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Author Topic: Zero and Q device  (Read 48044 times)

v8karlo

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Re: Zero and Q device
« Reply #105 on: January 11, 2019, 07:47:12 PM »
Hi v8karlo. If no one posts an in-depth analysis with measurements of your 'zero device' soon,
I will see if I have the parts to build it and test in the next few weeks.
Just need to find some time. All is good.  :)

No problem.

For Europe 220V grid:

1. You don't need diodes UF4007, it can be any diodes you have from your spare. It is 50Hz (60Hz) so it doesen't matter if they are slow.
    I used UF4007 because I have 200pcs of them. (on higher frequencies it does matter)

2. C1, C2, C3 capacitors can be any but don't use less than 20uF and they have to be at least 200V rated. I used 3 x 640uF, 200V and it was fine (watch your fingers, do not touch them while and after device is working). Be careful, I didn't. It hurts. That is why I used 22uF, 400V later.

3. Transistor can be any bipolar. Can be mosfet too, tried it with IRF840, it worked without driver. Prefer bipolar rather for this situation.

4. For CZero you can use 2 electrolytic 300V minus to minus (100uF, 300V perfect). I tried it works. It becomes bipolar that way. Don't use less than 20uF, rated at least 300V.

5. I tried 75W bulbs. If I remember correctly they had around 70-80ohms and they worked, so you don't have to search for specific bulb. 2 bulbs of same    ratings. I also tried 50W. Just 2 of a same kind. The 75W output was glowing less than 25W but the effect was visible and heat was there. With bulbs with more resistance effect is more visible, of course. Resistance has role.



You don't need special parts, grab something from trash.

citfta

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Re: Zero and Q device
« Reply #106 on: January 11, 2019, 08:46:55 PM »
Sorry to inform you v8karlo, but you have only reinvented the voltage doubler circuit that has been used in televisions and industrial machines for over 50 years.  As you add capacitors you raise the voltage and drop the current.  And you are incorrect about being able to raise the voltage only a little when you add more caps to keep the current up. You have to raise the voltage an equivalent amount to keep the current up that you can draw from the circuit. And you clearly claim this is a free energy device on page 4 of your PDF and again on page 22.  And on page 15 you refer to self looping the device.  You can NOT self loop this device.  As soon as you remove the grid power your device will discharge the caps and quit.

The reason L1 does not light is because the voltage drop across L1 is very low.  All the caps and diodes are taking the voltage so there is none left to force enough current through L1 to light it up.  As you explained in your PDF when you use larger bulbs then L1 will start to glow.  That is because L2 is now large enough to let enough voltage through to start to light L1.  What your circuit does is force the voltage to L2 and not leave enough for L1.  But it is not free energy and can not be looped.

Respectfully,
Carroll

v8karlo

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Re: Zero and Q device
« Reply #107 on: January 11, 2019, 09:01:00 PM »
As you explained in your PDF when you use larger bulbs then L1 will start to glow.  That is because L2 is now large enough to let enough voltage through to start to light L1.  What your circuit does is force the voltage to L2 and not leave enough for L1.  But it is not free energy and can not be looped

When you use larger bulbs, more watts, L1 does not start to glow, L2 is glowing less. (which is logical also)
When you use only 2 caps, C1 and C2 but without C3 then L1 starts to glow, but L2 is still glowing more. You are multiplying current only 2 times with 2 caps, not 3 times like with 3 caps.

On L1 is 220V directly from grid, butt less current is passing trough L1. On L2 is 175V, but 3 times more current. That means on L1 is more voltage than on L2, but still no glow.
If on L1 220V is passing less current then on L2 175V that means that power through L1 is less than trough L2.
So you have more power on output than on input.


The reason L1 does not light is because the voltage drop across L1 is very low.  All the caps and diodes are taking the voltage so there is none left to force enough current through L1 to light it up. 

The bulbs are the same. Both should expirience same voltage drop.

If there is no force (power), that means there is not enough power at L1. So we are talking about power here.
Voltage on L1 is higher (220V) than on L2 (175V), but L1 does not glow. L1 is directly on 220V grid input.


As I said before, try it, build it, don't talk just by looking at scheme.
There is 5 components (from trash) and 2 bulbs. 1 hour project.


I don't wanna defend it. If you say so, then it is.

