flows, Iin & Ifb, discussed here in this post


i didn't attempt to measure the transistor power dissipation - i measured Iin and calculated the total power dissipation for the whole of that path from +ve supply line to -ve supply line and subtracted main LED dissipation to arrive at the dissipation in the transistor and transformer windings - you can see this measurement in post #19

step (g) doesn't mention the transistor (because i thought that you mentioned that your reading of 43.38mW in post #31 was for that whole branch, between +ve & -ve supply lines, not just the transistor)

so step (g) refers to "the total dissipative load across supply rails (per cycle)" ie. the whole current path between supply +ve & -ve lines, entering the circuit into the Tr. emitter - this would include the transistor 'on' state, transformer winding losses, & main LEDs (ignoring the much smaller values of bias losses, elcap leakage etc.)


so, i see that your circuit has been updated with new readings - let's plug them into steps (c) to (h):-

c) the 'supply' current is [the average of] the RMS current [values for the 2 half-cycles]
     test example: 7.4mA

d) Input voltage is a DC voltage, essentially constant, negligible phase shift
     test example: 3.76V

e) the Power In (per cycle) is the product of the static supply voltage and the [RMS] 'supply' current [the mean current, as per your updated measurement after removing supply caps]
    = 3.76 × 7.37 = 27.3 mW

f) the [positive half-cycle] RMS current enters the circuit into the transistor branch
    test example: 9.2 mA

g) the total dissipative load across supply rails (per cycle) is the product of the static supply voltage, as above, and the positve half-cycle RMS 'supply' current
     = 3.76 × 9.2 = 34.6 mW

h)  the Efficiency n, for driving this load level is   ( dissipative load / power in)
      = 34.6 / 27.3  =  1.27   (or 127%)

[unit time, common to both terms to produce units of energy, cancels out]

(additional energy is dissipated in the feedback LED which, when included in the results, increases the efficiency of the circuit)


i hope this helps
Title: Re: To be deleted
Post by: itsu on December 07, 2018, 09:02:47 AM

NP,

thanks for the info, need to digest most of it still though.

We had an internet outage all morning, so need to catch up now.

I changed the feedback led now back to green, so we should almost be there :-)

Measured again the voltage across the now green feedback led with the scope to be 1.47V rms @ 2.54mA rms and 800uW
The DMM (Fluke 179) shows 0.792V, but it is operating out of its specs, so unreliable.

Removed the 2200uF and 0.1uF caps (i allways include them when working on pulsing systems).

The sharp negative green current pulse visible in my last screenshot above is the same as the current pulse through the (now again) green led.
So it seems that the feedback signal (current) through the green led is fed back into the battery pack now.

I have combined the supply input current (in white) with the current (in green) through the green led in the below screenshot.
There you see that the both sharp pulses (green is reversed for better comparing) are the same.

Rest of the post will be analyzed later today.

Itsu
Title: Re: To be deleted
Post by: Void on December 07, 2018, 10:55:53 AM
Hi nul-points. Going by Itsu's scope power measurement values posted on his diagram in reply # 37 above,
below is what the efficiency works out to. If the scope math function that Itsu is using
calculates all the instantaneous power points by multiplying all the corresponding current and
voltage waveform points and then takes the average (mean) of all those calculated instantaneous
power points, then those scope power calculations should be pretty close to actual. This method
would take into account any negative going parts of voltage and current waveforms as part of the
calculations.

Just taking an RMS measurement of a complex asymmetrical spikey waveform which goes both
negative and positive may not give an accurate reading, since the meter or scope may only
look at just the positive or negative half of the waveform to calculate the RMS value.
It would depend on how the meter or scope calculates the RMS values, so I wouldn't trust
such RMS readings on asymmetrical waveforms which go both positive and negative without knowing
exactly how a particular meter or scope determines the RMS values.

Based on Itsu's posted power measurements in his reply # 37:
-------------------------------------------------
Input power from battery: 28.4 mW

Total circuit power consumption: 1.4 mW + 2.3 mW + 21 mW = 24.7 mW
(power consumed by base resistor and base circuit of transistor should be fairly small)

Efficiency = (24.7 mW / 28.4 mW) x 100 = 86.97 = 87%

nul-points, I will have to take a closer look at your comments later this evening when I have time,
but the calculated efficiency above here looks about in the right vicinity for that type of circuit.
The power consumption of circuit components like transistors is not normally included in
efficiency calculations, but, anyway, I understand why you are doing that, and it doesn't change
things by that much anyway since the power consumption of the transistor is relatively low.

Itsu, with your white LED measured as consuming around 21 mW, was the white LED glowing very very bright,
like almost flashlight bright? At 21 mW, it seems to the me the white LED would be glowing super bright.
Title: Re: To be deleted
Post by: itsu on December 07, 2018, 01:14:35 PM

Void,

The way you describe the working of the scope i use is correct.
It takes 250M(ega) samples / sec across the whole display screen, then average (mean) them.

I also have to agree with your effeciency calculations (±87%).


The white load led is/was indeed very bright as to unable to look into.

Itsu
Title: Re: To be deleted
Post by: itsu on December 07, 2018, 01:33:24 PM

I received my BC327-25 transistor, so build a new setup using a PCB instead of the breadboard.
The breadboard is kind of unstable and i want repeatable results.

The setup is shown below.

Using:

the new BC327 transistor,
the same transformer,
the same 50K pot,
a green feedback led,
a white (10mm) load led,
a ceramic 47pF cap,
a 100uF electrolytic capacitor and
the 3.7V 3.1KWh (840mAh) battery pack

It has an on/off switch and elevated leads so to be able to put the current probe almost anywhere.
The transistor and potmeter are removeable.

Will do again the measurements this weekend.

Regards Itsu
Title: Re: To be deleted
Post by: Void on December 07, 2018, 02:35:01 PM
The new circuit layout looks good Itsu. Nice to have those raised wires
for easy access by your current probe. Ok on the white LED being very bright
in the previous test. That passes the sanity check. :)

For the scope math calculation, if you mean that the math function does
multiply each corresponding point for the current and voltage waveforms
to get the instantaneous power for each point set, and then takes the mean of
all those multiplied values, then from my understanding that is a quite accurate way to
determine the average power. I have used that same method in the past to analyze
joule thief circuits and similar, but I had to do the multiplication of the waveform points
and take the average using an Excel spreadsheet to do the calculations on all the points. :)

Title: Re: To be deleted
Post by: itsu on December 07, 2018, 03:10:22 PM

Thanks void,

Yes, this "multiply each corresponding point for the current and voltage waveforms
              to get the instantaneous power for each point set, and then takes the mean of
              all those multiplied values",
is for me the most (only) accurate way to determine average power on such waveforms.

I do not even want to think about having to do the multiplication of the waveform points etc, by hand  :o

Itsu 
Title: Re: To be deleted
Post by: Void on December 07, 2018, 03:43:28 PM
Hi Itsu. In my case it was not done manually. I exported all the points for a period of time
for the voltage and current waveforms into a CSV file, and then used the formula features of Excel
to do the calculations automatically. Works quite well. :)

Yes, for any asymmetrical waveform which goes both positive and negative, even using the RMS feature
of a scope is likely going to be inaccurate, as I think many if not most scopes only look at one half of
the waveform (postive or negative half) to do the RMS calculation. The method you are using which makes
use of a current probe and voltage probe and then uses a math function to calculate the average power consumption
is the only reliable way to make power measurements on those types of waveforms, assuming the probes are
reading reasonably accurately.

In the 'efficiency' calculation I did above, the discrepancy between the input power and the total 'output power'
may possibly be due to the current feeding back from the feedback LED into the battery not properly
being taken into account. If you measure the battery voltage with a scope probe and the battery
current with a scope current probe and use your scope math function to compute the average input power,
then that should properly take into account the feedback current from feedback LED, if that is not what you
did previously. Then the input power and total 'output power' should come out more closely, I would think,
as the power dissipation in the coils and ferrite core is probably not very significant.

I have done a lot of testing with these type of circuits and have never seen any signs of OU
in regards to them, and nor would I expect any OU there since there is really nothing
out of the ordinary included in those types of circuits, unless maybe someone does something very
unusual with a ferrite or iron core.

Title: Re: To be deleted
Post by: nul-points on December 07, 2018, 04:20:19 PM
Quote from: itsu on December 07, 2018, 01:33:24 PM
I received my BC327-25 transistor, so build a new setup using a PCB instead of the breadboard.
The breadboard is kind of unstable and i want repeatable results.

The setup is shown below.

Using:

the new BC327 transistor,
the same transformer,
the same 50K pot,
a green feedback led,
a white (10mm) load led,
a ceramic 47pF cap,
a 100uF electrolytic capacitor and
the 3.7V 3.1KWh (840mAh) battery pack

It has an on/off switch and elevated leads so to be able to put the current probe almost anywhere.
The transistor and potmeter are removeable.

Will do again the measurements this weekend.

Regards Itsu


nice build (as always!) Itsu, the parts list is looking good

i'm glad to see that you have confirmed a bidirectional supply current, consisting of both negative and positive pulses (one relating to the feedback path and the other to the main drive path)

i'll be meeting with family for a couple of days now, but should be back online later on Sunday

i hope your internet access has been restored - we've had outages on the cellphone network over the last couple days (that was fun!)

regards
np
Title: Re: To be deleted
Post by: Turbo on December 07, 2018, 04:26:31 PM
yes yes is that a ferite core ?
your almost there.
there is a specific mark and a space ratio.
fire the pulse train and vary the frequency until you get a response back in the space (null) time that the transistor is off, then when you got the sweet spot frequency right, fine tune the pulse train duration and adjust the space time lastly.
Title: Re: To be deleted
Post by: AlienGrey on December 07, 2018, 06:20:33 PM
The thread label is FLY back In order to achieve a true fly back you need a charge time on a tv its like 52usec and a return time (energy dump)
into the flyback of 12 microseconds but I presume you want to beat the magnetic vortex known as lens law in that case you might want to go
a lot faster than 64 usec and shouldn't the horn waveform be the other way round without the ringing oscillations try putting the scop on the
transistor base to see if that's got a clean cut off signal.
Title: Re: To be deleted
Post by: itsu on December 08, 2018, 04:51:56 AM

Void,


QuoteIf you measure the battery voltage with a scope probe and the battery
current with a scope current probe and use your scope math function to compute the average input power,
then that should properly take into account the feedback current from feedback LED, if that is not what you
did previously.

I use both DMM's and the scope to compare, but for calculating power i always use the scope to get consistent measurements over the circuit.
Concerning the use of DMM's, see also the video in a later post.



NP,

Internet is back up, so was able to catch up   :)



Turbo,

not sure who you are asking, but yes thats a ferrite pot core i use in the transformer.
I don't see to what screenshot you refer with your somewhat oversized   ;) screenshot, could you elaborate?



