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Solid States Devices => solid state devices => Topic started by: tinman on November 10, 2013, 02:34:54 PM
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This is a new type of transformer arangment im testing at the moment.
Some here wanted me to start a thread on it,so here it is.
The first video explains the setup,and the second video is of it running.
Would like an interpritation of the scope shots from those that are well versed reading scope's-as im still on my L plates.
https://www.youtube.com/watch?v=-sxM1eSaKmw
https://www.youtube.com/watch?v=zAKGIV_TdF8
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This is a new type of transformer arangment im testing at the moment.
So here wanted me to start a thread on it,so here it is.
The first video explains the setup,and the second video is of it running.
Would like an interpritation of the scope shots from those that are well versed reading scope's-as im still on my L plates.
https://www.youtube.com/watch?v=-sxM1eSaKmw (https://www.youtube.com/watch?v=-sxM1eSaKmw)
https://www.youtube.com/watch?v=zAKGIV_TdF8 (https://www.youtube.com/watch?v=zAKGIV_TdF8)
hi tinman,
I have seen some of your previous projects and i am impressed.
Upon seeing this current circuit of yours especially the waveform,peak voltage and sound.It just hit me you are actually building a battery desulfator unknowingly.
I had created a simpler circuit years ago the timing 56us on with 1ms off.Someone from instructable had built this circuit originally using 555. The current consumption for my circuit is the same around 11mA.
The back emf on old uncharged battery is measured around 44volts for my case and i have use it to recover dead battery <1 volt which i have not charged for 8 years.
My circuit can be modified by breaking input of D2 away from circuit to be the input positive power supply as well.
http://www.instructables.com/id/PIC12F629-Lead-Acid-Battery-Desulfator
(http://www.instructables.com/id/PIC12F629-Lead-Acid-Battery-Desulfator)
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@Magpwr
Any pulsed inductor can be made into a battery desulphator,it is quite easy.
Tomorrow i will make a video of a desulphator of the simplest kind-no transistor,reed switches or hall effect devices. It will cost you about $1.00 to make,and will work just aswell as any out there.
Below is a screen shot of the scope while the system is running-as per second video-along with my interpritation of what is happening.Please correct me if im wrong.
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Interesting design Tinman although I finally had to pull the earphones off before the end of the second video as the squeal of the resonance was a bit too much. Very hard to focus on what you are saying over that. May I suggest after showing the initial view of the coil to put something over it that would muffle the sound. I do like your concept on this but didn't get whether you have calculated yet how much power out vs. power in it is doing. BTW how do you like the Atten O-scope?
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Interesting design Tinman although I finally had to pull the earphones off before the end of the second video as the squeal of the resonance was a bit too much. Very hard to focus on what you are saying over that. May I suggest after showing the initial view of the coil to put something over it that would muffle the sound. I do like your concept on this but didn't get whether you have calculated yet how much power out vs. power in it is doing. BTW how do you like the Atten O-scope?
Yes,in the next video,i will put the muffler on it.
P/in v P/out is yet to be calculated accuratley. The blue trace on the scope shot is across the 18 ohm resistor/tank circuit.The P/in is accurate,as it is pure DC without ripple.My DMM also reads the same current and voltage as the power supply. The P/out is what needs more attention yet.
The tank coil is of very few turns(90 i think it was)and .5mm copper wire. It wasnt designed to be a power source,but more just an indicator coil to see if i could get the effect i wanted-and we certainly got that. The secondary core(wire wound core) was quick and nasty,and would be a very poor performer.Idealy we would slip another ferrite core over the tank windings,and have a lot more turns on that tank coil-aswell as the drive coil.
The idea was to create a magnetic wave that swept across the tank coil,and it seems that we have achieved that-looking at the current still flowing through the 18 ohm resistor/tank,even though the drive coil is in it's rest phase.
The Atten is great,although it's maths function has me baffled???.When useing the math function,i cant find anyware that i can add channel A and B to do the math-to get the sume of voltage and current over time-nor subtract. What the manual says,and what actualy is,is two different things.
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The Atten is great,although it's maths function has me baffled???.When useing the math function,i cant find anyware that i can add channel A and B to do the math-to get the sume of voltage and current over time-nor subtract. What the manual says,and what actualy is,is two different things.
Hi Tinman,
You surely can find the Math button and pressing it you would see Math Funtion Menu appear on the display and could perform addition, substruction multiplication and division of channel A with channel B. Personally I have not played with such a 'marvel' :) like the Atten ADS1102CAL but that is what is written in its manual ( http://micromir.ucoz.ru/Oscil/Atten/ADS1000_User_Manual.pdf (http://micromir.ucoz.ru/Oscil/Atten/ADS1000_User_Manual.pdf) Page 27, PDF page 36).
If you find those instructions do not work in practice on the scope, then I pass this question here to others to comment.
I have a much simpler DSO (a dual channel OWON 5022S and it also has a Math Menu button with 3 choices: CH1-CH2, CH2-CH1 and CH1+CH2. And each channel input has its own Menu too like yours and includes the Invert function should you need to flip the measured waveforms on A or B (to turn them upside down).
One more notice: you use a wire wound 18 Ohm 11W power resistor for the load (at least I see it as a wire wound type) and in this particular case when you use it in parallel with the output tank circuit which is in resonance at the output frequency, then the wire wound resistor is not an issue. In case there would be no resonant circuit involved with such load resistor in parallel, then its inductive loading effect would become an issue. (At least I gather from your comments the 18 Ohm is in parallel with the tank, maybe I am wrong.)
Gyula
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Hello Tinman,
can you please give us the link to the original circuit of the Flynn-Website ?
Do you know this development where only the magnets are doing the work ?
http://gap-power.com/index-2.html (http://gap-power.com/index-2.html)
A permanent magnet is influenced ( switched on an of ) best with the Coil on top of the magnet.
Taking in consideration gyulas comment about the inductive reactance of the 18 Ohm resistor and assuming a sinus at the output ( which it is not, we know this ) I calculated the power out about 32 mWatt.
Input-Power calculated according to the T of the frequency and the T_on of the MOSFET about 22 mWatt not considering the recycled power of the primary collapse-spike.
It looks promising I would however change the resistor-type with no inductive properties ( 1/4 Watt-metal-oxide-type) and tune the tank-circuit with different cap-values... I assume however you have already done that tuning.
Regards
Kator01
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Hi Tinman,
You surely can find the Math button and pressing it you would see Math Funtion Menu appear on the display and could perform addition, substruction multiplication and division of channel A with channel B. Personally I have not played with such a 'marvel' :) like the Atten ADS1102CAL but that is what is written in its manual ( http://micromir.ucoz.ru/Oscil/Atten/ADS1000_User_Manual.pdf (http://micromir.ucoz.ru/Oscil/Atten/ADS1000_User_Manual.pdf) Page 27, PDF page 36).
If you find those instructions do not work in practice on the scope, then I pass this question here to others to comment.
I have a much simpler DSO (a dual channel OWON 5022S and it also has a Math Menu button with 3 choices: CH1-CH2, CH2-CH1 and CH1+CH2. And each channel input has its own Menu too like yours and includes the Invert function should you need to flip the measured waveforms on A or B (to turn them upside down).
One more notice: you use a wire wound 18 Ohm 11W power resistor for the load (at least I see it as a wire wound type) and in this particular case when you use it in parallel with the output tank circuit which is in resonance at the output frequency, then the wire wound resistor is not an issue. In case there would be no resonant circuit involved with such load resistor in parallel, then its inductive loading effect would become an issue. (At least I gather from your comments the 18 Ohm is in parallel with the tank, maybe I am wrong.)
Gyula
Hi Gyula
Yes,if i push the math button,the math trace dose pop up on the screen. But no math function menu to be found as stated in the manual???. I even had poynt have a look at it once via webcam,but still couldnt work it out.
Yes the 18 ohm resistor is in parallel to the tank circuit.
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Hello Tinman,
can you please give us the link to the original circuit of the Flynn-Website ?
Do you know this development where only the magnets are doing the work ?
http://gap-power.com/index-2.html (http://gap-power.com/index-2.html)
A permanent magnet is influenced ( switched on an of ) best with the Coil on top of the magnet.
Taking in consideration gyulas comment about the inductive reactance of the 18 Ohm resistor and assuming a sinus at the output ( which it is not, we know this ) I calculated the power out about 32 mWatt.
Input-Power calculated according to the T of the frequency and the T_on of the MOSFET about 22 mWatt not considering the recycled power of the primary collapse-spike.
It looks promising I would however change the resistor-type with no inductive properties ( 1/4 Watt-metal-oxide-type) and tune the tank-circuit with different cap-values... I assume however you have already done that tuning.
Regards
Kator01
I dont have the original flynn circuit,as my setup was only based around an electromagnet canceling out the PMs magnetic field-same as the flynn type setup.However mine is solid state,while most of the flynn devices are mechanical devices-like that gap power site you posted.
My setup is probably more like the MEG,but differs in that i neutralize an already existing field around the coil,where the MEG diverts the field into the coil.
I am also lost with this comment-Quote:Input-Power calculated according to the T of the frequency and the T_on of the MOSFET about 22 mWatt not considering the recycled power of the primary collapse-spike.
P/in when dealing with pure DC is simply volts x amps. 12 volts x 11mA give us 132 mW.
How did you get 22mW?.
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Tinman:
There are issues with your experiment and I am going to make some comments. Don't let my comments knock the wind from your sails. Keep in mind that we are all better off getting input from all viewpoints, even when it is stuff that we might not like to hear.
You posted this in the other thread:
Well two days ago,i found one answer to that puzzle-after many failed attemps. Through a combination of another fellows video,and some of my own changes,we now have a transformer that has 0 lenz effect when we draw of the secondary-even the tank coil outputs more power when we draw from the secondary-with 0 reflection on the power input-all thanks to the permanent magnet.
So you are just starting out testing this setup and you make a posting that states that you have done it and you have zero "Lenz effect." That's a classic mistake and one you should learn to avoid. You can't just jump on your first set of measurements and pronounce them as being fact. I would not be surprised if, based on your statements, that you already have people that want to replicate what you are doing. The truth is that you have barely even scratched the surface and you simply can't make any definitive statements. For example, before you made the posting above, did you double check your measurements or try to make measurements in a completely different way to confirm your findings? I am willing to bet that you didn't.
You are claiming that you are drawing power from the secondary with no load reflected on the primary. Transformers don't work like that, if you draw power from the secondary then you have to put power into the primary. Just that fact should have stopped you in your tacks and make you say to yourself, "I am seeing something that doesn't seem to make sense, so I will have to really investigate and check and double-check and make measurements in alternate ways before I am sure enough to announce this." That is what the scientific method is all about.
Another big issue is using permanent magnets in any kind of transformer configuration. I probably have posted to you and for sure I have posted many times that using permanent magnets in transformer configurations is nonsensical. To put it harshly, it's junk. I take some liberty in being harsh here because it's so fundamental and I have repeated it many times to no avail. Transformers work with AC signals and a magnet is essentially a "DC" source of magnetic flux. The magnet is not even "seen" by the AC circuit. The magnet does absolutely nothing to improve the performance of the device and in fact can easily degrade the performance of the device. If you don't believe me then the challenge to you would be to build another transformer setup but with a normal non-magnetized ferromagnetic core, and then compare the two. I also looked at the "Flynn" page with all sorts of transformer configurations that include permanent magnets and it's pure junk.
This "permanent magnets in transformers" business is frustrating for someone like me. I will give you an analogy off the top of my head. Suppose you own a car body shop. When you paint a car it requires primer, then several coats of the right grade of paint, and then possibly a protective clear-coat on top of that. It's a serious business and paint jobs can be very precise work to get right look on an expensive sports car. Then you hire somebody for your shop, and he says, "I am going to the hardware store to get some latex paint on special." What? Latex house paint on a car, is the guy nuts? You tell him no, that doesn't make sense, but the guy doesn't believe you and he does the same thing two weeks later. A week after that he does it again. He seemingly can't understand how latex house paint can't be used to paint a $150,000 sports car.
This business of using permanent magnets when you are building a transformer is a good analogy to the paint story. It's nonsensical and ridiculous. It's just based on a blind belief and there is no evidence at all that it does anything. Anybody disagree with me? Then do an A-B comparison like I stated above.
MileHigh
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Tinman:
I will make one more comment and one more time don't let the comments take the wind out of your sails.
In your second clip you start by saying that you are going to check if the signal generator is injecting any power into the circuit. So to "test" for this you shut off the main power, turn on the signal generator, and then look at the transistor waveform. You see some tiny glitches and no real activity and then conclude that the signal generator is not injecting any power into the circuit.
Where did the logic for that test come from? Did you just decide that if you see nothing on the transistor output when the main power is off that that "must" mean that no power is being injected into the circuit from the signal generator? That's what it feels like to me at least. That doesn't make any sense whatsoever. Really think about what you are saying in the beginning of your second clip.
The only way to see if the signal generator is injecting power into the circuit is to actually make that measurement while the circuit is powered and in operation. You would have to measure the instantaneous signal generator voltage and the instantaneous signal generator current and then do the required crunching to calculate the average power supplied to the circuit by the signal generator.
You have to understand this is a generic comment and a generic issue. You seem to have made up a "test" for input power that is simply disconnected from the reality of the circuit. Electronics doesn't work like that, you can't simply make up things like like this on the fly. Everything you do when you are testing has to actually make sense and not just "sound like it makes sense." Do you see the error in what you did? It's a huge mistake.
Okay Tinman, you are off the hot seat. Look, you are starting to test something new, so why not get it right? Take the "we all know this is true" free energy experimenter assumptions and look at them with a critical eye. Don't just blindly assume things and end up repeating the same mistakes and just get stuck in an endless revolving door. Look at the case of the SMOT business. You put a magnet near a marble track and people start to swoon thinking they are seeing free energy. It's the paint story all over again. You can put one or 100 magnets along a track for a rolling metal ball and they will not contribute one millionth of a Joule of energy to the rolling ball.
For what it's worth there is a final mini issue. I bet not one of your peers on your forum or on this forum inquired about your "test" for the possible power contributed by the signal generator. As far as I am concerned that's a "brain logjam" and people have to start questioning the methods of their peers. It's a way for people to hone their skills and challenge each other in order to keep learning and improving.
MileHigh
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Tinman:
There are issues with your experiment and I am going to make some comments. Don't let my comments knock the wind from your sails. Keep in mind that we are all better off getting input from all viewpoints, even when it is stuff that we might not like to hear.
You posted this in the other thread:
So you are just starting out testing this setup and you make a posting that states that you have done it and you have zero "Lenz effect." That's a classic mistake and one you should learn to avoid. You can't just jump on your first set of measurements and pronounce them as being fact. I would not be surprised if, based on your statements, that you already have people that want to replicate what you are doing. The truth is that you have barely even scratched the surface and you simply can't make any definitive statements. For example, before you made the posting above, did you double check your measurements or try to make measurements in a completely different way to confirm your findings? I am willing to bet that you didn't.
You are claiming that you are drawing power from the secondary with no load reflected on the primary. Transformers don't work like that, if you draw power from the secondary then you have to put power into the primary. Just that fact should have stopped you in your tacks and make you say to yourself, "I am seeing something that doesn't seem to make sense, so I will have to really investigate and check and double-check and make measurements in alternate ways before I am sure enough to announce this." That is what the scientific method is all about.
