I keep getting the explanation that "the current is spiky" and "it's hard on the electrodes", but why exactly? I hear about harmonics as well, I have no idea what that is about either. Please enlighten me...

Thanks!
 

I have said already (if you read carefully ofcourse!) that combining discharge negative resistance with a capacitance of the ballast causes this. If you still is not convinced, you can run some natural experiments with a neon lamp or a fluorescent tube and a capacitor, looking at current waveform on the oscilloscope, or run a numerical simulation in Spice, or just read a good ol' book on lamps and ballasts after all!

Theoretically, the term is called 'instability'. You can easily be buried by tons of differential equations (nonlinear, for sure) describing this, the math is extremely complex for such a simple circuit! In the beginning of the line period just as the discharge is reignited, the current is not stabilized as in the case of the inductor or resistor in circuit, but keeps growing because negative resistance of the discharge means lower voltage with higher current, and for the capacitor increasing rate of voltage rise (as line voltage minus discharge voltage) means (a differentiating behavior) a higher current. So a current spike appears. Then as capacitor gets charged, this process goes in reverse, potentially dropping lamp current to zero.


Here is a current waveform captured many many years ago on that crappy kind of fluorescent nightlights consisting of a short crude fluorescent tube and a simple capacitive ballast. The ballast is not even pure capacitive (a combination of a resistor and capacitor in series actually) but still you see how bad the waveform is.



 

I've tried it with a 8w lamp and a capacitor calculated to have the same impedance as an inductive ballast, treating the lamp like a standard resistor. It is not a good idea. Both lamp and capacitor make scary sounds.

If you zoom in time wise to one half mains cycle: Current spikes can happen if the lamp ignites with the capacitor being empty. The cap won't charge if the tube is not conducting. But once it does, the lamp itself is practically a short circuit (see: negative resistance) and the capacitor will charge via the lamp with theoretically unlimited current (in practice there will be a limit, but it's way too much anyway). 

As for exactly why spikey current with high current crest is bad for lamp life one can imagine that many processes like sputtering and evaporation are not linearly proportional to current density, but more likely are a function of some high order or even exponent. So for the same average current, high current crest means significantly higher wear. Also, for fluorescents, high current crest will mean less efficiency again because of higher momentary current density. Interestingly, for some lamps the reverse is true - xenon dielectric barrier discharges definitely work better with spiky current.

This especially goes to delicate fluorescent lamps filaments with tiny structures with low thermal mass. Some massive electrodes of for example high power mercury lamps are significantly more robust and tolerate some parallel capacitance in the circuit and also high current crest of CWA circuits. Also it seems that HID lamps have some larger degree of gas ionization so show less extreme oscillations when used on capacitive circuits compared to low pressure fluorescents.

Interesting explanations!  I have never really considered it before, but what would be the consequences for a fluorescent lamp run on an LC series ballast?

For instance, the first British fluorescent tube was the extremely overloaded 5ft 80W, born out of the WW2 emergency situation that required such a short tube on the then smallest available ballast of an 80W high pressure mercury lamp. 

Very soon it became obvious that a higher power tube was required, and the length was extended to 8ft and the rating upped to 125W.  Contrary to popular belief that was not used on the 125W high pressure mercury ballast, which would have been impossible since the longer tube resulted in a lamp voltage whose reignition spikes exceed the available mains voltage.  The solution was to add a series capacitor and use the same old 80W ballast.  I suppose this must have introduced a phase shift to at least partially delay the lamp reignition until a higher mains voltage was available, closer to the peak of the sinus.  Or maybe instead the capacitor was by that point fully charged and its voltage would be added to the sinus to help the tube reignite.

Either way, what would have been the effect on the current waveform delivered to the tube and the consequences for its life, and the system efficacy?

With series inductor, the spikey problem of the capacitive ballast is gone - the inductance smooths out the rise times of the current, so the peak won't go that high.

For the series LC and zero cross reignition:
The important thing is, what is the ballast OCV at the instant when the current drops to zero, so the arc goes away. Because the arc went away, the circuit becomes interrupted so no voltage across the inductor. So with normal coil only ballast, the mains voltage present at that instant becomes present across the lamp. Because of the phase shift, the voltage uses to be close to the mains peak, usually above the voltage needed to reignite the discharge in the gas. And once reignited, the circuit continues as usual - current buildup when the mains is greater than the arc drop (there the energy gets stored within the magnetic field), current collapse when the mains voltage is below the arc drop (then the energy from the magnetic field gets transferred to the arc), until the current drops to zero and the same repeats.
The point of the series LC is, at the point of the current falling to zero, the capacitor is charged to a significant voltage. Until the discharge reignites, this voltage is added in series with the mains, so the sum of both becomes present across the lamp electrodes as a voltage available for reignition. And because it is a sum, it can easily be larger than the mains voltage, in fact (assume zero losses) there is no theoretical limit how high it could go, if the mains frequency is equal to the resonance frequency of the series LC.
The practical limit is imposed by the sensitivity to tolerances of the inductance, capacitance and the mains frequency, usually you may reach factor of 2 or so reignition OCV boost.
At the same time in order to pass at least any current, the mains still needs to be bigger than the 1'st harmonic of the lamp arc voltage.
That means the OCV is not anymore a limit to operate the lamp at a given mains voltage (and without any transformer), but just the arc voltage to be smaller than the mains and the ability for the LC to cover the tolerances of the difference. And with this tolerance, the inductance drop at higher currents is helping here - it detunes the circuit more at higher currents, so is able to compensate the rather large variation of the voltage drop available for the ballast.
So you can easily have a 180V arc lamp operating at 240V mains on a series LC, it still could have easily more than 600V available for the reignition, very comfortable even for things like a 180V arc lamp.

