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Author Topic: Slayer driven neon-producing DC via resistor ?.  (Read 15572 times)

Offline tinman

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Slayer driven neon-producing DC via resistor ?.
« on: August 21, 2014, 04:01:15 AM »
This is an effect i stumbled across about a year ago,and just revisiting it. Here i have a neon being driven via the slayers electric field,and then from the neon to a small AC cap(cap size not important)via a resistor on one leg. Seems some how that the resistor gives rise to an AC offset,so as we end up with an overall DC voltage in the cap.

https://www.youtube.com/watch?v=OWnLBrDkTTc&list=UUsLiBC2cL5GsZGLcj2rm-4w

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Offline MarkE

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #1 on: August 21, 2014, 02:01:08 PM »
With lots of RF around your meter could be reporting nonsense.  You should look at the capacitor with a scope.

Offline tinman

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #2 on: August 21, 2014, 02:11:10 PM »
With lots of RF around your meter could be reporting nonsense.  You should look at the capacitor with a scope.
Used the scope when i found this a year ago-same result. But what do you think the scope might show in way of trace Mark?

I will go make another video ,using the scope,and post it here asap.

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #2 on: August 21, 2014, 02:11:10 PM »
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Offline TinselKoala

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #3 on: August 21, 2014, 03:01:24 PM »
Used the scope when i found this a year ago-same result. But what do you think the scope might show in way of trace Mark?

I will go make another video ,using the scope,and post it here asap.
A little while ago you demonstrated that the NE-2 can act as a rectifier. Maybe that's what's happening here. If so I would expect that the maximum voltage you can get on the capacitor is going to be the neon's cut-off voltage, maybe 65 or 70 volts or so. Just guessing though.

Offline MarkE

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #4 on: August 21, 2014, 03:19:12 PM »
Used the scope when i found this a year ago-same result. But what do you think the scope might show in way of trace Mark?

I will go make another video ,using the scope,and post it here asap.
The scope will show whether or not you have a symmetrical waveform.  If the neon bulb is acting like a rectifier, even a very leaky rectifier, then the waveform will by asymmetric.

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #4 on: August 21, 2014, 03:19:12 PM »
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Offline tinman

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #5 on: August 21, 2014, 03:34:33 PM »
Video uploading now.
But my guess ,after looking at the scope-i see an AC offset of about 250-300mV from the neon/cap circuit. I would say that this offset is additive over a number of pulses,until a maximum voltage is obtained in the cap. But my only question is-why dose the voltage in the cap only rise when i add a resistor in series with the neon and cap? the resistor is non linear,and i turned it around to make sure of that.

I'll leave the last bit of the post as is,but i just went and swaped the neon around,and no difference-still a DC voltage in the cap of same polarity.

Offline MarkE

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #6 on: August 21, 2014, 03:41:30 PM »
Video uploading now.
But my guess ,after looking at the scope-i see an AC offset of about 250-300mV from the neon/cap circuit. I would say that this offset is additive over a number of pulses,until a maximum voltage is obtained in the cap. But my only question is-why dose the voltage in the cap only rise when i add a resistor in series with the neon and cap? the resistor is non linear,and i turned it around to make sure of that.

I'll leave the last bit of the post as is,but i just went and swaped the neon around,and no difference-still a DC voltage in the cap of same polarity.
DC amplifiers like the ones that are in your voltmeter are susceptible to RF noise.  Some meters do better than others. 

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #6 on: August 21, 2014, 03:41:30 PM »
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Offline tinman

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #7 on: August 21, 2014, 03:56:00 PM »

Offline MarkE

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #8 on: August 21, 2014, 04:13:17 PM »
A look across the cap with the scope.

https://www.youtube.com/watch?v=lxhIVwvTo2Q&list=UUsLiBC2cL5GsZGLcj2rm-4w
That shows that the noen is acting as a leaky diode.  When it is connected through the resistor, the resistor unloads the weakly rectified source and the average capacitor voltage rises.

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #8 on: August 21, 2014, 04:13:17 PM »
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Offline tinman

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #9 on: August 21, 2014, 04:24:31 PM »
That shows that the noen is acting as a leaky diode.  When it is connected through the resistor, the resistor unloads the weakly rectified source and the average capacitor voltage rises.
I thought that aswell,but as i wrote above,i went and turned the neon around,and no change in voltage polarity in the cap???.

So 1 thing i can think of,is that the cap itself can accept a charge in one direction better than in the other direction. So next i guess i go and swap the cap around,and see what happens. I dont think this is the case though,as i have tried many cap's,and get the same result.

The only other thing i can think of,is the effect that the resistor would have on the electric cables them self from the neon. Could the magnetic field around the cable with the resistor in series ,be weaken'd,while the other is not-resulting in a stronger field around that cable?.

