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Author Topic: Graham Gunderson Energy conference High COP demonstration  (Read 235559 times)
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It's turtles all the way down
Brad

Where are the probes placed for this measurement? If it is input to output phase shift, this is something different than what Poynt is asking.

Capacitance coupling between windings can relate to transfer of power.

So, are you measuring power  (current and voltage phase difference at the input to your transformer) or are you measuring input / output phase difference?



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Really?

How is that possible if the source is supplying an average power (real) of 0W?

Well, I'm sure you and others have seen my MEI experiments that were recently posted on my bench but I will attach a sample here with a schematic for others to view.

The example below is extreme in that the current lead is >90'. This is made possible by the physical coil structure with it's distributed capacitance and electrical connections that permit the magneto electric induction to take place. What you see is the result. These measurements were taken with extreme care and are relative to the individual points measured and do indicate that resistive power is able to be produced with reactive input and with gain.

With an output power of 33.2 watts rms, the VAR to watts average ratio ~11:1 for this example.

So, is this device OU when considering the power required to supply the class E amplifier? No.

pm

Edit: The DUT includes T1, RL and Rs. 
   

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It's not as complicated as it may seem...
What does, or would the sim say?


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"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
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What does, or would the sim say?

Dear poynt99,

Graham (at the conference) said that he used LT Spice to simulate his machine with the weird wave forms and found that the simulator actually displayed a positive output energy in excess of the input energy. Neither he nor I have an idea as to how a classical calculator can come up with a non-classical phenomena. I do know that LT Spice can process non-world components like negative resistance and negative transformer coupling factors and who knows what else..

I'm sure that Graham is not going to be going back to the simulator for a while.

Spokane1
   

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It's not as complicated as it may seem...
Spokane 1,

I am fairly confident that if the simulator is indicating OU, then either there is leakage from other components (as we've has postulated about GG's device), or the measurements were not done correctly.

Partzman had a sim circuit that was indicating OU, which later turned out to be a measurement error (once I helped him with the proper way to do the power measurement in SPICE).

Chances are that GG's sim was not being measured properly. If you could obtain the sim files, we can perhaps sort out the error.


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   

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Buy me some coffee
Brad

Where are the probes placed for this measurement? If it is input to output phase shift, this is something different than what Poynt is asking.

Capacitance coupling between windings can relate to transfer of power.

So, are you measuring power  (current and voltage phase difference at the input to your transformer) or are you measuring input / output phase difference?

It can be either way ION,as the voltage trace for the input is in phase with the output,but the current trace on the input is 90* out of phase with both voltage input,and output across the load-->or 370* out of phase with the load,depending how you wish to place your scope across the AC output-of course :)

So 1-the input voltage and current are 90* out of phase.
2-the output voltage and current are 90* out of phase with the input current
3-the output voltage and current are in phase with the input voltage.
4-the input current is 90* out of phase with the input voltage,and output voltage and current.


Brad



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Never let your schooling get in the way of your education.
   
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SPICE).

If you could obtain the sim files, we can perhaps sort out the error.


Dear poynt99

Thank you for the review. I doubt we will be getting those sim file for many moons if ever. By then we will be much more involved in more serious matters or this will be another dead horse.

Spokane1
   
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Let the Junk Box Replication Experiments Begin!
   
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First Step is to Decode the logic board. This is where the timing signals are derived and control both the H-Bridge and the Backend Diode. I think I have most of it figured out - now to draft it and get my head around how it works. I'm sure there are several other approaches that could be used.
   
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It's turtles all the way down
It can be either way ION,as the voltage trace for the input is in phase with the output,but the current trace on the input is 90* out of phase with both voltage input,and output across the load-->or 370* out of phase with the load,depending how you wish to place your scope across the AC output-of course :)

So 1-the input voltage and current are 90* out of phase.
2-the output voltage and current are 90* out of phase with the input current
3-the output voltage and current are in phase with the input voltage.
4-the input current is 90* out of phase with the input voltage,and output voltage and current.


Brad

I can only guess that there may be a sneak path through capacitance coupling between the windings and back to the scope ground on the secondary that is lighting the LED. If there is no ground path on the secondary then this does not apply, but since you are giving output phase shift with respect to input, I assume you have a scope ground also on the secondary.

