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Author Topic: Pierre's 170W in 1600W out Looped Very impressive Build continued & moderated  (Read 184961 times)

Offline d3x0r

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@gotoluc


you're making a miscalculation in your head...


input + recovered = energy that was used through the device.


recovered doesn't go nowhere; it goes back into power to be used through the coils...
100% recovery would take your input down to 0...


but if you had input of 10W and recovered 30W , that's 40W to the coils...
if you didn't recover any... you'd get 40W input and 0W recovered...


It's not input - recovered... though recovered does lower the input; the input is already lowered by it when you read the input.


so you could say 'The input is now 10W, and we're recovering 8, so the input *would be* 18W" so you're saving 8 watts of input by recovering the output... but it's still 10W input.

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Offline gotoluc

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@gotoluc


you're making a miscalculation in your head...


input + recovered = energy that was used through the device.


recovered doesn't go nowhere; it goes back into power to be used through the coils...
100% recovery would take your input down to 0...


but if you had input of 10W and recovered 30W , that's 40W to the coils...
if you didn't recover any... you'd get 40W input and 0W recovered...


It's not input - recovered... though recovered does lower the input; the input is already lowered by it when you read the input.


so you could say 'The input is now 10W, and we're recovering 8, so the input *would be* 18W" so you're saving 8 watts of input by recovering the output... but it's still 10W input.
That's what you think I'm doing but the fact is I'm just showing the differences as it gets more efficient when frequency goes up.
It has nothing to do with what you wrote.
Regards
Luc


Offline d3x0r

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That's what you think I'm doing but the fact is I'm just showing the differences as it gets more efficient when frequency goes up.
It has nothing to do with what you wrote.
Regards
Luc


so for input of 4.3 wats and rercovered 4.3 watts it's still 8.6 watts total used... and only 50% recovered.  not 90% like you claimed.


again, the input already has the recovered taken away from it when you read it; so you can't say it's any less than what it already is.


90% recovery owuld be if you had like 9 watts on the recovered side and 1 watt on the input.

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Offline gotoluc

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so for input of 4.3 wats and rercovered 4.3 watts it's still 8.6 watts total used... and only 50% recovered.  not 90% like you claimed.


again, the input already has the recovered taken away from it when you read it; so you can't say it's any less than what it already is.


90% recovery owuld be if you had like 9 watts on the recovered side and 1 watt on the input.

I don't dispute anything you wrote, all is correct. What's incorrect is you assume I don't know this.
If you're a trained EE then the way a say things won't sound right to you as I have no schooling.
However, it's good that you point this out for others that will assume differently.
Regards
Luc

Offline partzman

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so for input of 4.3 wats and rercovered 4.3 watts it's still 8.6 watts total used... and only 50% recovered.  not 90% like you claimed.


again, the input already has the recovered taken away from it when you read it; so you can't say it's any less than what it already is.


90% recovery owuld be if you had like 9 watts on the recovered side and 1 watt on the input.

Whaaat!?  Wait a minute, let's reason together.  We have a black box that we connect to a fixed power supply.  The box has three ports that is, an input, output, and common.  The input and output both connect to the positive terminal of said power supply and the negative lead of the supply connects to the common terminal.  We now attach a watt meter to the input and measure 1 watt of power drawn from the supply.  We also attach a watt meter to the output and we measure 9 watts returned to the same supply.  Our black box is somehow giving us an 800% gain not 90% recovery.

Regards,
Pm

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Offline d3x0r

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Whaaat!?  Wait a minute, let's reason together.  We have a black box that we connect to a fixed power supply.  The box has three ports that is, an input, output, and common.  The input and output both connect to the positive terminal of said power supply and the negative lead of the supply connects to the common terminal.  We now attach a watt meter to the input and measure 1 watt of power drawn from the supply.  We also attach a watt meter to the output and we measure 9 watts returned to the same supply.  Our black box is somehow giving us an 800% gain not 90% recovery.

Regards,
Pm
the measure isn't on the common.  It's not 800% gain.  It's 9W from one side + 1W on the other side, one can know that 10W is used on the input common.


The recover is going to the caps, the power supply is going to the caps, if the power supply doesn't have to apply that much power (1W) because most of what was used was returned by recovery....


very similar to the way a resonant circuit is; in a tank you can have KW of power for a few watts of input...


At initial power on in the 90% recovery case... the power supplied would have to be 100% (10W), after the first cycle of recovery, 9W returns from the coils, which leaves 1W required from the power supply.


There's no gain... it's still no more than the required 10W total.


@gotoluc :) just makin sure... I did hope you knew... but that's not exactly the words you were sayin; sorry to seem like a grammar nazi :)

Offline gotoluc

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@gotoluc :) just makin sure... I did hope you knew... but that's not exactly the words you were sayin; sorry to seem like a grammar nazi :)
No problem :)

So I decided to add a switch on the combined 30 diode output lead I use for the clamp amp probe to show the difference when we switch off the return but now I'm confused because I thought the input power would go up when I switch off the diode collection but it only has a small increase. Not sure what's up with that?

Have a look:  https://youtu.be/9ELVELKx4LU

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Offline listener192

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No problem :)

So I decided to add a switch on the combined 30 diode output lead I use for the clamp amp probe to show the difference when we switch off the return but now I'm confused because I thought the input power would go up when I switch off the diode collection but it only has a small increase. Not sure what's up with that?