Void

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Re: Zero and Q device
« Reply #108 on: January 11, 2019, 09:29:47 PM »
Sorry to inform you v8karlo, but you have only reinvented the voltage doubler circuit that has been used in televisions and industrial machines for over 50 years.  As you add capacitors you raise the voltage and drop the current.  And you are incorrect about being able to raise the voltage only a little when you add more caps to keep the current up. You have to raise the voltage an equivalent amount to keep the current up that you can draw from the circuit. And you clearly claim this is a free energy device on page 4 of your PDF and again on page 22.  And on page 15 you refer to self looping the device.  You can NOT self loop this device.  As soon as you remove the grid power your device will discharge the caps and quit.

The reason L1 does not light is because the voltage drop across L1 is very low.  All the caps and diodes are taking the voltage so there is none left to force enough current through L1 to light it up.  As you explained in your PDF when you use larger bulbs then L1 will start to glow.  That is because L2 is now large enough to let enough voltage through to start to light L1.  What your circuit does is force the voltage to L2 and not leave enough for L1.  But it is not free energy and can not be looped.

Respectfully,
Carroll

Hi Carroll. Yes, that all sounds correct to me. C1, C2, C3, Czero, and L1 are forming a voltage divider.
The (average) voltage across L1 therefore is not high enough to light up L1, except maybe just a little bit,
even though current is flowing through the bulb to charge up the capacitors. 

Warning to potential experimenters:
Build at your own risk.
The 'zero device' circuit is actually not a very safe circuit to experiment with when connecting to the mains
or the output of an inverter. Not only could someone get a lethal zap from that circuit if they touch the wrong thing,
but if they connect a scope probe ground lead to the hot side (phase) of the AC line, they will short the hot phase to ground
(big arc, blown fuse or breaker) since the ground on a scope is typically not isolated from the mains ground.   


v8karlo

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Re: Zero and Q device
« Reply #109 on: January 11, 2019, 09:36:25 PM »

Warning to potential experimenters:
Build at your own risk.
The 'zero device' circuit is actually not a very safe circuit to experiment with when connecting to the mains
or the output of an inverter.

It is not safe. I've got zapped with my left hand.
Do not touch capacitors while device is working and when it is off use 100ohm 2W resistor to empty capacitors before you touch them.
Be careful with capacitors.


But if you touch capacitor on your washing machine while it works, you will be zapped too.
Anything working with 220V is dangerous.
That is why I am experimenting with 12V first. Always.


So yea, you are right.

v8karlo

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Re: Zero and Q device
« Reply #110 on: January 11, 2019, 10:00:32 PM »

Hi Carroll. Yes, that all sounds correct to me. C1, C2, C3, Czero, and L1 are forming a voltage divider.
The (average) voltage across L1 therefore is not high enough to light up L1, except maybe just a little bit,
even though current is flowing through the bulb to charge up the capacitors. 


If voltage on L1 is so low, try to grab L1 with your hand.

Everything will be clear to you in split second.
L1 is directly on mains 220V.

But it can not push lot of current through bulb.
And in the second phase voltage is double on that line.

v8karlo

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Re: Zero and Q device
« Reply #111 on: January 11, 2019, 10:14:03 PM »

We can search for 1000 reasons why it won't work,
or why it can't work.


You have to build it only once in 1 hour.


We are talking here like we are going to build shuttle.
Any other scheme on this forum has more components than this.

lancaIV

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Re: Zero and Q device
« Reply #112 on: January 11, 2019, 10:40:52 PM »
Hi Carroll. Yes, that all sounds correct to me. C1, C2, C3, Czero, and L1 are forming a voltage divider.
The (average) voltage across L1 therefore is not high enough to light up L1, except maybe just a little bit,
even though current is flowing through the bulb to charge up the capacitors. 

