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Overunity Machines Forum



Effects of Recirculating BEMF to Coil

Started by gotoluc, July 02, 2009, 06:24:29 PM

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0 Members and 2 Guests are viewing this topic.

gotoluc

Quote from: poynt99 on July 05, 2009, 12:57:06 PM
Luc,

A DC-DC down converter is handy when you need to be able to vary your output voltage between the ranges of your supply and almost 0V. You do this of course by changing the pulse width. 100% will yield your full 170V, and 0% will yield 0V output. This is indeed what you've built. Try increasing the pulse width and watch your magnet rise even further. It should do this continuously until either your supply gives out due to current limitations, or your coil burns up.

As a test, connect your coil directly to your variable DC power supply and begin at 0V. Increase the voltage until the magnet levitates to the same height you had with the PWM supply you built, and I am sure you will find that the current and voltage from the DC supply is very close to the numbers I gave above, i.e. ~3VDC @ ~300mADC.

The only advantage gained from a switching power supply (i.e. the PWM supply you have built with your 555, switch and coil) and a linear variable DC power is efficiency, IF you need to vary the supply voltage. If for example you only needed a fixed 3V @ 300mA supply, it would be more efficient and practical to design a linear supply tailored to the application.

I am not sure what else you will want to do with your circuit, but if you want an efficient way of varying the strength of the magnetic field from your coil, then you have accomplished that. Linear supplies become inefficient when you turn the voltage quite low and draw high current, but at full voltage they are just as efficient as a switching power supply, or at least very close.

I've attached two scope shots matched to the shots wattsup posted, as it seems you have doubts that I did any testing. It's the best I can do with the time I have available, but I can assure you that the information is accurate. The yellow trace (coil voltage) scale is the far left, while the green trace (gate voltage) scale is to the right (-2V to 30V).

I encourage you to perform the measurements I did in addition to the DC power supply test. Measuring the coil voltage and current should be straight forward for you, and you will see that indeed you've converted 170V to about 3V or so, and 10mA to about 300mA or so.

Regards,
.99

Hi .99,

sorry you feel I was doubting you actually doing the test research since that was not my intent. I was just joking around since I'm not good at making reports like you have presented. It would take me a few pictures at least. You did a great job ;)

You are also 100% correct that this circuit is not making a coil perform magnet levitation height more efficiently then using strait DC as far as current and voltage are concerned. I have tested what you mention above before starting this topic and have found that strait DC to be slightly more efficient which demonstrates that the circuit has losses (heat at the switch). I'm not claiming either to have found the holy grail of electronic circuits as I don't believe there is one.

What I'm trying to communicate here is that an air core coil inductive kickback has more punch on a permanent magnet then the energy used to create it.

That's it :D

I'm also suggesting that using an electronic component to do the switching (even though I used it in the video test 2) is not the ideal switch to use in this application even tough it shows an interesting effect.

You are a fast and smart individual as I can see and it is good to be honest and looking at this from all angles as a good show can be deceiving if we don't look at it all.

So the question is, can we use this inductive kickback energy to power a motor more efficiently then using strait DC will remain unknown I would say until we build and test it.

Thanks for sharing

Luc

poynt99

Quote from: gotoluc on July 05, 2009, 04:00:23 PM

What I'm trying to communicate here is that an air core coil inductive kickback has more punch on a permanent magnet then the energy used to create it.

Luc

Luc, I've been around long enough to know when to "throw in the towel" so to speak, so I will graciously bow out of this one. I'm not here to aggravate anyone, and there comes a time when folks just need to find the facts themselves rather than hearing them from others, and that is fine :) .

I am not 100% certain of what you mean be the quote, but if it is what I think it is, it goes directly against my published numbers up at the top of this page in my findings. Either that or I just don't understand it  :D

Regards,
.99

PS. Always look forward to your new videos ;) They are top notch!
question everything, double check the facts, THEN decide your path...

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gotoluc

Okay .99,

I understand and I'm sure you understand what I'm talking about.

I'll try to build something that can somewhat work on this principle and you can enjoy the video show ;D when I post it.

Thanks for your time and willingness to help out.

Luc

petersone

Hi Luc
Your tests etc.are interesting,doe's the mag.behave the same if it's north or south pole on the coil?
peter

gotoluc

Hi Peter,

well, no since one side would be in attraction mode but yes! if I change the coil polarity.

Luc