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

[oscar]

Hi all,
please note the number of wires coming from Pierre's induced coil.
The attached image is an enlarged detail from his 1st video at ca. 6:30

Hello Pierre,
please consider how Linus Torvalds made an operating system for computers and open sourced it.
Today he has thousands of collaborators who provide a free operating system - an alternative to the products of the military industrial complex.

[listener191]

Hi Luc,

I don't have time today to make a drawing but a description should work as well.

Each coil has two leads, input lead A and output lead B, so e.g. Coil 1 has leads C_01A and C_01B, coil 2 has leads C_02A and C_02B, etc.... Coil 36 has leads C_36A and C_36B. The leading C_ stands for Coil. We will use H_ for the H-bridge and A_ for the Arduino.

The Arduino has 54 digital I/Os. We are going to use only 36 of them (for 36 coils): A_01 through A_36.


Each of the half bridge has 6 inputs: Inputs 1 through 4 and Enable A and Enable B. You can leave the Enable A and B jumpers in place (so they are connected to Vcc and HIGH). Input 1,2,3,4 should all be connected together as we are going to activate both the high (Vcc) FET and the low (GND) FET at the same time to activate a set of 6 coils. You already have all coils connected in series on the stator, so you can leave that in place.


So let's call the four inputs that are connected together H_01I for bridge 1, H_02I for bridge 2, etc. through H_36I. The I stands for input.

The H-bridge has 4 outputs. Output 1 and 2 are connected together and Output 3 and 4 are connected together (for double amps). Let's now assign these two resulting outputs as H_01A (for output 1-2) and H_01B  (output 3-4).

Then the connections you will need to make are:

A_01 output to input H_01I, and outputs H_01A to C_01A and H_01B to C_06B (so six coils in series)
A_02 output to input H_02I, and outputs H_02A to C_02A and H_02B to C_07B
etc.

A_35 output to input H_35I, and outputs H_35A to C_35A and H_36B to C_04B
A_36 output to input H_36I, and outputs H_36A to C_36A and H_36B to C_05B

That's it. Just label all the wires accordingly so you can easily find back wires later on.

I note a few things. Please also see the datasheet of the L298N:

https://www.sparkfun.com/datasheets/Robotics/L298_H_Bridge.pdf

Each output of the L298N can carry 2Amps of current max (maybe 2.5-3amps at 1/12 duty cycle but let's use 2amps for now). Since we hooked two outputs together, each of the outputs H_01A and H01B can supply or sink 4Amps of current.

Each of your coils has about 0.5ohms resistance. We are driving 6 coils in series, so that is about 3ohms total per coil.

However, we are activating 6 poles with 6 FETs (3 on the high (Vcc) side and 3 on the low (GND) side, see the image in reply #9).

So e.g. we would activate A_01, A_13 and A_25 on the Arduino which will activate H-bridge H_01I, H_13I and H_25I and coils C01A-C06B, C_13A-C_18B and C_25A-C_30B with forward current and coils C_07A-C12B, C_19A-C24B and C_31A-C36B with reverse current.

This makes the 6 poles. One each subsequent Arduino step, everything shifts by one:

We would activate A_02, A_14 and A_26 on the Arduino which will activate H-bridge H_02I, H_14I and H_26I and coils C02A-C07B, C_14A-C_19B and C_26A-C_31B with forward current and coils C_08A-C13B, C_20A-C25B and C_32A-C01B with reverse current[/size].
etc.

I note that because of this configuration (Which is the same as Pierre is using), the current that each FET supplies is split into two coilsets (6 series coils per set, let's call it a 6-coilset). So e.g. the FET at coil 1 (H-bridge output H_01A) provides current to two coils sets (6-coilset C_01A-C_06B and 6-coilset C_31A-C_36B). Since the resistance of each 6-coilset is 3 ohms, the current will be split equally, so 2Amps max per 6-coilset.

At each point in time there are 6 poles on (each pole has 2 6-coilsets), so total current draw will be 6x2amps=12amps max.

The max voltage over one 6-coilset is 3ohms x 2amps = 6 volts. So you should not be driving your coils with anything higher than 6 volts, otherwise you will burn out your FETs.

So this confirms Pierre's concern about the H-bridges not being able to provide enough current. But I think this is good for an initial try to see if we can make a rotating magnetic field.