AG,

not sure to who you are responding, but as my scopeshots show no ringing i guess you are responding to turbo his post.



Itsu
Title: Re: To be deleted
Post by: itsu on December 08, 2018, 05:10:08 AM

Here the first results of my measurements of the new setup.

Basically i measure at 4 points being across the battery (input) and the 3 power users (2x led and transistor).
I have taken screenshot of those 4 measurement points and put them in below.

The wattage of the transistor is in the 800uW range, so not 800mW as i mention in the video.

A (boring!) video of those 4 measurements can be seen here: https://www.youtube.com/watch?v=7uQ8BBxiU0s

I will repeat these measurements a few times to see if the results stay the same over a period of time.

Regards Itsu
Title: Re: To be deleted
Post by: Void on December 08, 2018, 11:45:50 AM
Quote from: itsu on December 08, 2018, 05:10:08 AM
Here the first results of my measurements of the new setup.

Basically i measure at 4 points being across the battery (input) and the 3 power users (2x led and transistor).
I have taken screenshot of those 4 measurement points and put them in below.

The wattage of the transistor is in the 800uW range, so not 800mW as i mention in the video.

A (boring!) video of those 4 measurements can be seen here: https://www.youtube.com/watch?v=7uQ8BBxiU0s

I will repeat these measurements a few times to see if the results stay the same over a period of time.

Regards Itsu

Hi Itsu. Nice job! Nice clear presentation. Something does not appear to be adding up in those power measurements
however, but I am not sure off hand why that might be. From your diagram:
Power In: 17.96 mW
Total circuit power consumption: 1.1mW + 770uW + 10.6mW = 12.47mW

I would expect that the total input power and the total circuit power consumption should compare at least fairly close,
but that is not what we have. Either something else in the circuit is consuming a significant amount of power that
is not being measured (I can't think what that might be at the moment), or the power measurements are off in one or
more places for some reason. Can you think of any reason for this discrepency? Unless the coil windings on the ferrite core
are consuming about 5mW? That seems quite high to me, as I wouldn't expect small coils of wire to consume anywhere
near that much power. What do you think?

I think a sanity check test setup of your scope power measurement method to see if the power measurements with
the scope are reasonably correct or not with complex waveforms would be a good idea here.
Title: Re: To be deleted
Post by: Hoppy on December 08, 2018, 12:02:31 PM
Quote from: Void on December 08, 2018, 11:45:50 AM
Hi Itsu. Nice job! Nice clear presentation. Something does not appear to be adding up in those power measurements

Pot, caps and Inductor power dissipation.
Title: Re: To be deleted
Post by: itsu on December 08, 2018, 12:49:38 PM

Hi Void,

Yes, there are some things that don't add up, like the current through the transistor (7.1mA) and below it
the current through the load led (3.7mA) and the current to the 100uF cap / feedback led (5.2mA) being 8.9mA.


Another point i see is the voltage across the transistor and the load led.
If you combine those you get more then what the battery pack (3.75V) delivers:
1.62 + 2.84 = 4.46V
1.72 + 2.89 = 4.61V

When scoping from emitter transistor to cathode load led it shows 3.75V rms again.


So its very hard to get accurate readings, even with my scope setup, especially the current probe needs constant calibration / degausing.

Therefor i am planning to take daily measurements and take the average after some time.

Today i did a second measurement and put the results in a spreadsheet, see below.
There you see that the difference between input and the power users is much smaller (about 1mW)



Itsu
Title: Re: To be deleted
Post by: Void on December 08, 2018, 01:21:00 PM
Hi Itsu. OK, I see. Maybe using an appropriate CSR and using a voltage probe would
give more stable current readings.

Title: Re: To be deleted
Post by: Void on December 08, 2018, 01:43:35 PM
Quote from: Hoppy on December 08, 2018, 12:02:31 PM
Pot, caps and Inductor power dissipation.

Hi Hoppy. Well, in my estimation those would not account for the approx. 5 mW discrepency
as their power consumption would be fairly low, but Itsu has pointed out that his current probe
may not be giving consistent readings (at those low current values). All the best.
Title: Re: To be deleted
Post by: itsu on December 08, 2018, 02:38:58 PM
Quote from: Void on December 08, 2018, 01:21:00 PM
Hi Itsu. OK, I see. Maybe using an appropriate CSR and using a voltage probe would
give more stable current readings.

I did use a 10 Ohm 1% inductionfree csr (1 Ohm is to noisy) in the beginning and compared the values with the current probe
and found them to be very similar, but they might influence the working of the whole circuit.

So therefor i opted for the current probe as it is much more convenient to use on a fixed PCB circuit.

I think i will be very close to accurate measurements when averaging like 10 measurements or so.


itsu
Title: Re: To be deleted
Post by: Void on December 08, 2018, 03:11:07 PM
Yes, sounds good Itsu.
Title: Re: To be deleted
Post by: itsu on December 09, 2018, 09:14:52 AM

Collected measurement data up till now:


Itsu
Title: Re: To be deleted
Post by: Void on December 09, 2018, 12:52:24 PM
Looks good Itsu.

My Siglent SDS1204X-E scope can do the same sort of math calculations
that you are using to compute the power consumption, so I tried a power measurement
comparison test with two test circuits using the exact same type of white LEDs in both test
circuits. One test circuit is DC, so the power measurement is straight forward. The other
test circuit is using a 10% duty cycle pulse train to power the LED. The LEDs in both circuits
were adjusted to be approximately the same brightness, but that is very hard to do by eye,
so their brightness is probably not a really close match, but they were in the same ballpark of
brightness anyway. :) The frequency of the pulse waveform was set to 10 kHz.

See the attached test circuit diagrams which I used in this test.
The LED power consumption in the DC test circuit came out to: 2.71V x 5.2 mA = 14.1 mW
The LED power consumption in my AC test circuit was measured using a product math function
and then taking the mean of the resulting waveform (White trace).
(mean of the product of the voltage and current waveforms), and is: 1.67W /100 = 16.7 mW

The difference between the two power measurements may be largely due to my eye not being able
to accurately compare the brighness of the white LEDs in the two test circuits. There may be other factors
in there that are affecting the power measurement a bit as well. The 100 Ohm CSR resistor value is close
enough to 100 Ohms, but off by a few Ohms. So using the scope product Math function and taking the
mean of the resulting waveform seems to give a reasonable result, in this test setup anyway.

Notice that in the attached Scope screenshot that the measured scope RMS values for the LED voltage and
LED current are not useful in doing any power measurements when measuring asymmetrical AC waveforms.
The current is being measured across a 100 Ohm CSR, so current must be divided by 100.

Ch 1 - Yellow trace - LED Voltage waveform
Ch 2 - Pink trace - LED current x100 (Set to Inverted)
White trace - Power waveform - Product of Ch1 x Ch2, with the Mean of this waveform indicated.

P.S. No jokes about Channel 2 having a pink trace. :D That's the way the scope is configured, and I don't
see any way to change the trace color. Why they chose pink instead of orange or red is beyond me. ;D

Title: Re: To be deleted
Post by: itsu on December 09, 2018, 01:30:37 PM

Very nice Void,

yes, eyeballing a led (pulsed or steady) is almost impossible, you need a black box / lux meter to measure properly, but still you came very close.

Indeed, it again shows that the measured scope RMS values for the LED voltage and LED current are not useful in doing any power measurements
when measuring asymmetrical AC waveforms.


 
Strange colors indeed, most scopes i know have fixed colors, so nothing to do about that i guess.


Itsu
Title: Re: To be deleted
Post by: Void on December 09, 2018, 02:45:49 PM
Hi Itsu. A few things I don't really understand with my scope is how the scope
is computing those RMS values. I would have thought the RMS values would come out
a lot lower considering I was only using a 10% duty cycle. My scope indicated 9.54 V RMS
for the voltage waveform and 12.8 mA RMS (1.28 / 100) for the current waveform. Not sure
how my scope is arriving at those values. The actual RMS voltage across the LED should be
around 2.7 V to 2.8 V, and the actual RMS LED current should be around 6 mA.  :o
All my scope's manual says for the RMS measurement is: "RMS: Root mean square of all data values."

Another thing that I think is odd is both Channel 1 and 2 were set to have the traces centered
vertically in the screen, but channel 1 has the waveform sunk down to its average DC value,
which I think is normal when displaying this kind of waveform on a scope using DC coupling, but
the waveform for channel 2 is displaying with the lower part of the waveform in the center of the screen.
Channel 2 is also set to DC coupling. I am not sure why channel 1 displays as sunk down to the average
while channel 2 is not. Can anyone explain why the 2 channels are displaying differently?

The scope is a fairly new scope, so it could have bugs I suppose. Not sure if the above mentioned results
are what should be expected, but they seem odd to me anyway. :)

Title: Re: To be deleted
Post by: itsu on December 09, 2018, 03:28:01 PM

Void,

i have the same (down offset) yellow voltage trace when using an AC pulse.
When using a DC pulse its like normal.

Also, do you use a grounded Signal Generator?  If so then the (grounded) black lead will short out
the 100 Ohm  resistor as very probably the both scope ground leads are also grounded.

Itsu
Title: Re: To be deleted
Post by: Void on December 09, 2018, 03:42:55 PM
Hi Itsu. Yes, I guess because the current is rectified by the LED,
the current waveform shows differently than the voltage waveform, which was AC with 0V DC offset. 

I was using a portable signal generator that was powered from my bench power supply.
I used my DMM and measured the resistance between the scope ground and the signal
generator ground across the 100 Ohm resistor, and it measures as 103 Ohms, so the
signal generator ground was isolated from the scope ground. If it was a short,
I wouldn't have measured any current across the 100 Ohm resistor.

Also, I think I may know why the scope is calculating the RMS values the way it is, by I am
still reviewing that. I will post something here if I figure it out. :)


Title: Re: To be deleted
Post by: itsu on December 09, 2018, 03:51:06 PM

I used a battery operated SG, but it has only 10Vpp output, see white trace when unloaded (10Khz, 10% duty, AC)


The setup used is the same as your right diagram above, white led, 100 Ohm resistor, probes setup the same.

Yellow is the (sunken) voltage across the led
blue is the current through the 100 Ohm csr (rms value adjusted to show correct value, so no /100 needed).
red is the math trace yellow x blue = power of the led.

So nothing wrong with your scope, its probably the led (diode) doing this.


Itsu 
Title: Re: To be deleted
Post by: Void on December 09, 2018, 03:55:46 PM
Thanks for that Itsu. Yes, the LED is rectifying the current waveform, so it
shows differently than the AC voltage waveform. Thanks for confirming that!

