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Slayer driven neon-producing DC via resistor ?.

Started by tinman, August 20, 2014, 10:01:15 PM

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tinman

Quote from: MarkE on August 21, 2014, 10:13:17 AM
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.

magpwr

Quote from: tinman on August 21, 2014, 10:29:06 AM
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.


TinselKoala

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.)

tinman

Quote from: TinselKoala on August 21, 2014, 02:00:26 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.)

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?.

TinselKoala

Quote from: tinman on August 21, 2014, 07:22:42 PM
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