Slayer driven neon-producing DC via resistor ?.

[tinman]

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

Well im not sure how using a DC reference in this case(the batteries in series)is the same as what we have here,as we are dealing with an AC source. So insted of having two batteries conected in series,we have a center taped transformer,say 12/0/12 AC. I can clearly see in my neon that both pins are glowing. Further testing shows it is the line with the resistor in series that has the +DC component,regardless of which side of the cap it is on.This can only mean that the resistor is offsetting the charge rate of one side of the cap-some how?.

What you are showing TK,is that the pin in the neon that has the negative polarity is the one that glow's-this we knew. What im showing is how a resistor some how offsets the charge rate to the plates in the cap. Now oddly enough,(and this seems ass about) is that the higher the value of the resistor,the higher the DC value in the cap is reached.

[MarkE]

The rectification effect is weak.  If it were strong you would see a good percentage of the neon hold voltage on the capacitor.  You can try scoping the neon bulb, but the probe is likely to distort the voltage.  You also want to be careful not to hurt your scope.  If you are going to try that, I would do it through a series 1Meg resistor.  That will cost some bandwidth.  At these frequencies that should not be a problem, and it is a lot better than frying your vertical amplifier.  You are looking for asymmetry in the waveform.

[tinman]

Below is a simple test i carried out. It would seem one pin in the neon is recieving a higher charge than the other. Although both wave forms are in phase,we can see a potential difference between the two.There dosnt seem to be any offset in either trace though.

[MarkE]

That's just where each of the wires are in the RF field.  The capacitor charges from the difference potential.  You want to measure from TP1 to TP2 or vice versa.  You also want the connections in your test set up from the bulb towards your scope to have as little exposure to RF as possible.  You could use a piece of coax from the bulb to get away from the field.  Or you could use nice short connections such as soldering the two bulb side resistors to the bulb with short leads and wrapping the ground clip lead around the probe hook to minimize the ground clip loop area.  The difference may still be hard to see because it is only about 1% of the signal.

[tinman]

That's just where each of the wires are in the RF field.  The capacitor charges from the difference potential.  You want to measure from TP1 to TP2 or vice versa.  You also want the connections in your test set up from the bulb towards your scope to have as little exposure to RF as possible.  You could use a piece of coax from the bulb to get away from the field.  Or you could use nice short connections such as soldering the two bulb side resistors to the bulb with short leads and wrapping the ground clip lead around the probe hook to minimize the ground clip loop area.  The difference may still be hard to see because it is only about 1% of the signal.

As all the cables run together,and are of same length,i dont see this as being the cause for the cap charging up. There is also the issue that it seems to be the pin in the neon that has the positive charge,is the one that light's up in my setup-this will be seen in the next video,as soon as it has uploaded. The resistor determonds as to which side of the AC cap has the negative potential-the side with the resistor is negative. resistor value also determonds the voltage potential reached in the cap.