Another big issue is using permanent magnets in any kind of transformer configuration. I probably have posted to you and for sure I have posted many times that using permanent magnets in transformer configurations is nonsensical. To put it harshly, it's junk. I take some liberty in being harsh here because it's so fundamental and I have repeated it many times to no avail. Transformers work with AC signals and a magnet is essentially a "DC" source of magnetic flux. The magnet is not even "seen" by the AC circuit. The magnet does absolutely nothing to improve the performance of the device and in fact can easily degrade the performance of the device. If you don't believe me then the challenge to you would be to build another transformer setup but with a normal non-magnetized ferromagnetic core, and then compare the two. I also looked at the "Flynn" page with all sorts of transformer configurations that include permanent magnets and it's pure junk.
This "permanent magnets in transformers" business is frustrating for someone like me. I will give you an analogy off the top of my head. Suppose you own a car body shop. When you paint a car it requires primer, then several coats of the right grade of paint, and then possibly a protective clear-coat on top of that. It's a serious business and paint jobs can be very precise work to get right look on an expensive sports car. Then you hire somebody for your shop, and he says, "I am going to the hardware store to get some latex paint on special." What? Latex house paint on a car, is the guy nuts? You tell him no, that doesn't make sense, but the guy doesn't believe you and he does the same thing two weeks later. A week after that he does it again. He seemingly can't understand how latex house paint can't be used to paint a $150,000 sports car.
This business of using permanent magnets when you are building a transformer is a good analogy to the paint story. It's nonsensical and ridiculous. It's just based on a blind belief and there is no evidence at all that it does anything. Anybody disagree with me? Then do an A-B comparison like I stated above.
MileHigh
"Anybody disagree with me?"
I do. ;D
"It's nonsensical and ridiculous. It's just based on a blind belief and there is no evidence at all that it does anything.
Another big issue is using permanent magnets in any kind of transformer configuration. I probably have posted to you and for sure I have posted many times that using permanent magnets in transformer configurations is nonsensical. To put it harshly, it's junk."
Tell that to Hitachi Magnetics Corp. ;) ;D
If your saying using a magnet as a core has zero effect, your wrong. There are some big companies out there that use magnets in the cores, and the effect is not the same as a core without a magnet. Just by biasing the core with a magnet increases the power handling of a switching supply transformer is one advantage. ;)
"This "permanent magnets in transformers" business is frustrating for someone like me."
Im sure it is. ;) ;D
"He seemingly can't understand how latex house paint can't be used to paint a $150,000 sports car. This business of using permanent magnets when you are building a transformer is a good analogy to the paint story."
Not the same thing at all, sorry to say. ::)
" I also looked at the "Flynn" page with all sorts of transformer configurations that include permanent magnets and it's pure junk."
Can you go into greater detail of why it is junk? ??? ;)
"You are claiming that you are drawing power from the secondary with no load reflected on the primary. Transformers don't work like that"
Some can operate like that. ;) I have a transformer that does just that, in that it actually lowers the input when drawing from the secondary. It is not OU, but it does what I said. ;) The PDF below shows how to do it. ;D
Mags
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Tinman:
I will make one more comment and one more time don't let the comments take the wind out of your sails.
In your second clip you start by saying that you are going to check if the signal generator is injecting any power into the circuit. So to "test" for this you shut off the main power, turn on the signal generator, and then look at the transistor waveform. You see some tiny glitches and no real activity and then conclude that the signal generator is not injecting any power into the circuit.
Where did the logic for that test come from? Did you just decide that if you see nothing on the transistor output when the main power is off that that "must" mean that no power is being injected into the circuit from the signal generator? That's what it feels like to me at least. That doesn't make any sense whatsoever. Really think about what you are saying in the beginning of your second clip.
The only way to see if the signal generator is injecting power into the circuit is to actually make that measurement while the circuit is powered and in operation. You would have to measure the instantaneous signal generator voltage and the instantaneous signal generator current and then do the required crunching to calculate the average power supplied to the circuit by the signal generator.
You have to understand this is a generic comment and a generic issue. You seem to have made up a "test" for input power that is simply disconnected from the reality of the circuit. Electronics doesn't work like that, you can't simply make up things like like this on the fly. Everything you do when you are testing has to actually make sense and not just "sound like it makes sense." Do you see the error in what you did? It's a huge mistake.
Okay Tinman, you are off the hot seat. Look, you are starting to test something new, so why not get it right? Take the "we all know this is true" free energy experimenter assumptions and look at them with a critical eye. Don't just blindly assume things and end up repeating the same mistakes and just get stuck in an endless revolving door. Look at the case of the SMOT business. You put a magnet near a marble track and people start to swoon thinking they are seeing free energy. It's the paint story all over again. You can put one or 100 magnets along a track for a rolling metal ball and they will not contribute one millionth of a Joule of energy to the rolling ball.
For what it's worth there is a final mini issue. I bet not one of your peers on your forum or on this forum inquired about your "test" for the possible power contributed by the signal generator. As far as I am concerned that's a "brain logjam" and people have to start questioning the methods of their peers. It's a way for people to hone their skills and challenge each other in order to keep learning and improving.
MileHigh
MileHigh
Lets do the math on the maximum power that could be put into the system from the signal generator-there is no rocket science here.
The SG is set at 4Vpp,so 2 volts on the forward side.
There is a 220ohm resistor on the base of the transistor.
The duty cycle is 23%.
So the maximum power/watts that can be achieved is .018 watts
.018 x 23%=.00414 watt's.
I think your splitting hairs there MH,concidering the P/in is 132mWatts.
But just for arguments sake,we will change the P/in to 136mWatts.
Second-you asume that the video's represent the only testing i have done on the device.
Could it be at all possable that i was starting from the begining,so as all those on my forum(which is the forum the video's were intended for)could follow right from the start?-or would i just be better off going straight to where i am now,so as to loose everyone in the dust?.
With all due respect MH (and i do have a lot for you),you seem to be so shut of when it comes to PMs being of any use at all in devices such as this-and others.
Well here is my challenge to you.Put together a bifilar inductor,and fire up one of the windings as i have.Then send the inductive kickback to a low ohm load as i have(the 12 volt SLA) Then with your second winding,make a tank circuit as i have,with an 18 ohm resistor (like i have)across the tank cap.Make sure you dont use any PMs in there anyware. Fine tune it as you will,and show us the waveform across the 18 ohm resistor. Ill bet you find(as i did)that when the primary winding is in it's rest phase,the bottom half of your wave form across the 18 ohm resistor will be flat,and almost 0 volts.
I am happy to do this again for you,if you cannot.
Now in regards to the scope shot i posted above-from that information,can you tell me how much power is being disipated across the 18 ohm resistor? Any help would be great.
In the next video,i will smooth the inductive output,so as we can get an accurate P/out from that part of the circuit. I have already done this ofcourse,but was starting at the begining,so as other could follow from the start-if they choose to replicate it.
My reference to the lenzless effect,was in relation to absolutly no power increase on the input was shown when we loaded one of the outputs-while remembering that the tank coil is already loaded(power yet to be determond).I am well aware that every transformer uses power at idle,but show me one where the power input dosnt go up when a load is placed on the output-while drawing a load on the second output.
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Magluvin,
You are doing it again. What's wrong with you? Your native language is English? Are you incapable of making your posting with out addressing me directly? What the fuck?
Let me remind you of something. You called me a "freak" and an "idiot" repeatedly and maliciously over three times that spanned several weeks each time and you enjoyed it. You never had the courage or the character to own up to it and apologize for your outrageous behaviour. They were arguments about the basics of electricity and magnetism, and here you are making argumentative points again that are outside the scope of the discussion. Also, you never followed through with your projects that were supposed to back up your case.
So here you are back again playing the "clarifier" and trying to score "points" with some stuff that is not even on the table.
What I said is technically true in a generic sense. I don't care that I did not qualify my statements because it's reasonable to talk in a generic sense when you are talking about basic electronics. No one doubts that a magnet can influence the operation of a transformer, in fact TK just sent me an email about that. But what I said still stands and is absolutely true.
Here is a thing that we see all the time: Someone makes a generalized statement about energy or electronics. Then somebody else chimes in and says, "What you are saying is not true because in special case XYZ it's not true." Well that statement is not true because we are NOT talking about some special esoteric case. That can be very frustrating and tedious.
TK commented that a magnet close to a JT core will change the way the JT operates. No kidding, the magnet will DEGRADE the performance of the JT core and thus affect the JT itself. But the real issue is that there is no reason to put a magnet near a JT core at all.
Everything I said in my postings to Tinman is absolutely true. That's the thing to appreciate. Stop looking for exceptions and actually try to understand the basics. The core in a transformer or in a JT or in Tinman's transformer is there to store magnetic energy or act as a conduit for the flow of energy from input to output. The core is not supposed to be DEGRADED in performance by the influence of an external magnet. If the core itself is a magnet, then the ability to store magnetic energy or act as a conduit for the flow of energy is degraded. So what the relative permeability of a magnet acting as a core will be different for flux going in one direction compared to the other direction. Big deal, it's not on the table right now. Tinman will not notice anything special about using a magnet as a core in his application. There is no rational reason to use a magnet as a core.
Your .pdf with all sorts of bizarre and unusual transformer configurations is not impressive. Plus there is no such thing as drawing power from a secondary without having to supply power into the primary, no matter how esoteric or strange looking the coupling is being done and no matter what that .pdf says. No transformer will output secondary power without drawing primary power. The fundamental principles for transformers will not change and that's the key point that people should appreciate.
TK pointed out that there is a part in a CRT TV consists of a magnet that biased a ferrite that is part of an inductor. Again, that is a specialized application and there is a logical reason for doing it. That's in stark contrast to making transformers for generic purposes.
Here is the challenge: If somebody is going to use a magnet as a core in a transformer, then they should explain exactly why they are doing it. Back it up with waveforms for both a magnet as a core and a regular ferrite core. In other words, make your case with logical arguments and with data to back it up. Prove that you actually have a reason for doing it and show your work and data. Now all of a sudden it's not so easy, is it?
This is all about understanding electronics and using your critical thinking skills and not using magnets in transformer cores by rote because somebody else did it.
MileHigh
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Magluvin:
This posing is dedicated to you.
Quoting myself:
Here is a thing that we see all the time: Someone makes a generalized statement about energy or electronics. Then somebody else chimes in and says, "What you are saying is not true because in special case XYZ it's not true." Well that statement is not true because we are NOT talking about some special esoteric case.
The above statement applies directly to you:
There are some big companies out there that use magnets in the cores, and the effect is not the same as a core without a magnet. Just by biasing the core with a magnet increases the power handling of a switching supply transformer is one advantage.
Yeah indeed you can increase the power handling because a switching power supply repeatedly charges and discharges an inductor and in this specialized case you can use the fact that you have to twist already biased magnetic domains and that means the magnetic domains can therefore store more energy. When you flip the core by 180 degrees then the effect will be the opposite and the magnetic domains will already be "occupied" and therefore store much less energy.
The problem is that we are not even talking about this specialized case, do you get it?
DO NOT ENGAGE WITH ME ON THE FORUM. Do you understand? You posting in Tinman's thread is just a mere two postings away from the Magluvin pissing contest and you will start up again and become maliciously demeaning and degrading towards me and I will not tolerate it. In the context of my point, "let's understand the basic of electronics," your posting was just gratituous and looking for an argument. I will not have it and I don't want you to post to me. Put your brain in gear and make your points, be them off base or on base, without engaging with me.
DO NOT ENGAGE WITH ME ON THE FORUM. Don't tell me that your language skills are so weak that you couldn't have responded to my positing and made your case without directly talking to me, and then move on. Surely you have the brainpower to do that.
Here is an example of a posting by you ad my response in email:
Magluvin: "What about HPV vaccinations that in some states are given to students without parental 'notification' let alone permission. Then the recent articles of how teen pregnancies are DOWN in ALL 50 STATES!!! Idiot."
My response: Are you trying to call me an "idiot" about something that I never even discussed? How utterly ridiculous and gratuitous and without merit.
DO NOT ENGAGE WITH ME ON THE FORUM.
MileHigh
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When a permanent magnet is brought near a coil conducting a AC wave, it (magnet) simply vibrates indicating that it is experiencing attraction during one half cycle and repulsion during other half cycle of AC wave. But the field of PM cannot affect the current flowing in the coil.
A permanent magnet's field can be considered as a standing AC wave with north pole as positive half of the cycle and south pole as negavitive half of the cycle. So a permanent magnet can affect the AC wave in the coil only if it is rotated and coupled (or synchronised) with the magnetic field produced in the coil by AC wave.
I think this coupling principle is made use of in rotary transformers but no PMs are used.
http://en.wikipedia.org/wiki/Rotary_transformer
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Tinman:
It's very late so I will just cover this one topic and respond more later.
Lets do the math on the maximum power that could be put into the system from the signal generator-there is no rocket science here.
The SG is set at 4Vpp,so 2 volts on the forward side.
There is a 220ohm resistor on the base of the transistor.
The duty cycle is 23%.
So the maximum power/watts that can be achieved is .018 watts
.018 x 23%=.00414 watt's.
I think your splitting hairs there MH,concidering the P/in is 132mWatts.
But just for arguments sake,we will change the P/in to 136mWatts.
Here is the big problem: You are shifting the goalposts. In your second clip you made a nonsensical test for checking for the possible power injected into the circuit by the signal generator. Do you agree with my point or do you have something to support what you did in the clip?
You are crunching numbers above but there was no discussion of crunching any numbers in the clip. You also make a mistake but that is not important. The only issue is what transpired in your second clip. I just commented on what I saw and you didn't respond to what went on in your clip.
MileHigh
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http://webpages.charter.net/dawill/tmoranwms/Elec_Magnetics.html
http://www.eevblog.com/forum/beginners/why-stick-a-magnet-to-an-inductor-permanently/
http://www.google.com/patents/US6710693
Note especially the waveforms shown in the first article.
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Useing the scope was to show the small signal coming from the SG,which is a visual way for me to show the small amount of power being supplied by the SG-and you have to agree,it is very small.
Also ,please feel free to point me out to my mistake-as this is how we learn. Im just as good at missing things as anyone else. If you are refering to how the transistor distributes that signal power,please remember i have a diode across the base/emitter junction.
I would like to know what you think is being disipated by the 18 ohm resistor,as far as the scope shot go's.
I am also drawing up a quick scetch for you and all to have a look at,and post your opinions on as to what happens throughout 1 cycle.It will be in 2D,so wont be anything fantastic,but it will show what im am asking.
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@ MH and all
Useing the picture below,i would like to ask the following questions.
Lets asume that the electromagnet has just enough power flowing through it to equal the strength of the field of the permanent magnet.Both magnets are of equal distance from the inductor core. So once the electromagnet is switched on,the field within the inductor is neutral throughout.
So first,what happens when we switch on the electromagnet?
Is there a BEMF or lenz force applied to our electromagnet coil?
Will the electromagnet still use the same amount of power ,with and without the inductor and PM being there?.
second-what happens when the electromagnet is switched off?-becomes open circuit.