Title: Re: To be deleted
Post by: Void on December 09, 2018, 04:25:51 PM
For the current waveform in my AC test, it is a DC pulse waveform with a 10% duty cycle,
so I believe the correct RMS formula for that type of waveform is: Vp x sqrt (duty cycle ratio).
Based on the current waveform in my scope screen shot, this would give:
4 x sqrt (0.1) = 1.26
The scope calculated the current RMS as 1.28, so that is pretty close. :)
1.28 /100 = 12.8 mA
However, the actual RMS current for the LED should be around 6 mA (estimated), so we can't
use that calculated RMS value by the scope.

For the voltage waveform, we have an AC pulse waveform which the scope is offsetting
with a negative offset. The RMS voltage value calculated by the scope is also not correct for our circuit,
because the LED is only conducting for 10% of the time, but the scope is calculating the RMS value for
the entire waveform cycle. I am still fiddling to see if I can figure out how the scope calculates that 9.54V RMS
value for the voltage waveform. :) The actual RMS voltage across the LED was probably about 2.75 V (estimated).

At any rate, the RMS values calculated by my scope obviously can't be used to calculate the power consumption
of the LED with a pulse type waveform being applied to the LED. :)

P.S.
After thinking about it, I guess the way the scope calculates RMS values is to square all the measured sample
values taken over a period of time, then calculate the mean of those squared values, and then take the square root
of the calculated mean. I am feeling too lazy right now to save waveform samples to a CSV file and compute it
all out using Excel, but I might try it at some point to confirm if that is how the scope calculates the RMS values.  ;D

P.P.S.
Since the voltage and current waveforms are pretty close to being a nice rectangular shape in my test setup,
another way to do a sanity check on the measurement of the power consumption of the LED in my AC test circuit
is to use the actual scope measured values (as shown in the scope screen shot I posted above).

If we do it this way we get (I think this is right, but maybe I am overlooking something :) ):
We are only concerned about the voltage waveform when the LED is conducting,
so, V measured = 0.65 divisions x 5 volts per division = 3.25V
I measured = ( 4  x 1 A per division) / 100 = 40 mA
Duty cycle is 10%, so the duty cycle ratio is 0.1
3.25V x 40 mA x 0.1 = 13 mW
The scope's measured/calculated value for the LED power consumption (using the Math function) was 16.7 mW.
Ballpark, but not exactly really close. Maybe this method of estimating the power consumption of the
LED is not right, or maybe it is not so accurate. Or maybe the power calculated using the Math function was off a bit.

I think this only works out (roughly) because the waveforms are quite clean rectangular waveforms.
However, thinking about it, that should actually be what the LED is doing. According to the scope measurements,
the LED has 3.25V across it and conducts 40 mA during its On time, which is 10% of the time.
That very possibly may not be so good for the LED however, if left running for a long time.  :D

This is why it is way easier to try to self-loop a circuit that you think might be OU, and see
if it will self-sustain itself.  No measurements or calculations required, and much less chance of
making an error somewhere.  ;D

Title: Re: To be deleted
Post by: itsu on December 10, 2018, 05:31:38 AM

Hi Void,

on this wiki there is some info on rms values / waveform:
https://en.wikipedia.org/wiki/Crest_factor

So every waveform needs its specific calculation to arrive at the correct rms value.
I can not see how a scope does that, so probably there is some generic way.

But i agree, (power) measurements are no easy task and marginal cop > 1 values should be treated with causion.


Itsu
Title: Re: To be deleted
Post by: nul-points on December 10, 2018, 06:16:34 AM
Quote from: itsu on December 08, 2018, 12:49:38 PM

Therefor i am planning to take daily measurements and take the average after some time.

Today i did a second measurement and put the results in a spreadsheet, see below.
There you see that the difference between input and the power users is much smaller (about 1mW)

Itsu


interesting observation, thanks Itsu - i'll look forward to seeing how those readings develop


np
Title: Re: To be deleted
Post by: Turbo on December 10, 2018, 06:17:28 AM
There is a function called Sound_Play() in mikroC library and i have had success with that.
The Prototype is void Sound_Play(unsigned freq_in_hz, unsigned duration_ms);
https://download.mikroe.com/documents/compilers/mikroc/pic/help/sound_library.htm (https://download.mikroe.com/documents/compilers/mikroc/pic/help/sound_library.htm)
So you can easily pass frequency and duration as parameters and so you can roll your own function that incorporates the mark duration, as well as delay for the space duration.

If you need higher frequency's you have to switch to classic PWM.
But who does not remember being a child and hearing the 20K whistle coming from the CRT Flyback tv when we was young ? and the resulting crackle from the electrostatic field on the tube when touched...
Of course now i'm old and i can't hear that high no more.
Title: Re: To be deleted
Post by: nul-points on December 10, 2018, 06:27:01 AM
time to recap on the original circuit we're looking at here (and which Itsu is kindly looking at replicating)

i'm attaching an overview of the circuit - it has 1 supply branch and effectively 2 branches (ignoring a higher impedance biasing branch with an effective impedance of approx 10k ohm at 130kHz)

one branch is entirely a current drain, Iin, on the supply and the other is entirely feeding current, Ifb, back to the supply

the net supply current, Isupply, is Iin - Ifb


np
Title: Re: To be deleted
Post by: Void on December 10, 2018, 09:01:03 AM
Quote from: itsu on December 10, 2018, 05:31:38 AM
Hi Void,
on this wiki there is some info on rms values / waveform:
https://en.wikipedia.org/wiki/Crest_factor
So every waveform needs its specific calculation to arrive at the correct rms value.
I can not see how a scope does that, so probably there is some generic way.
But i agree, (power) measurements are no easy task and marginal cop > 1 values should be treated with causion.
Itsu

Hi Itsu. I think the way a scope does it may be as follows:
The way a scope calculates RMS values is to square all the measured sample values taken over a period of time,
then calculate the mean of those squared values, and then take the square root of the calculated mean.
It's been a long time since I studied such things, but I think this may be how it is done in scopes (or similar).
As my scope's manual states: "RMS: Root mean square of all data values." 

Your measurements so far are showing a fair bit of variation, so not so super reliable, possibly because
the circuit is not so stable, but likely at least in part due to the current probe not giving highly
accurate measurements at those low current levels. Still it seems pretty clear already that there is no
indication of OU in that circuit arrangement when measurements are done properly. :)

Title: Re: To be deleted
Post by: Void on December 10, 2018, 09:18:12 AM
Quote from: nul-points on December 10, 2018, 06:27:01 AM
time to recap on the original circuit we're looking at here (and which Itsu is kindly looking at replicating)
i'm attaching an overview of the circuit - it has 1 supply branch and effectively 2 branches (ignoring a higher impedance biasing branch with an effective impedance of approx 10k ohm at 130kHz)
one branch is entirely a current drain, Iin, on the supply and the other is entirely feeding current, Ifb, back to the supply
the net supply current, Isupply, is Iin - Ifb
np

Hi Nul-points. Although Itsu's measurements are not showing as super consistent/accurate,
which may possibly be partly due to circuit instability and very possibly/likely an accuracy limitation of his
current probe at low currents, the measurement method Itsu is using appears to be correct (and it passes sanity checks), 
so Itsu's measurements should be close enough for general analysis purposes, so it should be pretty clear by now
that there are so far no signs of OU in your circuit arrangement.

My recent comments have been to demonstrate how tricky it can be to make really accurate measurements
on circuits that contain complex AC waveforms, and how important it therefore is to question and examine everything
very closely to see where a person might possibly be going wrong or overlooking something in their measurements.
Nul-points, you seem to have ignored this, and also ignored that your previous measurements and calculations
are way off compared to what Itsu has been measuring using an (apparently) correct measurement method (although
apparently not so super accurate with Itsu's current probe). Nul-points, what do you think about this? Do you acknowledge
that Itsu's measurement approach is correct, even if it may not be super accurate/consistent when using his current probe?
If you don't think Itsu's measurement approach is correct, please explain why.

P.S. Regarding your latest drawing posted above, efficiency is calculated as a ratio of average power out to average power in,
so you taking a ratio of currents is completely wrong. You also seem to be trying to analyze the circuit as if it is some
sort of DC circuit or simple AC circuit, which is not correct, as the circuit produces complex AC waveforms, and I have already
gone into much detail in this thread to demonstrate and explain what to watch out for and how to make proper measurements
in that type of circuit, and showed possible ways to do a sanity check on measurements to make sure the measurements are
reasonably in the ballpark of what should be expected. As best as I can tell, Itsu's measurement approach is correct, and is not
showing any signs of OU.

Title: Re: To be deleted
Post by: Void on December 10, 2018, 11:16:29 AM
Quote from: itsu on December 10, 2018, 05:31:38 AM
But i agree, (power) measurements are no easy task and marginal cop > 1 values should be treated with causion.

This bears repeating. :) At low power levels such as in the low mW or lower, 
a difference in measurement of even several mW can be well within the margin of measurement error,
so, yes, I would say jumping to conclusions when measuring on AC circuits with complex waveforms
at low power levels is not at all a good idea. There is just too much room for error or oversights there.

Now, if a person can scale up their circuit to much higher power levels and still measure a higher average
output power than average input power with a difference in the order of many Watts, then IMO that would bear a
much closer look, assuming the measurement methods stand up to close examination. The easiest of all is
just to self-loop a circuit and see if you can get it to self-sustain, as that way no measurements are required. :)

Title: Re: To be deleted
Post by: itsu on December 10, 2018, 04:21:24 PM

I added some more measurements onto the spreadsheet.
I also removed the very first one as that one seems wrong as i probably did not follow my measurement protocol.

But i don't think that like Void mentions it is inaccurate.
We are talking about milli-volts and micro-amps deviations, so pretty accurate if you ask me.

Anyway, here it is:

Itsu
Title: Re: To be deleted
Post by: itsu on December 10, 2018, 04:32:40 PM
Quote from: nul-points on December 10, 2018, 06:27:01 AM
time to recap on the original circuit we're looking at here (and which Itsu is kindly looking at replicating)

i'm attaching an overview of the circuit - it has 1 supply branch and effectively 2 branches (ignoring a higher impedance biasing branch with an effective impedance of approx 10k ohm at 130kHz)

one branch is entirely a current drain, Iin, on the supply and the other is entirely feeding current, Ifb, back to the supply

the net supply current, Isupply, is Iin - Ifb


np

NP,

thanks for the recap, i see what you are trying to say, but i think it is somewhat more complex.

The below screenshot shows the 3 currents:

Isupply grey/white R1
Iin     White R2
Ifb     green

The yellow trace is the voltage across the transistor to show when it turns on and off.

R1 is Isupply and includes the positive cycle Iin and the negative Ifb totaling 7.17mA
R2 is Iin and only has the positive cycle as during the negative Ifb the transistor is off, see yellow trace.
Green is Ifb and can only go left (transistor off) into the supply.