It is very difficult to eyeball an exact 90* shift when many cycles are displayed on the scope, much better resolution if only one cycle is shown, but still difficult to be exact.

Since you are varying the phase shift, this may also not apply.

Have you attempted to measure input and output power (with a non-inductive resistive load) to see if you have a COP> 1 or tried to loop the transformer for sustained stand alone operation? Maybe you have done all this on another thread.

Perhaps continue to pursue this if you think you have something.



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"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

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Buy me some coffee
I can only guess that there may be a sneak path through capacitance coupling between the windings and back to the scope ground on the secondary that is lighting the LED. If there is no ground path on the secondary then this does not apply, but since you are giving output phase shift with respect to input, I assume you have a scope ground also on the secondary.

It is very difficult to eyeball an exact 90* shift when many cycles are displayed on the scope, much better resolution if only one cycle is shown, but still difficult to be exact.

Since you are varying the phase shift, this may also not apply.

Have you attempted to measure input and output power (with a non-inductive resistive load) to see if you have a COP> 1 or tried to loop the transformer for sustained stand alone operation? Maybe you have done all this on another thread.

Perhaps continue to pursue this if you think you have something.

I did have a thread on this transformer,and if i recall correctly,Smudge was analysing  the results from some test we carried out,which did seem to show a negative resistance result.

I would need a 4 channel scope with isolated grounds to confirm this,but at the moment all my spare cash (+ some),is being spent on the issue mentioned in my last email to you,and i think you would agree that that is priority  ATM..

What i dont understand is why it would seem  so odd that real power cannot be produced from the output of the transformer,when the voltage and current are 90* out of phase on the input side?.
After all,it is current flow through the primary that produces the magnetic field-not the voltage across it,and this we can see with regard to super conductors. The only problem with this in my toroid transformer,is the current and voltage on the output is in phase with the voltage on the input-not the current,so not sure what is going on there?.

Perhaps this phase relationship thing only complies with your standard type transformer?.


Brad


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After all,it is current flow through the primary that produces the magnetic field-not the voltage across it
Brad,

It's a bit more complicated than that.  The magnetic field in a transformer is produced from both the primary current and the secondary current.  These two currents almost cancel each other out, so it is the difference between primary and secondary ampere-turns that actually creates the field, it is not just the primary current.  In fact with the transformer under load most of the primary current does not produce any field because of that cancellation, it is only the small so-called magnetizing current that creates the field.  And you can't divorce the voltage from the field since it is related to it.  If you do the math you find that for transformers connected to low impedance voltage sources, as is normally the case in power transformers, it is the voltage and frequency that determines the magnetizing current hence the magnetic field.  The much larger primary load current, that is opposed by the secondary load current, does not produce any field.  But it does present a mmf to the magnetic circuit, in the same way that the secondary current produces an opposing mmf.  It is that balance of opposing mmf's that is crucial to the way the transformer works, but unfortunately this isn't taught in electrical engineering courses.  They teach a transformer equivalent circuit that has at its heart a "perfect" transformer that somehow, magically, without the use of magnetic fields, transforms voltage according to the turns ratio and transforms current according to the reciprocal of the turns ratio.  Then the "imperfections" of a practical transformer, that includes the need for a magnetic field, are accounted for by adding components around the outside of that magical component.  Unfortunately this approach completely hides the inner workings of the magnetic domain.  So when you get unusual transformer designs, be it like your buried secondary or like Cobb's double winding, few people know how to properly analyze them. 

Smudge
   
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I would also like to know how it is possible or where does the real power come from?

Sorry for the delay ION but just got back to my computer.

In my MEI designs there is no outside source of energy as best I can determine at this time. The transformer construction and electrical properties allow real power to be developed from the reactive input.

Quote
If there are 500 VA at near zero power factor, you don't have to be off by much to get 5 to 10 Watts of real power hidden by some small non-zero portion of the power factor, enough to light a small lamp.

I agree totally. One design objective is to obtain as low a ratio of VAR/watts as possible.

Quote
BTW, does anyone know the type of lamp and how bright or dimly it was lit? This would allow an approximate back calculation of the actual power factor.