Offline tinman

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #10 on: August 21, 2014, 04:29:06 PM »
That shows that the noen is acting as a leaky diode.  When it is connected through the resistor, the resistor unloads the weakly rectified source and the average capacitor voltage rises.
Oh i must ask-how exactly can a neon act as a diode. We have two plate's and a chamber full of gas.I dont see how this could be any sort of diode.

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #10 on: August 21, 2014, 04:29:06 PM »
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Offline magpwr

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #11 on: August 21, 2014, 05:20:45 PM »
Oh i must ask-how exactly can a neon act as a diode. We have two plate's and a chamber full of gas.I dont see how this could be any sort of diode.

hi tinman,

Just look a look at history of diode.Especially the gas discharge diode.
 

Offline TinselKoala

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #12 on: August 21, 2014, 08:00:26 PM »
I thought we have been here before. Tinman, I think you have been corrupted by you-know-who!

Recall: Only the electrode of the neon which is experiencing +negative polarity+ actually glows. It is ionizing the neon in its vicinity by donating an electron, making negative neon ions, which migrate to the anode side which is dark. This is the rectification action! When connected to AC, of course which electrode is negative changes at the line frequency, so it looks like both electrodes are glowing. They are not.

So that is why reversing the neon has no effect on the rectification. You have just put the other electrode inside the envelope, in contact with the negative polarity of your voltage source.

Now think about this: Those neons below are in series. That means the glowing electrode of one is directly connected to the dark electrode of the other one, by a little bit of wire. One glows, the other does not. Why? (For a real laugh, ask TA to explain it.)

Offline tinman

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #13 on: August 22, 2014, 01:22:42 AM »
I thought we have been here before. Tinman, I think you have been corrupted by you-know-who!

Recall: Only the electrode of the neon which is experiencing +negative polarity+ actually glows. It is ionizing the neon in its vicinity by donating an electron, making negative neon ions, which migrate to the anode side which is dark. This is the rectification action! When connected to AC, of course which electrode is negative changes at the line frequency, so it looks like both electrodes are glowing. They are not.

So that is why reversing the neon has no effect on the rectification. You have just put the other electrode inside the envelope, in contact with the negative polarity of your voltage source.

Now think about this: Those neons below are in series. That means the glowing electrode of one is directly connected to the dark electrode of the other one, by a little bit of wire. One glows, the other does not. Why? (For a real laugh, ask TA to explain it.)

Well that makes no sence to me TK,as both electrodes would be negative at one point,as the source is AC,which would be equal and opposite to the other. So the net result of electron donating should be equal-unless the rise and fall of the electromagnetic field around the tower is not symmetrical???

In your picture above,are the neons being powered by a hard wired voltage,or wirelessly?.

Offline TinselKoala

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Re: Slayer driven neon-producing DC via resistor ?.
« Reply #14 on: August 22, 2014, 02:03:12 AM »
Well that makes no sence to me TK,as both electrodes would be negative at one point,as the source is AC,which would be equal and opposite to the other. So the net result of electron donating should be equal-unless the rise and fall of the electromagnetic field around the tower is not symmetrical???

In your picture above,are the neons being powered by a hard wired voltage,or wirelessly?.

This goes back to the issue of "what is AC and what is DC". Alternating means just that, alternating.

Take a neon and a resistor in series and connect them to an AC supply. Now consider the polarity of the voltage at one neon terminal. The voltage is positive during half the AC cycle, and it is negative during the other half of the cycle. Right? And the other electrode of the neon, connected to the other side of the mains supply line,  is negative while the other is positive, and is positive while the other is negative. Right? So the electrodes glow alternately, at the line frequency. One is on, the other is off. Then the one is off and the other is on. The negative-most electrode is the only one that glows, as the DC case shows.

Now... voltage is _relative_ and so is polarity. In the DC case, where two electrodes are connected together, but one is glowing and the other is not... they are both at the same _potential_ wrt some external reference like true Earth ground.... but with respect to the supply voltage to the stack, you can see that the "polarity" at any point is also relative. Consider two batteries connected in series. Look at the "middle terminal" where the positive of one battery is connected to the negative of the other battery. What is the polarity of this point? Is it negative, or positive? It is Positive wrt the Negative of the entire stack, but it is Negative wrt the Positive of the entire stack. In the neon stack, the two connected electrodes are at the same potential, but in one tube this is negative wrt the other electrode so it glows, and in the other tube that same potential is positive wrt the other electrode so it does not glow.

In the apparatus above, the neons are lit by the DC output of the HV Wireless Receptor, that I demonstrated in one of the microQEG videos.
http://www.youtube.com/watch?v=jcGTBA7NoVI

 

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