Once the electromagnet switches off,and the field of the PM becomes the field within the inductor core-where or what is the BEMF or lenz force between?.
If we load the inductive kickback from our electromagnet coil,as in charging a battery,or placing a low value resistor across the output,what happens to the magnetic field within the inductor?.
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Hello Tinman,
you say that you are lost with my calculation.
You need to calculate the rms-voltage. The period T of the frequency shown on your oscillosope:
T = 1 / 3.279 Khz = 304 µ . Because your mosfet is switched on for 48 µ sec, then then average voltage which drive the current is calulated roughly by : Urms = 13,2 V x 48 µ sec / 304 µ sec = 2.05 V
Assuming your current-measurement is correct and is representing average current i_rms then this gives you P = U_rms x i_rms = 22 mW.
In order to be sure about i_rms you need a current probe capable of measuring up to 10 KHz.
So this was just a rough calculation of me and certainly has to be verified.
Regards
Kator01
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@ MH and all
Useing the picture below,i would like to ask the following questions......
I think it is only the variation of flux that matters. If you keep permanent magnets near the inductor it is not going to affect the performance because PMs flux remains constant with respect to the inductor whether you ON or OFF the electromagnet. If at all it affects the performance, it might only change the wave form of output wave without any nett gain.
The figure drawn is same as keeping permanent magnets on induction coil which is not going to affect its performance in any way.
http://www.madteddy.com/indcoils.htm
http://www.physics.gla.ac.uk/~kskeldon/PubSci/exhibits/E4/
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Hello Tinman,
you say that you are lost with my calculation.
You need to calculate the rms-voltage. The period T of the frequency shown on your oscillosope:
T = 1 / 3.279 Khz = 304 µ . Because your mosfet is switched on for 48 µ sec, then then average voltage which drive the current is calulated roughly by : Urms = 13,2 V x 48 µ sec / 304 µ sec = 2.05 V
Assuming your current-measurement is correct and is representing average current i_rms then this gives you P = U_rms x i_rms = 22 mW.
In order to be sure about i_rms you need a current probe capable of measuring up to 10 KHz.
So this was just a rough calculation of me and certainly has to be verified.
Regards
Kator01
Hi Kator01
The current being measured is constant current,and i believe the current needed to do your calculations correctly ,is instantaneous current.(This is where the math function would come in handy-if it actualy was there as it should be?).This will be a lot higher that the constant current required to keep the caps at the supply voltage.When the coil switches on,i can asure you that the current draw will be much more than 11mA. To view this,we can place a 1ohm resistor in series with the driven coil.
Our DC P/in is 11mA @ 12 volt's. W=V x I. So our power consumption is 132mWatts
This is why i said i was a little confused with your power consumption result,as your sum of 22mWatts is far from my calculated 132mWatts.
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Hi Tinman,
agreed that you will have a bigger current in that time-period of 48 µ sec. Now how come that your scope shows 13,2 V when you drive your circuit with 12 Volt ? Anyway. lets assume 12 Volt , but than you can not use the 12 Volt as driving voltage over the full cycle of 304 µ sec. as the 12 volts are only swiched on and driving the unknown high current for 48 µ sec , you need the average voltage ( ever lower 1.89 Volt ) which is related to one cycle ( since you do not have the surge-current in that 48 µ sec window9 and multiplying it with the average current you have measured because you only have the average current of 11 mA.. see ?
Your scope shows the false rms-voltage at the top, it does calculate the high value-distribution ( 13,2 V and this value is wrong also ) , which is mainly composed out of the mosfet-off-state. It wrongly calculates 13.2 V related to the 304 - 48 µ sec = 256 µ sec.
You always have to relate the short on-time-voltage to the full cycle.
So again I insist: it looks good...at least not bad
Regards
Kator01
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Hi Brad,
Referring to your drawing shown above I would say the following comments (with assuming your inductor has an 'I' shape core (i.e. a straight open core) and your electromagnet core has indeed a horse shoe like shape like your permanent magnet):
when the electromagnet is off, most flux from the permanent magnet is directed and closed into the I shaped core of the inductor provided the I core is ferromagnetic i.e. conducts flux much better than air
when you switch on the electromagnet with the polarities shown, I assume two explanations may be valid, depending on distances:
a) either the flux of the electromagnet will join to the poles of the permanent magnet via the upper and lower edge parts of the I core (facing unlike magnetic poles tend to join even via a ferromagnetic core piece in-between), hence the earlier flux of the permanent magnet will move out from the I core lengthwise
b) or the flux from the electromagnet will be directed into the I shaped core just like that of the permanent magnet so that as you say the two opposing flux may neutralize each other lengthwise in the core, the result is again a flux decrease to near zero in the I core
when you switch off the electromagnet, the flux from the permanent magnet can again penetrate through the I core full in lengthwise due to the assumed good flux conducting properties of the I core
Now on your questions
(So first,what happens when we switch on the electromagnet?)
I discussed above what may happen when you switch on the electromagnet, cases a) or b).
(Is there a BEMF or lenz force applied to our electromagnet coil?)
The moment the current is on in the electromagnet coil (with the proper intensity) AND the R load is hooked to the inductor coil (perhaps together with your tank capacitor not shown) I think the Lenz law effect can manifest only in a smaller amount than in case of a normal Faraday induction because the flux which is causing the main induction in the I core comes mainly from the permanent magnet, especially if flux change really happens like in case a) above.
(Will the electromagnet still use the same amount of power, with and without the inductor and PM being there?)
You may surely have found that placing an I core near to the prongs of a C core changes the inductance of the coil wound onto the C core, how much the inductance changes depends on the air gap left between the I core and the prongs, highest inductance is received when you fully close the gap and smallest when you remove the I core and the prongs of the C core become an open magnetic circuit again. Now considering this, your actual setup already has a certain air gap I suppose which already established a certain L inductance value for the electromagnet coil and once you fixed the distances in a real setup the inductances for the coils are set.
Another factor to consider is the flux coming from the permanent magnet via the I core towards the electromagnet C core, it may influence the permeability of the C core, albeit it can be a small flux influence only, depending on mainly the thickness of the I core.
So the answer to your question I think is yes, the electromagnet would use very nearly the same amount of input power, with or without the inductor and the PM being there and allowing for the above reasonings. Here I assume that the cores of the I and that of the electromagnet are not driven towards saturation in any instant.
(second-what happens when the electromagnet is switched off?-becomes open circuit)
I discussed this above and I add that when you wish to collect the energy coming from the collapsing field of the electromagnet, you have to be careful with choosing the correct 'on time' for the electromagnet because loading the spike from the collapsing field may extend the 'on time' of the electromagnet. The 'on time' can be conveniently adjusted / compensted by the duty cycle in this case I believe.
(Once the electromagnet switches off,and the field of the PM becomes the field within the inductor core-where or what is the BEMF or lenz force between?)
As I mentioned, I assume the Lenz action-reaction force takes place mainly between the permanent magnet flux and the flux in the inductor core, the current taken by the R load surely creates a flux against that of coming from the permanent magnet, effectively reducing it, as if the original flux strength from the permanent magnet would have been weaker. This may set a certain limit on the amount of output power.
So Lenz law is still valid but mainly acts between the permanent magnet-output coil flux and in a much less rate between the input-output coil flux, this is how I see this, I may be wrong.
Greetings, Gyula
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One more time:
The presence of an external magnetic field, like that provided by a permanent magnet, effectively changes the permeability of the core material. This is a "nonlinear" effect. Take a look at the B-H curve of some core materials. The field from the PM can move the core closer or further away from being completely saturated and this can have a _strong effect_ on the inductor's (transformer's, whatever) behaviour depending on frequency, polarity, offset, etc of the signal applied to the inductor.
If there were no effect from a PM on an inductor's behaviour, then Please Tell Me why there are so many inductors and transformers manufactured with Permanent Magnets as part of the structure? Why does my 6-NE2 JT _require_ such a magnet-biased inductor to work? Why do core-effect pulse motors benefit so greatly from biasing the core to near-saturation using permanent magnets? It is because the PM can effectively increase or decrease the permeability of the core. One can even make a sort of inductive diode, where one polarity of the applied signal sees a core of low permeability (thus low inductance) and the other polarity of the applied signal sees a high permeability core resulting in correspondingly high inductance. It simply is not true that a permanent magnet has no effect on the AC behaviour of inductors and transformers!
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Tinman, TK:
Let me address this issue in the context of Tinman's actual clip, and in a generic context.
For starters, let's just use some arbitrary abstract units to make the discussion simpler. Let's assume that we have a transformer with an unmagnetized core. Let's say that the core can be magnetized by the windings to the "left" by -5 flux units before it saturates and to the "right" by +5 flux units before it saturates. Let's not discuss the BH curve and the hysteresis loop and all that stuff to keep things simple. Is that fair enough?
Now, lets assume that we drive the primary of the coil with an AC excitation that will polarize the core between +3 and -3 flux units.
Case 1:
If the core of the coil is not magnetized then there is no issue, and you never see saturation. The secondary only responds to the AC excitation, and the secondary sees two events: a) a positive slope in magnetic flux spanning 6 units, and b) a negative slope in magnetic spanning 6 units. I hope so far this makes sense to both of you. The point being that the secondary only responds to AC excitation.
Case 2:
Now let's suppose the core is partially magnetized to +2.
In this case the AC excitation will result in the core varying between -1 and +5 flux units. In other words, the permanent flux from the +2 magnetization is added to the external flux coming from the excitation of the primary.
In this case, from the perspective of the secondary, it sees exactly the same thing as in Case 1: a) a positive slope in magnetic flux spanning 6 units, and b) a negative slope in magnetic spanning 6 units.
Therefore the output from the secondary in Case 2 will be identical to Case 1. Do you guys get this?
I will very briefly discuss a Case 4 where the core is magnetized to +4 flux units. Then the AC excitation will quickly saturate the core in the positive direction, the apparent inductance will drop drastically, and you will be "clipping the core" which will result in the transformer not working properly.
The most important thing to grasp is that the output in Case 1 and Case 2 will be identical.
The issue is that experimenters what to believe that the magnet gives some "extra kick-back" and therefore you get more energy. It's total crap, just another myth that leads to bad design decisions, poor practices, and it's simply electronics voodoo bullshit that pollutes people's conception of how a transformer works. "Add a biasing magnet to the side of your transformer for more output" is TOTAL CRAP.
Look at Tinman's experiment. We know that he is pumping quite low power into his transformer. Relative to my simplified example above, perhaps his core is biased to +3. But his excitation range is perhaps only -0.2 to +0.2. In other words, with the setup he has in his clip he is far far away from saturating his core. Therefore, the secondary output from his transformer will be the SAME if his core is a normal unbiased core at zero, or if is core is biased at -4, -3.....0.... +3, +4.
Again, for emphasis, in the context of Tinman's experiment and in a general sense, magnetizing a transformer core is nonsensical and does nothing. Tinman is NOT playing with skirting at the edge of saturation of his core in search of some kind of non-linear response from his transformer.
This is basic electronics, and it's worth it to understand the basics. What most people should be thinking is "Why the hell would I ever want to magnetize the core of my transformer?" If you are going to magnetize the core of your transformer then you need a legitimate reason to do it. Blindly believing that you will get "extra kick" because you "stress the magnet" and then the "magnet pushes back" is complete and total nonsense. The secondary in the transformer never even sees the permanent magnetization in the core. The secondary in the transformer only responds to the changes in core flux caused by the primary.
MileHigh
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Tinman:
In your first clip, you clearly state the myth or whatever you want to call it when you say this:
"So we are basically rocking this magnetic field backwards and forwards and causing our tank coil (to be excited)."
You clearly believe that you are "rocking" or modulating the magnetic field of the magnetized core and the effect of this "rocking" is supplying power to your LCR circuit. I would not be surprised if all of your buddies on your forum believe it also.
The problem is that it's not true, at all. On the other hand, I am simply telling you the truth. It's a good thing to bust electronics myths. When it comes to electronics and free energy forum experimenters, there are myths abound everywhere.
MileHigh
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The magnetized core transformer is not an AC transformer as MH keeps suggesting. ::)
Neither is Tinmans input to his primary, in case some didnt look at his scope shot. ;)
When pulsing a primary of a magnetically biased core, the field of the coil should be opposing the cores field, not adding to it to make it stronger and saturate/over saturate the core. Building a field from the coil, working in opposition to the core magnet, gives the field of the coil a lot further to go before saturation than a non magnetized core of the same specs. So the amount of energy of the coil/magcore can be substantially more, up around twice the V/A capability compared to a non biased core.
So talking AC as an input is senseless. ::)
Tinmans project is a special case, as he is using a magnetically biased core. ;) ;D
Mags
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One more time:
The presence of an external magnetic field, like that provided by a permanent magnet, effectively changes the permeability of the core material. This is a "nonlinear" effect. Take a look at the B-H curve of some core materials. The field from the PM can move the core closer or further away from being completely saturated and this can have a _strong effect_ on the inductor's (transformer's, whatever) behaviour depending on frequency, polarity, offset, etc of the signal applied to the inductor.
If there were no effect from a PM on an inductor's behaviour, then Please Tell Me why there are so many inductors and transformers manufactured with Permanent Magnets as part of the structure? Why does my 6-NE2 JT _require_ such a magnet-biased inductor to work? Why do core-effect pulse motors benefit so greatly from biasing the core to near-saturation using permanent magnets? It is because the PM can effectively increase or decrease the permeability of the core. One can even make a sort of inductive diode, where one polarity of the applied signal sees a core of low permeability (thus low inductance) and the other polarity of the applied signal sees a high permeability core resulting in correspondingly high inductance. It simply is not true that a permanent magnet has no effect on the AC behaviour of inductors and transformers!
;) ;D
Mags
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http://www.youtube.com/watch?v=EQJFuKvrUEs
Of course since MH has narrowed the goalposts to talking about the completely linear, unsaturated behaviour of simple iron laminated cores ... he's right as well.
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TK:
Yes, thank you for putting it so succinctly. That's the territory Tinman is in and the principle I am trying to operate with is to understand the basics and give yourself a foundation.
If you are operating in the linear unsaturated region of a transformer and you add the influence of a magnet to the mix, then as long as you are still in the linear unsaturated region then the magnet will have no affect whatsoever on the operation of the transformer.
Tinman:
I am going to try to respond to your postings.
For starters, you asked about the power dissipation across the 18-ohm resistor in your RLC circuit. This is kind of a case of not seeing the forest for the trees. Your fancy scope is giving you a real-time Vrms so the answer is right there. If you want to do it right measure the actual resistance of the 18-ohm resistor with your best multimeter.
My reference to the lenzless effect,was in relation to absolutly no power increase on the input was shown when we loaded one of the outputs-while remembering that the tank coil is already loaded(power yet to be determond).I am well aware that every transformer uses power at idle,but show me one where the power input dosnt go up when a load is placed on the output-while drawing a load on the second output.