No idea here.

As a side note, I see from a post by Spokane on OUdotcom that Graham has decided not to open source his new technology.

pm


   
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Let the Junk Box Replication Experiments Begin!

Hi All!

Just a few of the things to understand and consider!  If you GOOGLE "H Bridge motor drivers"......thousands of articles. 
 
https://www.embeddedrelated.com/showarticle/424.php
http://www.modularcircuits.com/blog/articles/h-bridge-secrets/h-bridges-the-basics/

It gets complicated fast!  With the voltage ratings and current ratings of those H components.  I'm real curious which is blowing, the FET's or the Diodes in them!  Who knows.

We have to "forward, reverse and float inductor off and or maybe short to ground!  Too bad GG has decided to not open source!  Probably he was made an offer that he can't refuse! :-X

Ben K4ZEP

   
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I would also like to know how it is possible or where does the real power come from?

If there are 500 VA at near zero power factor, you don't have to be off by much to get 5 to 10 Watts of real power hidden by some small non-zero portion of the power factor, enough to light a small lamp.

BTW, does anyone know the type of lamp and how bright or dimly it was lit? This would allow an approximate back calculation of the actual power factor.

The bulb is an automotive brake light bulb, I believe, maybe a #1156.

From one of the images it looks like it is pretty bright. The one I have just tested on my bench draws almost exactly 2 amps when supplied with 12 VDC and it's pretty damn bright and white at that 24 watt power.  At 6 VDC it draws 1.4 amps (8.4 Watts) and is pretty bright but yellowish.

See the images below. There are some things to note in the face-on image showing the two Clarke-Hess power analyzers.

First, the bulb does not seem to be glowing in that image. Yet the "output" C-H is showing a reading. Second, the "input" C-H has its Current input sense bypassed. This would normally be in series between the DUT and its power supply. Instead of the series connection through the instrument the leads are simply shorted with a red cliplead. So naturally the instrument reads Voltage but no current, and therefore should naturally read 0 on the Watts (leftmost set of numbers), but is showing a slight Current indication. Next, both the voltage and current sense leads are fed through toroids, probably to suppress spikes.


The final image is from my cellphone camera showing the brightness of the #1156 at just under 13 Watts DC. I took this with the cellphone camera instead of the DSLR because it's comparable to the "end_view.jpg" image from the Gunderson demo.

   
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Hi All!

Just a few of the things to understand and consider!  If you GOOGLE "H Bridge motor drivers"......thousands of articles. 
 
https://www.embeddedrelated.com/showarticle/424.php
http://www.modularcircuits.com/blog/articles/h-bridge-secrets/h-bridges-the-basics/

It gets complicated fast!  With the voltage ratings and current ratings of those H components.  I'm real curious which is blowing, the FET's or the Diodes in them!  Who knows.

We have to "forward, reverse and float inductor off and or maybe short to ground!  Too bad GG has decided to not open source!  Probably he was made an offer that he can't refuse! :-X

Ben K4ZEP

You might find more relevant information if you look for "H-bridge Solid State Tesla Coil Drivers"  where you will also find thousands of articles.

I'm sure Gunderson's decision not to open-source is based on a desire not to be proven wrong by a bunch of amateurs who think they know how to measure stuff.

An offer he can't refuse? Not likely, but.... one had better be careful with those kinds of offers.... if you don't deliver what you've claimed you might end up sleeping with the fishes.

As to the exact failure mode of the mosfets... sometimes it can be hard to tell. Even if you _can_ find all the pieces.
   
