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Florescent lighting?

T

Tim

Jan 1, 1970
0
Is it feasable, or doable to dim an 18 watt Tri phosphor white tube ( 2
foot)?
Looking for about 50%intensity range.

I think they need a voltage across them to maintain ionization, but perhaps
current control?
Any pointers?
 
V

Victor Roberts

Jan 1, 1970
0
Is it feasable, or doable to dim an 18 watt Tri phosphor white tube ( 2
foot)?
Looking for about 50%intensity range.

You haven't given us the exact lamp type but almost all fluorescent
lamps are dimmable. The only exceptions I can think of are lamps that
have an integral glow switch starter, and almost all of these are
compact fluorescent lamps. (The reason dimming is a problem when the
lamp has a glow switch attached, is that the lamp voltage rises when
the current is decreased and this can trigger the glow switch for some
lamp types.)

Instant start fluorescent lamps, those with only one pin on each end
or those that have two pins on each end that are shorter together, are
dimmable, but the dimming range is limited since the electrodes cannot
be heated from an external source.
I think they need a voltage across them to maintain ionization, but perhaps
current control?

All discharge lamps, including fluorescent lamps, must be operated on
a ballast that controls the lamp current while providing the voltage
that the lamp wants to run at. To repeat - you choose the current. The
lamp chooses the voltage. Just the reverse of the more common voltage
source.
Any pointers?

Google is your friend. Also, International Rectifier, www.irf.com, has
a number of application notes on ballast design, including circuit
diagrams. Note that all their circuits are designed to use their
ballast control ICs, which are not necessary in other designs.

--
Vic Roberts
http://www.RobertsResearchInc.com
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T

Tim

Jan 1, 1970
0
Thanks for the feedback, I would be happy with being able to dimm from
100% brightness, to about 50% level.
The tube type is Sylvania, 18W/860 daylight with a diameter of about an
inch, and two pins at either end. I was wondering if I can add any other
components to a standard ballast, to vary/limit the current. I would like
to be able to vary the brightness to see what output is right for a custom
made light fitting, and then set the level at that setting pemamnently. I
guess I may need a variable current source, and some understanding of the
operation of the existing ballast, or is there a better, simpler way of
achieving what I need??
 
V

Victor Roberts

Jan 1, 1970
0
Thanks for the feedback, I would be happy with being able to dimm from
100% brightness, to about 50% level.
The tube type is Sylvania, 18W/860 daylight with a diameter of about an
inch, and two pins at either end. I was wondering if I can add any other
components to a standard ballast, to vary/limit the current. I would like
to be able to vary the brightness to see what output is right for a custom
made light fitting, and then set the level at that setting pemamnently. I
guess I may need a variable current source, and some understanding of the
operation of the existing ballast, or is there a better, simpler way of
achieving what I need??

Whet kind of ballast? If you are using a line frequency EM ballast
then you can dim the lamp to some extent by adding additional
impedance between the lamp and ballast - this was done in the early
days of the energy crunch in the 1970's, before low energy lamps and
ballasts were available. However, if the lamp is rapid start you still
need a way to heat the lamp electrodes. Those early low wattage lamps
with built-in series impedance also had a small 1:1 transformer added
to carry the electrode heating current around the additional series
impedance.

If you have an electronic ballast, the various designs are too
different to suggest an external solution. If you understand
electronics, you could go into the ballast and change the operating
frequency, which will change the lamp current. However, this requires
a substantial bit of knowledge. Messing with an electronic ballast is
quite difficult and risky, for both you and the ballast.

Have you considered purchasing a dimming ballast made for this lamp?
Or a ballast made for a lamp that runs at about 1/2 the current of
your lamp but the same voltage?

--
Vic Roberts
http://www.RobertsResearchInc.com
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A

Andrew Gabriel

Jan 1, 1970
0
Thanks for the feedback, I would be happy with being able to dimm from
100% brightness, to about 50% level.
The tube type is Sylvania, 18W/860 daylight with a diameter of about an
inch, and two pins at either end. I was wondering if I can add any other
components to a standard ballast, to vary/limit the current. I would like
to be able to vary the brightness to see what output is right for a custom
made light fitting, and then set the level at that setting pemamnently. I
guess I may need a variable current source, and some understanding of the
operation of the existing ballast, or is there a better, simpler way of
achieving what I need??