You are in Thane Heinz territory here. You know that the transformer is drawing power, but you don't have a clear picture of where that power is going. So when you add a small load to the output of the transformer, what's happening is that the way the power is being split up on the output side changes. The net result is no increase in input power and you still can't account for precisely where the input power is going. At least you can say when you add a load and make a measurement that you know were some of the input power is going. That's all fine, but you should not have made your original statement the way you made it. You can always qualify statements after the fact but my suggestion is to just be straight from the start. If you said something like, "I made a power input measurement and when I added a load to the output the power input measurement did not change. However, I haven't accounted for where all the input power is going therefore I can't make any conclusions," that would have been the real deal.
but show me one where the power input dosnt go up when a load is placed on the output-while drawing a load on the second output.
Your request for someone to do this is not valid. Someone else could have a transformer that is drawing input power and when they add a load to the output the "balance of output power" changes with no extra current draw on the input. There is nothing remarkable about that.
Time to see if there will be a next posting....
MileHigh
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Tinman:
Okay I will try to tackle this one:
Useing the scope was to show the small signal coming from the SG,which is a visual way for me to show the small amount of power being supplied by the SG-and you have to agree,it is very small.
Also ,please feel free to point me out to my mistake-as this is how we learn. Im just as good at missing things as anyone else. If you are refering to how the transistor distributes that signal power,please remember i have a diode across the base/emitter junction.
I would like to know what you think is being disipated by the 18 ohm resistor,as far as the scope shot go's.
I am also drawing up a quick scetch for you and all to have a look at,and post your opinions on as to what happens throughout 1 cycle.It will be in 2D,so wont be anything fantastic,but it will show what im am asking.
I will just repeat again that if you want to measure the power that the signal generator is possibly pumping into your setup then you have to do it when the setup is live and running. Shutting the setup off will invalidate any attempts to measure the signal generator power and looking at the small glitches on the scope display while the setup was switched off was meaningless.
One more time, I am being conservative because I don't have a schematic of your setup. If the signal generator is only switching on an NPN transistor through a base resistor and an inductor, then for sure you can "eyeball" the input power and make an estimate without really having to make a measurement at all. But I absolutely refuse to make assumptions like this about your circuit.
Looking forward to seeing your schematic.
Moving on, in post #19 I am sorry to say that your diagram doesn't make much sense. You have magnetic fields from horseshoe magnets travelling radially across an inductor. So by definition even if you change the strength of those fields nothing will happen to the inductor coil. If I can offer you some serious advice, hunker down and look at this guy's videos and watch them from the beginning.
http://www.youtube.com/user/lasseviren1/videos
If it's too much work then just focus on the electricity and magnetism related videos. Don't be fazed by the math and the guy also talks a lot in plain English. If you get those clips and go learn stuff then you will be in a better position to experiment.
I will still try to answer your questions:
So first,what happens when we switch on the electromagnet? >>>>>>>> The current will increase in the coil and then level off. If it's "against the grain" of the magnetic flux in the horseshoe magnet it will take a longer amount of time.
Is there a BEMF or lenz force applied to our electromagnet coil? >>>>>>>>>>> There is standard BEMF just like when you energize a regular coil.
Will the electromagnet still use the same amount of power ,with and without the inductor and PM being there?. >>>> Yes.
second-what happens when the electromagnet is switched off?-becomes open circuit. >>>> You get an inductive spike that ionizes the air at the switch contacts and the coil discharges its stored energy.
Once the electromagnet switches off,and the field of the PM becomes the field within the inductor core-where or what is the BEMF or lenz force between?. >>>>>>> I don't understand your question.
If we load the inductive kickback from our electromagnet coil,as in charging a battery,or placing a low value resistor across the output,what happens to the magnetic field within the inductor?. >>>> The magnetic field will collapse. The horseshoe magnet is "not really there" because it's a magnet with a constant unchanging field and the coil only responds to changing fields - the same old story.
MileHigh
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Hi Brad,
Referring to your drawing shown above I would say the following comments (with assuming your inductor has an 'I' shape core (i.e. a straight open core) and your electromagnet core has indeed a horse shoe like shape like your permanent magnet):
when the electromagnet is off, most flux from the permanent magnet is directed and closed into the I shaped core of the inductor provided the I core is ferromagnetic i.e. conducts flux much better than air
when you switch on the electromagnet with the polarities shown, I assume two explanations may be valid, depending on distances:
a) either the flux of the electromagnet will join to the poles of the permanent magnet via the upper and lower edge parts of the I core (facing unlike magnetic poles tend to join even via a ferromagnetic core piece in-between), hence the earlier flux of the permanent magnet will move out from the I core lengthwise
b) or the flux from the electromagnet will be directed into the I shaped core just like that of the permanent magnet so that as you say the two opposing flux may neutralize each other lengthwise in the core, the result is again a flux decrease to near zero in the I core
when you switch off the electromagnet, the flux from the permanent magnet can again penetrate through the I core full in lengthwise due to the assumed good flux conducting properties of the I core
Now on your questions
(So first,what happens when we switch on the electromagnet?)
I discussed above what may happen when you switch on the electromagnet, cases a) or b).
(Is there a BEMF or lenz force applied to our electromagnet coil?)
The moment the current is on in the electromagnet coil (with the proper intensity) AND the R load is hooked to the inductor coil (perhaps together with your tank capacitor not shown) I think the Lenz law effect can manifest only in a smaller amount than in case of a normal Faraday induction because the flux which is causing the main induction in the I core comes mainly from the permanent magnet, especially if flux change really happens like in case a) above.
(Will the electromagnet still use the same amount of power, with and without the inductor and PM being there?)
You may surely have found that placing an I core near to the prongs of a C core changes the inductance of the coil wound onto the C core, how much the inductance changes depends on the air gap left between the I core and the prongs, highest inductance is received when you fully close the gap and smallest when you remove the I core and the prongs of the C core become an open magnetic circuit again. Now considering this, your actual setup already has a certain air gap I suppose which already established a certain L inductance value for the electromagnet coil and once you fixed the distances in a real setup the inductances for the coils are set.
Another factor to consider is the flux coming from the permanent magnet via the I core towards the electromagnet C core, it may influence the permeability of the C core, albeit it can be a small flux influence only, depending on mainly the thickness of the I core.
So the answer to your question I think is yes, the electromagnet would use very nearly the same amount of input power, with or without the inductor and the PM being there and allowing for the above reasonings. Here I assume that the cores of the I and that of the electromagnet are not driven towards saturation in any instant.
(second-what happens when the electromagnet is switched off?-becomes open circuit)
I discussed this above and I add that when you wish to collect the energy coming from the collapsing field of the electromagnet, you have to be careful with choosing the correct 'on time' for the electromagnet because loading the spike from the collapsing field may extend the 'on time' of the electromagnet. The 'on time' can be conveniently adjusted / compensted by the duty cycle in this case I believe.
(Once the electromagnet switches off,and the field of the PM becomes the field within the inductor core-where or what is the BEMF or lenz force between?)
As I mentioned, I assume the Lenz action-reaction force takes place mainly between the permanent magnet flux and the flux in the inductor core, the current taken by the R load surely creates a flux against that of coming from the permanent magnet, effectively reducing it, as if the original flux strength from the permanent magnet would have been weaker. This may set a certain limit on the amount of output power.
So Lenz law is still valid but mainly acts between the permanent magnet-output coil flux and in a much less rate between the input-output coil flux, this is how I see this, I may be wrong.
Greetings, Gyula
Gyula
It is good to see some of us can understand a simple drawing,and answer with responces that make sence. From the many test i done with a setup like that shown in the picture above,i can say you are right. When i stated a transformer that showd a 0 lenzz effect when the output was loaded,was in regards to no reflection shown on the input. As you stated,there is a lenz force created between the PM and the inductor,but we dont need to worry about that,as it comes at no cost to us,as it's now the PM that is doing the work !fancy that,a PM doing useful work!
This is where experimenting kills text book physics-actual results befor your eyes,insted of blind faith that all has been looked into and accounted for.
My actual test setup(as pictured above in previous post) showed exactly what you said Gyula.
With or without the inductor and PM there,the electromagnets power draw remaind the same.
Without the PM there,the electromagnets power draw went up,and a voltage was produced across the resistor (standard transformer action)
With the PM in place,the electromagnets power draw went down,and a voltage was produced across the resistor of the same value.
The trick is getting the right amount of current at a set voltage going to the electromagnet,so as it only neutralises the PMs field in the inductor core. This can be done useing a resistor in series with the electromagnets coil,and then measureing the voltage across that resistor. Then we just have to set our SG's signal strength to the transistor so as we get the same voltage across that same resistor during our on period. Then that is when i set up the tank circuit,so as to be able to adjust the frequency until i gained maximum amplitude across that tank circuit.
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The magnetized core transformer is not an AC transformer as MH keeps suggesting. ::)
Neither is Tinmans input to his primary, in case some didnt look at his scope shot. ;)
When pulsing a primary of a magnetically biased core, the field of the coil should be opposing the cores field, not adding to it to make it stronger and saturate/over saturate the core. Building a field from the coil, working in opposition to the core magnet, gives the field of the coil a lot further to go before saturation than a non magnetized core of the same specs. So the amount of energy of the coil/magcore can be substantially more, up around twice the V/A capability compared to a non biased core.
So talking AC as an input is senseless. ::)
Tinmans project is a special case, as he is using a magnetically biased core. ;) ;D
Mags
Hi Mags
Yes,there is no AC in the drive coil,so not sure where the AC bit is coming from. AC is alternating current-not alternating voltage.The current flow remains in 1 direction when the drive coil is switched on and off,AC is current flowing one way,and then switches and flows the other way-this never happens in the drive coil.
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Hi Mags
Yes,there is no AC in the drive coil,so not sure where the AC bit is coming from. AC is alternating current-not alternating voltage.The current flow remains in 1 direction when the drive coil is switched on and off,AC is current flowing one way,and then switches and flows the other way-this never happens in the drive coil.
Hey Brad
"so not sure where the AC bit is coming from."
It comes from "moving the goal posts" like TK said, in order to put the magnetic biased core in the realm of 'junk'. ;) lol
Just keep doin what you do. ;) Your doing fine. ;D
Mags
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Many people are confused about what "AC" really means. But "transformer effect" doesn't depend on true alternating current in the primary of the transformer. Recall that induced voltage in a secondary is proportional to the time rate of change of... the magnetic field in the primary. And the magnetic field is proportional to the current (amp-turns). So if the current rises and falls, that's all that is needed for transformer induction effect to happen. The current doesn't have to actually reverse, it can keep going in the same direction, as long as the amplitude of the current rises and falls. This of course produces a change in the magnetic field. The secondary output will be true AC though, even if the input to the primary is pulsed DC or "AC" with enough DC offset so that the current doesn't actually reverse.
"dB/dt", the time rate of change of the B field, can be thought of as the instantaneous slope of the curve describing the B field strength over time. So while the B field is increasing, dB/dt is positive, and when the B field peaks dB/dt is zero, and when the B field is decreasing, dB/dt is negative. So the induced voltage in the secondary flips sign at the peaks (and valleys) of the current which is causing the primary B field to wax and wane.
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Hi Tinman,
agreed that you will have a bigger current in that time-period of 48 µ sec. Now how come that your scope shows 13,2 V when you drive your circuit with 12 Volt ? Anyway. lets assume 12 Volt , but than you can not use the 12 Volt as driving voltage over the full cycle of 304 µ sec. as the 12 volts are only swiched on and driving the unknown high current for 48 µ sec , you need the average voltage ( ever lower 1.89 Volt ) which is related to one cycle ( since you do not have the surge-current in that 48 µ sec window9 and multiplying it with the average current you have measured because you only have the average current of 11 mA.. see ?
Your scope shows the false rms-voltage at the top, it does calculate the high value-distribution ( 13,2 V and this value is wrong also ) , which is mainly composed out of the mosfet-off-state. It wrongly calculates 13.2 V related to the 304 - 48 µ sec = 256 µ sec.
You always have to relate the short on-time-voltage to the full cycle.
So again I insist: it looks good...at least not bad
Regards
Kator01
Hi Kator
If you look at the scope shot,you will see that the 13.1 volt peak is from the inductive kickback spike.But you will also see that it is quickly brought down by the cap and battery-more so the battery,as its internal resistance would only be around 1-1.5 ohms-provideing its a healthy battery.
You can see in the rest phase ,that the supply voltage is indeed 12 volts.
If we look at the scope shot,we can see that the efficiency is quite good-as far as the inductive kick back charging go's.
But as far as our tank circuit go's,well thats no so exciting. From that part of the circuit,we have only around .033 watts-but every bit helps.
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Hey Brad
"so not sure where the AC bit is coming from."
It comes from "moving the goal posts" like TK said, in order to put the magnetic biased core in the realm of 'junk'. ;) lol
Just keep doin what you do. ;) Your doing fine. ;D
Mags
Experimenting is never junk Mag's lol.
Here is a good example,and it seems that his results are the same as what i get with the setup i posted in a previous post.
http://www.youtube.com/watch?v=2kosQIrDCWM
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Tinman:
I can assure you that your transformer input is being excited by "AC." That is a generic term used in electronics. It means "changing with respect to time." If there was no AC input to your transformer then the output wound be dead and the RLC circuit would be dead. Talking about "moving the goalposts" with respect to the standard definitions for electronics terms is a non-starter. I am asking you to not go there because that would be another "contest." I won't even argue the point beyond what I just stated because it would be silly.
With respect to the "other" moving of the goal posts, I already covered that earlier. The moving is done when you make a general statement about something that relates directly to your setup, and then someone else responds stating "not true" because special case XYZ is not true. That's the real moving of the goal posts. I discussed your particular setup and what you are doing and my explanation about the futility of the permanent magnet core is because that's where you start - at the beginning - within the context of your actual experiment. You learn that non-moving magnets have no affect on transformers because there is no changing magnetic flux with respect to time. Then you go on the bench and view and measure it for real and confirm it for yourself. These are the basic building blocks of electronics. You don't believe me? That's fine, feel free to do some bench experiments and prove it for yourself. Also, don't make the mistake of alleging that I am "all theory" because I have done all of this stuff on the bench for real.
From my perspective you are playing with a transformer setup where your excitation is a switching transistor. That bangs the primary coil with a regular stream of pulses (which is AC) and that excitation is at the resonant frequency for your RLC circuit. The fact that your core is a magnet has zero affect on your setup. If you don't believe me then why not put in an equivalent non magnetized core and compare waveforms and make some good measurements? It would be blind faith on your part to believe that the magnetized core of your transformer does something useful, because it doesn't. Nor are the magnets doing any work, they are as dead as proverbial doornails. That's more blind faith on your part, "One day someone will prove that magnets are a source of energy," it will never happen.
That's the real deal Tinman. If you are going to make measurements on the input and all of the outputs and compare total input power to total output power that would be great. You won't find anything special but you will deserve a lot of credit for making good measurements.
The thing that may be throwing you off is that the magnetic core does nothing. It's a fact. The best thing that you could do for yourself is do the research and create some good test procedures to prove it for yourself. That's the challenge, find the real truth. If you refuse and just "want to believe" that the magnetized core does something beneficial and the "magnets are doing work" then it would be a shame and you won't benefit from the project. I'll put it another way, if you do the project and intentionally hunt for something to confirm your beliefs and prejudices and expectations, then you are just cheating yourself to your own detriment.
