PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2024-05-05, 18:05:46
News: Registration with the OUR forum is by admin approval.

Pages: 1 ... 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 [31] 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 ... 48
Author Topic: Akula0083 30 watt self running generator.  (Read 932669 times)

Group: Professor
Hero Member
*****

Posts: 3366
Here a screenshot of pin 3 (purple trace) and pin 2 (blue trace) together again with the gate voltage (yellow trace)
I am still not sure if the signal at pin 3 can be trusted.  The spikes on it become negative (EMI?) and the outputs of TL494's error amplifiers just cannot go below ground naturally.
Also, these spikes appear to go only 500mV above ground, and for them to have any effect on TL494's internal PWM comparator, they'd have to be at least 600mV above ground.
So my verdict is that the outputs of the error amplifiers (pin 3) are not getting activated and your feedback loop is not functioning.
To troubleshoot this you need to scope the "feedback takeoff point".

the L2 current (green trace)...2e screenshot is with same settings as first one, except the green (current) trace now is from L1.
The comparison between L1 and L2 current reveals that they flow at the same time like in a regular transformer.
I don't know if this is what the author had intended.

I was expecting a current flow like in a flyback transformer (alternating between L1 and L2).  My expectation was that L2 skims off the energy from C3 that the LEDs were not able to use up, because D6 conducts only when C3 is overcharged, precisely meaning VC3 + VL2 > VC11.   In the present configuration VL2 = +3*VL1 when the MOSFET is ON because of the transformer's 1:3 turn ratio and L2 polarity.
For my expected circuit behavior, L2 must be reversed making VL2 = -3*VL1 when the MOSFET is ON, so the current in D6 & L2 cannot flow when the MOSFET is ON.  L2 current can flow through D6 only when C3 had been overcharged by the flyback pulse via D5.
See this simulation for what I mean.

« Last Edit: 2014-04-18, 16:30:37 by verpies »
   
Group: Guest
I believe you,  but it is a secondary side effect, not a primary effect of the "rail voltage".
The R5 & C5 LC current has more direct causal connection to the feedback loop and pin 3 behavior, than the "rail voltage"



It is nonetheless a feature that Akula may have intended in his design and makes sense insofar as limiting current to the LED array.

Once the PWM output has ceased, there may still be an amount of DC current through the LED array depending on the combined forward voltage of the series connected LED elements in the array and the supply voltage level. The typical forward voltage for my LED array is 3.6V x 5 LED elements = 18V for my array. However, current conduction will start as low as 2.6V per element, so I do see an increasing level of illumination as the supply voltage is increased beyond a PWM cut-off setting of 13V. Those builders using ordinary 12V LED lamps either singly or in parallel, will be forward conducting at a much lower voltage level.

Hoppy
« Last Edit: 2014-04-18, 09:03:11 by Hoppy »
   

Group: Professor
Hero Member
*****

Posts: 3366
Do you still believe in the intended PWM cutoff directly related to the rail voltage level after fixing the C4 error on Groundloop's PCB ?
   
Group: Guest
Do you still believe in the intended PWM cutoff directly related to the rail voltage level after fixing the C4 error on Groundloop's PCB ?

Yes, the cap fix makes no difference to the PWM cut-off. I can see no observable affect of the cap being completely removed on the PWM output waveform.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4102
I am still not sure if the signal at pin 3 can be trusted.  The spikes on it become negative (EMI?) and the outputs of TL494's error amplifiers just cannot go below ground naturally.
Also, these spikes appear to go only 500mV above ground, and for them to have any effect on TL494's internal PWM comparator, they'd have to be at least 600mV above ground.
So my verdict is that the outputs of the error amplifiers (pin 3) are not getting activated and your feedback loop is not functioning.
To troubleshoot this you need to scope the "feedback takeoff point".

The comparison between L1 and L2 current reveals that they flow at the same time like in a regular transformer.
I don't know if this is what the author had intended.

I was expecting a current flow like in a flyback transformer (alternating between L1 and L2).  My expectation was that L2 skims off the energy from C3 that the LEDs annot not use, because D6 conducts only when C3 is overcharged, precisely meaning VC3 + VL2 > VC11.   In the present configuration VL2 = +3*VL1 when the MOSFET is ON because of the transformer's 1:3 turn ratio and L2 polarity.
For my expected circuit behavior, L2 must be reversed making VL2 = -3*VL1 when the MOSFET is ON, so the current in D6 & L2 cannot flow when the MOSFET is ON.  L2 current can flow through D6 only when C3 had been overcharged by the flyback pulse via D5.
See this simulation for what I mean.



I will concentrate on the feedback signal to see what is happening there tonight.

