I expect whats below will cover this well enough.
Not really - see below.
that is not the circuit of a fl lamp though, nor is it comparable.
They are actually very comparable as far as the negative incremental
impedance problem is concerned, which is the issue here. Why do you
think that are not comparable?
I should have written dv/dt, my mistake. Cap current depends on dv/dt
across the cap, so is controlled through the cycle, except at high
frequencies.
At hf:
firstly hf makse up a minority percentage of the lamp current
second, lamps care primarily about temperatures, and temp change takes
time.
The current runaway in fluorescent lamps is not related to lamp
temperature, it is a result of increasing electron density. And, the
electron density can increase quite rapidly. If you were to connect
your 60-volt lamp to the 240-volt power line the current would rise to
very high values in less than 1 msec, probably less than 100 usec.
Other than the free electrons, which have so little mass that they
don't add to the "temperature" of the lamp as we conventionally think
of temperature, nothing would get hot except for lamp electrodes which
would be vaporized by the high current.
I understood this was evened out to a good extent at 50 or 60 Hz
though.
It is not "evened out." The current varies as a distorted sine wave,
while the voltage is almost a square wave, so the conductance is
changing constantly over a 50 Hz half-cycle. And, the only reason why
the current follows the shape of the power input is because a proper
ballast is being used.
The lamp conductivity is still there during the <60v portions
of the mains cycle, if it were not the lamp would have to be restarted
100/120 times a second on ac.
Near the zero crossings the conductivity has decreased quite a bit due
to electron-ion recombination. This can be seen from the fact that the
lamp voltage is still quite high while the current is almost zero. At
the zero crossing itself the lamp voltage finally collapses. Even on
an inductor ballast there is a "reignition" spike on the lamp voltage.
One advantage of an inductive ballast is the voltage kick that is
generated at the zero crossing of the lamp current that helps to
restart the lamp.
It never needs that, even on a cap
ballast that keeps the i at zero for part of every half cycle.
If you had voltage waveforms you would see that the lamp voltage is
much higher than ~ 60 volts at the start of each cycle when you are
operating from a C ballast. And, the longer you keep the lamp off, the
higher the reignition voltage spike will be. The reason your lamp
restarts without intervention from you (such as reheating the
electrodes) is because your peak operating voltage is more than 5.5
times the operating voltage of the lamp, and the electrodes have a
reasonable thermal mass.
thats for how much change though? Conductance may reduce, but it doesnt
cease altogether, even while i does. Excuse my ignorance on this point,
but isnt there more going on in there than just e- clouds, like ionised
Hg and Ar?
There is a classic paper by PC Drop and J Polman titled "Calculations
on the Effect of Supply Frequency on the Positive Column of a
Low-Pressure Hg-Ar AC discharge," J. Phys. D: Appl Phys., Vol. 5,
1972. Figure 2 of that paper shows the various quantities in the
fluorescent lamp discharge as a function of time over a 50 Hz
half-cycle. The electron density generally follows the current
waveshape. At the start and end of the 50 Hz half-cycle the electron
density is about 25% of the peak value. There is a clear "reignition"
peak in the electric field at the start of each half-cycle which is
about 1.4 times the mean value of the electric field.
There is no significant amount of ionized Ar in the positive column.
Any ionized Hg must be balanced by free electrons since the plasma
must be neutral, so the Hg ion density varies as the electron density.
yes, for lf components, which make up the bulk of the waveform.
yup, i is proportional to dv/dt
no, really. Current is proportional to dv/dt. The xenon flash dumps its
load without delay, the capacitor ballast doesnt.
I don't understand your comment that the xenon lamp dumps its load.
The energy is stored in the capacitor, not the lamp. The energy is
"dumped" from the capacitor into the lamp. The flash is intense and of
short duration because of the nature of a capacitor and the high
conductance of the xenon discharge.
In a fl lamp the
mains waveform controls the current through the cycle, along with other
factors.
The lamp current waveform will follow the mains waveform only with a
proper current limiting ballast.
yes, but
a) only to a limited extent
b) only for a small limited time, not enough to cause destructive
thermal rise
See discussion above about "thermal" effects. Thermal inertia as you
normally think of it does not control current runaway in a low
pressure discharge.
Re that limited extent, what is the delta v on a 2' 60v lamp during
this increasing conduction time? Youve got a ballast running on 240v
ac, which is 330v peak, with a 60v lamp of just 20w. So really, how
much extra power is one putting through it for how long?
I don't understand. Using the data from Drop and Polman, if you had an
inductive ballast the voltage at the start of each cycle would be
about 1.4 times the average operating voltage. If the discharge is off
for longer than the short period of time than is typical of an
inductive ballast, the voltage spike at the start of each cycle would
be higher.
Filament lamps have their power turned on and off 100 or 120 times a
second, and they dont mind a bit. And fl lamp filaments have more
thermal intertia, more temperature headroom, and only see a fraction of
the total lamp power.
As discussed above, this is not about the thermal inertia of a
tungsten filament.
yes and no. Having done it I know it works, with a 60v lamp and a 240v
supply.
But you still haven't said that you ever looked at the lamp current
waveform. Certainly, with a low enough lamp voltage the current will
be the same shape as the line voltage. When a capacitor is placed
directly across an AC power line, the current through it matches the
shape of the voltage waveform. So if we imagine that we have a lamp
whose operating voltage is, say, 1/1000 of the line voltage, then the
current distortion caused by this very low voltage non-linear load
placed in series with the capacitor would be small and the current
would still be substantially sinusoidal. As the voltage of the
non-linear lamp becomes a larger fraction of the line voltage, the
amount of current distortion will increase. I believe that your case
will have significant current distortion, but until I model it or
measure it I cannot tell you how much distortion.
If OTOH youre running a 100v lamp on a 110v supply, the
situation will be way worse. I'm fairly sure this is why we have had
quite different results.
would be interesting. I dont know what sort of capacitor ESR one would
need to figure into it, that might be relevant. But I'll bet you the
60v lamp 240v mains results will look very different to a 100v lamp on
110v.
I am assuming a perfect capacitor, that is zero ESR. If you want to
add a partial resistive ballast later that is fine, and it will
improve the waveform. And, yes a 60-volt lamps on 240-volt mains will
have lower CCF than a 60-volt lamp on 120-volt mains. A 100-volt lamp
will not operate on 120-volt mains with a capacitor ballast. In fact a
100-volt lamp will not run properly on 120-volt mains with an inductor
ballast, but that doesn't explain how those cheap LC ballasts manage
to operate 100-volt lamps from 120-volt power lines. .
--
Vic Roberts
http://www.RobertsResearchInc.com
To reply via e-mail:
replace xxx with vdr in the Reply to: address
or use e-mail address listed at the Web site.
This information is provided for educational purposes only.
It may not be used in any publication or posted on any Web
site without written permission.