Warning to potential experimenters:
Build at your own risk.
The 'zero device' circuit is actually not a very safe circuit to experiment with when connecting to the mains
or the output of an inverter. Not only could someone get a lethal zap from that circuit if they touch the wrong thing,
but if they connect a scope probe ground lead to the hot side (phase) of the AC line, they will short the hot phase to ground
(big arc, blown fuse or breaker) since the ground on a scope is typically not isolated from the mains ground.


http://web.archive.org/web/20040610181420/http://www.theverylastpageoftheinternet.com:80/menu/main.htm

http://web.archive.org/web/20040606131649/http://theverylastpageoftheinternet.com:80/forsale/plans/earthbattery/ebpage1.htm
http://web.archive.org/web/20040617011947/http://theverylastpageoftheinternet.com:80/forsale/plans/earthbattery/ebpage8.htm

super-low or super-fast Ohm conductor
Physics law and order modelling : the conditioning
https://translate.google.com/translate?hl=de&sl=de&tl=en&u=https%3A%2F%2Fde.m.wikipedia.org%2Fwiki%2FOhmsches_Gesetz
  a. not- b. steady , temperature, .......
     ..... EINZIG DIE KONSTANZ .... IST DIE KERNAUSSAGE DES OHMSCHEN GESETZES.
 
freI nach Stein : DIE KONSTANZ IST DIE KONSTANZ IST DIE KONSTANZ ( und nicht immer am Bodensee)

 constant up and constant down, constant up and sharp down, sharp up and sharp down = Dirac surges

                    macro-/ nano-scale : friction/ Reibung, material delay actio/reactio time 
                                             

Hoppy

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Re: Zero and Q device
« Reply #113 on: January 11, 2019, 11:16:27 PM »
 My explanation for the operation of the 'Zero' circuit: -

Using my 25W bulb having a cold resistance of 204R, the bulb in series with a 10uf cap and diode, the bulb does not visibly flash on connection to a 240V mains supply. However, it will flash with a 20uf or higher value cap. The difference is due to the reactance of the capacitor allowing sufficient current to heat the bulb filament to a glow in the case of the 10uf cap. The combined series capacitance of C1 to C4 in the zero circuit is around 5uf, thus no flash.
 
When the transistor is switched ‘ON’ and conducts on one phase of the mains supply, caps 1 to 3 effectively discharge in parallel to a much lower resistance output circuit, via steering diodes, to common output rails to which L2 is connected. The combined currents through the lower resistance, discharged from caps 1 to 3 are sufficient to heat the filament of L2. However, L1 cannot light on this phase as the diodes in series with the caps are reverse biased. The caps are now discharged ready for re-charge on the next phase when the transistor is switched ‘OFF’.
 
Altering the power rating of the bulbs and cap values will alter the impedance of the input loop to an extent whereby LI may be seen to glow together with L2.
 

v8karlo

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Re: Zero and Q device
« Reply #114 on: January 12, 2019, 12:02:45 AM »
My explanation for the operation of the 'Zero' circuit: -

Using my 25W bulb having a cold resistance of 204R, the bulb in series with a 10uf cap and diode, the bulb does not visibly flash on connection to a 240V mains supply. However, it will flash with a 20uf or higher value cap. The difference is due to the reactance of the capacitor allowing sufficient current to heat the bulb filament to a glow in the case of the 10uf cap. The combined series capacitance of C1 to C4 in the zero circuit is around 5uf, thus no flash.
 
When the transistor is switched ‘ON’ and conducts on one phase of the mains supply, caps 1 to 3 effectively discharge in parallel to a much lower resistance output circuit, via steering diodes, to common output rails to which L2 is connected. The combined currents through the lower resistance, discharged from caps 1 to 3 are sufficient to heat the filament of L2. However, L1 cannot light on this phase as the diodes in series with the caps are reverse biased. The caps are now discharged ready for re-charge on the next phase when the transistor is switched ‘OFF’.
 
Altering the power rating of the bulbs and cap values will alter the impedance of the input loop to an extent whereby LI may be seen to glow together with L2.

It is very scientifically said.

Can not see any errors here.

And when grid line comes in serial with CZero it raises that line voltage to 525V, which explain 175V on each cap, C1, C2 and C3.
I also tried with 640uF caps and 2 x 150uF (minus to minus). Combined capacitance is little bit more in that case (3 x 640uF + (150uF / 2)) / 4 and effect was same as when I used 22uF. With bigger caps I was creating only bigger buffer. That means when I plug off the L2 continues to light for a second or two.

You are right Hoopy, it sounds ok to me!