Good to see that your simulation also shows ± 10A being pulled when closing the 12V switch.

A short video about the present circuit and the way it behaves when powering off, especially pin3 signal is going up/down shortly after power off.
Also some temperature measurements are done on the severall components.

https://www.youtube.com/watch?v=ofRW4OeFExk&feature=youtu.be

Regards Itsu
   

Group: Professor
Hero Member
*****

Posts: 3366
A short video about the present circuit and the way it behaves when powering off,
I little girl just run up to me with her hands in the air, yelling:  "Aaaaa, go, go... Itsu has made another video!" :)
   

Group: Professor
Hero Member
*****

Posts: 3366
OK, C11 once charged will supposedly discharge slowly thru L1 (Choke).  The purpose of the  L1 (choke) would be to act a surge suppressor for the transients so the 7812 doesn't burn up.  
Yes

This feedback loop as you stated is the major power loop for the OU system, right?  
In my posts I use two phrases containing the word "loop"
The "major power loop"  consisting of L2, SB1, R3, C11, R1, L1, D5, C3, D6 in that order (with Q1 as a catalyst).
The small signal "feedback loop" consisting of the "feedback takeoff point", R7, C15, pin 2 & pin 15, culminating in pin 3.

So, as C11 charges up at the same time it would be discharging into the L1 (choke) to feed the 7812 to power the TL494.
Yes, I like to think about the VR1 as the origin of the auxiliary power supply.

But until the system is free running, the input voltage/current must pass thru L1 (choke) from opposite side to charge C11
Yes, in other words it appears that the author had intended to charge the major energy tank (C11) from the external battery via L3 in the bootstrap mode and from L2 via D6 in the self-running mode.

the then thru L1 Primary to the Drain of the MOSFET (when switching).
That is the major discharge path for C11.

...So C11 is held @ Vcc as a minimum voltage thru L1 (choke).
What is Vcc in your analysis?  Voltage across C11 or C6 or C3 ?

As stated C11 can charge to a higher voltage when L1  EMF (primary) and C3 voltage exceed Vcc.
IMO only D6 and L2 provide the recharge path for C11.
I don't count the series LC tank formed by C11 & L1 during the power-up transient.

My idea is this:  What if a small boost/buck converter (1-3 amp variety) was substituted in place of the 7812 volt reg.  When the voltage dipped below the set value, the boost converter would raise it, and when it went above, the buck would lower it....  This *could* give more leeway to the operation of the control circuitry to get it free running.  Usually these small boost/bucks converters have decent efficiency ratings.
Yes, a buck/boost converter would be able to maintain the auxiliary voltage with wider input voltage range on C11 via L3.
« Last Edit: 2014-04-18, 16:33:30 by verpies »
   
Group: Guest
I little girl just run up to me with her hands in the air, yelling:  "Aaaaa, go, go... Itsu has made another video!" :)

Sweet.  :)
   

Group: Moderator
Hero Member
*****

Posts: 2740


Buy me a cigar
Temporarily remove one side of R6 on Groundloop's PCB.  (R7 on Akula's diagrams).
This will disable the feedback loop and pin 3 of TL494 will stay low.

Please make a short video about what happens as the power supply voltage is increased slowly above the critical voltage while the Ch2 is connected to the junction of R6 and R7.

Dear Verpies.

As you have since discovered the C4 problem would you like me to show a before, and after, the capacitor fix, with relation to R 6 removal ??

Cheers Grum.


---------------------------
Nanny state ? Left at the gate !! :)
   

Group: Professor
Hero Member
*****

Posts: 3366
As you have since discovered the C4 problem would you like me to show a before, and after, the capacitor fix, with relation to R6 removal ??
Removing R6 (R7 on Akula's diagram) disables the feedback loop as stated in this message.

After you do the C4 fix (C8 on Akula's diagram) you might find out that pin 3 does not go high for higher supply voltages anymore, and disabling/troubleshooting the feedback loop is not necessary.

A short video about scoping the "feedback takeoff point"* on Ch2 while the power supply voltage is increased slowly, is still called for.
You may trigger your scope on Ch1 connected to pin 10 so you are far away from EMI generating components (TC4428, MOSFET, L1 & L2)



* "feedback takeoff point" is the junction of R7 and R6 on Groundloop's PCB  (R5 & R7 on Akula's diagrams)
« Last Edit: 2014-04-18, 11:49:19 by verpies »
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4102
I little girl just run up to me with her hands in the air, yelling:  "Aaaaa, go, go... Itsu has made another video!" :)


I am glad you guys are having fun about my video's, it won't prevent me making them though  ;D

What about increasing R5 (Akula diagram) to say 10 Ohm? Will it not increase the feedback signal to a level where it does affect the duty cycle?