You should be able to buy an electronic diming ballast with a
0-10V control voltage. When using just a single ballast, with
some of these you can just connect a small potentiometer to the
dimming input as they contain a low current internal 10V supply
just for this purpose.
 
J

JM

Jan 1, 1970
0
pemamnently.

ok, situation clear now. If youre using the standard British glow start
circuit (flashes during start up), an extra impedance in series with
the choke will reduce lamp run power. But... if you drop the power by
more than 20% you'll get falling lamp life, and possibly difficulty
starting.

This is fine for a lashup test, and might even be ok for long term use
in some cases if youre willing to accept the downsides.

The best series impedances are inductors and capacitors, but caps must
be calculated with care, since adding the wrong vaalue of cap can cause
destruction to the equipment by opposing the inductive impedance
element, resulting in destructive currents and voltages.

Get rid of the starter, and use a low voltage transformer to supply the
cathode heaters. Now just dim the lamp, while keeping the cathode heaters at
full power. Or, otherwise get a regular dimming ballast. I'm sure they are
available as the 18w lamps are "their" equivalent to "our" 17w lamps.
 
V

Victor Roberts

Jan 1, 1970
0
The best series impedances are inductors and capacitors, but caps must
be calculated with care, since adding the wrong vaalue of cap can cause
destruction to the equipment by opposing the inductive impedance
element, resulting in destructive currents and voltages.

Using a capacitor at line frequency is a bad idea unless the
calculations are done carefully. If the capacitive reactance is
smaller than the inductive reactance, the capacitive reactance will
always subtract from the inductive reactance, reducing the 50 Hz or 60
Hz impedance, instead of increasing it, and increasing lamp current
instead of decreasing it. Once the capacitive reactance is greater
than the inductive reactance, the capacitor will dominate, but the
inductive reactance will subtract from the capacitive reactance.
Better to use an inductor, which will always decrease current for a
simple series inductive ballast.

--
Vic Roberts
http://www.RobertsResearchInc.com
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V

Victor Roberts

Jan 1, 1970
0
Using an added inductor is probaly the better idea if the OP doesnt
have the calc abilties that a cap requires. If you do, the cap has the
advantages of higher energy efficiency and lower cost. It is also
smaller and lighter. Also, less critically, it produces a capacitive
power factor, resulting in less pfc in a multi lamp setup, where some
are inductive and some capacitive.

Also tends to produce terrible lamp CCF at line frequency, though it
might not if placed in series with an inductor with sufficient
inductance to ballast the lamp by itself.

--
Vic Roberts
http://www.RobertsResearchInc.com
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A

Andrew Gabriel

Jan 1, 1970
0
A CL ballast is not a problem crestwise, which is what the OP would be
doing. A C only ballast is problem on 110, but not 240

No, a C only ballast very quickly wrecks a lamp on 240V 50Hz -- I've
tried it.

CL is fine -- Thorn Lighting had a patent for many years for their
lead-lag twin tube fittings getting near unity power factor.
 
V

Victor Roberts

Jan 1, 1970
0
A CL ballast is not a problem crestwise, which is what the OP would be
doing. A C only ballast is problem on 110, but not 240

We're getting repetitive. Typical CL ballasts on 120-volt mains have
terrible CCF. They are used in very cheap "shoplight" fixtures and
lamp life is very short. As stated earlier, if the L in a CL ballast
is a large as it would need to be in an L-alone ballast, then I
suspect the CCF would be OK, but have not done any calculations or
measurements. I can't comment on 240-volt circuits.

--
Vic Roberts
http://www.RobertsResearchInc.com
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V

Victor Roberts

Jan 1, 1970
0
Andrew said:
I've tried it too, and got full rated lamp life and more.

Do you have an lamp current waveforms? What was the operating voltage
of the lamp you used? A capacitor cannot prevent current runaway at
either 50 or 60Hz, but perhaps if the lamp voltage is a small fraction
of the supply voltage you can get away with a capacitor.
What lamp were you using?

What type of lamp were you using?

--
Vic Roberts
http://www.RobertsResearchInc.com
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T

TKM

Jan 1, 1970
0
Tim said:
Is it feasable, or doable to dim an 18 watt Tri phosphor white tube ( 2
foot)?
Looking for about 50%intensity range.

I think they need a voltage across them to maintain ionization, but
perhaps current control?
Any pointers?

If you don't need dynamic dimming, just sleeve the lamp with a neutral
density filter (theatrical gel works). It's not energy efficient, of
course.

