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Author Topic: Pierre's 170W in 1600W out Looped Very impressive Build continued & moderated  (Read 254092 times)

Offline listener192

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Not only are metal contacts bi-directional they also allow for simultaneous events to pass through them, and they can offer almost no resistance.

If you use  a  MOSFET as the low side switch, and set the RDS the same as the relay contact resistance (which you can do in the latest version of Falstad),
you get a quiet a different recovery current waveform, through the two diodes on the end of the switched series of coils.

Not sure if the simulator assumes a body diode in the MOSFET.

Regards
L192

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Offline listener192

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Thanks Cheors,


I've attached the loading sequence video to review for you as well. Play in loop.   This is what I'm seeing and hearing before his loop sequence begins with the two high steps which is the beginning of his loop code.



pin.high(20,21) '' RELAY (39, 52) AND (41 AND 54)
short delay
pin.high(01)      '' RELAY 01 AND 14 ON
short delay
pin.high(14)      '' RELAY 25 AND 38 ON
long delay
pin.low(14)       '' RELAY 25 AND 38 OFF
long delay
pin.high(14)      '' RELAY 25 AND 38 ON
long delay
pin.low(14)       '' RELAY 25 AND 38 OFF
regular delay


Then....we get the loop sequence with the double high. 
_____________________________________________________ 


I believe he is priming the coils, this has to be important. Pin 14 is turned on/off twice while Pins 1,20, and 21 are held.  WHY TWICE?

LOOK AT THE PATTERN.

52-39 =13
14-01 =13
54-41 =13
38-25 =13

We can confirm that each PIN runs two switches 13 apart. 

Jerdee

The naming convention appears to  take the clockwise slot...
junction 36-1 =1
junction 6-7   =7
junction 12-13 =13

Same polarities spaced 13 apart, ties up with the original scheme shown.

L192

Offline d3x0r

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Okay so there is a skip then between the first and last coils... (in original code)




This is what I thought at first glance... and that the single north went from (right to left) when the cycle restarted...

so - wouldn't 6 poles have short fields?  except maybe where the pickup coil core is?




(1 cycle, 1 coil (and it's mating coil) )



Edit : (Added 'Overlap Tangle" )
And I think maybe, the pitch on 36 should be 5, or on 30, should be 4, so that you don't overlap driving coils... otherwise you get a small tangle that doesn't do anything except add impedance....


If you had a 4 skip, then 2 coils on is 5, and will never directly overlap a south that'[size=78%]s also on... (not illustrated)[/size]

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Offline onielsen

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Hi seaad,

5A through the resistor makes it dissipate (5A)2 x 4ohm = 100W as heat with 4ohm x 5A = 20V across it. 8A through the resistor dissipates (8A)2 x 4ohm = 256W as heat with 8A x 4ohm = 20V across it.

Assuming that the first column of numbers is the voltage across the left capacitor and the second column is the voltage across the right capacitor making the third column in the box the voltage across the resistor. The last column then can't be the current through the resistor! Are you sure that this isn't the current  on the other side of one of the capacitors?

Next assumption is that the capacitors are super capacitors. Super capacitors mustn't be charged to above their breakdown voltage (as all other capacitors). Thus super capacitors in series have a charge balancing circuit to protect each cell in them and this probably also protects for over voltage of the total series of cells. If the voltages and currents are without any load the input power is dissipated in the resistor but also in the protection circuit of the capacitors. If a load is connected it also becomes a source in the second row where the current changes direction to a negative value. Here the energy comes from the right hand side. As the bridge rectifier prevents the current from going into the transformer the current must be dissipated in the left hand side capacitor as well as in the right hand side.

If the voltages in the third column are the voltage across the resistor the current through the resistor isn't as shown. The current (I) is given as I = U / 4ohm where U is the voltage. The fourth column then becomes:
@ 5V:  1.25A
@ -7V: -1.75A

@ 12V: 3A
@ 0V:   0A

@ 25V: 6.25A
@ 13V: 3.25A

 The power dissipation P in the resistor is P = U2 / 4ohm where U is the voltage across the resistor. The power dissipation with the above values becomes:
@ 5V:  6.25W
@ -7V: 12.3W

@ 12V: 36W
@ 0V:   0W

@ 25V: 156W
@ 13V: 42.3W

Perhaps the current is measured before or after the capacitors and thus also includes the current into the protection network of the capacitors. Just make sure to check how much power the protection network can dissipate before activating it.

Regards
Ole

Offline seaad

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Hi Ole, I have revised my pic in my previous post. I hope this below is less baffling.

