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Author Topic: Graham Gunderson Energy conference High COP demonstration  (Read 151347 times)

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So do either of you believe that it is possible to extract _more_ energy from a PM than was applied to the base material in order to magnetize it in the first place?

What happens to the magnet when you extract energy from it? Does it become less and less magnetic? If you extract energy from "nuclear spin interaction" does the material change properties, become more brittle, evaporate, emit radiation, or what? Is it channelling dark energy from the expansion of the cosmos?

Your rolling ball experiment should answer that for you TK.

Dose the magnet on a flywheel of a small engine loose it's stored energy to the ignition coil after millions of cycles ?.
What about the PMs in a PM DC motor--do they loose there stored energy to the rotor after millions of revolutions ?.


Brad


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Your rolling ball experiment should answer that for you TK.

Dose the magnet on a flywheel of a small engine loose it's stored energy to the ignition coil after millions of cycles ?.
No, because it isn't doing any work. Say the magnet is closing and opening a small reed switch on every cycle. The work to do this is coming from the flywheel, not the magnet. Replace the magnet with a cam and replace the reed switch with a microswitch that is hit by the cam on every cycle. Does the cam lose energy, or is the work coming from the flywheel? If the flywheel isn't powered, will it eventually slow down? If you break rocks with a hammer, is the hammer doing work... or are you just using the hammer to _transmit_ the work that you are doing?
Quote
What about the PMs in a PM DC motor--do they loose there stored energy to the rotor after millions of revolutions ?.

No, because they aren't doing any work. The work to drive the rotor comes from the electrical power applied to the coils.




   

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It's not as complicated as it may seem...
Poynt,

All you had to say was "PM, you need to reverse or invert the current thru L2 because the reference point is inverted" without all the condescending verbiage. Yes, you are correct in that no OU is present and I am sure with that you are pleased.

pm

Partzman,

In my opinion, my posts below were all that needed to be said, but apparently they were dismissed without due consideration. As such, and since it appeared you were still convinced of your original assertion, I ventured to explain it in greater detail (much to my chagrin).

Am I happy about it, not really. Do I have tact, not really, especially after I've already made my point clear enough. :)

I think Pin is being measured across the input fet/diode/cap, not the primary. How much difference it will make I won't know until we see Pin done with the voltage across the primary.

Yeah, I figured it was -2.81W.

But no big deal it being negative (it all depends on probe polarity or how/where you subtract the 20V). The fact is this accounts for the missing power and indicates the true Pin for the transformer.

Sorry, but the sim is not showing OU on the transformer, and thankfully, ION won't have to eat his hat!
« Last Edit: 2016-08-08, 02:55:01 by poynt99 »


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No, because it isn't doing any work. Say the magnet is closing and opening a small reed switch on every cycle. The work to do this is coming from the flywheel, not the magnet. Replace the magnet with a cam and replace the reed switch with a microswitch that is hit by the cam on every cycle. Does the cam lose energy, or is the work coming from the flywheel? If the flywheel isn't powered, will it eventually slow down? If you break rocks with a hammer, is the hammer doing work... or are you just using the hammer to _transmit_ the work that you are doing? 
No, because they aren't doing any work. The work to drive the rotor comes from the electrical power applied to the coils.

So replace the magnet on the flywheel with a block of steel,and see how your motor runs--with there be spark-will the block of steel produce an electromagnetic field?

In regards to the DC motor-the same applies. Replace your PMs with steel keepers,and see how well your motor runs. You will find that it consumes a lot of power,and has very little torque--but why?. Well now you do not have the generating effect of the PMs to reduce the current flowing through the rotor coils--the BackEMF just is not there.

In time,i will be carrying out some calorimetry tests,and we shall see if PMs can actually do useful work  O0


Brad


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It appears to me that what partzman and ION are describing is a loss mechanism, dissipating power in the core and H-bridge components, rather than one which could produce any gain in energy.

Oh, where is MarkE when he's needed. I'm sure he would have some very interesting things to say about this.

While MarkE was a great man,and was well versed in EE-->we have Poynt  O0

It would be a tough bet,but my money would have been on the man that still walks among us.


Brad


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Ah..... It seems Brad beat me to it!

Should I open a topic? " Do permanent magnets DO work "

Consensus?

Cheers Grum.


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Ah..... It seems Brad beat me to it!

Should I open a topic? " Do permanent magnets DO work "

Consensus?

Cheers Grum.

I think it is time Grum,to answer this question once and for all.

For every action,there is an equal and opposite reaction--and then there is the counter reaction  ;)

Open the topic,and lets get ION on the scene ,as he knows his way around calorimetry setup's,and i believe this is the way to show PMs doing useful work.


Brad


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No, because it isn't doing any work. Say the magnet is closing and opening a small reed switch on every cycle. The work to do this is coming from the flywheel, not the magnet. Replace the magnet with a cam and replace the reed switch with a microswitch that is hit by the cam on every cycle. Does the cam lose energy, or is the work coming from the flywheel? If the flywheel isn't powered, will it eventually slow down? If you break rocks with a hammer, is the hammer doing work... or are you just using the hammer to _transmit_ the work that you are doing? 
No, because they aren't doing any work. The work to drive the rotor comes from the electrical power applied to the coils.

With a magnet on the flywheel passing a coil to generate current, the magnetic field of the magnet is always there, it is not the movement of the flywheel, the flywheel just brings that field into proximity of the coil for a time thus generating current in the coil. That is the flywheel did not create the energy, the magnetic field yes, the flywheel was expended energy only to move that field too and from the coil.

Another way, a magnet held above a piece of steel and the steel rises up to the magnet when close enough, what created the energy to rise the steel? the movement of the magnet closer to the steel or the movement itself? ;) Or is it the magnetic field?