Let me know if you have any questions or if things are not clear.

PmgR
====
Help end the persecution of Falun Gong * www.faluninfo.net * www.stoporganharvesting.org

Hi PmgR,

You have to use the EnA & EnB inputs as when you take these low both upper and lower transistors are then turned off, otherwise the In1/In2 In2/In3 just determine which transistor is turned on.

The half bridges not in use have to be  completely switched off.

Regards

L192

[listener191]

Hi Luc and everyone
Just my two cents and disregard if you "don't think so"(no problem)  but it scares me a bit with those modules with their "built in" recovery diodes....as I am thinking where do they go to?

Is there an external "collector" capacitor that collects the backemf/recoil/flyback energy via those steering diodes  and this collector capaciyor  is then pulsed out to load (for example)s after 4 pulses filling them, the...load in this case would be the supercaps

To fill up super-caps "straight" from backemf/recoil/flyback steering diodes usually does not work very well, since the caps being so large of UF will act like a resistive load to the backemf/recoil energy steered out through the  steering diodes....and so this immediately reflects back to lurch up the primary input to much larger draw and it just becomes a big loss to system...I am sure any of you working with flyback/backemf energy know this - and what needs to be done is proper size UF cap needs to be filled, and this cap filled all by itself, with no resistive load over it too.....then this cap is periodically discharged to a load (so cap does not fill up and stay filled up otherwise being filled up it will stop collecting the backemf/recoil energy!)

And important this collector cap must be disconnected from the steering diodes and/or switching whenever it does discharge, or the discharge reflects back to the primary, and it goes up in draw terrible ....(anyone who has worked with backemf/flyback energy also knows this)
This is called a two-stage cap discharge output...pretty common stuff really....

Anyways I cannot believe those built in diodes in those modules will do anything at all but to suppress the backemf/flyback energy and snuff it out so it will not disturb the electronics in the modules and this would be the normal thing to do - ususally the designer will steer the energy straight back into the inductor, snuffing out the :"destgructive transients" that way - and I will suppose that is all those modules will do too, with their built-in diodes...

So I think how Pierre has external steering diodes mounted to his relays to recover the backemf/recoil energy into the supercaps will be actually recovering that backemf/recoil energy and helping the super caps stay stocked up in joules, keeping the system looped and self-sustaining....
(at least helping it to)
While I think those modules will only suppresss and snuff out the backemf/recoil energy, adding nothing to the supercaps...
The supercap UF value might not be such a factor, since Pierre pre-charges those caps, and the caps being already high in voltage will make them more receptive to recieveing and capturing the backemf/recoil energy - that is all I can think of as to how it can actually work OK

One more thing - having modern "ultra fast" or even better "hyper fast" diodes as the backemf/recoil recovery diodes could make the system even better....at least I hiope those are shottkys but there are even better diodes out nowadays....look up the hyper fast ones....

So that is my two cents sorry!
Please continue everyone how you are all doing great!!!!
I like this moderated board it was quite ridiculous the arguments and accusations and eventual sabotage of this project by the usual dickweeds ....

Hi Konehead,

There is nothing wrong with a half bridge arrangement and diodes as shown, except you are correct the super caps will present much impedance to charge which will likely overheat the diodes. If a half bridge with MOSFET's were used with a rating of say 20A or greater, the body diodes would have the same rating and this would not be an issue. I think the L298N's can only be run at very low current in this arrangement.

Regards

L192

[listener191]

Hi all,
please note the number of wires coming from Pierre's induced coil.
The attached image is an enlarged detail from his 1st video at ca. 6:30

Hello Pierre,
please consider how Linus Torvalds made an operating system for computers and open sourced it.
Today he has thousands of collaborators who provide a free operating system - an alternative to the products of the military industrial complex.

Hi Oscar,

Yes,he is using two parallel windings to provide two 115V outputs.

Regards

L192

[pmgr]

Hi PmgR,

You have to use the EnA & EnB inputs as when you take these low both upper and lower transistors are then turned off, otherwise the In1/In2 In2/In3 just determine which transistor is turned on.

The half bridges not in use have to be  completely switched off.

Regards

L192

Yes, you are right. Luc, remove the jumpers to Vcc in and instead connect ENA to ENB to arduino. Connect in1/in2 to Vcc and in3/in4 to GND.
I have updated the original post.
PmgR