Regards Itsu
 
   
Group: Guest
The components arrived this morning but is missing the Toroidal Choke.....and the large heat sink


« Last Edit: 2014-04-18, 19:13:51 by TutorialFE »
   

Group: Professor
Hero Member
*****

Posts: 3366
The components arrived this morning but is missing the Toroidal Choke.....
Disassemble on old computer power supply. 
You will find a suitable choke inside.
   

Group: Professor
Hero Member
*****

Posts: 3366
What about increasing R5 (Akula diagram) to say 10 Ohm? Will it not increase the feedback signal to a level where it does affect the duty cycle?
It could.
Also, changing R5 to an inductive wirewound resistor will dramatically change the signal at the "feedback takeoff point".
   
Group: Guest
Disassemble on old computer power supply. 
You will find a suitable choke inside.

Don't tell him that, he might find at least half the components he needs inside.   :)
   
Group: Guest
@ Verpie Disassemble on old computer power supply. 
You will find a suitable choke inside.


Hy Verpies I don't have the device to measure uH ...

This are about an year of serching ferrite in the litter....

But I don't know which one is good... Do you have any suggest?
   
Sr. Member
****

Posts: 375
Hi,

There is new video from akula (obviously it was attempted to delete very quickly, so no original link) - https://www.youtube.com/watch?v=ogMcs6iBJwY

The control circuit in the left is creating main transformer driving frequency and resonant frequency to the secondary primary.
The box in the right is impulse power supply (unless it was modified).

To me it is almost same what we did in Lithuanian Yoke experiment, just we did not self loop on the time.

Good luck!
   
Group: Guest
Could someone tell to Akula to do a TutorialFE seriously with all information (All) to replicate in all the World with electronic components......

I'm very glad to see that Him is good constructor about Free Energy but his knowledge never go on market.... so He is wasting time to do movie without the ALL information...  ^-^

More video more probably he will begin a Slave.....  >:(

Tell him also open a Paypall account to receive money from everyone..  ;)
   
Group: Guest
No spark gaps, no ground wire and not even heatsinks on the MOSFETs.  Looks like a typical core from a flyback transformer.  Good stuff T-1000.  Maybe we can get lucky and see how close the schematic matches what we are working on.
   
Sr. Member
****

Posts: 336
@ Verpie Disassemble on old computer power supply. 
You will find a suitable choke inside.


Hy Verpies I don't have the device to measure uH ...

This are about an year of serching ferrite in the litter....

But I don't know which one is good... Do you have any suggest?


Use the core that has approximately the same size as the drawing on the PCB.
If the core has more than one winding, then remove the thinnest winding.

GL.
   

Group: Administrator
Hero Member
*****

Posts: 3867


Buy me some coffee
   
Group: Guest
   
Group: Guest
@GL

Ok It has got only one wire thikness 0,7 mm...

Thanks
   

Group: Professor
Hero Member
*****

Posts: 3366
But I don't know which one is good... Do you have any suggest?
The one Groundloop suggested is fine  or the one to the left of it.
   

Group: Administrator
Hero Member
*****

Posts: 3867


Buy me some coffee
Thanks Matt, thats interesting so i wonder if the Russian named who owns the most shares is actually Akula.
This all reminds of the the Return of the Hungarian

Quote
The basic principle of deriving free energy from wave-fields can be understood even without higher mathematics. If we would have to summarize the essence of Mr. Vajda?s discovery in few sentences without maths, the following explanation could be given:

When two waves with identical polarization, frequency, phase and amplitude propagate in the same direction and meet (and merge) in free space, then their amplitudes will add together and the amplitude of the resultant wave will be double that of a single input wave. This physical phenomenon is called superposition or interference of the waves, when (under the above conditions) the
amplitude of the resultant wave is calculated by simply adding together the amplitudes of the incoming waves.
The energy content of a wave is directly proportional with the square of its amplitude. This fact has a profound impact on the energy balance of the wave-fields.
Calculating the energy balance of the above example, we get that if two units of energy enter the system, then the energy of the output resultant wave will be (calculated as the square of the
resultant?s amplitude, that is) four times that of one single input wave (and not only double). As we see, two units of energy enter the system and four units leave, that means we have gained two times more energy then what we have feed into it. If we take two units of energy from the output and feed it back into the input, then there are still two units remaining for utilization and the process can go on continuously.

http://www.overunity.com/3068/the-return-of-the-hungarian-free-energy-from-wave-fields/#.U1FF-aJrFZM
   
Pages: 1 ... 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 [31] 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 ... 48
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2024-05-05, 18:05:46