Well, I am definitely in need to borrow a nice modern hi-rez 12 bit 4-trace scope at work to capture some quality waveforms of such circuit. Stay tuned.

In my understanding how the things works, LC leading ballast for fluorescents forms a series LC tank tuned to a frequency slightly above line frequency. So at the fundamental (50Hz) the behavior is capacitive, for lamp current/voltage harmonics it is still inductive. 

Capacitor voltage is lagging behind lamp current in a series circuit. That means some charge (voltage) is carried over from a previous line half-period and then added to line voltage in the next. Adding to inductive kick still present from the inductance that is in the circuit too, leading circuit eases re-ignition and allows for lower line voltage to arc voltage ratio while still keeping the discharge stabilized. Additional useful behavior is likely a ferroresonant regulation meaning that increasing current saturates the choke core to more degree decreasing its inductance but increasing the overall circuit impedance. So the ballast exhibits some degree of constant current regulation, more so than of pure inductive (lagging) one.

Inductive part of the circuit isolates the capacitor at line frequency harmonics from the lamp, suppressing most ill effects of pure capacitive circuit. But the resulting higher current crest from leading current behavior is still evident, and there is a penalty on tube life in 20-30% range.

The same is likely true for HID CWA circuits, though these are extremely rare where I live.

The 20..30% life penalty on tube life with preheat circuits and standard switching pattern (3h ON/ 0.5h off) comes mainly from insufficient preheating current in the lead branch.
Plain choke (lag) ballast tends to increase the current when the load is effectively shorted (preheat phase, partly because of the missing lamp drop, partly because of slight core saturation) and lamp filaments are designed for this elevated preheat current (so a 0.37A F18T8 is designed to get about 0.5A during preheat). But in the lead circuit the ferroresonant current stabilizing mechanism does not allow this current increase during preheat, it stays at the normal (0.37A for the F18 example), so about 20% lower than the tubes are actually designed for. Plus the extra capacitor voltage added on top of the mains causes the lamp to ignite easier, so with electrodes less preheated. So the bottom line, the starts happen at cold cathodes, so with extra sputtering wear.

In the past this was overcome by using "starter compensation coils" in series with the starter, shifting the resonance so the preheat current becomes closer to its rated value, But this adds quite significant cost and as second, it causes extra voltage spike seen just by the starter, so causes the ignition voltage spike to get reduced. It was not a problem with T12 tubes which tend to have rather low ignition voltage, but it would become problematic with T8's requiring higher ignition spike. This ignition spike reduction could be fixed by some extra MOV or capacitor parallel to the starting compensator, but that complicates the circuit even further.
Because this compensator fixes only starting wear and brings other problems and complexities, while most fluorescent installations were operated for way longer than 3h/start so the starting wear became less relevant, its use has stopped and the industry just accepted the extra starting wear from the insufficient preheat.

I have said already (if you read carefully ofcourse!) that combining discharge negative resistance with a capacitance of the ballast causes this. If you still is not convinced, you can run some natural experiments with a neon lamp or a fluorescent tube and a capacitor, looking at current waveform on the oscilloscope, or run a numerical simulation in Spice, or just read a good ol' book on lamps and ballasts after all!

Theoretically, the term is called 'instability'. You can easily be buried by tons of differential equations (nonlinear, for sure) describing this, the math is extremely complex for such a simple circuit! In the beginning of the line period just as the discharge is reignited, the current is not stabilized as in the case of the inductor or resistor in circuit, but keeps growing because negative resistance of the discharge means lower voltage with higher current, and for the capacitor increasing rate of voltage rise (as line voltage minus discharge voltage) means (a differentiating behavior) a higher current. So a current spike appears. Then as capacitor gets charged, this process goes in reverse, potentially dropping lamp current to zero.


Here is a current waveform captured many many years ago on that crappy kind of fluorescent nightlights consisting of a short crude fluorescent tube and a simple capacitive ballast. The ballast is not even pure capacitive (a combination of a resistor and capacitor in series actually) but still you see how bad the waveform is.

I've tried it with a 8w lamp and a capacitor calculated to have the same impedance as an inductive ballast, treating the lamp like a standard resistor. It is not a good idea. Both lamp and capacitor make scary sounds.

If you zoom in time wise to one half mains cycle: Current spikes can happen if the lamp ignites with the capacitor being empty. The cap won't charge if the tube is not conducting. But once it does, the lamp itself is practically a short circuit (see: negative resistance) and the capacitor will charge via the lamp with theoretically unlimited current (in practice there will be a limit, but it's way too much anyway). 

Thank you both so much! This actually helps a lot and makes total sense.