Terry McGowan
 
V

Victor Roberts

Jan 1, 1970
0
If you don't need dynamic dimming, just sleeve the lamp with a neutral
density filter (theatrical gel works). It's not energy efficient, of
course.

Gee Terry - Just like you to think of a simple solution that
works instead of the Rube Goldberg solutions the rest of us are
suggesting :)

--
Vic Roberts
http://www.RobertsResearchInc.com
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T

TKM

Jan 1, 1970
0
Victor Roberts said:
Gee Terry - Just like you to think of a simple solution that
works instead of the Rube Goldberg solutions the rest of us are
suggesting :)

Well, I was going to suggest how we really do it when fluorescent lamps are
too bright -- say behind a piece of diffuse plastic that is part of a
transilluminated sign. We wrap the lamps with black electrical tape. But
that's pretty crude, the tape gets messy with the heat and it's probably
unsafe; so I don't recommend it.

Terry McGowan
 
V

Victor Roberts

Jan 1, 1970
0
Victor said:
Andrew Gabriel wrote:


no. One can estimate them, but thats all.

Then please provide an estimated waveform and describe your means of
estimating.

OK - I see the lamp type at the bottom of your reply. A 2-foot 20-watt
lamp should have an operating voltage of about 60 volts more or less.
of course it can, current is limited by C and dv/di, and charge per
cycle is limited by Vpk and t.

I assume you know how an electronic flash works. A charged capacitor
is connected across a xenon discharge tube. Due to the characteristics
of the capacitor the energy in the capacitor is dumped into the lamp
in a very short period of time, limited only by the resistance of the
xenon discharge and the value of the capacitor.

I don't understand your "dv/di" above.

The basic problem is that discharge lamps have negative incremental
impedance, which means that their operating voltage drops as the
operating current increases. The current of any discharge lamp will
increase explosively if the lamp is operated directly from a voltage
source. The cause if the increasing current is a rapidly increasing
electron density which causes a rapidly increasing discharge
conductance. The electron density, and therefore the conductance and
the current, however, can be controlled by a series inductor placed
between the lamp and the voltage source.

The time scale for changes in the electron density of a fluorescent
lamp is far less than 100 usec, less than 10 usec under some
conditions, so for all practical purposes a 50 Hz or 60 Hz power
source has a fixed output voltage over a 100 usec time period.

Consider a simple series circuit consisting of a power source, an
inductor or capacitor and the lamp. Based on the discussion in the
previous paragraph, the voltage of the power source can be considered
to be constant over the time period of interest.

Inductors tend to resist changes in current. The transfer function for
an inductor is V = L* di/dt. As the lamp current tries to increase due
to increasing electron density and the resultant increase in discharge
conductance, a voltage is generated across the inductor that reduces
the voltage applied to the lamp. The reduced lamp voltage reduces the
internal electric field that is responsible for creating free
electrons and thereby limits the increase in electron density. The
faster the current tries to increase, the greater the reduction in the
electric filed, and this brings the system into balance.

Capacitors on the other hand tend to resist changes in the voltage
across them, and have little impact on current until they are fully
discharged. The transfer function for a capacitor is I = C* dV/dt.
Using the same series circuit, but this time with a capacitor instead
of an inductor, we now see that as the lamp voltage drops due to
increasing current, the increasing difference between the supply
voltage and the lamp voltage appears as an increasing voltage across
the capacitor and now tends to increase the capacitor and lamp current
since dV/dt is positive. So, while the series inductor has a negative
feedback effect on the lamp current, a series capacitor has a positive
feedback effect and the current will increase quickly and then finally
decrease as the lamp voltage approaches its minimum value and the
voltage across the capacitor can not increase any further. This rapid
increase in lamp current, followed quickly by a drop to zero current
creates the high lamp CCF that destroys lamp electrodes.

I number of years ago I developed a PSPICE model of a fluorescent lamp
that works quite well from 60 Hz to high frequency (as high as you ant
to go) using either inductive or resistive ballasts. I've never run it
with a capacitor ballast. I'll try to dig out the model and run a
capacitor model simulation and then post the waveforms at various
points in the circuit.
precisely, this is why the difference between 240v C ballasts and 120v.
We've been over this one before.

You never told us your lamp voltage. I assumed it was a 2 to 4-foot
lamp which would make the voltage 60 to 100 volts, and only the
60-volts lamp (which I see from below you are using) approaches what I
consider to be a small fraction of 240 volts.
2' 20w 3500K

Thanks. Your capacitor ballast would not work as well with a 4-foot
lamp even with a 240-volt supply.