We know that the transformer "takes" 2 Amp when Pierres apparatus is running (from the wall) and we know that the voltage across the supercaps  (feeding his unit directly??)  is about 20 Volt.

 The transformer has to deliver 8 Amps with its voltage ratio if it is a normal transformer with normal losses.

The 4 Ohm series resistor consumes 32 Volt at 8A.

The transformer after the diode bridge and filter capacitor (left) delivers some 25 to 36 Volts which enables ONLY a deliverance of 5 to 12 Volt accross the series resistor.
 Which consumes as mentioned above = 32Volt !!
???

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Offline pmgr

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Can somebody solve this riddle?

Pierre; "the resistor is really hot"

1]  Check in first film at 2min. 40sec. Start the charging of the supercaps (4.6 Volt). About 0.5 Ampere from the 'wall' goes to the transformer.

2]  Check in film at 3min. 40sec. The supercaps 25.1Volt with 0 to 0.1 Amp in to the transformer.

3]  Check in film at 5min. 06sec. Arduino have just Start the program, The supercaps <25Volt while 1.5--2 Amps is Now feeding the transformer!!
Hi seaad,


1) This is indeed a bit strange, if the supercaps are low in voltage, I would expect a large inrush current, but maybe the supercap boards have some current limiting.


2) This would mean this 25.1V is the max voltage that the transformer can put out after rectification and the caps are charged up to that so current draw goes to zero. This would be the peak voltage of the secondary.


3) I believe Pierre states in his first video that the transformer is home made. Mains is 170Vpp, secondary is 25Vpp, so winding ratio is 6.8, or 1.5A at primary becomes 10A at secondary which would cause 40V voltage drop over resistor... so indeed this doesn't make any sense... ??? Only thing I can think of is that the primary current meter is not reading correctly.


PmgR

Offline seaad

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Hi seaad,
1) This is indeed a bit strange, if the supercaps are low in voltage, I would expect a large inrush current, but maybe the supercap boards have some current limiting.

2) This would mean this 25.1V is the max voltage that the transformer can put out after rectification and the caps are charged up to that so current draw goes to zero. This would be the peak voltage of the secondary.

3) I believe Pierre states in his first video that the transformer is home made. Mains is 170Vpp, secondary is 25Vpp, so winding ratio is 6.8, or 1.5A at primary becomes 10A at secondary which would cause 40V voltage drop over resistor... so indeed this doesn't make any sense... ??? Only thing I can think of is that the primary current meter is not reading correctly.
PmgR

From first film: 26 Volt only not p-p? , 30 A 

That current limiting seems to not work when arduino up and running!   2Amps  in to the transformer then.

Regards Arne

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Offline onielsen

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Hi seaad,

The numbers on the figure doesn't add up.

Quote
3]  Check in film at 5min. 06sec. Arduino have just Start the program, The supercaps <25Volt while 1.5--2 Amps is Now feeding the transformer!!
I agree with PmgR that the meters aren't correct calibrated for the waveform. The output of the device isn't a pure sine wave and it is of low frequency. Cheap meters usually measure mean value and thus are calibrated to show the correct RMS value for a sine wave. As the form factor of the wave isn't the same as that for a sine wave such meters will show the wrong value. Even RMS meters and true RMS meters may not be correct calibrated. See the following articles about the problem: https://meettechniek.info/opinion/true-rms-or-a-true-lie.html and https://meettechniek.info/multimeter-avo/measurement-deviation.html.

An oscilloscope would be preferable for the measurements or the current could be measured through the resistor which is put between the two filter capacitors. At this point the waveform must be close to DC.

Regards
Ole

Offline jerdee

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The naming convention appears to  take the clockwise slot...
junction 36-1 =1
junction 6-7   =7
junction 12-13 =13

Same polarities spaced 13 apart, ties up with the original scheme shown.

L192


Funny that the image shows that he wants to solve the problem with 4 poles instead of 6.  In order for him to do this he would have to replace each coil to a 9 pole bridge instead of his 6.  Always look at the pattern.  For a 4 pole to work correctly, you would need to trigger 9 coils in series with a bridge gap of 9. Do you see his pattern?   This means we might have problems doing our setup with the same bridged poles as the 36 pole.  It should be 5 to match his same configuration.  [size=78%]I beginning to realize why the overlap in coil windings will be a problem.  I can see how his new configuration will work better!

I also believe we could do this method of pulsing on a toriod completely solid state. 