I have overlayed green (and inverted) on R2 to show it is similar as the Isupply waveform R1.

The rms value of Isupply is higher (7.17mA) with the negative cycle, compared to Iin without
the negative cycle (6.5mA), but this is due to how AC rms voltage/current works, it is equivalent to the same
DC value, but it uses both the positive and negative parts to do work (if you can call it that here).


Itsu
Title: Re: To be deleted
Post by: Void on December 10, 2018, 05:04:04 PM
Quote from: itsu on December 10, 2018, 04:21:24 PM
But i don't think that like Void mentions it is inaccurate.
We are talking about milli-volts and micro-amps deviations, so pretty accurate if you ask me.

Hi Itsu. It's all relative. You have now removed one measurement entry that had wide deviation,
so with that entry removed the measurement deviation seems not so wide now. Looking
at the load LED power measurements, you have one measurement at 10.3 mW and one
measurement at 11.27 mW. That is a variation of about 1mW, or about 9% variation,
so not so small a deviation in measurements. Part of the problem could be due to circuit
instability however, as mentioned previously.

There is no point at all in comparing input and output currents in regards to trying to analyze
efficiency of the circuit. As I have already explained, efficiency of a circuit is determined from the ratio
of average output power to average input power. Comparing currents alone gives no indication whatsoever
on the efficiency of a circuit. I have also explained and demonstrated why looking at RMS readings alone in
circuits of this type with complex AC waveforms will very likely lead a person completely astray. :)

At any rate, it should be clear already that this circuit configuration gives no indication at all of OU.

Title: Re: To be deleted
Post by: itsu on December 11, 2018, 04:10:39 AM

Void,

thanks, we get it now.

I have to agree, all the power is almost being accounted for (95%), the rest is very probably consumed like Hoppy mentioned in the potmeter, transformer, cap esr, etc.

Left is now how to correctly interpret the data and the energy flows.


Itsu
Title: Re: To be deleted
Post by: AlienGrey on December 11, 2018, 04:19:41 AM
Oh yes very nice triangle waveforms of loss, the problem with AC and magnetics is you get loss as your magnetising a transformer
core and then you destroy that magnetic flux by reversing it and doing it again in the other direction!
What you really need to do is recover that energy before it's lost in heat.
If you used only positive or negatively charged pulses that didn't reverse in polarity and ironed out the back 'ringing' in your magnetic drive
the results you obtain might turn out far better, also better magnetic devices might produce better results.
Title: Re: To be deleted
Post by: nul-points on December 11, 2018, 05:35:51 AM
Quote from: Void on December 10, 2018, 05:04:04 PM
[...]

There is no point at all in comparing input and output currents in regards to trying to analyze efficiency of the circuit.
As I have already explained, efficiency of a circuit is determined from the ratio of average output power to average input power.
Comparing currents alone gives no indication whatsoever on the efficiency of a circuit

[...]


the supply and the 2 significant current branches share a common voltage (let's call that voltage Vsupply)

the generic equation for a measure of Energy = Volts * Amps * Time

say we measure our energy for a period of 20 cycles (let's call the total time t1)

Our energy supplied for this period is

  Esupply = (Vsupply * Isupply * t1)

Our energy converted for this period is:

  Ein = (Vsupply * Iin * t1)


therefore
Efficiency, n,  = Ein / Esupply = (Vs * Iin * t1) / (Vs * Isupply * t1)

the voltages cancel (because they are common to Numerator and Denominator)


so now Efficiency, n,  = (Iin * t1) / (Isupply * t1)

the time values cancel (because they are common to Numerator and Denominator)


and now Efficiency, n,  = (Iin) / (Isupply)


this is the calculation i show in my overview diagram (and i also used in some previous posts)

so - we ARE able to calculate Efficiency using only current values WHEN the current branches share a common voltage (which they do) AND all the current readings are averaged over a common measurement period (which they are)


in my example above (overview diagram), Vsupply = 3.7V, t1 = 20 cycles (approx 60uS) - these cancel and leave us with:

  Efficiency, n, = (6.93 mA) / (5.7 mA) = 1.2  (= 120%)


Quote from: Void on December 10, 2018, 05:04:04 PM[/size]
[...]

At any rate, it should be clear already that this circuit configuration gives no indication at all of OU.

lol



np
Title: Re: To be deleted
Post by: Void on December 11, 2018, 06:10:44 AM
Sigh...  :o

Title: Re: To be deleted
Post by: itsu on December 11, 2018, 08:30:02 AM

Come on guys, lets agree to disagree for the time being.

I need to make some 4 more measurements to get a decent average and then i can make some duration tests.
See how long it takes to drain the battery.

I can also change the frequency to see if there are major changes in the severall power users, perhaps there is a sweet spot somewhere.

But step  by step.


Thanks,   itsu
Title: Re: To be deleted
Post by: nul-points on December 11, 2018, 09:00:57 AM




hi Itsu


everybody is entitled to an opinion (that includes all of the posters to this thread) - i hope that no opinions have been restricted on this thread?


maths is maths however, and if two sets of values, eg. Current and Energy (under the limits explained), are proportional then you can certainly use one set as validly as another when calculating a ratio/proportion or percentage

if i see someone claiming otherwise** as justification for denying a possibility then i am going to call it out

...and i would hope that ANY other well-intentioned and informed person would do the same


(**does anyone here think that we should not flag up maths issues like this?)



the duration tests sound good - as one part of my investigation with these tests i've been logging battery terminal voltage vs time - it''ll be interesting to compare our results and try to get a better idea of the characteristics of this circuit



np


Title: Re: To be deleted
Post by: Void on December 11, 2018, 09:30:59 AM
Quote from: itsu on December 11, 2018, 08:30:02 AM
Come on guys, lets agree to disagree for the time being.

Hi Itsu. Anyone is free to do as they wish. It is already very clear that the circuit
is not producing OU however, and that is fair and honest assessment. All the best.

Title: Re: To be deleted
Post by: nul-points on December 11, 2018, 09:45:42 AM
as i said, everyone is entitled to their own opinion...  using bad maths to try & justify it, however, is not acceptable
Title: Re: To be deleted
Post by: Turbo on December 11, 2018, 12:28:02 PM
It's all bad math based on theory and assumption and we tend to stick to the formulas and models that work best but the truth of the matter is that we don't have a clue about what is really going on.

Title: Re: To be deleted
Post by: nul-points on December 11, 2018, 01:25:55 PM

the operating cycle for the circuit, based on a pnp blocking oscillator, is as follows:-

-  drain 'half-cycle':
at power on,  the transistor switches on, the primary voltage rises and its current starts to charge the elcap via the coil and also to drive the main LEDs;

the antiphase secondary voltage drives the transistor harder with its low-level output;

when the primary winding saturates, the secondary voltage rises sharply, and turns off the transistor;

- feedback 'half-cycle':
the resulting output current pulse from the coil-field collapse is fed via the feedback LED and elcap, back into the supply battery;

after several cycles, the elcap charges to a steady DC voltage across the main LEDs, so that those LEDs are continuously lit throughout the cycle, not just pulsed once each cycle;

the elcap receives a current pulse via the primary each cycle, to restore the charge which is passed back to the battery during the feedback pulse;

although the feedback current is approx. 20% of the drain current, there is imperceptible ripple on the elcap, compared to its DC voltage


due to the feedback current opposing the drain current, the net supply current is lower than the drain current by approx 20% (in this circuit -YMMV!)


hopefully this is a factual and non-contentious description of the circuit operation

this is a simple circuit - you don't need to believe either the claims, or the counter-claims for its performance - i'm sure you're all well able to build, test & decide for yourselves


np
Title: Re: To be deleted
Post by: itsu on December 11, 2018, 03:33:13 PM

NP,

thanks for your description of your circuit.

I am terrible at math so i won't go there, but about this statement above:

Quotedue to the feedback current opposing the drain current, the net supply current is lower than the drain current
by approx 20% (in this circuit -YMMV!)

Are you sure the feedback current is opposing the drain current?
I mean, yes its of opposite direction, but it is shifted in time, see the screenshot of the supply current R1
in my post #81 here:
https://overunity.com/16384/flyback-data/msg527979/#msg527979

To me it seems that the supply current = drain current and then after the transistor is off, the supply current
stops, followed by the FeedBack current in the opposite direction charging the battery.

This FB current is part of the supply/drain current but time shifted, so one should not subtract it from the
supply current to get a new net supply current.

just my 2 cents,  

Regards Itsu

Title: Re: To be deleted
Post by: nul-points on December 11, 2018, 03:52:41 PM
hi Itsu


thanks for your question, i understand what you're saying - we don't get a lot of experience of AC current with battery sources  :)


think of current as an amount of charge which could flow (in either direction) - if, say, two lots of the same value of charge flowed out of, and back into, a battery then technically it doesn't matter if those 2 flows of charge occurred in 2 sequential periods of 30 minutes (eg. driving a motor, then recharging the battery), or 2000 sequential mS periods of alternate flow direction

the important thing in our calculations here is to choose a regular period which contains the same number of each flow direction (it could be 1 of each, for one cycle; or many cycles each containing 1 of each flow direction). where each flow direction occurs within each cycle is not important

our 2 flow directions here have different average current value, but if you could make them equal then the net charge which has flowed would be zero (regardless of the total time period for the exchange) and the battery would be in the same state of charge after the test as it was before (except maybe temperature!)


i hope this helps - i don't think i've explained it well !


np
Title: Re: To be deleted
Post by: nul-points on December 11, 2018, 03:54:55 PM
[Edit: i'm going to remove this post comparing battery run-times until i can confirm that i'm comparing use of the same battery and not 2 from the same batch]

i'll repost later


np
Title: Re: To be deleted
Post by: nul-points on December 11, 2018, 06:46:17 PM

Quote from: itsu on December 11, 2018, 03:33:13 PM
[...]

This FB current is part of the supply/drain current but time shifted, so one should not subtract it from the
supply current to get a new net supply current.

[...]

hi Itsu

i don't think i fully addressed your question earlier - i see your concern that the feedback current should only be considered as part of the 'drain' current, Iin

yes, you're right - all charge must have come from the initial drain current, Iin, for each cycle

but if, say, 2 Coulombs of 'charge' flow into the circuit, as Iin, and 0.5 Coulombs of charge flows back out, as Ifb, in the same cycle, then a net 'charge' of 1.5 Coulombs has flowed into the circuit, and the effective current to achieve that is (Iin - Ifb), regardless of the sequential timing of those 2 different charge-transfer directions (within the same cycle)

i hope this answers your question


np
Title: Re: To be deleted
Post by: itsu on December 12, 2018, 04:59:29 AM

NP,

thanks for trying to explain  :)

i understand what you are saying, but in my mind there must be a flaw somewhere otherwise we would have excess energy in this
circuit and the battery would not drain down.