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Brad,

It's a bit more complicated than that.  The magnetic field in a transformer is produced from both the primary current and the secondary current.  These two currents almost cancel each other out, so it is the difference between primary and secondary ampere-turns that actually creates the field, it is not just the primary current.  In fact with the transformer under load most of the primary current does not produce any field because of that cancellation, it is only the small so-called magnetizing current that creates the field.  And you can't divorce the voltage from the field since it is related to it.  If you do the math you find that for transformers connected to low impedance voltage sources, as is normally the case in power transformers, it is the voltage and frequency that determines the magnetizing current hence the magnetic field.  The much larger primary load current, that is opposed by the secondary load current, does not produce any field.  But it does present a mmf to the magnetic circuit, in the same way that the secondary current produces an opposing mmf.  It is that balance of opposing mmf's that is crucial to the way the transformer works, but unfortunately this isn't taught in electrical engineering courses.  They teach a transformer equivalent circuit that has at its heart a "perfect" transformer that somehow, magically, without the use of magnetic fields, transforms voltage according to the turns ratio and transforms current according to the reciprocal of the turns ratio.  Then the "imperfections" of a practical transformer, that includes the need for a magnetic field, are accounted for by adding components around the outside of that magical component.  Unfortunately this approach completely hides the inner workings of the magnetic domain.  So when you get unusual transformer designs, be it like your buried secondary or like Cobb's double winding, few people know how to properly analyze them. 

Smudge

Most useful insight.

Spokane1
   
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The bulb is an automotive brake light bulb, I believe, maybe a #1156.

From one of the images it looks like it is pretty bright. The one I have just tested on my bench draws almost exactly 2 amps when supplied with 12 VDC and it's pretty damn bright and white at that 24 watt power.  At 6 VDC it draws 1.4 amps (8.4 Watts) and is pretty bright but yellowish.


Dear TK,

Your assessment of the load lamp seems pretty accurate. The second photo you posted is the one I took at the convention. Graham had just shut the machine off about 10 seconds prior, so we are not going to get any useful lamp brightness information from that one.

Spokane1
   

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Frequency equals matter...


Buy me a drink
In regards to Heisenberg's quote:
And you will find the enemy too or he will find you.
Most assuredly if one peeps, the fox will find the henhouse.
The technology is not allowed here, at all.

Any argument is easily beaten down by the history of prior prominents. They were all intelligent and esteemed at one point.
Tesla, Schauberger, Keely, E.V. Grey, Stanley Meyers, Moray. The list goes one.
I wonder who the next prairie dog will be?


---------------------------
   
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In regards to Heisenberg's quote:
And you will find the enemy too or he will find you.
Most assuredly if one peeps, the fox will find the henhouse.
The technology is not allowed here, at all.

Any argument is easily beaten down by the history of prior prominents. They were all intelligent and esteemed at one point.
Tesla, Schauberger, Keely, E.V. Grey, Stanley Meyers, Moray. The list goes one.
I wonder who the next prairie dog will be?

Tesla would roll in his grave if he saw your list. The inventions of Nikola Tesla, or derivations of them, are all around you, from the radios that connect you to the internet, the logic gates that run your computer, the AC motors that run your appliances, and many more. Those others.... frauds, fakers, deluded old men, none of whom has contributed a single _real_ thing in all their lifetimes of tomfoolery.
   
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Dear TK,

Your assessment of the load lamp seems pretty accurate. The second photo you posted is the one I took at the convention. Graham had just shut the machine off about 10 seconds prior, so we are not going to get any useful lamp brightness information from that one.

Spokane1

That makes sense. The bulb is still connected to the capacitor bank and has drained it down to about 1 volt according to the C-H analyzer and looks like it is drawing about a quarter of an amp. My own test #1156 draws about 0.6 A at 1 volt, so maybe Gunderson's bulb is a different type, but in the same package type. Perhaps it is a #1157 and he is using the higher-resistance filament in that dual-filament bulb. I'll have to dig around to see if I can find a  #1157 in my bulb stash.

Now.... what about the "input" C-H setup? Did Gunderson conduct the demonstration with it wired the way it is shown in the photo? I mean, with the Current side not connected and bypassed with the red cliplead? If that's how he got his "0.000 W" readings..... well, just go ahead and insert a ROFL here.  Just kidding.... he would not have done that. Surely not.... still, the setup does look rather suspicious.
   
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That makes sense. The bulb is still connected to the capacitor bank and has drained it down to about 1 volt according to the C-H analyzer and looks like it is drawing about a quarter of an amp. My own test #1156 draws about 0.6 A at 1 volt, so maybe Gunderson's bulb is a different type, but in the same package type. Perhaps it is a #1157 and he is using the higher-resistance filament in that dual-filament bulb. I'll have to dig around to see if I can find a  #1157 in my bulb stash.