Have a look:  https://youtu.be/9ELVELKx4LU


Hi Gotoluc,


How can you turn off recovery the upper and lower switch diodes are the MOSFET body diodes you can’t turn them off, all you can do is look at the bridge board supply with the current clamp to see the current pulses returned to the caps when the MOSFETs turn off.


Regards


L192


Offline partzman

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the measure isn't on the common.  It's not 800% gain.  It's 9W from one side + 1W on the other side, one can know that 10W is used on the input common.


The recover is going to the caps, the power supply is going to the caps, if the power supply doesn't have to apply that much power (1W) because most of what was used was returned by recovery....


very similar to the way a resonant circuit is; in a tank you can have KW of power for a few watts of input...


At initial power on in the 90% recovery case... the power supplied would have to be 100% (10W), after the first cycle of recovery, 9W returns from the coils, which leaves 1W required from the power supply.


There's no gain... it's still no more than the required 10W total.


@gotoluc :) just makin sure... I did hope you knew... but that's not exactly the words you were sayin; sorry to seem like a grammar nazi :)

Who said anything about measuring the common?  How do you account for 10W total input?  Refer to the attached schematic.  This is an example of my understanding of Luc's measurements but using your values.  Using the current flows shown, we consume 1 watt from the supply via the input, and we return 9 watts via the output thru the recovery diode to the supply.  Thee black box is magically generating
8 watts over and above the input consumption of 1 watt.

Regards,
Pm   

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Offline gotoluc

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Hi Gotoluc,

How can you turn off recovery the upper and lower switch diodes are the MOSFET body diodes you can’t turn them off, all you can do is look at the bridge board supply with the current clamp to see the current pulses returned to the caps when the MOSFETs turn off.

Regards

L192

Yes, I forgot about the mosfet body diode. So basically when I shut off my 30 external quality diodes the mosfet's body diode kick in which is not quite as good quality and why we see a slight increase in input current.

Thanks for the reminder

Regards
Luc

Offline cheors

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"réfléchissez 36 bobine peuvent être relier avec une autre"

My translation as i understand Pierre:
Think again, 36 coils could be (possibility, probability) connected to one more coil (a 37th coil inside the core ?)
Remember the wire(s) going directly to the core not through the 5 green connectors !
And  remember 37 transistors !!

just a thought

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Offline gotoluc

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"réfléchissez 36 bobine peuvent être relier avec une autre"

My translation as i understand Pierre:
Think again, 36 coils could be (possibility, probability) connected to one more coil (a 37th coil inside the core ?)
Remember the wire(s) going directly to the core not through the 5 green connectors !
And  remember 37 transistors !!

just a thought

Thanks cheros,

you're right, looking at Pierre's French writing again, the word "réfléchissez" is the action of thinking over something which could be translated as "Think about it" or as you wrote "Think again"

You're also right that the word "peuvent" may be more related to the word "could" instead of "can" as I wrote

So I translated :

Fr. et réfléchissez 36 bobine peuvent être relier avec une autre

to En. and reflect 36 coils can be connected with another

but it could also be:  En. and think about it, 36 coils could be connected to one more

Notice I did not add the word coil at the end as he did not write that.

Thanks for bringing this up and you or anyone else is free to point out translation options as it can or should I write could change the outcome,

Regards
Luc


 

Offline hanon

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Maybe you could get some good ideas from the next link:


https://figueragenerator.wordpress.com/my-interpretation/


According to that, the key is to distort the magnetic fields in order to bend the magnetic lines and create flux cutting induction. Creating two fields in repulsion is one possible way to distort their magnetic lines and get the wires cut by those magnetic lines.

Offline d3x0r

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Who said anything about measuring the common?  How do you account for 10W total input?  Refer to the attached schematic.  This is an example of my understanding of Luc's measurements but using your values.  Using the current flows shown, we consume 1 watt from the supply via the input, and we return 9 watts via the output thru the recovery diode to the supply.  Thee black box is magically generating
8 watts over and above the input consumption of 1 watt.

Regards,
Pm


In1 + In2 = InT... as Recovey goes up, (never to exceed in1 form the powesupply) ... nevemrind just rehashing that point...


The measuements ae the common voltage top of the capcitor, current from power supply, and currrent from recovery diodes.  Power at InT Is not erally measurable because it's thorugh a multitude of bus lines and not from a single source.

Offline pmgr

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Yes, I agree and this is what I was trying to establish and test: https://www.youtube.com/watch?v=WxFxnVld9-8  before winding the coils but I made errors and was way off in my calculations and no one helped since they probably didn't understand what I was trying to do.
RegardsLuc
Luc, we discussed this before. Based on the measurements you did, your coils have a resistance of 0.53ohm and an inductance of about 0.56mH. When in registration with the rotor, this increases with about a factor of 4, so about 2.1mH.

The time constant of a coil is L/R and cut-off frequency (3dB point) is fc=R/(2*pi*L). Filling in the above values gives fc=0.53/(2*pi*2.1e-3)=40Hz.

Considering your stator steel was probably designed for 50-60Hz, the fc of 40Hz that you have for your coils is not bad. In fact it is in the correct ballpark.

The questions is why we are not seeing more voltage on the output coil. There can be two answers to that:

1. You don't have enough windings on the output coil. Solution: try and output coil with more windings, like a factor of 5 or so.
2. The output rotor is too thick and shorting part of the two neighboring poles when they pass the rotor. Solution: use a thinner rotor.

Or do a combination of both. Hope this helps.

PmgR

 

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