Not sure though what this flaw is, perhaps its NOT in the same cycle (supply and feedback).

Anyway, the duration tests could shed some light on that i think.

Today i will finish my last two measurements for a total of 10 so i should have a decent average.

Then i will remove the feedback led connection and do some measurements to see how this new situation influence
the severall current / energy flows.

Itsu
Title: Re: To be deleted
Post by: nul-points on December 12, 2018, 02:52:59 PM

hi Itsu

i understand your doubts about the drain and feedback currents alternating in direction within each cycle, but the green LED wouldn't illuminate if there wasn't a flow of current from the circuit to the battery, and that current can only be generated by the coil-field collapse current (which occurs once at the end of each cycle, as i described in post #92)

the value of the current returning to the battery is only about 20% of the current that the battery supplied to the circuit, so all that will happen is that the circuit runtime will be extended.  we can't create extra energy (because energy is conserved) - but what we can do is store & then convert some of the original energy again to do some extra work (because work is not conserved)

each cycle, the circuit stores some energy in the coil field and in the elcap; the coil field collapse causes some current to flow in the secondary and this current can draw on energy stored in the elcap to flow through the feedback LED and battery (this is the extra work done)

the only way for the battery to not drain down is if we could return more current to the battery than the battery supplies to the circuit (it would need to be more because the battery recharge efficiency is only about 70%)

we're only returning about 20% of Iin, so we're nowhere near a self-runner (and i guess we never will be); but we are increasing the runtime of the circuit for the same initial energy stored in the supply

i look forward to seeing the results from your battery rundown tests


np
Title: Re: To be deleted
Post by: nul-points on December 12, 2018, 03:03:00 PM
Quote from: nul-points on December 11, 2018, 03:54:55 PM
[Edit: i'm going to remove this post comparing battery run-times until i can confirm that i'm comparing use of the same battery and not 2 from the same batch]
i'll repost later
np


i've supplied my circuit from a 1F capacitor to do some comparative run tests

i charged the cap to 4+ V and started the run as the voltage slowly approached 4V

i stopped the run as the cap voltage just crossed the 3.4V

i measured the time between the crossing at the 4V & 3.4 cursors



the runtime for the circuit WITH NO feedback is approx 125s

the runtime for the circuit WITH LED feedback is approx 140 s

(i also measured the runtime for a circuit variant WITH SCHOTTKY DIODE feedback at approx 224 s
- but i think the nearly doubled time was mostly due to a change in drive conditions)


np
Title: Re: To be deleted
Post by: itsu on December 12, 2018, 04:20:39 PM

NP,

thanks for the run down graphs, how did you make those?

Good idea to use a supercap, else run down times would be endless.
So the feedback (led) provides for about 15s longer runtime which is 18.75% (your 20% feedback current?)

Anyway, i have completed my 10 runs for the normal circuit, see below spreadsheet top part.
The 10x average is on the bottom of this first colored part underscored.

Then i quickly today ran 4 runs without the feedback (led) attached, to show that we have a limited increased power
drawn from the battery 15.28mW v 13.65mW (+1.63mW).
See lower part of the spreadsheet, again 4x average on the bottom of this lower part underscored.
Guess some power is lost somewhere as the ratio "pwr users / input" drops from average 94% to 85%.


Will start with some run down tests now,               looking for my super caps......


Itsu
Title: Re: To be deleted
Post by: nul-points on December 12, 2018, 05:19:55 PM

hi Itsu

you've been BUSY!!!!

those power user results look pretty comprehensive


its a shame that you haven't been able to replicate the true net supply current value (ie net flow of charge into & out of the circuit) - your scope traces show the 2 parts, so they are there (and your Green LED current value shows that charge is flowing back to your supply)


yes, large value caps seem to be the way to go when you have many tests to do

my 1F cap is physically large as well as large value - i think they were designed for crazy-loud car audio systems, so presumably very low ESR

i just replaced the battery with the 1F cap, charged it slightly above my required start voltage, then watched the voltage approach the cursor and started the circuit when the voltage was slightly above. i stopped logging just after the end cursor was crossed and subtracted the end time from the start time


good luck with your next tests
np
Title: Re: To be deleted
Post by: nul-points on December 14, 2018, 06:47:14 AM
Quote from: itsu on December 12, 2018, 04:20:39 PM
Then i quickly today ran 4 runs without the feedback (led) attached, to show that we have a limited increased power drawn from the battery 15.28mW v 13.65mW (+1.63mW).

hi Itsu

i'm not sure about your input power calcs...

on your spreadsheet, you make them:
15.28mW (with feedback)
13.65mW (no feedback)

from your data, i make them:
3.75V * 6.97mA = 26.14mW (with feedback)
3.75V * 6.45mA= 24.19mW (no feedback)


thanks
np
Title: Re: To be deleted
Post by: itsu on December 14, 2018, 07:13:28 AM

Hi NP,

indeed, you make them:
3.75V * 6.97mA = 26.14mW (with feedback)
3.75V * 6.45mA= 24.19mW (no feedback)


but i did not make them, i had the Tektronix TDS3054B scope calculate them to be:
15.28mW (with feedback)
13.65mW (no feedback)

Your formula is for DC signals only, while we are dealing with a complex AC like current waveform.
The scope is able to deal with those complex AC like waveforms using its elaborate math functions.

Itsu

Title: Re: To be deleted
Post by: AlienGrey on December 14, 2018, 08:21:27 AM
I don't know its just another blocking oscillator like all the rest but its upside down!
Sling the ferrox core out and replace it with as NANOCRYSTAL one and tune for best output.

Not trying to be clever but that stuff the nanoparticles stay charged and if you just pulse the device
you're bound to get again.
Title: Re: To be deleted
Post by: itsu on December 14, 2018, 08:30:53 AM

Finaly got some drain duration results.

Using 3x 10F caps in series = 3.3F

Charged till 4.15V, activate the circuit, at 4V start the timer, at 3.4V stop timer.

Circuit with feedback      shows drain time from 4V => 3.4V to be average (10X) 10 minutes 14 seconds = 614s
Circuit with no feedback shows drain time from 4V => 3.4V to be average (10x)   9 minutes                    = 540s

So the remove of the feedback shorts the draining time by 1m14s = 74s = 12%.

The graph shows the both situations:

Regards Itsu
Title: Re: To be deleted
Post by: nul-points on December 14, 2018, 09:34:30 AM
thanks Itsu, your cap draw-down results seem comparable with mine


here is a way to visualise what is occurring with this circuit in terms of average quantity for the transfer of charge (ie. current flow):

say you have 3 watertight containers (2 larger ones, Cb & Cc, and 1 smaller cup Ct) and a timer

(NO FEEDBACK)
after using Ct to fill Cb with a quantity of 20 cupfuls of water,
start the timer, and then at every new minute use Ct to transfer a quantity of 5 cupfuls of water from Cb to Cc;
(so a quantity of water is transferred in one direction only, each minute, from Cb to Cc)

Q1) after the 4th minute of water transfers, how many cupfuls of water remain in Cb?
A1) 0 cupfuls

Q2) what is the average quantity of water transferred per minute?,
A2) (5 cupfuls * 4 transfers) / 4 minutes = 5 cupfuls per minute


now we repeat the test with a slight difference to the water transfer procedure:

(WITH FEEDBACK)
after refilling Cb with 20 cupfuls of water,
start the timer, and then every minute use Ct to transfer 5 cupfuls of water from Cb to Cc, and also transfer 1 cupful back from Cc to Cb
(so 2 different quantities of water are transferred, one in each direction, each minute, between Cb and Cc)

Q3) after the 4th minute of water transfers , how many cupfuls of water remain in Cb?
A3) 4 cupfuls

Q4) what is the average quantity of water transferred per minute?
A4) ((5-1) cupfuls * 4 transfers) / 4 minutes = 4 cupfuls per minute

note that you still have 4 cupfuls of water left in Cb - it would take another minute's transfer event to empty Cb. (this extra runtime behaviour matches what we're seeing with our capacitor draw-down tests)

so, in the posted circuit, if the voltage waveform across a CSR (or equivalent sensor) at the battery terminal shows 2 pulses of different polarity, then 1 pulse is caused by current, Iin, flowing INTO the circuit from the battery to power the components, and the other is caused by current, Ifb flowing OUT of the circuit back into the battery, recharging it.  the average supply current, Isupply = (Iin - Ifb)

in my circuit, the average supply current, Isupply = (Iin - Ifb) = (6.5mA - 1.2mA) = 5.3 mA. (DMM reading was 5.4mA)


np
Title: Re: To be deleted
Post by: nul-points on December 14, 2018, 09:48:01 AM
Quote from: itsu on December 14, 2018, 07:13:28 AM
indeed, you make them:
3.75V * 6.97mA = 26.14mW (with feedback)
3.75V * 6.45mA= 24.19mW (no feedback)

but i did not make them, i had the Tektronix TDS3054B scope calculate them to be:
15.28mW (with feedback)
13.65mW (no feedback)

Your formula is for DC signals only, while we are dealing with a complex AC like current waveform.
The scope is able to deal with those complex AC like waveforms using its elaborate math functions.

ah, ok

my 'formula' Vsupply * Isupply is the standard calculation for input power calculated from the average Voltage & Current supplied to the circuit

i understand that you're reporting the values provided by the scope, but i thought that those values would have some similar meaning (ie.  that you could multiply the voltage reading by the current reading and that the result would produce some meaningful value for the input power)

in that case, it's not clear to me what is the relationship between the data provided by the scope, and what is the meaning of any individual reading in terms of standard power calculations - eg. is there any reading on the spreadsheet which represents 'average input current' in the conventional meaning of that phrase?


thanks
np
Title: Re: To be deleted
Post by: itsu on December 14, 2018, 10:15:34 AM

NP,

as said, the scope uses its elaborate math function to compute the data found on the traces it is processing
(here CH1, yellow, voltage times CH4, green, current).
The red math trace presents this data averaged over many samples.

Besides what the math function is doing with the, in this case, data found on CH1 and CH4, i can display the
data on those channels to my likings, like in rms, mean etc.

Normally i put the data on those channels in rms, but if i would put the data in mean, nothing would change on
the math trace as it is still doing its thing with the data.

In post #56 of this thread:
https://overunity.com/16384/flyback-data/msg527867/#msg527867

you can see in the first screenshot that i have the input power being calculated and present the (DC) voltage
yellow in rms and the complex AC current green in both rms and mean (together with the math red calculation in mean).

If you think that the complex AC current in green presented in mean would be more meaningfull, then i could use
that for next calculations, but i still consider the calculated power via the red math trace more reliable. 