Now.... what about the "input" C-H setup? Did Gunderson conduct the demonstration with it wired the way it is shown in the photo? I mean, with the Current side not connected and bypassed with the red cliplead? If that's how he got his "0.000 W" readings..... well, just go ahead and insert a ROFL here.  Just kidding.... he would not have done that. Surely not.... still, the setup does look rather suspicious.

Dear TK,

I have wondered about that as well. I didn't notice it at the time but there is no doubt that there is no current inputs to the Wideband Power Analyzer for the input to the H Bridge. My photo was taken just after the machine was turned off so this is how it operated during the entire presentation.

That is going to be the first question I ask Graham the next time we go out for Pizza and Chicken Wings.

I know this one presumed omission might be a game stopper for anyone reviewing this material and that would be a legitimate call.

I am going to take the path that assumes that Graham knows what he was doing and had some technical work around to account for the missing connections and why his approach provided more accurate data. If there was no current data then the instrument would have been unable to calculate a 1.53 watt input value when the demonstration first started. I would say that the instrument only registered the 0.000 reading during the last 5 minutes of the show.

This is a good catch. Did any one else besides yourself notice it?

Spokane1
   
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Doesn't the "input" C-H monitor the _output_ of the H-bridge going to the input of the transformer?

I have no idea if anyone else noticed it. I'm fairly familiar with the use of the Clarke-Hess 2330 myself, though, so I know what to look for on the instrument. I str that .99 poynted out the toroid filters on the patch wiring. He probably noticed the Current bypass too.

I'm not sure what effect this filtering would have on the overall measurement problem. Since the Current patch lead is making a series connection with the C-H, going from the H-bridge to the transformer, the filtering would keep spikes out of both the C-H and the transformer. At first I thought maybe the spikes could be sneaking energy into the transformer that was deliberately being filtered out of the C-H. But since the circuit is a series one for the Current, that's probably not the case.

If, that is, the C-H is being patched in properly to begin with, which I can't tell from the photo.

Performed the whole demonstration with the "input" C-H current input bypassed/disconnected? That's hardly kosher IMHO.  Under certain conditions the C-H display will hold the last value sensed, which might explain the 1.53 W starting reading you saw.

Here's the brochure with specifications for the C-H 2330.
http://www.clarke-hess.com/2330.html

I recently saw one listed, manual included, for 300 dollars US on Ebay. I have a friend about 100 miles away who has one I can borrow but it costs me about 30 dollars to make the round trip to pick it up, and the same to return it.
   
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It's pretty hard to tell from the low-resolution photo I have to work with but here are my best "guesses" as to the settings on the two C-H analyzers.

"input" (RH) instrument settings, from R to L:
Voltage Range 200V, showing 131.1 V
RMS/PEAK in RMS mode, Auto range OFF
Current (disconnected) Range 50 mA, showing 00.01 mA
Function: PWR, W range, 00.000 W indicated

"output" (LH) instrument settings, from R to L:
Voltage Range 2V, showing 0.961V
RMS/PEAK in RMS mode, Auto range ON
Current Range 500 mA, showing 254.3 mA
Function: Px10, mW range, showing 243.18 mW


   
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You might find more relevant information if you look for "H-bridge Solid State Tesla Coil Drivers"  where you will also find thousands of articles.

I'm sure Gunderson's decision not to open-source is based on a desire not to be proven wrong by a bunch of amateurs who think they know how to measure stuff.

An offer he can't refuse? Not likely, but.... one had better be careful with those kinds of offers.... if you don't deliver what you've claimed you might end up sleeping with the fishes.

As to the exact failure mode of the mosfets... sometimes it can be hard to tell. Even if you _can_ find all the pieces.
Morning TK,

I see you have let the smoke out of a few devices yourself.  Soon after I do that wife says "OK, what you burning up now!"
Your direction to search "H-bridge Solid State Tesla Coil Drivers" is very good, .  Looking at some of
the schematics there from experienced coil builders looks very much like what I think he has done on the H driver board.
One of many sites, http://www.stevehv.4hv.org/SSTCindex.htm,  really shows what we can expect to have to do in the future!

Ben K4ZEP

 
   
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