|
@verpies SORRY-SORRY AND SORRY AGAIN.
First, let's now put the main skeleton on the table. Our only reference of this circuit working is via x-name41. No video of Akula was ever shown with this circuit working. If x_name41 made his video based on the original Akula 30 watts circuit diagram, then it did not work in loop and that is why he faked his video probably because he was so pissed off in following the circuit to the letter and not showing looping. So he spiked his circuit with two feed wires from underneath. I know he faked it and there is solid proof. So x_name41 was not able to self-run it, only fake it. I am convinced that the problem is R3 and he could not figure it out himself. Probably if he is reading this forum, he will do the change himself on his circuit. This does not mean the Akula device is fake, only that there are mistakes in the diagram and we need to work them out.
I really appreciate your extra effort on my behalf and am sorry to be so repetitive. A real pain I know, but then you or anyone else would need to answer these questions just to make quadruple sure when looking at your original diagram. I am sure some others are asking themselves the same thing.
1) Why would you accept to pass dc + directly through R3 - then L2 - then R2- then to load, continuously? This is like a straight line to load and since load is at the end of the line, the DC+ entering R3 is still positive when it enters L2 and still positive when exiting L2 and still positive when entering R2 so still positive when exiting R2 and when hitting the load. L2 is always positive going to load.
You can pulse L1 as much as you want and L2 bleed will not change. You can switch the R2/R3 around with diode and cap (like on your second diagram you made for me) but that does not change this continuous DC bleed to load at all times. @Itsu said his R3 was really hot? That's not a good path to OU by wasting energy in a resistor, especially when it is not required.
2) Then how do you expect to refill C11 if L2 is so busy passing DC to load? You are saying D6 is passing a high voltage spike to L2 then R3 then to C11. But why would you want to more resistance in this path when it is not required. In any case any spike through D5 will hit the load (path of least resistance) before it can pass L2 and R3. So that logic is not fluent enough to work.
3) Then why do you think L1 is not only going to the Mosfet Q1, but just before that it is bypassed by D5 and sent to load as well? This part is OK and for me this is the only main out-of-the-box feature of the Akula circuit. Instead of feeding the load with L2, he is feeding the load with L1 via D5. That's all he needs to do to feed the load. He then must be using the free L2 to replenish the C11. There is no other way. But L2 cannot replenish C11 and feed the load at the same time because it cannot respond to the L1 pulse off the E-core because L2s' internal polarity is always held positive by R3.
4) How do you think the mosfet is influencing what L1 is sending to the load? The pulsed mosfet is useless because pulsing L1 is useless as per the diagram. You pulse a coil to impart to the another coil otherwise you are stuck with overflux in the core, but if the other coil is already full of DC, then how do you expect it to react? It cannot.
5) What adverse effect would you anticipate if R3 was replaced by a diode? Is such a change going to risk blowing the circuit? I can't see how it could.
For me, when I consider all of these conditions as the diagram is draw presently, none of it makes sense if R3 is not replaced with a diode pointing to C11. Nothing else, just a diode at that point prohibits DC+ from entering the top of L2 but provides a clear and simple path so L2 can refill C11 ASAP. If you fill C11 fast enough you can have a self-runner but you need an available and receptive L2 to accomplish that.
The diode D6 can still be there and there is no problem having two diodes one on each side of L2 because they both react differently depending on the polarities and intensities present in L2. My opinion on the D6 diode is it is never really used except if L1 really created some very high spikes that could pass the D6 and enter the L2 but if the coils are well tuned, that should never occur. The other reasons is if D6 is held in reverse threshold mode by extreme changes in L2. But those extremes cannot occur if R3 is holding it hostage.
Just having that one D6 is not the same thing as also replacing the R3 with a diode because the source of polarity changes is supposed to originate from inside L2 and not something that can be controlled by D6. If you check D6 temperature, it is probably not hot at all because it is not being used for much.
Look I know I may risk ("risk" huh more like "for sure") coming off as the resident shmuck of the forum, being such a pain and all. But I think everyone should stop and carefully answer the above questions. There is no point to continue if this is not resolved in a logical manner. We seem to think it is OK to force feed through L2 then to load, I do not think, well actual I know, that if L2 is feeding the load as it is doing, then impulses from L1 to L2 are useless. L2 is just a conductor. Add the fact that some are using the T-1000 wind and this is compounded even greater.
Maybe asked again, why is R2 present in the circuit. For me it is if R3 was a diode then when C11 is full and cannot take anymore, if the stress in L2 increases further it will dump more and more to load instead of over saturating L2 and killing the overall effect. It's a question of balance but you cannot achieve that balance if one side is tipped way over the scale as R3 is doing now.
I won't bring this up again but I will make my own circuit "eventually" and do it my way.
wattsup
PS: This post took 3 hours to prepare (last night and this morning) because I wanted to make sure I did not vex you since I hold you in top regard and I consider this part of my duty to point out something so wrong. Akula is not infallible and I would not be surprised if he put R3 there (instead of a diode) on purpose to see who will figure it out. That fact that he is not coming forward any more openly is his way of saying "There is no free lunch man". Use your brains and figure it out. His way of saying, "if you can find the problem, then you have proven to me that you have finally learned well what the circuit is doing".
---------------------------
|
Author
Topic: Akula0083 30 watt self running generator. (Read 974990 times)