--
Vic Roberts
http://www.RobertsResearchInc.com
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V

Victor Roberts

Jan 1, 1970
0
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
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V

Victor Roberts

Jan 1, 1970
0
it is precisely about that. It is because of that that the fl lamp is
happy on a c ballast. This point is the central one to understanding
why it works fine. I'm going to put this para at the top.

The reason that high lamp CCF damages electrodes is that for short
periods of time each cycle the lamp current rises to values higher
than those that can be provided by the electrode via thermionic
emission. When thermionic emission is not sufficient to provide the
required number of electrons, the cathode fall voltage rises, which
increases the ion impact energy. The impact of high velocity ions on
the surface of the electrode releases the additional electrons
required by the discharge, but these impacts also sputter the electron
emissive coating off the electrode, shortening its life.

You are correct that the thermal inertia of the electrode keep it from
physically melting during these cyclic short-term high current events,
but it is that same thermal inertia that makes it impossible for the
electrode temperature to rise quickly enough to provide sufficient
thermionic emission during the high current portions of a high CCF
waveform - and this is when the real electrode damage takes place.
the difference is that with fl the energy dump on strike is small
compared to the overall working power of the lamp. It does not produce
enough temp rise to be destructive. With xenon flash OTOH the energy
dump is large, and produces the full working temp rise in the lamp.

It's not about temperature. See above. As for how fast the energy from
the capacitor is dissipated in the lamp, I suggest you either measure
or model the current waveform.
We're at cross purposes here. My point is that the energy dump on
restrike is small enough not to take anything upto destructive temps.

It is not about temperature.
Bear in mind that the final purpose of a ballast is temp control rather
than current control: meaning the current is controlled in order to
control the temps in the lamp. The lamp temps are the real final say on
what works and what doesnt.

I suggest you read good text on low pressure discharges. The role of
the ballast is to control lamp current, and the current can rise to
destructive values in far less time than it takes any part of the
"lamp" to heat up. The free electrons heat quickly to about 11, 000 K,
but their total mass is so small that if you placed a thermometer in
the lamp it would measure the temperature of the Hg and rare gas
atoms, which are only slightly above room temperature, not the
temperature of the electrons.

I can understand your "final purpose" comment in that the final result
of current runaway is often the physical destruction of parts of the
lamp - usually the lamp electrodes - and the time constant for the
electrodes to melt is limited by their thermal inertia. But, as was
discussed above, the electrodes are damaged by short-term high current
events long before their temperature rises.

In the end a ballast must be designed to control lamp current on the
time scale of electron creation and destruction process, not the
thermal time constants of the electrodes or any other part of the
lamp. It the ballast cannot control current on the time scale of the
electron creation/destruction process, the lamp discharge will run
away.
The cap controls the current well enough, if not perfectly, with a 20w
lamp on 240v. The restrike energy dumps are not large enough to take
anything to destructive temp.

The electron emissive coating is damaged by a high CCF lamp waveform.
For me this point is almost beyond question, for the simple reason that
I've already had 26,000 hours off a (probably already used) 2' tube
running on nothing but a series cap on 240v. It will thus be a tough
job to convince me that didnt happen :)

So what remains now is for us to explain and understand it all.

Can you post a picture of this lamp? Are the ends clean or dark? It is
possible that your electrodes have been sputtered down to the bare
tungsten but you lamp continues to run since your supply voltage is so
high compared to the normal lamp operating voltage that you can
support the higher cathode fall voltage of bare tungsten compared to
barium coated tungsten. This sometimes happens in the US when lamps
operating on instant start ballasts lose their electron emissive
coating. In fact, lamps have been known to run on IS ballasts after
the electrode has been broken or even after the electrode has fallen
off the support wires - in that case the support wires become the
electrodes. This is a dangerous mode of operation since the high
cathode fall voltage heats the glass near the electrodes to high
temperatures. This can cause the glass to break which can, in turn,
allow the lamp to fall out of the socket.
indeed. Both those points are significant. It is the thermal mass that
stops the electrodes vapourising or being damaged at each restrike
energy dump. And it is the high voltage ratio of supply/lamp that keeps
the energy dump low enough.