Itsu
Title: Re: To be deleted
Post by: nul-points on December 14, 2018, 10:32:57 AM
ok, i understand what you're doing




thanks
np
Title: Re: To be deleted
Post by: itsu on December 15, 2018, 05:30:04 AM

As the percentage "power users versus input" drops from 94% with feedback to 85% without feedback (see post #100),
i was trying to find out where that missing 15% in the no feedback situation went.

I was doing that by monitoring the temperature of some components like the transformer, the potmeter and the
electrolytical cap using my laser temp. meter.

But it seems that this method is too rough for these small currents as i do see some temp. changes, but in a
irregular way, so probably due to room temp changes.

Guess i have to scope the current paths to these components to get a more accurate view.


Itsu
Title: Re: To be deleted
Post by: nul-points on December 15, 2018, 06:50:42 AM
that's a shame about the resolution issue of a low-powered circuit, Itsu - it would have been a very interesting view of what's going on where


i've repeated the 1F (nominal) cap draw-down test swapping out the original coil with a larger coil (both turns and ferrite, also a toroid rather than solenoidal)

the average runtimes (for same draw-down voltage range) are as follows:-

NO FEEDBACK: t = 68.4s

LED FEEDBACK: t = 89.6s

ratio = 89.6 / 68.4 = 1.31


thanks
np
Title: Re: To be deleted
Post by: itsu on December 15, 2018, 03:21:07 PM

I started measuring the current through the potmeter with and without feedback
and the current to the electrolytical cap with and without feedback.

We are dealing with minute currents through the potmeter, but we see a big increase of the current when feedback
is removed from 402uA with, to 636uA without,  see screenshot 1.
White is the current through the potmeter with feedback, green without.
Also to notice is that the frequency went up when feedback is removed.


The current to the elco is much bigger, but behaves the other way around, less without feedback, from 6.22mA with, to
5.04mA without, see screenshot 2.


I don't think though that this will account for the missing power in the system with feedback removed.


Itsu
Title: Re: To be deleted
Post by: nul-points on December 15, 2018, 08:11:49 PM
yes, good call Itsu


the frequency does increase with 'no feedback' (and i think this is because in the feedback case, the f/b pulse increases the cycle duration whilst the coil collapse energy performs work, charging the battery somewhat & lighting the feedback LED - in the 'no feedback' case the coil collapse energy dissipates wherever it can)

interesting about the resistor pot - after i used the pot to select a suitable operating drive level i replaced the pot with a fixed resistor

the elco behaviour seems reasonable -

with feedback, the elco is drained in 2 ways:
1) buffering the LEDs
2) completing the coil collapse current circuit back thro' the Feedback LED

without feedback, the elco is only drained as in 1) above


thanks
np
Title: Re: To be deleted
Post by: nul-points on December 15, 2018, 08:15:06 PM
QUESTION:  Is the average supply current, Isupply, equal to the difference between the current drawn, Iin, and the feedback current, Ifb?

ANSWER:  Yes it is

SCOPE TEST:  measure Isupply at a mid-range voltage (3.75V)

(With Feedback)
signal cycle = 70.4kHz, full cycle = 14.1us

DC average, across 1 ohm CSR, using scope math from traces shown below

Isupply: 8.0mA 
(full cycle)

Ifb: -5.24mA  *   (4.7us / 14.1us) = -1.67mA  (averaged over full cycle)

Iin: 14.6mA * (9.4us / 14.1us) = 9.7mA  (averaged over full cycle)

Difference between Iin & Ifb = (9.7 - 1.67) = 8.03mA
Title: Re: To be deleted
Post by: itsu on December 16, 2018, 05:39:58 AM
Quote from: nul-points on December 15, 2018, 08:11:49 PM
yes, good call Itsu


the frequency does increase with 'no feedback' (and i think this is because in the feedback case, the f/b pulse increases the cycle duration whilst the coil collapse energy
performs work, charging the battery somewhat & lighting the feedback LED - in the 'no feedback' case the coil collapse energy dissipates wherever it can)
interesting about the resistor pot - after i used the pot to select a suitable operating drive level i replaced the pot with a fixed resistor

the elco behaviour seems reasonable -

with feedback, the elco is drained in 2 ways:
1) buffering the LEDs
2) completing the coil collapse current circuit back thro' the Feedback LED

without feedback, the elco is only drained as in 1) above


thanks
np

NP,

you mention: "in the 'no feedback' case the coil collapse energy dissipates wherever it can".

So that i should be able to measure, but up till now (temperature tests, current tests) i was not able to detect.
Perhaps i should check with the Spectrum Analyzer if the spectrum is more polluted without feedback.


Itsu



Title: Re: To be deleted
Post by: itsu on December 16, 2018, 05:58:54 AM
Quote from: nul-points on December 15, 2018, 08:15:06 PM
QUESTION:  Is the average supply current, Isupply, equal to the difference between the current drawn, Iin, and the feedback current, Ifb?

ANSWER:  Yes it is

SCOPE TEST:  measure Isupply at a mid-range voltage (3.75V)

(With Feedback)
signal cycle = 70.4kHz, full cycle = 14.1us

DC average, across 1 ohm CSR, using scope math from traces shown below

Isupply: 8.0mA 
(full cycle)

Ifb: -5.24mA  *   (4.7us / 14.1us) = -1.67mA  (averaged over full cycle)

Iin: 14.6mA * (9.4us / 14.1us) = 9.7mA  (averaged over full cycle)

Difference between Iin & Ifb = (9.7 - 1.67) = 8.03mA

NP,

i cannot confirm your above question as i did not record the average (mean) current.
My voltages and currents are in rms, and then you see that Isupply is almost Iin while Ifeedback is way more then the difference.

Itsu   
Title: Re: To be deleted
Post by: nul-points on December 16, 2018, 08:09:55 AM
Quote from: itsu on December 16, 2018, 05:39:58 AM
you mention: "in the 'no feedback' case the coil collapse energy dissipates wherever it can".

So that i should be able to measure, but up till now (temperature tests, current tests) i was not able to detect.
Perhaps i should check with the Spectrum Analyzer if the spectrum is more polluted without feedback.

Itsu

hi Itsu
yes, i was thinking energy radiated away from the circuit - the waveform across a supply CSR with no feedback shows a burst of HF and also some LF oscillations

thanks
np
Title: Re: To be deleted
Post by: nul-points on December 16, 2018, 08:22:32 AM
Quote from: itsu on December 16, 2018, 05:58:54 AM
i cannot confirm your above question as i did not record the average (mean) current.
My voltages and currents are in rms, and then you see that Isupply is almost Iin while Ifeedback is way more then the difference.

Itsu

that's ok, thanks Itsu - my general post about the average supply current was just provided as supporting evidence relating to the results from my circuit, for other interested viewers of the thread

thanks
np
Title: Re: To be deleted
Post by: itsu on December 16, 2018, 08:42:02 AM

NP,

i just redid the basic power measurements now including the average (mean) current through the severall paths.

Picture is probably to wide for the thread, but decreasing it would make it unreadable i guess.

I looked at the Spectrum with my SA, but only a very slight increase in overall bottom level in the first 33MHz
is noticed, no specific peaks or noise bands, both with or without feedback.

By the way, just to confirm, i remove the green led cathode from the plus rail to measure without feedback.


Itsu
Title: Re: To be deleted
Post by: nul-points on December 16, 2018, 10:22:08 AM
thanks for the update, Itsu

average readings (apart from their lower values) still suggest that our circuits are different (ie. your Isupply and Iin are similar, whereas my values show that Ifb as the difference between Isupply and Iin)

ok, so it doesn't look like the missing energy is being radiated significantly

yes, i use the same method for 'NO feedback' tests


np
Title: Re: To be deleted
Post by: itsu on December 16, 2018, 10:31:05 AM

NP,

Hmmm,  i would of thought that you would be pleased with my average results as according to me they present
exactly what you were saying (or do i understand you wrong?).

My average values are:

Isupply = 4.07mA,
Iin        = 4.74mA,
Ifb       = 0.80mA

so indeed like you say, "Ifb as the difference between Isupply and Iin"  well, almost.

Itsu
Title: Re: To be deleted
Post by: nul-points on December 16, 2018, 12:43:40 PM

hi Itsu

ah ok, i must have seen that the feedback current RMS value looked like the previous table (approx 2mA) where the math wasn't identified (just labelled as current(mA) and thought that value hadn't changed - my apologies!

yes, those values are a closer fit to the pattern of the circuit behaviour


thanks
np
Title: Re: To be deleted
Post by: nul-points on December 16, 2018, 07:00:43 PM
QUESTION2: Do the capacitor supply draw-down tests support the general comparison seen on this circuit when using another method (eg. scope math)?
ANSWER: Yes they do...

the 1F (nominal) supply capacitor was identified as having an actual capacitance of 1.111F, using the draw-down time for a 681 ohm resistor (and then using online calculation)

the start and end values for charge on the 1.111F cap for the tests was as follows:-
4.0V  4.444Coulombs
3.4V  3.777Coulombs

The total value of charge transferred was:
  (4.444 - 3.777) = 0.667Coulombs = 667mCoulombs


runtime for circuit WITHOUT FEEDBACK: 68.4s
667mC => 667mAs / 68.4s = an average Isupply of 9.75 mA for 68.4s

runtime for circuit WITH FEEDBACK: 90s
667mC => 667mAs / 90s = an average Isupply of 7.4mA for 90s


the CSR traces show that charge transferred from the supply is comparable for both cases:
WITH FEEDBACK: positive pulse (14.6mA for 9.6us) = 137.24uC
WITHOUT FEEDBACK: positive pulse (10.6mA for 12.72us) = 134.8uC

BUT in the 'WITH FEEDBACK' case, 24.4 uC are ALSO transferred BACK to the supply to be re-used;
whereas in the 'WITHOUT FEEDBACK' case, NO charge is recycled

the result is that the runtime of the 'WITH FEEDBACK' lasts around 20-30 % longer, with the same average power delivered by the supply to the circuit throughout - ie. some of the total energy converted is recycled to perform approx 20-30% more work

it appears that this 'WiThFeedback' circuit is able to gather energy which would otherwise have been dissipated and lost, and re-use it to do extra work

energy is conserved - work is not!


np
Title: Re: To be deleted
Post by: itsu on December 17, 2018, 04:47:55 AM

NP,

nice analysis, i like it.

Quoteit appears that this 'WiThFeedback' circuit is able to gather energy which would otherwise have been dissipated and lost, and re-use it to do extra work


But, to my understanding, considering your: "gathered energy which would otherwise have been dissipated and lost"  this dissipated energy also did work to get dissipated.

Not sure yet in which form (heat, light, radiation), but it was doing work then.


So would you not say that you utilized this "dissipated and lost energy" by using feedback and so improving on the overall efficiency (input versus work done) of the circuit?