So it's up to you. You can increase your knowledge and look for the truth or just look for confirmation of your preconceptions in your measurements.
MileHigh
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Many people are confused about what "AC" really means. But "transformer effect" doesn't depend on true alternating current in the primary of the transformer. Recall that induced voltage in a secondary is proportional to the time rate of change of... the magnetic field in the primary. And the magnetic field is proportional to the current (amp-turns). So if the current rises and falls, that's all that is needed for transformer induction effect to happen. The current doesn't have to actually reverse, it can keep going in the same direction, as long as the amplitude of the current rises and falls. This of course produces a change in the magnetic field. The secondary output will be true AC though, even if the input to the primary is pulsed DC or "AC" with enough DC offset so that the current doesn't actually reverse.
"dB/dt", the time rate of change of the B field, can be thought of as the instantaneous slope of the curve describing the B field strength over time. So while the B field is increasing, dB/dt is positive, and when the B field peaks dB/dt is zero, and when the B field is decreasing, dB/dt is negative. So the induced voltage in the secondary flips sign at the peaks (and valleys) of the current which is causing the primary B field to wax and wane.
Hi TK
Yes,i know what your saying.Just like it only takes a north field passing over an inductor to produce an AC output.This is why we have an AC output on the tank circuit. But the input is DC,and the primary coil only has a DC current flowwing through it at all times.If we put a DC current into an inductor,a magnetic field will be produced around that inductor. When we switch that coil off,the field collapses-but it never reverses polarity.Now if we load that kickback in the way of charging a battery,or place a resistive load across it,then a BEMF is produced from the coil that apposes that collapsing magnetic field. And as you can see in the scope shot,the current always flows in the same direction.
Definition of AC current:-In alternating current (AC, also ac), the flow of electric charge periodically reverses direction.
Well as we see in the scope,there is no change in direction-only a change in amplitude.
The voltage potential is ofcourse reversed,due to the way the circuit is-much the same as the SSG circuit.This is why the positive of the P/in become's the negative for the P/out,and the collector that use to be our negative side,now becomes our positive potential-thus the diode on the collector.
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MH
Once again,you are clutching at straws,and being far to picky.
Lets have a look at a couple of scope shots from the video-below.
Dose this not clearly show an indication,or give you a visual reference as to how little of the total power supplied to the system,is being supplied by the signal generator?
Also there is no AC current going ito the drive coil,nor coming out of it.Once again you are deviating from the true meaning of AC,and getting it mixed up with something else.
Definition of AC: In alternating current, the flow of electric charge periodically reverses direction. In direct current, the flow of electric charge is only in one direction.
The current flow through the driven coil remains in the same direction at all time's-only the amplitude of that current flow changes !! that is not AC current in the true sence of the definition !!
Edit-well now it seems that i cant post attachments-no !ADD ATACHMENT BUTTON!?-only clear attachments
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http://www.youtube.com/watch?v=36W4UdMKVwU
Can somebody explain what is he doing? Magnetic standing columnar waves?? :'(
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http://www.youtube.com/watch?v=36W4UdMKVwU
Can somebody explain what is he doing? Magnetic standing columnar waves?? :'(
Yes-he is playing with his balls ::)
A standing wave is just a stationary wave,which means it remains in a constant position.
Not to sure where the columnar part come into it, in regards to his video?.
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@MH
Scope shots regards to post 41
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@Magpwr
Any pulsed inductor can be made into a battery desulphator,it is quite easy.
Tomorrow i will make a video of a desulphator of the simplest kind-no transistor,reed switches or hall effect devices. It will cost you about $1.00 to make,and will work just aswell as any out there.
Hi tinman,
did you already made a short video of battery pulsing for low cost? Would appreciate much!
Thank you
magneto_DC
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Some information regarding the scope shots below.
As my scope,SG and power supply all have a common ground,i could not place my scope across the coil to measure voltage,nor across the .1 ohm 5 watt series resistor to measure current to the coil.
So i dug out my old home made, stand alone signal generator,and traded the power supply for a 12 SLA. this now gave me the means to measure the voltage across the drive coil,and the current going to it.
The one draw back is my home made SG is set at a 50% duty cycle,but this just makes the results a lot clearer.The frequency i can adjust,and got it as close as i could,as the home made SG is touchy on the pot adjustment.
So the yellow trace(ch1) is across the coil,and we can clearly see that we have an alternating voltage,although in a square wave.-but nothing out of the ordinary for this kind of setup.
NOTE-CH1 is also lifted 2 divisions
to get it clear of ch2.
The blue trace(ch2)is across the .1 ohm 5w series resistor,that is hooked to the positive input of the driven coil.
At no point can you see the blue trace show an alternating current-only a change in current amplitude
The change in voltage polarity is because of the way the circuit is set up-nothing more.It is the same as the SSG circuit-and remember that the positive of our power supply,is also the negative for our charge battery.
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Hi tinman,
did you already made a short video of battery pulsing for low cost? Would appreciate much!
Thank you
magneto_DC
Hi Magneto
I have made the device,but as yet havnt made the video. I will have it up for you within the next 24 hours-posted here,so as you can find it.
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TK reply #36:
Many people are confused about what "AC" really means. But "transformer effect" doesn't depend on true alternating current in the primary of the transformer. Recall that induced voltage in a secondary is proportional to the time rate of change of... the magnetic field in the primary. And the magnetic field is proportional to the current (amp-turns). So if the current rises and falls, that's all that is needed for transformer induction effect to happen. The current doesn't have to actually reverse, it can keep going in the same direction, as long as the amplitude of the current rises and falls. This of course produces a change in the magnetic field. The secondary output will be true AC though, even if the input to the primary is pulsed DC or "AC" with enough DC offset so that the current doesn't actually reverse.
"dB/dt", the time rate of change of the B field, can be thought of as the instantaneous slope of the curve describing the B field strength over time. So while the B field is increasing, dB/dt is positive, and when the B field peaks dB/dt is zero, and when the B field is decreasing, dB/dt is negative. So the induced voltage in the secondary flips sign at the peaks (and valleys) of the current which is causing the primary B field to wax and wane.
Very clear, thank you. As Tinman says, like passing a north pole over a coil, one gets AC output (as I've done many times.)
Tinman -- did I understand correctly that you have a steady DC input, making P-input easy to measure?
If so, how do you achieve steady DC input?
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TK reply #36:
Very clear, thank you. As Tinman says, like passing a north pole over a coil, one gets AC output (as I've done many times.)
Tinman -- did I understand correctly that you have a steady DC input, making P-input easy to measure?
If so, how do you achieve steady DC input?
By useing large caps to store power.The amp meter go's between the power supply and caps.
At these high frequencies,and small current draw,the amp meter sees no ripple.The higher your current draw,the larger the caps you must use,and high current caps are better for this aswell.
Tomorrow i was going to test the accuracy of my meters reading pulsed DC current.
So i will video it aswell,and post it here,so as you can see what caps can do to smooth out pulses-both on P/in and P/out.
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OK, makes sense. Now you can measure the input energy using E = 1/2 CV^2.
Same formula for output cap (seems then you would need a FWBR to feed the output cap.)
Power = energy/time interval, same time interval for Pin and Pout.
AC output can also be dumped on a simple resistor immersed in water, and determine Eout (=Qout) using calorimetry. This is quite easy to do actually, IMO.
... AS LONG AS your measurement techniques do not alter/diminish the effect you are trying to measure!
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Tinman:
Your interpretation of your scope shots is spot on. In the second scope shot where you see the increasing energizing current for the coil it looks to me to be about one-half of a time constant before the transistor switches off.
The issue of why you have a "magneformer" where you use a magnet as the core of your transformer remains. I didn't see you explain at the beginng of the thread why you chose to use a magnet. People can describe specific examples about the use of magnets in certain applications but they have nothing to do with your setup. Is it for better performance? If yes, what does that mean? Now that you have done the build what kind of tests are you going to do to confirm (presumably) that the magnet in the core is doing what it is supposed to be doing?
The point here is that if you made a design decision to use a magnet, there is supposed to be a valid reason for it. If the "valid reason" is that "'people know' that using a magnet does something special," but you have no specifics, then simply state that.
You know there is a peer pressure on the forums to always agree and almost never question your builder friends. Like I have pointed out in the past, that just leads to stagnation. People don't learn and next year and the year after the pattern repeats itself and they repeat the same mistakes over and over and no progress is made.
In the real world of electronics design you have to have valid reasons for your design choices that you can explain to your peers. So you can address this issue or sweep it under the rug, it's your choice. In the best-case scenario you would really think about this and do testing and arrive at the correct conclusion for your setup.
MileHigh
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Tinman:
Your interpretation of your scope shots is spot on. In the second scope shot where you see the increasing energizing current for the coil it looks to me to be about one-half of a time constant before the transistor switches off.
The issue of why you have a "magneformer" where you use a magnet as the core of your transformer remains. I didn't see you explain at the beginng of the thread why you chose to use a magnet. People can describe specific examples about the use of magnets in certain applications but they have nothing to do with your setup. Is it for better performance? If yes, what does that mean? Now that you have done the build what kind of tests are you going to do to confirm (presumably) that the magnet in the core is doing what it is supposed to be doing?
The point here is that if you made a design decision to use a magnet, there is supposed to be a valid reason for it. If the "valid reason" is that "'people know' that using a magnet does something special," but you have no specifics, then simply state that.
You know there is a peer pressure on the forums to always agree and almost never question your builder friends. Like I have pointed out in the past, that just leads to stagnation. People don't learn and next year and the year after the pattern repeats itself and they repeat the same mistakes over and over and no progress is made.
In the real world of electronics design you have to have valid reasons for your design choices that you can explain to your peers. So you can address this issue or sweep it under the rug, it's your choice. In the best-case scenario you would really think about this and do testing and arrive at the correct conclusion for your setup.
MileHigh
Your answers starts at post 19 MH,and guyla hit the nail on the head on post 24.
My idea is to make a transformer that is as efficient as it can be made here at home,and to show that PMs improve efficiency.
What im trying to do now,is to make sure all my meters are reading accuratly,and this is tricky,as my scope shares a common ground with my SG,and power supply. My home made SG dosnt have pulse width adjustment,only frequency adjustment. I am trying to make up a transformer based around the same priciples,but where i can remove the magnet,and replace it with a ferrite core of same dimentions-while the system is running. All these things take time,and i am working on them
My biggest problem is getting around this common ground between my three needed devices(scope,SG and PS)
Please remember-no one is shouting OU here,but the P/in and P/out so far,seem a little to close when we look at how roughly the device is made.
But a quick summery as to what i hope i have achieved.
The driven coil is to neutralise the PMs field in the tank coil.
When the driven coil is switched of,two things happen.
1-we collect the inductive kickback,and charge a battery with it.
2-the PMs field once again is induced into the tank coil. The lenz force is now between the PM and the tank coil. To me this is the PM doing the work,and is the reason we get a continual current flowing through the 18 ohm resistor-even when the driven coils energy has been depleeted.This can be clearly seen in the first scope shot i posted.
How is it that the primary coil is no longer pumping current into the charge battery,but the tank coil is still producing a current across the 18 ohm resistor? A bifilar coil would not do this,as i checked that today. A standard bifilar coil with an iron core reacts compleetly different. The bottom half of the tank circuit wave is flat.
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I am trying to make up a transformer based around the same priciples, but where i can remove the magnet,and replace it with a ferrite core of same dimentions-while the system is running.
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Hi Brad,
Please remember that the permeability of a permanent magnet is very close to that of the air (a PM is almost a 100% saturated piece of material) this means that you should not replace the magnet with a ferrite core because by doing so your coil would have a much higher self inductance than with the magnet inside. The best approach would be not to insert any core into the same coil when you remove the permanent magnet. Permeability figures for ceramic (i.e. ferrit) magnets is around 1.1 to 1.3 and for Neo magnets it is around 1.05 to 1.1 or so. If you have an L meter and some air core coils around (or you can remove the core from a multiturn coil), you could see how small the inductance changes (increases a few percent) with a permanent magnet inserted into an air cored coil.
On your Atten scope: perhaps the best would be to contact the Atten service people to have the software in the scope check and reload once the Measure Menu does not have any content in it to choose from.
Gyula
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Hi Brad,
Please remember that the permeability of a permanent magnet is very close to that of the air (a PM is almost a 100% saturated piece of material) this means that you should not replace the magnet with a ferrite core because by doing so your coil would have a much higher self inductance than with the magnet inside. The best approach would be not to insert any core into the same coil when you remove the permanent magnet. Permeability figures for ceramic (i.e. ferrit) magnets is around 1.1 to 1.3 and for Neo magnets it is around 1.05 to 1.1 or so. If you have an L meter and some air core coils around (or you can remove the core from a multiturn coil), you could see how small the inductance changes (increases a few percent) with a permanent magnet inserted into an air cored coil.
On your Atten scope: perhaps the best would be to contact the Atten service people to have the software in the scope check and reload once the Measure Menu does not have any content in it to choose from.
Gyula
Hey Gyula
If the pulsed primary input field is opposing the permanent magnet, the core wont be saturated till the field of the magnet is flipped/reversed and then pushed further into saturation, if the input can deliver that much opposition to the cores permanent field. ;D A ferrite mag can be demagnetized, or even reversed eventually, that is again if the input is enough. Brad isnt putting that much power in so saturation shouldnt be an issue and the magnets should last for quite some time.
Mags
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Hi Brad,
Please remember that the permeability of a permanent magnet is very close to that of the air (a PM is almost a 100% saturated piece of material) this means that you should not replace the magnet with a ferrite core because by doing so your coil would have a much higher self inductance than with the magnet inside. The best approach would be not to insert any core into the same coil when you remove the permanent magnet. Permeability figures for ceramic (i.e. ferrit) magnets is around 1.1 to 1.3 and for Neo magnets it is around 1.05 to 1.1 or so. If you have an L meter and some air core coils around (or you can remove the core from a multiturn coil), you could see how small the inductance changes (increases a few percent) with a permanent magnet inserted into an air cored coil.
On your Atten scope: perhaps the best would be to contact the Atten service people to have the software in the scope check and reload once the Measure Menu does not have any content in it to choose from.
Gyula
Hi Gyula
I will not be changing the core in the electromagnet,but only the generating inductor core-the tank coil. It will be set up the same as the picture i posted with the two horse shoe shaped magnets-1 electric,and 1 PM.
In regards to the scope-it seems that all who have this scope,have the same problem. This i have found out by browsing some reviews on the scope. Some seem to be loading software from another scope onto the atten(same scope,different name) ,but it comes at a risk.If the software dosnt take,then you just have a box full of parts,and it cant be reversed. I have contacted atten,and there is no software update for my modle yet.
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Tinman:
I have to retract what I said in my posting #32 about your diagram and questions in your posting #19. The reason is because your diagram confused me. You use the exact same symbol for the permanent magnet and the electromagnet, right down to the red and blue rectangles. So I thought the "electromagnet" was a second permanent magnet with a coil wrapped around it.
I read through Gyula's comments and pretty much agree with them. I also believe that I see what your train of thought is. You see the permanent magnet on the left as being the agent that is responsible for the change in flux through the inductor when the electromagnet switches off. So the logic is that the permanent magnet is doing the work to change the flux through the inductor which then creates current to flow through the resistor.
