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

[Jeg]

Hi Jeg,


The only hole in this theory is that the system still functions off the super caps, although this could just be less than OU as observed. Why have a frequency control at all? Perhaps it is really being used as a crude phase control when running the transformer from the line supply. When running looped the phase will be the same.

Well, looks like that caps are just for absorbing the spikes. It should be less than OU as "Spike return" never showed something promising. At least by itself.

I don't really see why the phase will be the same when running lopped. As i am thinking it, a maximum value of ripple voltage (120Hz), should be presented when we have a full registration as you call it during both cycles.

Some pics for the records

   

[pedro1]

J'ai voulue faire cette méthode mais je me suis rendue compte a mi-chemin que cela ne fonctionnera pas très beau travail quand même et cela pourra donner un excellent transfo pour le reste j'ai quelque reticence

En. I wanted to use this method but I realized half way through the build it won't work as I thought but nice job anyways and it could make a great transformer! but for the rest I have some reservations.

[Jeg]

Hello Pierre
Thanks for your comments. Yes it is a different approach and it will not work by driving it the same way as your first prototype.

Regards
Jeg 

[pmgr]

L192 et al,

With respect to the diodes and how Pierre has them hooked up, please see the attached images (I note that Pierre does not have the ground side diodes).

In the attached images, Vdd is connected to the rectifier bridge (small red cable that drives the relays). Vc is connected to the capacitor bank.

Since Vdd is always larger than Vc, the diode (D1_p in the attached image) that is connected to the relay that is turned on (S1_p in attached image) will always sink current to the capacitor bank through that relay.

So if the capacitor bank is not fully charged (or empty), it will be mostly charged directly through the relay and the diode and not through the 4 ohms resistor.

So it is very important that the capacitor bank is fully charged up to Vdd before starting the Arduino, otherwise it will be very likely that too much current will go through the relay (and diode) and overload the relay and maybe even the diode if its current rating is too low. The 4 ohm resistor is not going help much (if at all) in limiting the current to the capacitor bank in this case as it is parallel with the relay/diode.

And once one relays fails, the other switched on relays are going to get the current of this relay in addition to what they are already carrying and so they will fail as well. And since the Arduino is switching all relays one after another, eventually all relays will quickly fail.

And I note that this will also be the case if you are using H-bridges or MOSFETs. A capacitor bank can sink many many amps of current, so it is crucial to put a fuse in the main Vdd line, between rectifier bridge and relays, and make sure the capacitor bank is fully charged up.

Pierre, this is probably what caused all your relays to burn out.

There is a simple solution to this problem: make sure the capacitor bank is fully charged before starting the Arduino and use a fuse.

Lastly I note that if your output coil is sinking too much current and draining the capacitor bank to a voltage too low, this problem will also occur once the capacitor bank gets discharged to a value a lot lower than Vdd.

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

[r2fpl]

L192 et al,

With respect to the diodes and how Pierre has them hooked up, please see the attached images (I note that Pierre does not have the ground side diodes).

In the attached images, Vdd is connected to the rectifier bridge (small red cable that drives the relays). Vc is connected to the capacitor bank.

Since Vdd is always larger than Vc, the diode (D1_p in the attached image) that is connected to the relay that is turned on (S1_p in attached image) will always sink current to the capacitor bank through that relay.

So if the capacitor bank is not fully charged (or empty), it will be mostly charged directly through the relay and the diode and not through the 4 ohms resistor.

So it is very important that the capacitor bank is fully charged up to Vdd before starting the Arduino, otherwise it will be very likely that too much current will go through the relay (and diode) and overload the relay and maybe even the diode if its current rating is too low. The 4 ohm resistor is not going help much (if at all) in limiting the current to the capacitor bank in this case as it is parallel with the relay/diode.

And once one relays fails, the other switched on relays are going to get the current of this relay in addition to what they are already carrying and so they will fail as well. And since the Arduino is switching all relays one after another, eventually all relays will quickly fail.

And I note that this will also be the case if you are using H-bridges or MOSFETs. A capacitor bank can sink many many amps of current, so it is crucial to put a fuse in the main Vdd line, between rectifier bridge and relays, and make sure the capacitor bank is fully charged up.

Pierre, this is probably what caused all your relays to burn out.

There is a simple solution to this problem: make sure the capacitor bank is fully charged before starting the Arduino and use a fuse.

Lastly I note that if your output coil is sinking too much current and draining the capacitor bank to a voltage too low, this problem will also occur once the capacitor bank gets discharged to a value a lot lower than Vdd.

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

PmgR

Yes: 4 ohms they depend on the voltage on which the voltage capacitors are. It allows current from either the bridge or capacitors and it is important to fully charge the capacitors before switching on.