Hello all. I know this has been discussed before on this web site, but be darned if I can find any examples, so I figure I would make an official inquiry on the forum.

I have concluded that the beat up Coleman lantern I acquired a couple of months back is not worth restoring. The SLA battery is shot, and it's missing a part required to work which is impossible (and not worth the money) to find. However, because the ballast is fully functional, I figure as a project I can harvest it to make an instant start 8 watt fixture.

Consider this ballast, or any simple two-wire instant start ballast that runs on low voltage DC. Would it be possible to "convert" it, per se, to rapid start by simply connecting the low-voltage power supply directly to each filament? Or is it more complex than that because the low voltage supply will essentially be connected in parallel the high voltage output? If it's not feasible in a simple manner, would connecting the power supply to only one filament be?

As well, consider this idea; many moons ago I owned an LOA F17T8 undercabinet light, which utilized an instant start ballast, and the other two pins of the lamp were coupled with a capacitor to make a faux rapid start setup. Would this trick work with the given ballast? Would it actually do a good job of heating the cathodes given the right capacitor? Might it cause problematic stress on the ballast?

The capacitor trick is definitely simpler to set up. However, I can only assume connecting the power supply directly to the cathodes would result in better heating. So I'm just wondering how feasible both of these scenarios are.

Thanks for any thoughts.

The filament heating idea would be difficult since you can only heat one filament to avoid the H.V side of the ballast from interacting with the battery and even so you'd have observe the polarity of the H.V to avoid shorting the output of the ballast to ground. In commercial DC ballasts with preheating only one side of the tube is preheated so unless you have two SLA batteries on each side of the tube to provide electrical isolation it would be complex I think. The best and easiest way is to use separate heating transformers on each side of the tube hooked up to a dimmer to adjust filament voltage to suitable level. But to have rapid start effect the transformer has to switch on with the battery no certainty it will have the effect or not I have not tried it. 

As for the capacitor idea I did it before but the ballast I used was a 12v DC for 20w T8 two wire output ballast to run a 8w T5 tube. The capacitor I used was the yellow high voltage capacitor I forgot the values it was many many years ago. When switched on there is no light from the tube but after about a second or more it slowly brightens up so I guess there is some preheating until the tube conducts. But it made the ballast very hot since it was running a 8w T5 so the excess power was absorbed by the ballast as heat. But for your case I don't know since your ballast is designed for 8w T5 and the power from the ballast may be insufficient to heat the filaments for the tube to conduct.

The capacitor could be used only as a part of the resonance circuit, so it is used only in ballasts using the same concept as the mains powered electronic: Generate ~100..200V HFAC (in the mains electronic this is done by the halfbridge) and feed it via an inductor to the tube, with resonant capacitor parallel to it. With this concept, when the tube is not lit, the inductor and capacitor form a series LC, connected in series with the filaments. So when excited close to the resonance frequency, a high current will form, heating up the electrodes, and at the same time high voltage appear across the capacitor to ignite the arc. Once the arc ignite, it's low impedance damp the LC so, it become effectively only an R (the arc) in series with the inductor, connected to the AC source (the inverter). Now the low voltage differ from the mains ballasts only in the way, how they generate the 100..200V HFAC. The mains rectify the mains and use the halfbridge. The low voltage use a transformer between to boost the voltage from the 12..24V to the required 100..200V, while the primary could be connected either as a full bridge, half bridge (24..48VDC), or a center-tapped primary push-pull (24V and below). But otherwise these ballasts operate exactly as the mains ones.

The inverter you have use different concept, flyback, to generate the high voltage. This lead to the simplest circuits, but it have their drawbacks:
As the lamp is cold started, the only thing, what heat up the electrodes, is the ion bombardment. Beside of the lamp life problems, it mean only one electrode get heated, because the high voltage is present only in one polarity. That cause the lamp to act as a rectifier, so effectively operate the lamp at pulsed DC (the current flow only when the transistor is OFF). It have the advantage to provide quite wide room to adopt toward varying input voltage ("6V" battery mean operating voltages in the range of 3.6..7.5V), but it tend to cause mercury migration problems in the lamp, so not usable for longer, more efficient tubes.

To power the electrodes, you may add additional secondary windings. The turns should correspond to ~3.5V from the primary winding.
It will do two things: The filament resistance parallel to the transformer will limit the ignition voltage, mainly when filaments would be cold.
Second, when the lamp is not yet ignited, the inverter generate way higher rms voltage across all windsings, what mean the heaters will receive elevated  voltage for heating before the lamp ignite. Once it ignite, the arc reduce the voltage, so the filament voltage reduce as well.
Overall, you will get a rapid start ballast...

But never connect any capacitor to the secondary on a flyback inverter, it will overload the transistor before it would be able to heat up the filaments.


And other option is to use the Royer oscillator ballast, again auxiliary filament windings. There the capacitors could be calculated so, the filament current disappear, once the lamp ignite.

Thanks guys, that's good info. So suppose I went the route of heating only one side directly from the power supply - it can only be done on one end? Is the high voltage output with respect to ground, so I would just have to find which end is connected to ground? Or would the correct side have to be determined another way?

In many circuits the lamp is not isolated from the primary, so connecting the filament to the battery will disturb the circuit (sometimes it is connected to the feedback - just to suppress the battery voltage variation). Generally there are many variants of the flyback circuit in use and even when the principle is the same, the details do matter a lot.
So without knowing exactly, how the circuit is connected and how it works, better do not alter it in any way.

Okay, so I'm resurrecting this thread because I just had another idea. So we've established that cathode heating can't be rigged up as follows (first picture), because the output side of the ballast isn't isolated from the power source. But what if something like diodes or a rectifier were introduced, seen in the second picture, which allowed current to flow from the power source to the cathode(s), but not the other way around? Or wouldn't this work because the output from the ballast is AC?

The last was the an answer: It won't work, because the output is an AC voltage.
Don't get confused by many observations and descriptions stating the lamp is fed by "pulsed DC", as these descriptions mean the lamp current (so what is actually generating the light), but not the voltage.
The voltage is always AC with no DC component across the lamp, the complete secondary could be sometimes DC shifted, but this shift is very little compare to the AC component (when connected somewhere inside of the inverter circuit).
It is the lamp itself, what converts the AC voltage to the pulsed DC current, because initially the voltage is highly asymmetrical, so heats just one cathode, so after run up the lamp itself become the asymmetrical component, so the rectification remains, even when the voltage becomes about symmetrical with many designs.
And the rectification effect is very important there for the current stability: It ensures, the lamp draw current only when the transistor is OFF, so the voltage could be freely dictated by the arc. During the ON time the voltage is tied to the input, so it would be a voltage source directly on a discharge, what is everything but stable.

Yes, I was assuming a ballast with an ideal AC output. Okay, that answers my question.