The L1 flash for a moment at the beginning while C1, C2 and C3 are filling up. When they are filled L1 diminish and L2 starts to glow.
That means that C1, C2 and C3 must not be emptied all the way. Effect vanish. That is why resistance of load comes to play.At the beginning if the C1, C2 and C3 are bigger L1 will flash longer. It needs more time for caps to fill up.The L1 will flash only when all caps are emptied and then device plugged into grid.

It is best to build it and then observe it.
The fact is that L1 does not glow and L2 glows with heat.
Can you use this principle, adopt it with some other circuit or develop further?

lancaIV

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Void

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Re: Zero and Q device
« Reply #116 on: January 12, 2019, 12:34:33 AM »
My explanation for the operation of the 'Zero' circuit: -

Using my 25W bulb having a cold resistance of 204R, the bulb in series with a 10uf cap and diode, the bulb does not visibly flash on connection to a 240V mains supply. However, it will flash with a 20uf or higher value cap. The difference is due to the reactance of the capacitor allowing sufficient current to heat the bulb filament to a glow in the case of the 10uf cap. The combined series capacitance of C1 to C4 in the zero circuit is around 5uf, thus no flash.
 
When the transistor is switched ‘ON’ and conducts on one phase of the mains supply, caps 1 to 3 effectively discharge in parallel to a much lower resistance output circuit, via steering diodes, to common output rails to which L2 is connected. The combined currents through the lower resistance, discharged from caps 1 to 3 are sufficient to heat the filament of L2. However, L1 cannot light on this phase as the diodes in series with the caps are reverse biased. The caps are now discharged ready for re-charge on the next phase when the transistor is switched ‘OFF’.
 
Altering the power rating of the bulbs and cap values will alter the impedance of the input loop to an extent whereby LI may be seen to glow together with L2.


Hi Hoppy. Sounds right to me.
All the best...


Void

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Re: Zero and Q device
« Reply #117 on: January 12, 2019, 12:56:00 AM »
If voltage on L1 is so low, try to grab L1 with your hand.
Everything will be clear to you in split second.
L1 is directly on mains 220V.
But it can not push lot of current through bulb.
And in the second phase voltage is double on that line.

Hi v8karlo. I did not say the voltage 'on' L1, I said the voltage *across* L1, i.e. the average voltage drop across L1.
Just because you don't understand the basics of electric circuits very well, it doesn't mean other people don't or can't
understand the circuit well, even just by looking at the schematic.  :)

Carroll's and Hoppy's analysis seems to be correct to me. The caps charge up from the mains through L1 in one half
of the AC cycle. Since the caps and L1 are in series, the voltage from the mains will divide across them.
There is not enough average voltage drop across L1 to light it up brightly, even though current is passing through it
to charge the caps which are in series. In the second half of the AC cycle, the caps (in parallel) discharge through L2
and light it up. Given this analysis, there is no reason at all that I can see to think this circuit arrangement might be 'free energy'.
You are just charging up caps from the mains in one half of the AC cycle and discharging those caps through L2
in the second half of the AC cycle. Sorry mate. It appears there is no magic to find there. This is not surprising at all since
IMO any given setup would have to draw in extra energy from outside the circuit to be able to achieve COP >1.
I hope this is beginning to sink in now, but I won't hold my breath waiting for that moment to happen.  :)
All the best...


v8karlo

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Re: Zero and Q device
« Reply #118 on: January 12, 2019, 01:09:23 AM »
Sorry mate. It appears there is no magic to find there.
I know what you mean by voltage drop across element.

So, how you explain extra light and heat on output while input is cold and without light?
Hoopy didn't explained that, neither do you. He did not explain 175V on each cap?
Do you have your explanation of extra light and heat?


Try to remove CZero and plug it that way.
You will see how magic vanish and there is no 175V on caps.
Try that.

Power on output will be the same or less than input now.
Why is that?

You did not change anything else.
Your circuit stays the same,
you still have your voltage divider,
but the power from caps is gone now.

Something is not right here, does it?

Now you have ordinary, simple useless circuit.
More input than output.

v8karlo

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Re: Zero and Q device
« Reply #119 on: January 12, 2019, 01:56:12 AM »

At the end, maybe CZero is doing what it is supposed to do?
Raise the voltage on the caps.

With voltage the power in caps is also raised.
And the power on output is going up.

What you think about that?

Magic? I don't think so.

Some things are very hard to accept. Obviously.