Itsu
Title: Re: To be deleted
Post by: nul-points on December 17, 2018, 05:18:28 AM
thanks Itsu, appreciated

i understand what you're saying


before i reply, i should make it clear to other visitors that although i'm saying this whole arrangement is > 100% efficient, that relates to total work done (which is one or more kinds of conversion of energy) - all the usual losses apply (I^2R loss, heat loss, etc)

no laws of physics are broken here  - energy is not (and cannot be) created

god might disagree!  :)


i agree that the dissipation of the 'packet' of energy which was temporarily stored within the circuit (without feedback) is doing work (ie. it is still converting that part of the energy which was supplied) - this part of the energy, however, is not being converted into work as part of the circuit - and, at the moment, we don't know by what route it is leaving the system. my guess is that it is converting to heat, or other radiation

the WithFeedback circuit is also storing this part 'packet' of the total energy supplied, just the same, but because the coil field collapse is trying to maintain its original current flow when the transistor switches off the supply, it finds a suitable current path which just happens to be arranged to feedback this current, from the elcap, thro' the secondary, thro' the feedback LED, thro' the battery and back to the elcap

this reverse current is just a standard recharge current for the battery and so some of the recycled energy gets stored in the battery to be available for re-use by the circuit

the value of total work which is done by the 'mysterious', disappearing energy packet, will be the same value as the work done by that packet within the feedback path of the WithFeedback circuit, in the elcap, coil winding. feedback LED & battery

i guess its a bit like using a water-fuelled fuel cell - the fuel supply is water, the cell converts energy from water into Work, Heat, and... water

we could collect the water output and feed it back in with the water input supply and increase the efficiency of the whole system


...but hey, i'm just this guy called nul-points - what do i know?  :)

np
Title: Re: To be deleted
Post by: Void on December 18, 2018, 01:28:22 PM
Quote from: nul-points on December 17, 2018, 05:18:28 AM

before i reply, i should make it clear to other visitors that although i'm saying this whole arrangement is > 100% efficient, that relates to total work done (which is one or more kinds of conversion of energy) - all the usual losses apply (I^2R loss, heat loss, etc)

Hi nul-points. I have already explained that based on the proper measurement approach that Itsu took
with his scope, it is clear that this circuit arrangement is not showing any indications of OU.
Your circuit analysis is incorrect due to your misunderstandings of AC circuits.
Just want to make sure no one is mislead here. All the best. :)

Title: Re: To be deleted
Post by: nul-points on December 18, 2018, 02:53:15 PM
Quote from: Void on December 18, 2018, 01:28:22 PM
Your circuit analysis is incorrect due to your misunderstandings of AC circuits.
Just want to make sure no one is mislead here. All the best. :)

hi Void, welcome back

i'm happy to let the members here make up their own minds about the behaviour of this circuit

i find it intriguing that you think this unbalanced biphasic waveform, where the 1st pulse is generated by the draw of the circuit on the DC battery supply, and the 2nd pulse is generated by the action of the coil field-collapse and stored energy in the connected elcap (generating recharging current back through the battery) would be best treated by AC circuit analysis

you're welcome to your opinions, of course - and everyone else is welcome to theirs, too


np
Title: Re: To be deleted
Post by: Void on December 18, 2018, 03:15:03 PM
Quote from: nul-points on December 18, 2018, 02:53:15 PM
i find it intriguing that you think this unbalanced aphasic waveform, where the 1st pulse is generated by the draw of the circuit on the DC battery supply, and the 2nd pulse is generated by the action of the coil field-collapse and stored energy in the connected elcap, generating recharging current back through the battery, would be best treated by regular AC circuit analysis

Hi nul-points. I don't think that at all. I said Itsu's measurement approach is correct (as far as I can see anyway). 
Itsu's measurement approach takes into account what is happening dynamically in the
primary power consuming components in the circuit and should be giving a fairly reasonably
accurate indication of the average power consumption of those components, as long as his current probe
is measuring reasonably accurately that is. It doesn't matter if there are feedback currents or very spiky
wave forms with positive and negative swings or whatever else, this measurement approach will take them into account,
based on my understanding and quite a lot of prior experience in making my own measurements on similar types of circuits
with similar types of complex waveforms.

Title: Re: To be deleted
Post by: nul-points on December 18, 2018, 04:40:29 PM
Quote from: Void on December 18, 2018, 01:28:22 PM
Your circuit analysis is incorrect due to your misunderstandings of AC circuits.
Just want to make sure no one is mislead here. All the best. :)

this is the quote of yours i was addressing - don't try to change the subject please!

you said my circuit analysis is incorrect due to my misunderstandings of AC circuits

i say this circuit generates its own unsymmetric biphasic pulses

- its not driven by an externally generated AC waveform

- its not necessary, or appropriate, to use AC analysis

but then you ought to know that, based on your "understanding and quite a lot of prior experience in making [your] own measurements on similar types of circuits with similar types of complex waveforms"
Title: Re: To be deleted
Post by: Void on December 18, 2018, 04:51:04 PM
Sigh...  ::) :o  :D  ;D   

Title: Re: To be deleted
Post by: nul-points on December 18, 2018, 05:47:05 PM
Quote from: Void on December 18, 2018, 04:51:04 PM
Sigh...  ::) :o :D ;D


...and don't think you're going to get round me with that crazy, whispering in my ear, trick!
Title: Re: To be deleted
Post by: nul-points on December 18, 2018, 05:50:55 PM
Ladies andddddddd Gentlemen....   it's the 'nul and Void' Show






(apologies - *somebody* had to say it)
Title: Re: To be deleted
Post by: Void on December 18, 2018, 08:42:00 PM
Quote from: nul-points on December 18, 2018, 05:50:55 PM
it's the 'nul and Void' Show

Good one. :)

Title: Re: To be deleted
Post by: NickZ on December 18, 2018, 08:49:42 PM
   Efficiently conserving energy is one thing, but, most of the time it's not enough to self run a device,
though. Right?
   But isn't the idea to do more "work" than is provided for? In some type of "open circuit", that allows for tapping into extra energy, from wherever? Yet, the illusive source of extra juice seams to be very difficult to obtain, and harder yet to prove. In any case that's what I'm still working on, on my own, no direction home. Hate to be alone...   Using the Dr. Stiffler " double diode loops" ideas, and seeing where it wil take me. Tricky road to say the least.But, it does concern giving a portion of the output, back to the input (24v battery inputs). To see what you have mentioned about efficiency. However, that is not my goal, until a device can self run, with no outside man made source, at all. An ongoing work in progress for me, still.
Title: Re: To be deleted
Post by: nul-points on December 19, 2018, 04:11:41 AM

hi Nick
i agree, i think you're right on both counts-

- conserving energy is good, but just that on its own wouldn't be the road to a self-runner

- re-cycling energy to do additional work (ie. extra to what you would have expected to do, on the usual single-pass conversion of energy) appears to be opening up the n > 100% door for us

'overunity' can apply to work regardless of any limitations on energy, and i guess that we'll find plenty of ways to re-cycle packets of energy if we start looking


good luck with your Stiffler double-diode ideas

energy is more slippery than jello on a hot plate - make sure you have a good bucket handy

np
Title: Re: To be deleted
Post by: nul-points on January 04, 2019, 06:19:02 PM
The 2 battery-rundown graphs below provide further confirmation that a proportion of the energy being stored in this circuit is then getting recycled back into the supply storage (rechargeable battery or capacitor) - this recycled energy becomes available to do extra work, eg. by extending the battery runtime at the same power rate

the WiThFeedback circuit has extended the active runtime from the battery to approx 120%, compared to the same circuit when the feedback (read 'Looped') connection is removed

the useful battery voltage range here is taken to be 3.7V down to 3.4V (at which point the Output DC drive level starts to reduce sharply, slightly delayed from the battery voltage decrease

the unlooped circuit runs for  48s  and the WiThFeedback circuit runs for  58s,  providing 20% more work at a comparable level


for these tests, i'm using a pancake coil to generate the pulses (having already successfully tested the circuit with solenoidal, toroidal & multi-core transformers) - the pancake coil is centre-tapped (to form the common connection between primary & secondary), basket-weave, approx 30 turns of 0.45mm copper (aka magnet wire), approx 20mm ID, 75mm OD

it appears to give similar results with or without an adjacent ferrite 'core', but these results are from the tests with ferrite


i suspect that this WiThFeedback circuit concept is not novel...

as usual, Tesla got there first with his Oscillator-Shuttle-Circuits, 'bleeding' or 'siphoning off' a proportion of the total energy being shuttled between 'floating' grounds (see Barrett's paper, submitted to the Louis de Broglie Foundation in the 1990s, discussing Telsa's OSC theory)


TAKE NOTE
i'm happy to continue discussing other member's attempts (with real hardware, not Sim) to replicate the results i'm getting - i think i've provided more than enough detail already - you have now got way more info than i had when i started out on this journey!

i will NOT be entering into discussion with anyone who offers cliches, strawman arguments, appeals to authority, unsupported opinions, etc, etc in place of 'terse & technical' comments related to specific results reported in this thread


"The person who says it cannot be done, should not interrupt the person doing it"

np
Title: Re: To be deleted
Post by: tinman on January 07, 2019, 10:02:10 AM
Hi NP
Hope you dont mind me posting this here,but it is in regards to some flyback testing i have been doing on a coil for another project.

This coil differs from most,as it has a PM for the core,and is giving some very interesting results.

The measurements taken below do not include the power dissipated by the LED,but only the power flowing into C1 from the source,and then the power flowing out of C1 into the coil.

The calculated power from C1 is only calculating the power delivered by C1 to the coil when the transistor switches on-->duty cycle is 5%.
This is far greater than the power being delivered to C1 by the source.

So there is more power flowing through the coil,than there is being delivered by the source.
This means that the magnetic field being built by the coil is far greater than that that could be built by the supplied power from the source. And as i stated above,this is not including the power being dissipated by the LED.

I am not sure whether RMS or average values are used to calculate power in this case?.
Maybe void could shed some light on this.


Brad
Title: Re: To be deleted
Post by: nul-points on January 07, 2019, 10:54:05 AM
Quote from: tinman on January 07, 2019, 10:02:10 AM
Hi NP
Hope you dont mind me posting this here,but it is in regards to some flyback testing i have been doing on a coil for another project.

This coil differs from most,as it has a PM for the core,and is giving some very interesting results.

The measurements taken below do not include the power dissipated by the LED,but only the power flowing into C1 from the source,and then the power flowing out of C1 into the coil.

The calculated power from C1 is only calculating the power delivered by C1 to the coil when the transistor switches on-->duty cycle is 5%.
This is far greater than the power being delivered to C1 by the source.

So there is more power flowing through the coil,than there is being delivered by the source.
This means that the magnetic field being built by the coil is far greater than that that could be built by the supplied power from the source. And as i stated above,this is not including the power being dissipated by the LED.

