Going back to my comments about adding a resistor in parallel with the two diodes in series with the gap. A perfect silicon diode would have close to 0.6 volts across is even with only a few microamps of current. So with no parallel resistor you could have about 1.2 volts due to leakage current through the dielectric. With a parallel resistor of say 100 ohms the voltage would be almost zero. When the dielectric breaks down the current would be many tens or hundreds of amps so the voltage across the two diodes would rise to around 1.5 volts. It is this change that you are trying to detect to step your logic onto the next step.
Hi Les, The reason I wasn't going with a resistor with the two gap diodes was to keep the full voltage (open circuit voltage) from being sent across the gap. Since the cap bank voltage will be somewhat smaller than open circuit voltage. Doing this will make the finish of the burn more consistent, or at least that's my thinking. I am adding a resistor to the other diode on that comparator.The diode you are using to set the difference between the lower and upper thresholds of the window detector may not have any current trough it (And therefore no voltage across it) This is because the input leakage current to the op amp/ comparator is very small and can be positive or negative. I think you are hoping for about 0.6 volts across this diode.
Most of the DIY machines are just relaxation oscillator based. So some of the voltage/current from each discharge into the gap is used to ionize the dielectric. By keeping the Q1 on for the time it takes to ionize the gap, it should then give the cap bank it's full energy store into the burn. Again that's my thinking behind this.
I guess I should call this circuit a digital relaxation oscillator?
I wasn't trying to disparage your circuit, just trying to be funny. I really do appreciate all of your help with this. I just have a weird sense of humor.My circuit is quite simple when you break it down to blocks.