In reply to Jones Beene's message of Sun, 3 Aug 2008 19:32:00 -0700 (PDT): Hi, [snip] >http://www.youtube.com/watch?v=wlMELbK8zDQ [snip] Initially the air in the spark gap in non-conducting, and the spark gap is a very large resistance. So the voltage over the spark gap follows the voltage on the output of the coil as it rises. During this phase, the diode is connected such as to be non-conducting. However when the breakdown voltage is reached, and a plasma starts to form in the spark gap, everything changes. The resistance of the spark (plasma) suddenly drops to a fraction of an ohm, and without the diode, the current would be limited by the ability of the coil to supply it. This aint much (because coils don't like their current changing rapidly). However when the diode is present, and the plasma is shorting out the spark gap, the "high" voltage side of the coil is effectively shorted to ground through the plasma, hence the voltage at that point is actually quite low (nearly ground voltage). This means that the diode is now free to conduct current from the "low" voltage positive side of the circuit to ground through the plasma of the spark gap. If the output side of the low voltage circuit has a decent electrolytic cap. across it, then this can discharge through the diode and the plasma, resulting in a lot of energy being dumped into the plasma very rapidly. That's why the diode makes such a difference (and it is by far the largest difference to the circuit).
The water vapor may just provide extra ions, reducing the resistance of the plasma, or there may be an additional kick due to e.g. Hydrino formation, but this is by no means certain. Regards, Robin van Spaandonk <[EMAIL PROTECTED]>