That's all true but you have to look at the whole story as it unfolds in time. This is in a way related to the SMOT thread where people are not comfortable with the concept of magnetic potential energy and how it "came from somewhere else" by virtue of position and it is being invisibly stored like a compressed invisible spring.
When you first switch on the electromagnet you have to pump an amount of energy into the electromagnet to generate the field to counteract the flux going through the inductor from the permanent magnet. That will also cause changing flux with respect to time in the inductor and some energy will also flow through the resistor.
Then when you have DC current flowing through the electromagnet maintaining zero net flux through the inductor core (there are equal and opposite flux streams in the one-half core slices for net zero flux), you have magnetic potential energy stored in the stressed magnetic fields. Then when you switch off the electromagnet the collapsing field from the electromagnet allows the permanent magnet to "take over" putting flux through the inductor core, and that also causes the current to flow and energy is burned off in the resistor.
I am just giving you a general description of what is happening. The key thing is that the permanent magnet is not "doing work," rather, it's being stressed and storing some magnetic potential energy, and then when it is unstressed it's releasing that magnetic potential energy. There are no gains here.
Even though your diagram is relatively simple, if you had this problem in an electronics class, you would make a timing diagram where you have the switch closing and then opening, and you track all of various variables. If you constructed a timing diagram you should see how everything balances.
Another way to describe what is happening is as follows: When the electromagnet is off, the permanent magnet is responsible for the flux through the inductor core. Let's call that the unstressed condition of the core. When the electromagnet switches on, it "pushes out" half of the downwards flux inside the core due to the permanent magnet and sets up a counter stream of flux going in the opposite direction. Here we have the rare occurrence of a real true to life Bloch wall going down the center of the core. This is the stressed condition. It took work to create the stressed condition, and that's now stored magnetic potential energy. So the permanent magnet is being stressed and storing energy that the electromagnet put into it. When you cut the current to the electromagnet, the permanent magnet pushes back with its stored potential energy and "reoccupies" the inductor core with its flux and you are back to the unstressed condition.
It's all about accepting the notion that any permanent magnet can only appear to be a source of energy because someone or something else did some work to store potential energy in the magnet in the first place and the magnet can only give back as much as you put into it.
MileHigh
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Dammit Janet... I love you! lol
My idea is to make a transformer that is as efficient as it can be made here at home,and to show that PMs improve efficiency.
Got it.
For getting around the common ground issue you could get isolation transformers for your scope, signal generator and use batteries or get a isolation transformer for your power supply. They might be expensive, I don't know. When it comes to stuff like that safety is a priority and I would not be comfortable using my own home brew isolation transformers. I would recommend using batteries and making your own current measurements. I think it's safe to assume that your multimeter is better at measuring current than what you see on the power supply display.
The driven coil is to neutralise the PMs field in the tank coil.
When the driven coil is switched of,two things happen.
1-we collect the inductive kickback,and charge a battery with it.
2-the PMs field once again is induced into the tank coil. The lenz force is now between the PM and the tank coil. To me this is the PM doing the work,and is the reason we get a continual current flowing through the 18 ohm resistor-even when the driven coils energy has been depleeted.This can be clearly seen in the first scope shot i posted.
The reason you get the continuous current flowing through the resistor is because of the LC tank circuit. The voltage across the 18-ohm resistor is the tank circuit voltage. Your transistor pulse going through the primary is continually hitting your LC tank circuit with an injection of energy, so I don't see why you would find it significant that the current is always flowing through the resistor. The average power being burned off in the resistor plus the average power being burned off in the wire is exactly equal to the average power being supplied by the secondary, which also happens to be the 'L' in the tank circuit. You might have a Fluke true-RMS multimeter and I would use its measurement as opposed to your scope's measurement for the RMS voltage. You should be able to make a quite accurate power measurement for that part of the circuit. Since you know the pulse frequency you know the amount of energy per pulse for what it's worth.
Anyway, your quest is a great idea and a great exercise. I can see that you are doing filtering to make good current measurements and stuff like that. It should be interesting to see what your results are.
MileHigh
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Brad:
As Gyula said your scope may need a firmware upgrade and you should check their web site. You typically download a file and put it on a flash drive. Then when you power up the scope while pushing on one or two buttons and that triggers the firmware update.
Gyula:
Please remember that the permeability of a permanent magnet is very close to that of the air (a PM is almost a 100% saturated piece of material) this means that you should not replace the magnet with a ferrite core because by doing so your coil would have a much higher self inductance than with the magnet inside. The best approach would be not to insert any core into the same coil when you remove the permanent magnet. Permeability figures for ceramic (i.e. ferrit) magnets is around 1.1 to 1.3 and for Neo magnets it is around 1.05 to 1.1 or so. If you have an L meter and some air core coils around (or you can remove the core from a multiturn coil), you could see how small the inductance changes (increases a few percent) with a permanent magnet inserted into an air cored coil.
I didn't know that a magnet would have such a low relative permeability. You still might be able to take advantage of the polarization though as has already been stated. Note the excitation from the primary coil is unidirectional. This assumes the relative permeability is radically different depending on the direction of the external field. It would be an interesting and easy test. You just have to look at the slope of the current rise in the coil for same-direction and opposite-direction magnetic field generation by the coil wrapped around the magnetic core. You cross your fingers and hope that you don't ruin the magnet.
I also wonder if the L-meter will be thrown off by the introduction of a magnet into the coil. I assume they sample or sweep low-level AC frequency excitation for the coil under test and check the response to measure the inductance. So if the magnet does indeed radically change in relative permeability depending on direction it may have a small heart attack (throw off the measurement algorithm). It probably will read out as a high inductance - my guess.
Another point is that this is a transformer setup, not an inductance. So assuming the core (any core) is coupling the energy properly, you _don't_ see inductance on the primary, you see the load, which is an LCR circuit. So you see a wobbling resistance! lol Note since you are approximately at resonance, you are pretty much seeing the resistive component of the LCR circuit as the load of the primary. So that means that Brad should see the voltage and current going into the primary winding as mostly in phase, assuming that the core/coupling is doing it's job properly to transfer the power.
MileHigh
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Tinman:
I have to retract what I said in my posting #32 about your diagram and questions in your posting #19. The reason is because your diagram confused me. You use the exact same symbol for the permanent magnet and the electromagnet, right down to the red and blue rectangles. So I thought the "electromagnet" was a second permanent magnet with a coil wrapped around it.
I read through Gyula's comments and pretty much agree with them. I also believe that I see what your train of thought is. You see the permanent magnet on the left as being the agent that is responsible for the change in flux through the inductor when the electromagnet switches off. So the logic is that the permanent magnet is doing the work to change the flux through the inductor which then creates current to flow through the resistor.
That's all true but you have to look at the whole story as it unfolds in time. This is in a way related to the SMOT thread where people are not comfortable with the concept of magnetic potential energy and how it "came from somewhere else" by virtue of position and it is being invisibly stored like a compressed invisible spring.
When you first switch on the electromagnet you have to pump an amount of energy into the electromagnet to generate the field to counteract the flux going through the inductor from the permanent magnet. That will also cause changing flux with respect to time in the inductor and some energy will also flow through the resistor.
Then when you have DC current flowing through the electromagnet maintaining zero net flux through the inductor core (there are equal and opposite flux streams in the one-half core slices for net zero flux), you have magnetic potential energy stored in the stressed magnetic fields. Then when you switch off the electromagnet the collapsing field from the electromagnet allows the permanent magnet to "take over" putting flux through the inductor core, and that also causes the current to flow and energy is burned off in the resistor.
I am just giving you a general description of what is happening. The key thing is that the permanent magnet is not "doing work," rather, it's being stressed and storing some magnetic potential energy, and then when it is unstressed it's releasing that magnetic potential energy. There are no gains here.
Even though your diagram is relatively simple, if you had this problem in an electronics class, you would make a timing diagram where you have the switch closing and then opening, and you track all of various variables. If you constructed a timing diagram you should see how everything balances.
Another way to describe what is happening is as follows: When the electromagnet is off, the permanent magnet is responsible for the flux through the inductor core. Let's call that the unstressed condition of the core. When the electromagnet switches on, it "pushes out" half of the downwards flux inside the core due to the permanent magnet and sets up a counter stream of flux going in the opposite direction. Here we have the rare occurrence of a real true to life Bloch wall going down the center of the core. This is the stressed condition. It took work to create the stressed condition, and that's now stored magnetic potential energy. So the permanent magnet is being stressed and storing energy that the electromagnet put into it. When you cut the current to the electromagnet, the permanent magnet pushes back with its stored potential energy and "reoccupies" the inductor core with its flux and you are back to the unstressed condition.
It's all about accepting the notion that any permanent magnet can only appear to be a source of energy because someone or something else did some work to store potential energy in the magnet in the first place and the magnet can only give back as much as you put into it.
MileHigh
Hi MH
Well by now,you will know im not so good at drawing diagrams or schematic's,but do my best to show something that resembles what im trying to show.
In regards to your last paragraph Quote: It's all about accepting the notion that any permanent magnet can only appear to be a source of energy because someone or something else did some work to store potential energy in the magnet in the first place and the magnet can only give back as much as you put into it.
Well some where i have a device that shows different,and i have been trying to find it for the last 3 days. But as we not long ago shifted house,most of my junk stuff is still in boxes,and which box my setup is in?-i have no idea yet,as i havnt found it.
But where im at now was based around the results i got from that little setup. It was based around a mixture of Tom's MEG and the flynn device.
My results were as follows,and the setup was like the one i posted the sketch of.
With just the inductor(electromagnet)on it's own,it would consume !lets say! 300mWatts.
From our inductive kickback ,we could get back !lets say! 150mWatts.
Now with the secondary coil in placeTank coil),with a resistor acorss it.
The P/in to the inductor(electromagnet) rose to say 350mWatts.
Our p/out from the inductive kickback droped down to 130mWatts
And the power across the resistor from the secondary (tank)coil was say 20mWatts
So the total in P/out remained about the same,but the P/in increased.
Now with the horse shoe magnet on the other side of the secondary(tank)coil-as per diagram.
P/in droped back down to 300mWatts.
P/out from inductive kickback rose to 170mWatts.
P/out from the secondary (tank)coil rose to 35mWatts.
So from this we can see that by adding the secondary coil with a load across it,and the magnet,the P/in didnt change.
But we also see that the total P/out rose,and that was only possable with the PM in place.
By replacing the PM with a horse shoe shaped ferrite core,the P/in would rise once again to 350mWatts-infact it was actualy a higher value than that.
(Note-the above figures are to give an indication of the effect in the experiment,and may not reflect the actual values i had back the-memory just isnt that good.)
This told me that the PM could be the only source for the extra P/out,as the P/in was the same with or without the PM and secondary combination there.
Now when i say the PM was the only source for the extra P/out,this can be seen in two ways.
1st-the PM provided the extra energy.
2nd-the PM was the only thing that could make the transformer become more efficient.
But no matter which way you look at it-the PM was the source of that energy gain Wether is was providing energy,or makeing the transformer more efficient dosnt realy matter. What matters is that it dose work- the PM dose make the transformer more efficient than it would be with just the ferrite core there.
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@MH-and all
I am going to start from the begining,so as all can see what it is im trying to show with the magneformer. As much as i have looked,i cant find the original device,so i have made a new one from scratch. Im guessing that the original is in one of my junk boxes out on the farm,as there just wasnt enough room in my new work shop to bring everything to our new home.
The core halves im useing are from an old flyback transformer,from either a TV or cpu monitor.
I now will make the pulse circuit for it,and it will be driven by my home made(kit)SG,and P/in will be in way of an SLA.This is so as the scope ground can be placed anywhere on the device for measurements.
Pictured below is the transformer part of the device ,that i made today. Pic 1 shows the transformer in its seperate pieces,and pic 2 shows how it all fits together. This way we can test it with and without the PMs in place,which are strong N48 neo's.
Once up and running,the first video will be about checking our DMM's for accuracy,and show how to smooth out the pulses across the DMMs-so as they read accurate.
In pic 1,the core looks like it is broken and is bent inward toward the open end's. But that is just because it is lying on an angle,and it is actualy straight. The cores are ofcourse ferrite ! grade unknown!.
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Some results of the first run.
The neo's had to be swaped out for ferrite magnets,as they were far to strong.
Measurements are as follows after some fine tuning,and the resistor across the secondary coil is 81.7 ohms. Tank cap is not yet in place.
Without magnets
P/in=12 volts@ 7.79mA=.0934watts
P/out(flyback)=12.43 @ 4.45mA=.0553watts
P/out(secondary/resistor)=400mV/81.7 ohms=.00196watts
P/out total=.0553+.00196-.05726watts.
Efficiency is 61.3%
With magnets
P/in=10.4mA @12volts=.1248watts
P/out(flyback)=12.44 @6.8mA=.08459watts
P/out(secondary/resistor)=920mv/81.7ohms=.01036watts
P/out total=.08459=.01036=.09495watts
Efficiency is 76.08%
Both meters are of the same type,and were set at the mA scale,and large smoothing caps used on input and output.
Meters were then swaped over,and test ran again. The results are an average of the two test.
The two DMMs used have a .02mA difference.
The smoothing caps on the input were 1x 10 000uf high current cap,and two 4200uf caps.
Cap used on the flyback output was 1x 10 000uf high current cap.
No ripple detected across the two DMMs.
The first scope shot shows the device running without the magnets in place.
The second scope shot shows operation with magnets in place.
Scope shots taken befor fine tuned.
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Hey Gyula
If the pulsed primary input field is opposing the permanent magnet, the core wont be saturated till the field of the magnet is flipped/reversed and then pushed further into saturation, if the input can deliver that much opposition to the cores permanent field. ;D A ferrite mag can be demagnetized, or even reversed eventually, that is again if the input is enough. Brad isnt putting that much power in so saturation shouldnt be an issue and the magnets should last for quite some time.
Mags
Hi Mags,
Yes it is all okay what you wrote, I was mistaken in that I had believed literally the permanent magnet was really inserted into the coil while in fact the magnets were only attached to the cores but not inserted into the coil :)
Gyula
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Gyula:
I didn't know that a magnet would have such a low relative permeability. You still might be able to take advantage of the polarization though as has already been stated. Note the excitation from the primary coil is unidirectional. This assumes the relative permeability is radically different depending on the direction of the external field. It would be an interesting and easy test. You just have to look at the slope of the current rise in the coil for same-direction and opposite-direction magnetic field generation by the coil wrapped around the magnetic core. You cross your fingers and hope that you don't ruin the magnet.
I also wonder if the L-meter will be thrown off by the introduction of a magnet into the coil. I assume they sample or sweep low-level AC frequency excitation for the coil under test and check the response to measure the inductance. So if the magnet does indeed radically change in relative permeability depending on direction it may have a small heart attack (throw off the measurement algorithm). It probably will read out as a high inductance - my guess.
Another point is that this is a transformer setup, not an inductance. So assuming the core (any core) is coupling the energy properly, you _don't_ see inductance on the primary, you see the load, which is an LCR circuit. So you see a wobbling resistance! lol Note since you are approximately at resonance, you are pretty much seeing the resistive component of the LCR circuit as the load of the primary. So that means that Brad should see the voltage and current going into the primary winding as mostly in phase, assuming that the core/coupling is doing it's job properly to transfer the power.