A perminent magnet does create energy, the problem is to extract that energy you may have to expend more than you get back.

Regards

Mike 8)


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Dear All,

Here is a drawing of the Synchronous Diode Component Layout to help facilitate discussions about this sub assembly.

Enjoy!

Spokane1
   

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Has anyone looked at the gate current injection into the drain circuits of MOSFETs in this device?
   
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Has anyone looked at the gate current injection into the drain circuits of MOSFETs in this device?

Is that a data sheet parameter?

Spokane1
   

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Is that a data sheet parameter?
No, it is a particular circuit's parameter.

The current supplied to charge the MOSFET's gate can make its way into the drain circuit through the gate-drain Miller's capacitance and sometimes through the gate-source capacitance.  It is an often overlooked input power path in high frequency circuits that is capable of injecting Watts of power into the drain circuit.
   

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It's not as complicated as it may seem...
Has anyone looked at the gate current injection into the drain circuits of MOSFETs in this device?
Not that I know of, but I already brought it up here over 2 weeks ago when this all started.


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Not that I know of, but I already brought it up here over 2 weeks ago when this all started.
It is a real concern.  See a simple calculation here.
For example, most MOSFET RF power amplifiers leak the input energy into the output stage even when that output stage is unpowered !.
« Last Edit: 2016-08-10, 13:57:50 by verpies »
   

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It's not as complicated as it may seem...
It is a real concern.
For example, most MOSFET RF power amplifiers leak the input energy into the output stage even when that output stage is unpowered !.

Agreed. I see this whenever the PA is blown. The exciter power gets to the output/antenna making the radio appear to still be working albeit with very limited range.

btw, in the video there is a short section (only about 6 minutes unfortunately) when Gundy talks about the power measurements. At one point he has the input voltage and current traces up. It is easy to tell by eye alone that the measurement makes sense, i.e. 2V x 2A / 0.33 = 1.33W, assuming the half-sine parts are perfectly 90 degrees and can be ignored. I'm just not convinced that the wave forms being presented to the scope and power analyzer are truly correct and not skewed by phase shift, especially when he says it can be tuned for negative power.

The output side is troubling as well. Although I have not analyzed it like you are going to do, I am suspicious of the MOSFET Gate bleeds and by the fact changing one d-g cap can apparently kill the "effect".

Anyway, looking forward to your analysis.


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Agreed. I see this whenever the PA is blown. The exciter power gets to the output/antenna making the radio appear to still be working albeit with very limited range.
Just a week ago that happened to me with this RF PA.

It was excited with 3W at its input and after the main 50V power supply went up in smoke, 1.8W was still leaking out of its output.
   
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I could not find anything listed under the mosfet part number on Spokane1's diagram (C2M002510D).

However, looking at the photos I cannot see a full part number but I can tell that the part number starts with C2M and ends with 120.

Cree and DigiKey list the C2M0025120D mosfet, 1200V 90A 34mOhm Rdson, for only 69 dollars and 80 cents _each_. Data sheet attached below.

Free Energy is getting more and more expensive every time I look!   :D
   

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It's not as complicated as it may seem...
Just a week ago that happened to me with this RF PA.

It was excited with 3W at its input and after the main 50V power supply went up in smoke, 1.8W was still leaking out of its output.
1.2kW?  :o

That ought to get you around the earth a few times.


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Just a week ago that happened to me with this RF PA.

It was excited with 3W at its input and after the main 50V power supply went up in smoke, 1.8W was still leaking out of its output.

Ahhh TK, the best things in life are free!!!!! ;D

Ben K4ZEP
   
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1.2kW?  :o

That ought to get you around the earth a few times.

For the life of me, I can't figure out what a blown line fed PA amp. and leakage through burnt circuits has to do with GG's design??? ;)
Anymore dust and dirt get thrown up in the air and we will forget what we are discussing on this thread!

Ben K4ZEP
   

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It's about the energy used to drive GG's MOSFETs, which are connected to the output of his device, reaching the secondary winding of his transformer and being accounted there as this transformer's energy output.

If the RF energy leaks through MOSFETs' gates in RF amps, then it must be leaking through GG's MOSFETs' gates, as well.


The question is "How much" ?
   

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However, looking at the photos I cannot see a full part number but I can tell that the part number starts with C2M and ends with 120.
With some pseudo-color mapping. I could recognize ...25120 or ...26120 at the end of the number .
   

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@Itsu

Could you put together a quick video showing the gate leakage of any N-Ch Power MOSFET in the circuit below ?
No power supply. Just program your FG to 1MHz square wave at 10V amplitude. Use a red LED (observe its polarity).
   
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@Itsu

Could you put together a quick video showing the gate leakage of any N-Ch Power MOSFET in the circuit below ?
No power supply. Just program your FG to 1MHz square wave at 10V amplitude. Use a red LED (observe its polarity).

Beautiful demonstration!

I did it with  IRF530n and IRF3205 mosfets and a white LED, and tested a wide range of frequencies using a 0 to +9V pulse train of 50 percent duty cycle. 1 MHz produced a brilliant light. Down at 50 kHz the light was still fairly bright. Going over 1 Mhz and the light dims progressively. Also reducing the amplitude of the square wave causes the LED to go out somewhere between 4 and 5 Vp-p.
With 9V peak pulse train, the IRF3205 produced peak light at about 450 kHz. The IRF530n peaked at about 800 kHz.

No video from me, my camcorder battery is kaput.
   
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Tinsel
Quote

No video from me, my camcorder battery is kaput.
 -------

not good

we have to fix that...

please post a link for a replacement battery
   
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