The thermal mass only keeps the electrodes from melting. It does not
prevent damage to the electron emissive coating.
If our mean v is 60v, that would give a restrike v of 84v. This gives
us an energy dump coming from 24v across IIRC 2.4 uF. Of that energy
dump, some is dissipated in the discharge, some at electrode 1 and some
at electrode 2.

As I said, the 1.4 factor was for an inductive ballast that had
minimum off time between half-cycles. For a capacitor ballast that
produces longer off time, the voltage at the start of each half-cycle
will be significantly higher.
Ballast apx 2.4uF
Capacitor delta V during energy dump: 84v-60v = 24v.

Q=CV
= 2.4 micro x 24
= 57.6 microjoules

The fl electrodes have thermal inertia just like any other filament.
This 57.6 microjoules amounts to around 1.5w when averaged over the
whole cycle. Filament bulbs do not have a problem with this level of
power variation through the cycle, and fl filaments are operating at
much below 2700K, giving them a lot more temp rise headroom.

We know from observation that the periodic temp rises this causes are
not enough to cause damage.

It is not about temperature.

[snip]
The current waveform doesnt follow the mains v waveform in any of the
common ballasts. With a c ballast and high ballast v drop, the majority
of the waveform follows the mains v waveform roughly. There is dead
time and there is restrike energy dump. During this dump the ballast
gives energy control rather than current control. That works ok.

Yes, a capacitor does have finite energy each cycle and does provide
"energy control", but it also allows very high peak currents which
will exceed the thermionic emission capability of the electrode and
thus damage the electrode electron emissive coating as described at
the top.


[snip]
I never scoped it, didnt even think about the issue at the time. There
is no record of the waveform.

I suggest you do look at the waveform.

[snip]
indeed. And a poorer pf than an L ballast. But since the overall grid
load is lagging, this is not such a problem. Especially when used with
another inductor ballasted lamp.

The small advantage to the grid is a big disadvantage to the lamp.
The flip side is better energy efficiency, since the c dissipation is
essentially zero.

Not necessarily. Fluorescent lamps are less efficient when operated at
high peak currents with long off times.

This will be my last reply to your messages in this thread. I have
explained lamp current runaway and its consequences to the best of my
ability.

I did dig out my PSPICE lamp model last night and reran old
simulations for both resistive and inductive ballasts at 60 Hz just to
make sure it ran on this relatively new computer. If I have time I
will run a C ballast simulation and post the waveforms to the group.
I might even make some real measurements :)

--
Vic Roberts
http://www.RobertsResearchInc.com
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R

R.Lewis

Jan 1, 1970
0
<snip>

There seems to be a fair amount of 'mumbo jumbo' coming out here.
for example:-
Ballast apx 2.4uF
Capacitor delta V during energy dump: 84v-60v = 24v.
Q=CV
= 2.4 micro x 24
= 57.6 microjoules

What has the charge (in coulomb, not joule) got to do with the price of
cheese?
Provided we know how fast this charge is passed through the 'tube we may
well be able to glean something about the (electron) energy density of the
discharge as this time.

That, per se, is not a major failure mechanism of commercial fluo tubes.
Pulling metal atoms off the cathodes is.
 
A

Andrew Gabriel

Jan 1, 1970
0
There seems to be a fair amount of 'mumbo jumbo' coming out here.

Nope, none of the text you quoted was from me.
 
D

Don Klipstein

Jan 1, 1970
0
OK. All I know for sure is what you say doesnt match what has been
observed. Beyond that we both search in the dark.

Perhaps its to do with the electron cloud, I've no idea. ISTR reading
something abot thermionic emitters being able to use their electron
cloud for current peaks.

I expect any electron cloud around a fluorescent lamp filament to supply
high peak currents for only a small fraction of a microsecond, probably
less than a nanosecond. How much capacitance does that electrode cloud
have - a few picofarads? The excessive peak current that results from a C
ballast lasts several microseconds.
I took the lamp out of use after apx 26,000 hours, and it was still in
good health. Ends were not blackened any more than usual, the tube ends
were not running hot, etc. It all still worked perfectly, I just didnt
need it any longer. If I ever find it I'll take a pic, its probably
still around somewhere. But where I dont know, if its anywhere it'll be
at another address, and Im seriously busy, so it wont be soon.

Was that lamp run on a purely C ballast or one with both C and L in
series? If the ballast was purely C, can you tell us the capacitance
value, capacitor type, and lamp type so that one can attempt to duplicate
these results?

- Don Klipstein ([email protected])
 
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