I am not sure whether RMS or average values are used to calculate power in this case?.
Maybe void could shed some light on this.


Brad


mind you cr@pping all over my thread again?  no, of course i don't...

...as long as you don't mind me returning the favour on one of yours


If your results have any merit, then i recommend that you start your own thread about them - that way your friend Void can go there and comment about them all he likes...

...and he won't be getting a hard time again here


thanks for asking

np
Title: Re: To be deleted
Post by: tinman on January 07, 2019, 09:17:52 PM
Quote from: nul-points on January 07, 2019, 10:54:05 AM

mind you cr@pping all over my thread again?  no, of course i don't...

...as long as you don't mind me returning the favour on one of yours


If your results have any merit, then i recommend that you start your own thread about them - that way your friend Void can go there and comment about them all he likes...

...and he won't be getting a hard time again here


thanks for asking

np

Enjoy
Title: Re: To be deleted
Post by: nul-points on January 08, 2019, 01:24:52 AM
Quote from: nul-points on January 04, 2019, 06:19:02 PM
TAKE NOTE
i'm happy to continue discussing other member's attempts ... to replicate the results i'm getting
...
...

np


Quote from: tinman on January 07, 2019, 10:02:10 AM
Hi NP
Hope you dont mind me posting this here,but it is in regards to some flyback testing i have been doing on a coil for another project.
...
...
I am not sure whether RMS or average values are used to calculate power in this case?.
Maybe void could shed some light on this

Brad

i suggest you go and dump your testing results from "another" project onto a thread started by Void, where he's sharing with us all an OU-related project he's made...

there must be plenty of those, right?
Title: Re: To be deleted
Post by: tinman on January 08, 2019, 08:45:39 AM
Quote from: nul-points on January 08, 2019, 01:24:52 AM

i suggest you go and dump your testing results from "another" project onto a thread started by Void, where he's sharing with us all an OU-related project he's made...

there must be plenty of those, right?

I know you dont like input from those that know,such as void,as it doesn't agree with your mistakes.

I just thought you might like something with substance on your thread,as you seem to be here all alone now  ;D .

But you enjoy your journey of solitude.


Brad
Title: Re: To be deleted
Post by: Void on January 08, 2019, 02:11:41 PM
Quote from: tinman on January 07, 2019, 10:02:10 AM
Hi NP
Hope you dont mind me posting this here,but it is in regards to some flyback testing i have been doing on a coil for another project.

This coil differs from most,as it has a PM for the core,and is giving some very interesting results.

The measurements taken below do not include the power dissipated by the LED,but only the power flowing into C1 from the source,and then the power flowing out of C1 into the coil.

The calculated power from C1 is only calculating the power delivered by C1 to the coil when the transistor switches on-->duty cycle is 5%.
This is far greater than the power being delivered to C1 by the source.

So there is more power flowing through the coil,than there is being delivered by the source.
This means that the magnetic field being built by the coil is far greater than that that could be built by the supplied power from the source. And as i stated above,this is not including the power being dissipated by the LED.

I am not sure whether RMS or average values are used to calculate power in this case?.
Maybe void could shed some light on this.

Brad

Hi Brad. Average power is calculated using RMS values for the voltage and current,
however it is not that straightforward. Coils and capacitors produce reactive currents
which are out of phase with the voltage across them. This is sometimes referred to as
'reactive power' but it is not really power. When you are looking at efficiency of circuits
you are only concerned with real power, which is power consumption that depletes energy in
a power source, or power consumption in a load that consumes power, such as resistors
and bulbs. Power calculations for the purpose of determining circuit efficiency applies
to the input power from your power source and the output power for real power consuming loads.
Looking at current flowing in coils and capacitors does not tell you about the efficiency of
the circuit. That is an interesting setup with the coil around a magnet core however.
All the best...
Title: Re: To be deleted
Post by: nul-points on January 08, 2019, 02:21:03 PM
Quote from: tinman on January 08, 2019, 08:45:39 AM
I know you dont like input from those that know,such as void,as it doesn't agree with your mistakes.

I just thought you might like something with substance on your thread,as you seem to be here all alone now  ;D .

But you enjoy your journey of solitude.


Brad


awwww, that's so SWEEEEEET, it's almost like you really cared about me being lonely!

how could i possibly be lonely when i have you here, Big Boy?!?

i don't care what the others say about you, you'll always have a special place in this thread now


...let's close the door quietly  now and give these two lovers some privacy  ;)
Title: Re: To be deleted
Post by: nul-points on January 11, 2019, 04:09:39 AM


...
Title: Re: To be deleted
Post by: synchro1 on January 12, 2019, 02:05:21 PM


Dragone and Zaev.
A few contributors have offered theories about the magnet core coil gain; mentioning field compression etc. These guys are reinventing the wheel.

Tinman and Itzu need to take caliometric readings of the coil's core magnet. They will find their magnets are cooling.

The PM field is not compressed it is annihilated by a dealignment of polarized electrons caused by the electrical pulse.

The Neo-magnet material has to work on the quantum plane to reorganize its alignment and consumes heat in the process. Electricity is generated by the work. 

These are the theories of Dr. Leon Dragone and Nicoli Zaev.

Title: Re: To be deleted
Post by: synchro1 on January 12, 2019, 03:45:23 PM

When the current is interrupted, the magnetic field is removed from the Neo magnet, an (MCE) material. This causes the electron spins to become random and the material cools.

"All magnetic materials exhibit MCE, although the intensity of this effect depends on the properties of each material. Generally, for a simple ferromagnetic material near its Curie temperature, when a magnetic field is applied, the spins tend to align parallel to the magnetic field, which lowers the magnetic entropy. To compensate for the loss in the magnetic entropy in an adiabatic (isentropic) process, the temperature of the material increases. When the magnetic field is removed, the spins tend to become random which increases the magnetic entropy and the material cools".


That's all they tell us in refrigeration science. The problem for them is that when the magnetic caloric material begins to work to re-polarize it's domains, it generates electrical power. This is an unwanted byproduct they simply run into an Earth ground. Tinman is running a D.C. motor with MCE waste current.
Title: Re: To be deleted
Post by: nul-points on January 12, 2019, 04:20:42 PM
Quote from: synchro1 on January 12, 2019, 02:05:21 PM

Dragone and Zaev.


would have loved to see the expressions on Leon's staff-colleagues' faces when he handed them his lovingly hand-typed theory paper for discussion

only saw Zaev's 'h' type current waveform once, in one of my pulsed motor experiments - real - but elusive!

ciao
Title: Re: To be deleted
Post by: synchro1 on January 14, 2019, 08:37:46 AM
Tinman or Itzu need to measure the temperature of their Neo magnet core to see if it's dropping while under operation. This will prove wether we're witnessing an (MCE) or not.
Title: Re: To be deleted
Post by: synchro1 on January 16, 2019, 08:53:08 AM
2SGen Episode 6: The 2SGen principle: The energy from the core magnetization/demagnetization process
In the scope pictures below:

the yellow curve is the pulse sent by the controller to build up the magnetic field of the core (magnetization phase),
the blue curve is the pulse measured across the output coil connected to a lamp as a load.

                               (http://jlnlabs.online.fr/2SGen/images/2sg1b.jpg (http://jlnlabs.online.fr/2SGen/images/2sg1b.jpg))


Look at the blue curve, the first part (negative curve) is the magnetization phase of the core (building up of the magnetic energy), you may notice some Barkhausen effect (http://jnaudin.free.fr/spgen/barkhausen.htm) bumps.
The second part (positive curve) is the demagnetization phase of the core. The excess free energy is tapped during the demagnetization process and not during the magnetization process due to the blocking diode connected at the output coil. To get more free energy from the 2SGen device, the clock pulse must be as short as possible (during the magnetization process).
The process of free energy generation from magnetization/demagnetization of a ferromagnetic core has been fully explained in the Nikolay E. Zaev paper "Ferrites and Ferromagnetics Free Energy Generation (http://jlnlabs.online.fr/2SGen/images/demag.pdf)" published in New Energy Technologies Issue #5 Sept-Oct 2002.
Title: Re: To be deleted
Post by: synchro1 on January 16, 2019, 09:24:59 AM

Neodymium as a magnetic refrigerant:

"AbstractThe adiabatic temperature change upon magnetization, ΔTs, and the field dependent heat capacity of Nd have been measured between 5 K and 40 K in applied magnetic fils up to 7 T in order to assess its potential for use as a magnetic refrigerant in an active magnetic regenerator (AMR). The ΔTs of Nd is proportional to temperature between 4 K and 10 K, with a maximal value of 2.5 K at 7 T and 10 K. Nd thus shows some applicability to AMR use".
Title: Re: To be deleted
Post by: synchro1 on January 16, 2019, 09:30:01 AM
The material generates electrical output as JLN shows as it works to realign it's domains after the eternally applied magnetic field is removed. Polarity makes a difference.

Wikipedia:

"Adiabatic magnetization: A magnetocaloric substance is placed in an insulated environment. The increasing external magnetic field (+H) causes the magnetic dipoles of the atoms to align, thereby decreasing the material's magnetic entropy (https://en.wikipedia.org/wiki/Entropy)and heat capacity (https://en.wikipedia.org/wiki/Heat_capacity). Since overall energy is not lost (yet) and therefore total entropy is not reduced (according to thermodynamic laws), the net result is that the substance is heated (T + ΔTad). Isomagnetic enthalpic transfer: This added heat can then be removed (-Q) by a fluid or gas—gaseous or liquid helium (https://en.wikipedia.org/wiki/Helium), for example. The magnetic field is held constant to prevent the dipoles from reabsorbing the heat. Once sufficiently cooled, the magnetocaloric substance and the coolant are separated.


Adiabatic demagnetization: The substance is returned to another adiabatic (insulated) condition so the total entropy remains constant. However, this time the magnetic field is decreased (H = 0), the thermal energy causes the magnetic moments to overcome the field, and thus the sample cools, i.e., an adiabatic temperature change. Energy (and entropy) transfers from thermal entropy to magnetic entropy, measuring the disorder of the magnetic dipoles".


















Title: Re: To be deleted
Post by: synchro1 on January 17, 2019, 07:51:07 AM
Imagine the "Magnet Core Coil" positioned in the diameter of a ferrite toroid that is wrapped with copper windings to collect MEG output; Rather then pulsing a rotor with the alternating magnetic flux.







Title: Re: To be deleted
Post by: lancaIV on January 17, 2019, 08:23:13 AM
Imagine  some becomes true - in future :

https://worldwide.espacenet.com/searchResults?submitted=true&locale=en_EP&DB=EPODOC&ST=advanced&TI=&AB=&PN=&AP=&PR=&PD=&PA=Hanna+awad&IN=&CPC=&IC=