harry wrote in
:



So most of the volt drop changes from being across the tube at the
instant of switch on (and the starter as it is in parallel) to being
across the ballast.



But if the tube hasn't "struck" yet, then wouldn't there be an absence of
current-flow through it until it strikes from the HT surge? Or does curent
begin to flow during the ionization period?



So when the tube is running it "shorts out" the starter, there is
insufficient voltage to cause a discharge.


I guess because the tube offers a path of less resistance than the now-open
starter?



The tube has a virtually nil resistance when running due to the
mercury vapour in the ionised arc and it has a relatively large cross
sectional area.
The starter needs a higher voltage than is available during running to
initiate the discharge as it has no mercury and the gas pressure in it
is higher plus it is a different gas.

All very simple in theory but it took years of experiment to make it
work.



I'll bet. Like a lot of seemingly-simple technology.

The system seems to be basically two incandescent bulbs in the same housing
with their light being produced by the gases between their filaments
instead of from the filaments themselves. This seems to be the reason why
fluorescents consume less electricity than incandescents: it takes much
less current to keep gases glowing than to keep filaments glowing.


--
Tegger