I was also able to place the priming sequence of coils to visualize the arrangement.  My original post had a wrong pin number, 14 should have been 13.  The reason for this PIN change is because all odd relays are on the outside and even relays are on the inside.  See image.  I still don't understand why he is priming the entire stator prior to sequence.  Maybe this image will help others.   

Jerdee

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Online r2fpl

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Rotate 6 coils = 6 poles

https://youtu.be/EtuZJFKu3uQ

Offline d3x0r

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Hi seaad,


1) This is indeed a bit strange, if the supercaps are low in voltage, I would expect a large inrush current, but maybe the supercap boards have some current limiting.


2) This would mean this 25.1V is the max voltage that the transformer can put out after rectification and the caps are charged up to that so current draw goes to zero. This would be the peak voltage of the secondary.


3) I believe Pierre states in his first video that the transformer is home made. Mains is 170Vpp, secondary is 25Vpp, so winding ratio is 6.8, or 1.5A at primary becomes 10A at secondary which would cause 40V voltage drop over resistor... so indeed this doesn't make any sense... ??? Only thing I can think of is that the primary current meter is not reading correctly.


PmgR


(Videos are gone from Pierre)


(You couldn't have > 6A ... E=IR (I=E/R) not I > E/R )


so 25V input across 4 Ohms (count 0 Ohms for capacitors?)  is max 6.25A


140V:25V = 1.1A : 6.25A.  that is input has to be at least 1.1A doesn't mean it couldn't be more...


suppose it's a lossy transformer? ...

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Offline seaad

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Hi seaad,

1) The numbers on the figure doesn't add up.

2) I agree with PmgR that the meters aren't correct calibrated for the waveform. The output of the device isn't a pure sine wave and it is of low frequency. Cheap meters usually measure mean value and thus are calibrated to show the correct RMS value for a sine wave. As the form factor of the wave isn't the same as that for a sine wave such meters will show the wrong value.

3)
An oscilloscope would be preferable for the measurements or the current could be measured through the resistor which is put between the two filter capacitors. At this point the waveform must be close to DC.

Regards
Ole

1)   The numbers on the figure doesn't add up. Correct. That was the purpouse of my post.   I. e.  An eye  opener.

2)   I agree with YOU and PmgR that the meters (maybe) aren't reliable. At this time in the film as I'm refferring to  Pierre is still using the 60Hz wall outlet with a good sine wave. Both when charging the super caps and running his unit.

3) Fully Agree.

@d3x0r Quote: (Videos are gone from Pierre)      Strange?
                           
                           (You couldn't have > 6A ...    Certainly.  Pierres transformer: 26V (30 A)  Peak value 35.7 Volt maximum rectified level.


Regards Arne
« Last Edit: April 22, 2018, 12:58:22 PM by seaad »

Offline listener192

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Funny that the image shows that he wants to solve the problem with 4 poles instead of 6.  In order for him to do this he would have to replace each coil to a 9 pole bridge instead of his 6.  Always look at the pattern.  For a 4 pole to work correctly, you would need to trigger 9 coils in series with a bridge gap of 9. Do you see his pattern?   This means we might have problems doing our setup with the same bridged poles as the 36 pole.  It should be 5 to match his same configuration.  [size=78%]I beginning to realize why the overlap in coil windings will be a problem.  I can see how his new configuration will work better!

I also believe we could do this method of pulsing on a toriod completely solid state. 

I was also able to place the priming sequence of coils to visualize the arrangement.  My original post had a wrong pin number, 14 should have been 13.  The reason for this PIN change is because all odd relays are on the outside and even relays are on the inside.  See image.  I still don't understand why he is priming the entire stator prior to sequence.  Maybe this image will help others.   

Jerdee


Yes, a larger stator with more slots, so you can reduce the polarity overlap would improve efficiency, still his scheme produces OU irrespective of its faults.

The priming issue is reminiscent of the permanent magnets in an alternator rotor, which are there to provide some output from the stator coils, which in turn energize the rotor with DC, starting the whole current generation process.

Regards


L192   

Offline listener192

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This was the FWBR used by Pierre.

Crydom EFG13C  125A continuous, 2500V

Somewhat overkill due to the 4ohm current limit resistor.

L192

Offline d3x0r

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Falstad Sim


current through 2 coils in series (with simulated 10 ohm resistor) is 2x the current per coil  (4x total, since both get doubled)
It's also 4x the power disappated across each coil (8x total)


Magnetic field is L * I * coilParams; so for L being the same, being able to increase current(I) will increase the field.

 

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