MileHigh
Hi MileHigh,
Yes permanent magnets are almost fully saturated magnetically, many FE tinkerers are not aware of that and when they use magnets in a "closed" magnetic circuit where permanent magnets are used to "close" the magnetic path, they actually build an "air gap" into the circuit at places they insert the permanent magnet(s).
Well, L meters mainly use oscillators inside which excite the coil to be measured with saw-tooth like waveform so there are no abrubt amplitude change across the coil. Indeed you have to be careful when testing coils with magnets when using an L meter and move the magnet slowly in or out of the coil to avoid harmful induced amplitudes, to save the inside oscillator circuit. I did check air core coils with inserting different magnets into them and never had any malfunction in my L meter. The permanent magnet in itself when plugged into an air core coil does not change any of its original properties its permeability also stays the same. The L meter normally use a low power oscillator to drive the coils to be measured.
If you use a coil with ferromagnetic core and attach a permanent magnet to the core or just approach it closely with a magnet, then the permeability of the core can change a lot (it normally goes down) so the L meter shows a decreasing inductance value too. Taking sudden movements with the magnet causes the L meter to go out of range for some moments, then normally it returns to the new L value.
It is okay what you wrote in your last paragraph above, I would add that the load resistance (earlier the 18 Ohm, now the 81.7 Ohm) establishes the loaded Q of the output tank circuit, and if this load does not change there is no 'wobbling resistance'. If there is no load resistance then indeed the LC tank has its own parallel resonant impedance and it is a high value (several kOhm or even higher) wrt the 18 or 82 Ohm loads, and it is also a constant value (if the switching frequency is stabil).
Gyula
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Some results of the first run.
The neo's had to be swapped out for ferrite magnets,as they were far to strong.
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Hi Brad,
My comment would be that the magnets still reduce the permeability of the C core on which the electromagnet coil is wound, this explains the higher input power taken when the magnets are in place.
If you have an L meter, then you can check the inductance values first with the magnets in place (the setup is unpowered of course and leave the cores and the magnets possibly in the fine-tuned positions) and then removing the magnets you could adjust the core distances to arrive at the same L values for the coils like with the magnets and then test the input output.
Of course this may sound as an obsolote test now, the goal is to get more output with a reducing input, by using the magnets.
One more thing: if you happen to find with the L meter that the reducement of coils inductance (due to the magnets) is not the cause for the increased input power, then the explanation for the latter is what Magluvin also mentioned: more input is needed to flush out the PM flux from the core. It is also possible that both the decrease in inductance and the extra flux from the magnets are indeed the cause for the extra input. It would be good you would find your original setup of course and could check it again.
Gyula
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Magnetics and transformers and materials, it's a pretty big subject if you are hard core. I am too tired to comment on previous postings right now, but let me mention something interesting. I saw this thread on a science forum but didn't try to read it, it looked like way like too much work and my interest level was not high enough.
The subject was what happens in the magnetic core of a transformer under normal operation. It's not so obvious. Think of a good quality fairly large 1:1 transformer. You put a nice pure 60 Hz sine wave into it and there is a resistive load. We don't really need to worry about values.
Here is the thing: The primary current creates magnetic flux in the core. But the secondary also reacts in perfect synchronicity and its current also creates an equal and opposite magnetic flux in the core. Since we know that flux in one direction can be cancelled by equal flux in the opposite direction, then what is going on in the core? If there is no net flux in it, what's happening? What are the magnetic domains doing? Are they flipping or not flipping?
We know that each coil in the transformer is creating a "blast of flux." But between the two blasts there is a kind of mutually assured destruction going on and there is no flux. It's almost like electrons and holes in a diode smashing into each other and self-annihilating (in a figurative sense).
I honestly have never read up on this subject at all. I just saw the subject line and it got me thinking.
MileHigh
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Magnetics and transformers and materials, it's a pretty big subject if you are hard core. I am too tired to comment on previous postings right now, but let me mention something interesting. I saw this thread on a science forum but didn't try to read it, it looked like way like too much work and my interest level was not high enough.
The subject was what happens in the magnetic core of a transformer under normal operation. It's not so obvious. Think of a good quality fairly large 1:1 transformer. You put a nice pure 60 Hz sine wave into it and there is a resistive load. We don't really need to worry about values.
Here is the thing: The primary current creates magnetic flux in the core. But the secondary also reacts in perfect synchronicity and its current also creates an equal and opposite magnetic flux in the core. Since we know that flux in one direction can be cancelled by equal flux in the opposite direction, then what is going on in the core? If there is no net flux in it, what's happening? What are the magnetic domains doing? Are they flipping or not flipping?
We know that each coil in the transformer is creating a "blast of flux." But between the two blasts there is a kind of mutually assured destruction going on and there is no flux. It's almost like electrons and holes in a diode smashing into each other and self-annihilating (in a figurative sense).
I honestly have never read up on this subject at all. I just saw the subject line and it got me thinking.
MileHigh
MH
The flux in the secondary wouldnt be equal or opposite. Not equal due to ohmic losses,nor opposite. It would be a weaker field,and of the same field-apposing yes-opposite no. This is where transformer loss comes from.If one end of the transformers primary builds a north field,then the secondary field would be a north field aswell at that end-this is an apposing field,as opposite would be south. There is no equal in the magnetic fields either,as there is heat produced.If it was a 1 to 1 transformer,and was equal,then we would get out what we put in. But as we know we dont get out from the secondary what we put into the primary,we know the magnetic field isnt equal in both winding's.The output is less,and if we add the heat energy created by ohmic resistance to the output,we then have an equal amount of energy to that of what we put in.
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I think this paper is fairly close to the mark for the layman's needs.
Although it's not second nature to most of us it's not rocket surgery either, if we look in the right places the info is around.
I recommend that anyone interested in transformers and not already trained to read the entire paper from start to finish a few times and use it for reference.
Of course transformers that vary from the efficient power transformer he talks about will behave differently it's a good starting point especially if we want to design our own transformers to use.
Transformers - The Basics (Section 1)
http://sound.westhost.com/xfmr.htm
Preface
One thing that obviously confuses many people is the idea of flux density within the transformer core. While this is covered in more detail in Section 2, it is important that this section's information is remembered at every stage of your reading through this article. For any power transformer, the maximum flux density in the core is obtained when the transformer is idle. I will reiterate this, as it is very important ...
For any power transformer, the maximum flux density is obtained when the transformer is idle.
The idea is counter-intuitive, it even verges on not making sense. Be that as it may, it's a fact, and missing it will ruin your understanding of transformers. At idle, the transformer back-EMF almost exactly cancels out the applied voltage. The small current that flows maintains the flux density at the maximum allowed value, and represents iron loss (see Section 2). As current is drawn from the secondary, the flux falls slightly, and allows more primary current to flow to provide the output current.
It is not important that you understand the reasons for this right from the beginning, but it is important that you remember that for any power transformer, the maximum flux density is obtained when the transformer is idle. Please don't forget this .
3. How a Transformer Works
At no load, an ideal transformer draws virtually no current from the mains, since it is simply a large inductance. The whole principle of operation is based on induced magnetic flux, which not only creates a voltage (and current) in the secondary, but the primary as well! It is this characteristic that allows any inductor to function as expected, and the voltage generated in the primary is called a 'back EMF' (electromotive force). The magnitude of this voltage is such that it almost equals (and is effectively in the same phase as) the applied EMF.
Although a simple calculation can be made to determine the internally generated voltage, doing so is pointless since it can't be changed. As described in Part 1 of this series, for a sinusoidal waveform, the current through an inductor lags the voltage by 90 degrees. Since the induced current is lagging by 90 degrees, the internally generated voltage is shifted back again by 90° so is in phase with the input voltage. For the sake of simplicity, imagine an inductor or transformer (no load) with an applied voltage of 230V. For the effective back EMF to resist the full applied AC voltage (as it must), the actual magnitude of the induced voltage (back EMF) is just under 230V. The output voltage of a transformer is always in phase with the applied voltage (within a few thousandths of a degree).
For example ... a transformer primary operating at 230V input draws 150mA from the mains at idle and has a DC resistance of 2 ohms. The back EMF must be sufficient to limit the current through the 2 ohm resistance to 150mA, so will be close enough to 229.7V (0.3V at 2 ohms is 150mA). In real transformers there are additional complications (iron loss in particular), but the principle isn't changed much.
If this is all to confusing, don't worry about it. Unless you intend to devote your career to transformer design, the information is actually of little use to you, since you are restrained by the 'real world' characteristics of the components you buy - the internals are of little consequence. Even if you do devote your life to the design of transformers, this info is still merely a curiosity for the most part, since there is little you can do about it.
4. Interesting Things About Transformers
As discussed above, the impedance ratio is the square of the turns ratio, but this is only one of many interesting things about transformers ... (well, I happen to think they are interesting, anyway ).
For example, one would think that increasing the number of turns would increase the flux density, since there are more turns contributing to the magnetic field. In fact, the opposite is true, and for the same input voltage, an increase in the number of turns will decrease the flux density and vice versa. This is counter-intuitive until you realise that an increase in the number of turns increases the inductance, and therefore reduces the current through each coil.
I have already mentioned that the power factor (and phase shift) varies according to load, and this (although mildly interesting) is not of any real consequence to most of us.
A very interesting phenomenon exists when we draw current from the secondary. Since the primary current increases to supply the load, we would expect that the magnetic flux in the core would also increase (more amps, same number of turns, more flux). In fact, the flux density decreases! In a perfect transformer with no copper loss, the flux would remain the same - the extra current supplies the secondary only. In a real transformer, as the current is increased, the losses increase proportionally, and there is slightly less flux at full power than at no load.
MIT Lecture - Inductance.
http://www.youtube.com/watch?v=UpO6t00bPb8
A transformer designed to be efficient can be made to behave like a Thane Crimes transformer just by increasing the applied frequency. :) And it will behave woefully. See link.
http://www.youtube.com/watch?v=Zxde9qga79c Please bear in mind I am no expert and I get things wrong and say the wrong thing at times, but the video shows some stuff in my opinion. I made it quite some time ago. I know a little bit more now. It still makes me laugh, I think I am a funny guy, no ?
..
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About the issue of the relative permeability of a magnet, it may be possible to have a magnet that still retains a high relative permeability if it is only partially magnetized.
Yes permanent magnets are almost fully saturated magnetically, many FE tinkerers are not aware of that and when they use magnets in a "closed" magnetic circuit where permanent magnets are used to "close" the magnetic path, they actually build an "air gap" into the circuit at places they insert the permanent magnet(s).
Note the attached graphic showing how you can control how strong the nominal flux in the magnet is if you travel up and then down the BH curve and choose your path carefully. You can back off on the externally applied field strength and then slide down to a flux density level that is about 50% strength. The trick is to know what your maximum externally applied field strength is to then back off and settle around the 50% flux density level. Where I am uncertain is about the choices and associated properties for the core material. How well will the material retain its partial magnetization if you wrap a coil around it and pulse it? But at least in theory there is a recipe for doing it.
When it comes to the various types of commercial magnets you play with, I would assume that when they "bang" them to magnetize them, the flux density in the magnet is nearly or is 100% strength - they are fully saturated. So does that imply that the relative permeability is very low in both directions because the magnetic domains are 100% "occupied?"
With a lot of care I would assume that you could demagnetize a commercial magnet and then give it a flux density level of 50% by yourself. You would have to carefully tip toe up and then down the BH curve.
If anyone wants to do some reading "BH curve" and "tape head demagnetiser" would be a good launching pad.
MileHigh
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When it comes to the various types of commercial magnets you play with, I would assume that when they "bang" them to magnetize them, the flux density in the magnet is nearly or is 100% strength - they are fully saturated. So does that imply that the relative permeability is very low in both directions because the magnetic domains are 100% "occupied?"
Yes it does imply. Ferromagnetic materials intended for permanent magnets are magnetically "hard" materials with very high coercivity value, this means that once magnetized to a certain strength they tend to stay at that magnetized level, you cannot control its magnetization as readily and easily as in case of a soft ferromagnetic material. For such hard materials, when the H field is reduced back to zero (or even to an opposite value), the B field in the material does not get reduced but it remains 'remanent' at a certain level.
Here is a link to show the BH curve for a strong and a not so strong permanent magnet, I refer to Figure 3: http://www.electronenergy.com/magnetic-design/magnetic-design.htm (http://www.electronenergy.com/magnetic-design/magnetic-design.htm)
By the way the formula for a BH curve is B=u*H and u (u=permeability) has to be close to unity in case of a material with very wide hysteresis curve.
With a lot of care I would assume that you could demagnetize a commercial magnet and then give it a flux density level of 50% by yourself. You would have to carefully tip toe up and then down the BH curve.
Yes that would be a hard and arduous task for sure. I would suggest a much easier solution using the so called electro-permanent magnet which includes a soft and a hard magnetic core combination, the soft core serves for a normal electromagnet and the hard core is in fact a normal permanent magnet. By changing the current in the electromagnet the resultant combination of the PM and that of the electromagnet gives a variable and strong magnetic field: http://en.wikipedia.org/wiki/Electro-permanent_magnet (http://en.wikipedia.org/wiki/Electro-permanent_magnet)
You surely remember the Hildenbrand or the Flynn setups etc: all such magnetic circuits add magnetic flux from permanent magnets to that of electromagnets, albeit not always in a linearly controllable way but by switching. See these products:
http://www.magnets2buy.com/acatalog/Electro-magnets.html (http://www.magnets2buy.com/acatalog/Electro-magnets.html)
Gyula
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@MH-and all
I am going to start from the begining.....
I you wind the wire in a zig - zag manner as seen in your picture, you won't get good results. The flux produced in individual turns may not add up properly. Better if you remove a coil from old small generator or a 6 Watts / 12 watts eleminator (rectifier) which will have a small step down transformer in it. You can see how neatly the individual turns sit one above the other in correct alignment.
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I you wind the wire in a zig - zag manner as seen in your picture, you won't get good results. The flux produced in individual turns may not add up properly. Better if you remove a coil from old small generator or a 6 Watts / 12 watts eleminator (rectifier) which will have a small step down transformer in it. You can see how neatly the individual turns sit one above the other in correct alignment.
Hi Newton
For the purpose of the demondstration,the coils dont have to be neat,as the same loss(if any) will be encountered with and without the magnets in place.
Some time ago,a member here that knows his stuff,said you would only get back 50% at best !on the kickback! to what you put in. As you can see from the result's(which are very accurate),we are getting back 76.08% from the device-even with the messy hand wound coils. But even so,i would have to disagree that neat windings make for a better performance-in regards to these type of systems.I have unwound many inductors,and wound the wire back on neat,and never have i seen a performance increase. Infact i believe messy wound coils are better in this type of system,due to the increased capacitance of the coil.
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Hi Newton
For the purpose of the demondstration,the coils dont have to be neat,as the same loss(if any) will be encountered with and without the magnets in place.
