Pierre's 170W in 1600W out Looped Very impressive Build continued & moderated

[listener192]

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

[listener192]

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

[d3x0r]

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]

[onielsen]

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

[seaad]

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 !!