Seems legit to me

But i think 1uF is way too much and 100 ohm is way too low. You dont need any significant current or capacity there, just to make tiny initial discharge in the lamp. Big capacity there means blowing off more emitter from tube electrodes at initial breakdown (when the capacitor discharge into the tube with the cold electrodes), low resistance means high current so you will have to take care about power dissipation - where you dont reallly need to dissipate any significant power in the 1st place....

Would be good to add some few Mohm resistor to bleed the capacitors down when not in use - that for safety when you mess with the circuit

In this circuit the D2 and R3 with their resistors do absolutely nothing, because they are plain parallel to the diodes within the bridge rectifier (and that is connected without any resistance at all, so will take all current in that direction).

The capacitors directly connected like this mean there is 0.5uF directly parallel to the lamp - that is quite too high to cause problems with current spikes like with capacitor only ballast. It does not matter that much, if the capacitor is in series with a low impedance voltage source, or parallel to the lamp. In both cases the lamp sees the capacitor as the impedance and that alone causes those current spikes.

So better connect the resistors in series with each of the capacitors. It will be still sufficient in the doubler mode for ignition, because there is no current (so no voltage drop), but once the lamp ignites, these resistors will separate the capacitors from the discharge.

In this circuit the D2 and R3 with their resistors do absolutely nothing, because they are plain parallel to the diodes within the bridge rectifier (and that is connected without any resistance at all, so will take all current in that direction).

Those relations between components are exactly the ones I don't notice, as I have some hobby-experience in electronics, but not in designing circuits, thanks. However, I'm not sure I understand you here. The bridge rectifier does not conduct until the tube strikes, right? So, the D2 and D3 with their resistors (although in parallel to the bridge) would be the only path for current at first, if I understand it right. But, I see that putting resistors in series with the capacitors is a better option, and will use that arrangement. I would gladly replace the bridge with a single diode for simplicity and low cost (as one is enough to create a high voltage barrier in direction of the bulbs) but I guess the tube would flicker badly and probably have reduced light intensity - both of which I don't want.

Regarding what Ash noted, what would be "the most optimal" values for capacitors and resistors, do you have any suggestions? He is right noting I need the lowest possible capacity and a highest possible resistance, as the ignitor has a rather simple task to do requiring very little current.

The "bridge rectifier" is no magic, it is nothing else than just a 4 regular diodes, only integrated into one package for the mechanical construction convenience. If you connect the four 1N4007, you will get electrically exactly the same, but it would mean four physical components on the board instead of one.

So with the structure you have drawn one diode from the bridge would have parallel connected other diode with a series resistor. So when any current would start to flow there, it will seek for the lowest resistance path and that would be through the diode inside of the rectifier bridge, as there is no resistor in the path.


For the ballast you really need full bridge, so 4 diodes. Even with that the time, where the mains voltage is not able to feed any current into the lamp (because the mains voltage is lower than the lamp arc voltage) is rather long to already cause significant overshoots (same as the resistive ballast I have shown with the oscilloscope pictures; just you will have all the time the same polarity), using just one diode will make it more than 10ms long, that would extinguish the arc...