Hello everyone,

i am looking to start my own Plasma experiments. But i have some basic questions i would like to ask. I have been looking at other forums and groups, and their seems to be lots of different way to get started.

Super God: your schematics refer to using 110v and 40kv.....

Tero: uses 110vac into a bridge rec, then its similar to Super God's

Capacitor70: 230v into transformer to 12v then into a transformer to 230v (for isolation i guess)

S1r's: Relay with HV (but know one knows where to get the relays..) and does not use a bridge rec.

Remoi: Just like Tero, but more up to date.

mdbreedi: FANTASTIC video on youtube.

So i guess what i am getting at is..what direction to take... i would say that a 230v system is not for me, i would like to keep it to a 110v system. SO do we want straight 110v with low amps or use 110v to make it DC with high amps? I am surrounded with HV, i work on arcade monitors, i used one to make a jacobs type ladder.

thanx!
-bill

I think the s1r9a9m9 set-up uses HV to switch the relay on/off which passes 110v to the plugs, however because of the high voltage, i bet it jumps from inside the relay to the plug which is why he was getting readings of 24,000v with 4amps or whatever.   You dont need relays as the spark gap theoretically completes the circuit,  protecting the inverter is the hard part.

HV Diodes are available on ebay for fairly high prices, id just really like to know how many amps the ign coil produces with its (so called) 30,000v spark.  Based on calculations it would be 0.002 amps or 2mA, therefore a 1amp diode is overkill and 10ma 20kv diodes are very affordable.   Anybody have any idea how to measure the amps and voltage of an ignition coil?  Ive noted a CRT flyback rated at 30kv makes a far more powerful spark than ign coils suggested to have the same volts.

A car the  ignition  system  can work well for the  high  voltage and  timing  control
A common  110 V inverter  is used for the  extra current .

Once  the  high  voltage  jumps the   gap in the spark plug  it   creates  a  plasma " channel " that is  relativly  low resistance .........  the much  lower voltage  can   follow this   path as long as the high  voltage maintains  the  channel
The high  voltage is in effect the switch  for  the lower  voltage .

It is the lower  voltage  ( high current )   that   flashes  the water  into  steam .

gary

Quote from: capacitor70 on June 13, 2008, 12:53:50 AM
Once  the  high  voltage  jumps the   gap in the spark plug  it   creates  a  plasma " channel " that is  relativly  low resistance .........  the much  lower voltage  can   follow this   path as long as the high  voltage maintains  the  channel
The high  voltage is in effect the switch  for  the lower  voltage .

It is the lower  voltage  ( high current )   that   flashes  the water  into  steam .

Well I agree almost completely with you, except for one point. The High-Voltage usually doesn't have to be maintained to keep the arc, for the plasma channel, which is a good conductor is fed by the current from the low voltage source, and usually will remain if the current is strong enough and the voltage not too small. Here in the water explosion this could be different for I think the water explosion shockwave could be so strong as to extinguish the arc.
BTW: In my opinion this is nothing else than a Trigatron (http://en.wikipedia.org/wiki/Trigatron) where one does just connect the high and the low voltage source to the same electrode. But then you have to make sure, that the HV-Trigger doesn't destroy your low voltage source. This could be a problem in my eyes, for if you take a diode then it has to be rated HV, but at the same time, it needs to withstand a lot of amps (the more the better) for the conduction of the LV-Current. The more current you can deliver in a short time, the stronger should be the bang...
Well, Britt's AEROPS-Engine is quite the same, but he doesn't use water, but inert gases. But except that, he uses the same principle. But there he uses a "normal" Trigatron Scheme.

BTW, I just calculated for myself how much energy one could use for a working car engine. Maybe some are interested, therefore here are my calculations:

* Let's assume the engine runs at 3000 RPM = 50 RPS = 25 Firings per Second for each cylinder
  (a 4-stroke engine needs 720? per cycle)

* Let's assume the car has 6 cylinders, that makes 6*25 = 150 Firings per Second in total.

* Let's assume the alternator has a power of 3kW (= 4HP). Thats makes 3000W : 150Hz = 20 J per Firing

* Let's assume we are in Europe ( ) and have an european inverter which gives 230 VAC -> rectified -> 320 VDC

* Then the capacitor size for one firing should be C=2E/U^2=2*20J/320V^2=390uF

This is just for a rough estimate. The capacitor would not become completely discharged in reality, but on the other side one needs power for a HV Spark...

BTW, just an idea.

It could be interesting to test if making a magnetic Plasma Trap during the discharge would increase the Bang.This could be easily done by wiring a coil around the plasma region and letting the LV-Discharge first through this coil, prior to the water. This should compress the plasma and if the coil is wound correctly could even trap the plasma so that it gets hindered at going out of the spark plug, while the discharge is in progress. Maybe this could increase the efficiency even more.
I think I will test this tomorrow...

Another idea:

If one puts a high voltage on the water, (connect only one side of the HV source, so that no current is flowing) the water gets ionized and will then be repelled from the plug. I think if the HV is high enough and the amount of water small it will immediately spray out due to the repellant electrostatic forces. If then one would fire the HV (let current flow) and afterwards the LV (a lot of current) it could also be beneficial for the plasma explosion, if the water is finely dispersed in the room when the plasma is built.