Some time ago,a member here that knows his stuff,said you would only get back 50% at best !on the kickback! to what you put in. As you can see from the result's(which are very accurate),we are getting back 76.08% from the device-even with the messy hand wound coils. But even so,i would have to disagree that neat windings make for a better performance-in regards to these type of systems.I have unwound many inductors,and wound the wire back on neat,and never have i seen a performance increase. Infact i believe messy wound coils are better in this type of system,due to the increased capacitance of the coil.
Hi Tinman. Boost converters use the flyback or inductive discharge and can be over 90% efficient, so I don't get how a knowledgeable person could say we would get 50% back at most, I have a boost converter which can be over 90% efficient. In my opinion neat wound coils are better because they are more uniform and repeatable to closer values.
The trick is to reduce losses by keeping DC resistance to a minimum Thick wire and as least as is needed, also using good core material and diodes with proper clean switching.
Generally I use 1 mm magnet wire on iron powder cores for high frequency inductors, I parallel it for transformer primaries side by side to make flat conductors. For a primary coil or an RF inductor I never use less than 1 mm of wire, sometimes I use multiple strands of 0.5 twisted.
As the paper I linked shows, more turns does not mean more magnetic flux intensity so I don't buy into the more turns are always better for magnets argument, I would argue less losses are the key, as little loss as is possible. When building a transformer the primary coil needs to have a certain amount of turns for the core size for it to work so that it can idle with small input. Cancelling he self induction and so forth is good but it cannot nullify DC resistance losses.
A very interesting phenomenon exists when we draw current from the secondary. Since the primary current increases to supply the load, we would expect that the magnetic flux in the core would also increase (more amps, same number of turns, more flux). In fact, the flux density decreases!
I do agree if the one coil is used it can be wound however, it stays the same so it does not matter. If I was winding coils for a multi coil boost converter I would wind them as neat as is possible for me for practical purposes, so they can be made the same.
An electro-magnet could be excited by an all positive or all negative "Sine" type or sine looking wave couldn't it ?
My toroid with four transformers on it has thousands of tuns side by side neatly wound by hand, I put a layer of wax paper between each layer of turns.
Cheers
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Brad:
Looking at your scope shots in reply #61 it looks like your current waveform is upside down. Not a big deal, sometimes inverting a signal on your scope compensates for the "backwards" voltage you might see because of the where the scope ground reference is relative to the probe.
From the top level view you can see that the impedance of the circuit drops and the power consumption goes up when you add the magnet. So there is more power to go around.
Also, the coupling to your secondary may explain the relatively low efficiency of the circuit. Ideally you would mate end-on to the laminations for the secondary for both the primary and the magnet. Instead, you are connecting to the flat tops of the laminations and there are small insulating gaps between each layer. So the magnetic circuit to the secondary could be relatively poorly coupled.
If you are curious enough, and perhaps for your own satisfaction, with the aid of your digital scope you could make a full timing diagram for the two cases, without the magnet and with the magnet. If you are good with an image editing program it would be fairly easy to do. You could take screen captures of all of the signals and then paste them into a large composite image, one on top of the other, everything lined up in time. You do it for the voltages and the currents on the primary and the secondary, and for the flyback energy collection, with and without the magnet.
You are only looking at less than one-half of the picture right now. If you were to consider doing it, do it for the resistor only as the load on the secondary. Forget about the capacitor and keep it simple.
Once you have a complete set of timing diagrams for the two cases, then look at the waveforms and explain and understand what they mean. It may sound tedious and like it's a lot of work and it is.
MileHigh
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Just a few more comments about the setup. Again let's start with a resistor as the secondary load to keep things simple.
Everything I discuss below can be considered to be duplicated so you test for the two cases, without the magnet and with the magnet.
For starters, you could put a 100 Hz, and a 1 KHz sine wave into the primary and scope the unloaded secondary to measure your turns ratio.
Note when you energize the primary with the square wave driving the transistor input this is what is considered to be a pulse circuit. You pulse the primary on and then disconnect from it and then collect the remaining back-EMF energy in the primary coil.
So if you first try with no load on the secondary, the overall circuit load relative to the power supply looks like an inductance. You should see a relatively slowly rising current waveform. That's the rising part of the triangle wave. The slope of that waveform tells you the inductance of the circuit.
Interestingly, when you add a load resistor to the secondary. That also causes a linearly rising current waveform on the primary. So that would tend to imply that your actual observed slope is some combination of the slop due to the inductance plus the slope due to the load resistance. That suggests that the lower the load resistance on the secondary, the steeper the slope of the current waveform on the primary.
Naturally your transformer will have a coefficient of coupling associated with it, but I don't know a quick and easy way to test for that.
If you look at the circuit and make tests and measurements with some kind of strategy like I state above, it can all be figured out and understood. To understand what is really going on, you need a suite of test vectors, and not just one isolated measurement. As suite of measurements allow you to see trends as variables change and stuff like that.
MileHigh
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Magnetics and transformers and materials, it's a pretty big subject if you are hard core. I am too tired to comment on previous postings right now, but let me mention something interesting. I saw this thread on a science forum but didn't try to read it, it looked like way like too much work and my interest level was not high enough.
The subject was what happens in the magnetic core of a transformer under normal operation. It's not so obvious. Think of a good quality fairly large 1:1 transformer. You put a nice pure 60 Hz sine wave into it and there is a resistive load. We don't really need to worry about values.
Here is the thing: The primary current creates magnetic flux in the core. But the secondary also reacts in perfect synchronicity and its current also creates an equal and opposite magnetic flux in the core. Since we know that flux in one direction can be cancelled by equal flux in the opposite direction, then what is going on in the core? If there is no net flux in it, what's happening? What are the magnetic domains doing? Are they flipping or not flipping?
We know that each coil in the transformer is creating a "blast of flux." But between the two blasts there is a kind of mutually assured destruction going on and there is no flux. It's almost like electrons and holes in a diode smashing into each other and self-annihilating (in a figurative sense).
I honestly have never read up on this subject at all. I just saw the subject line and it got me thinking.
MileHigh
MileHigh
Are you interested ? Really ? Here I proposed experiment which will prove what is going on ! Nobody is interested....I have done my tests with 1:1 isolation transformer when JackNoSkills posted thread , but lack of experience and analog scope would not allow me to get final conclusions...
I'm tired writing again and again when nobody sees the importance of the experiment. I thought maybe gotoluc would be interested because his reactive power setup is very close but unnecessary complicated.
Ok, in short : it is about resonance power and transformer. If we have pure resistor (I assumed bulb , but that's not ideal however easy to measure lumens) - we can power it in parallel tank circuit down to almost zero input real power. That's easy part. Now what if tank circuit is on primary side of transformer and resistor (bulb) alone is on secondary ?
Please tell, why it is SO IMPORTANT to measure such circuit precisely !!!??????
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A NEW ATTACK ON LENZ: WARDFORCE:
What do you make of this?
http://www.energeticforum.com/renewable-energy/5173-new-magnetic-forces-steve-ward.html
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Forest:
Resonance is not a form of power that can be applied to the input of a transformer. In very general terms, resonance is often misunderstood by beginning experimenters and they attribute "amazing" properties to it when the truth is that with more understanding and knowledge, then it's not so special.
Please tell, why it is SO IMPORTANT to measure such circuit precisely !!!?
Because that's how you learn. Another problem is that people sometimes make two measurements and then they conclude that they have over unity or something that "conventional electronics can't understand or explain." It's simply not true. They only make two measurements and they ignore what is going on in the rest of the circuit which is a mistake.
If you are beginner and you really want to understand how a circuit works, then with your scope and some current sensing resistors and by making multiple measurements where you change certain parameters (eg: a load resistor) you can understand how your circuit really operates. If you are serious and determined you will look at every single voltage and every single current in the entire circuit. You will construct a timing diagram for the circuit and show how an event on one signal creates a response event on another signal. Nobody ever does it and that's just the way things are. The collective group progress is very very slow.
I will give you one example: How may times have you seen a discussion about selecting the correct base resistor value for a transistor on the forums? For me the answer to that question is zero. On the other hand, every time you use a transistor in a circuit, you can simply blindly do what someone tells you to do or you can blindly copy the value for the resistor in the schematic. What you should be doing is understanding how you select a base resistor value for your transistor.
Farmhand:
About flux in transformers, the document you linked to says this:
For example, one would think that increasing the number of turns would increase the flux density, since there are more turns contributing to the magnetic field. In fact, the opposite is true, and for the same input voltage, an increase in the number of turns will decrease the flux density and vice versa. This is counter-intuitive until you realise that an increase in the number of turns increases the inductance, and therefore reduces the current through each coil.
I am not comfortable with this statement. No matter what, the flux density is proportional to the ampere-turns and the relative permeability and the current. The author is implying a set of initial conditions - AC excitation of the primary and more turns equals higher inductance equals less current. However, more turns with less current is still multiplied out to give you ampere-turns.
If you just look at DC excitation for a second, then for the same DC current more turns gives you more flux density.
One thing for certain is that your document clearly reminds people that unchanging (call them "DC") magnetic fields do not affect an inductor or a transformer. So there are hundreds and hundreds of experimental clips out there that get this basic fact wrong. And I am NOT referring to certain special cases here, I am referring to the generalized case and the vast majority of the video clips fall into that category.
MileHigh
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Tinman:
The flux in the secondary wouldnt be equal or opposite. Not equal due to ohmic losses,nor opposite. It would be a weaker field,and of the same field-apposing yes-opposite no. This is where transformer loss comes from.If one end of the transformers primary builds a north field,then the secondary field would be a north field aswell at that end-this is an apposing field,as opposite would be south. There is no equal in the magnetic fields either,as there is heat produced.If it was a 1 to 1 transformer,and was equal,then we would get out what we put in. But as we know we dont get out from the secondary what we put into the primary,we know the magnetic field isnt equal in both winding's.The output is less,and if we add the heat energy created by ohmic resistance to the output,we then have an equal amount of energy to that of what we put in.
I am assuming that you saw in Farmhand's links that the discussion backs up what I had to say. There is no flux inside the core of an operating transformer while AC currents are flowing through the primary and secondary. However, power is clearly flowing through the device from input to output. A perfect mechanical analogy for a transformer is a transmission - two interlocked gears of the same or different sizes. As long as you are looking at the correct mechanical variables that correspond to the voltages and the currents then you can see how an electrical transformer and a mechanical transmission function the exact same way.
Also, to be more precise, if you factor in losses, then the losses are responsible for the small amounts of net flux inside a transformer core while it is in operation.
Think about a long straight wire carrying current and you know that there is a magnetic field around the wire. Where is the North? Where is the South? The answer is that there is no North or South. North and South are convenient terms to use when discussing a bar magnet, but the actual truth applies in the same way as for the wire. We can say that a bar magnet has a North end, but there is no such thing as a "North" magnetic field in reality.
When it comes to transformer cores, it's a similar deal as a long straight wire. There is no such thing as the primary winding building a North field in the core. What you can say as an example is that the primary winding will produce a clockwise then a counter-clockwise magnetic field (or magnetic flux) in the core as the current changes direction. There is no "North" part of the core, ever. There is no beginning or end to a circle or a closed loop.
MileHigh
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MileHigh
With my old analog 10Mhz scope I cannot do much without big experience and to get more experience I should have good tools and teacher who can shown and ANSWER questions. Sorry, I can't follow that path due to various reasons.
What I tried to do is to attract more experienced person , preferably with extended knowledge and having proper tools to answer some important questions.
AS it failed I will explain WHY I did it , talking is cheap. :-\
What everybody was told after some time of learning about resonant tank circuits is this :
1.
"it is not free energy , it just accumulate the impulses , so if impulses has enough energy the energy level accumulated in tank circuit rise . Sure, energy is oscillating in unloaded tank circuit, and with proper input frequency related to resonant frequency the input energy required to sustain oscillations is very low while the apparent power circulating in tank circuit is high. You however cannot tap this energy, because doing so will destroy resonant condition"
that is first sentence I would like somebody with better then my experience would correct....please
The next sentence is about transformer , and it is also told to newbies as one of first "laws":
2.
Transformer is electro-magnetic device transforming electrical energy from primary circuit to the secondary circuit through the magnetic field. Then it is explained how to electric energy in primary is converted into magnetic field which induce current in secondary winding.It is stated that magnetic field is produced by primary and then "converted" into current in secondary - I mean energy of magnetic field is used up to create secondary current. So while the electrons from primary do not pass to secondary then energy is passed.
I'm not sure if I described that concept clearly so also please correct me.
Now the essential point. We know we can re-use the same current in tank circuit to heat up resistance for example, with carefully matched resistance placed in the path of oscillating current of tank circuit at resonance much lower input energy is required to get the same heat output in resistor then when powered by simply DC. Of course assuming temperature stable resistance and not disturbed resonant condition.
Let's now place tank circuit on primary side of transformer when transfor,er primary winding is a coil for tank circuit and ca matched capacitor placed in shunt (in parrallel to the primary), when resistor is placed across seondary winding.
If we could match input power supply frequency to the tank circuit to minimalize input power (real) and get the same heat on the resistor on secondary we would have kind of contradiction.
According to 1 tank circuit do not create energy from vaccum and according to 2 current in secondary was created by passing energy from primary via magnetic field link. Both cannot be true in the same time.
That's why it is important to experimentally test such condition.
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Think about a long straight wire carrying current and you know that there is a magnetic field around the wire. Where is the North? Where is the South? The answer is that there is no North or South. North and South are convenient terms to use when discussing a bar magnet, but the actual truth applies in the same way as for the wire. We can say that a bar magnet has a North end, but there is no such thing as a "North" magnetic field in reality.
When it comes to transformer cores, it's a similar deal as a long straight wire. There is no such thing as the primary winding building a North field in the core. What you can say as an example is that the primary winding will produce a clockwise then a counter-clockwise magnetic field (or magnetic flux) in the core as the current changes direction. There is no "North" part of the core, ever. There is no beginning or end to a circle or a closed loop.
MileHigh
I agree and raised this point some time ago, when describing the magnetic actions in his rotating magnetic field converters and motors Tesla used the words or term "magnetic intensities" and not magnetic poles or polarities, I think it is the magnetic intensities that rotate in the annular core. It depends on which way the rotation happens and how the coils are wound in relation to that as to the effect of the induction.
As for the turns and magnetic field intensity, if in both cases the wire is a standard size and the supply is a fixed voltage AC then more turns and more inductance means less current. I did mention that he was referring to regular efficient power transformers and to vary from that scenario will see different needs, wants and results.
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And a related comment: In the Bedini enthusiast world there is a debate about the rotor magnets. Should they be North facing out or South facing out? Some people claim that one is better than the other. I believe that even Bedini himself has made comments expressing favouritism for one pole over the other. It's assumed that when people do experiments and observe differences it's attributable to other factors, including the human factor. If you are pushing yourself too much and expect to see a difference then you risk being like the guy that was convinced that his horse could do additions and subtractions.
There is no difference between North facing out and South facing out beyond the expected differences. For example, the EMF induced in your coil will be opposite. Some beginners will rewind the coil from clockwise to counter-clockwise to compensate not knowing that all that they have to do is cross the drive coil wires.
MileHigh
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Actually TK and I, and some others, had done some experiments a few years back that did show some differences between N and S poles. Maybe he still has a video up on that subject. ;)
Mags