I have PL adapter (containing magnetic ballast) that allows to connect the PL to 240V socket. It is for 10 / 13W PL-D

I also have 125W mercury fixture

I tried to install the PL with adapter in the fixture, right on the mercury ballast, so that there are 2 ballasts (mercury fixture + PL adapter) in series. The PL lamp strikes normal and works at what looks close to normal brightness, maybe a bit dimmer but not too noticable

What are the ballasts values in mH ? (240V 125W 1.15A mercury and 240V 10/13W PL)

Will slightly underpowering the PL shorten its life ?

Don't worry, the lamp current would not change much, it would be about 10..15% lower, what should be still OK...

I let it run several hours like that, and here he results :

Mercuriy ballast - cool

PL adapter - warm

PL - warm, but you can touch it - unlike when it is running on full power

And now what i am unsure is good - the capacitor (polyester 250V 14uF) is warm. Should a capacitor connected across 240V heat up noticably at all ?



What are the ballast values, currents, and lamp arc voltges for 125W mercuy and 13W PL ?

I let it run several hours like that, and here he results :

Mercuriy ballast - cool

Should be expected

PL adapter - warm

Should be expected

PL - warm, but you can touch it - unlike when it is running on full power

It may be the 15% power difference: 15% in lamp current mean ~30% in ballast losses. So between 50 and 60degC is the "border" between "possible" and "too hot" to touch.

And now what i am unsure is good - the capacitor (polyester 250V 14uF) is warm. Should a capacitor connected across 240V heat up noticably at all ?

I would say not good at all, the capacitor is likely to fail soon... But this is not influenced by connected lamp/ballasts, as the capacitor see always the same 230V.
Maybe on long wires it may boost up the voltage of an unloaded circuit (Ferranti effect), where such elevated voltage could cause higher load of the capacitor. But if it heats, it will for sure not survive this condition for too long...

What are the ballast values, currents, and lamp arc voltges for 125W mercuy and 13W PL ?

When assuming loss-less linear (no saturation,...) ballast and sinewave arc voltage (both cause some error when neglected, but comparable between lamps)
     L=sqrt(Vmains^2-Varc^2)/(c*Pi*f*Iarc)
MV 125W: Varc = 120V, Iarc = 1.15A, so about 543mH
PL-C10W Varc ~ 70V, Iarc = 0.16A
PL-C13W Varc ~ 96V, Iarc = 0.15A
  In both cases about 4.4H.

So when connected in series, the total impedance would correspond to the sum, so about 5.15H, what mean about 4.4/5.15 = 86% of nominal current.

I would say not good at all, the capacitor is likely to fail soon... But this is not influenced by connected lamp/ballasts, as the capacitor see always the same 230V.
Maybe on long wires it may boost up the voltage of an unloaded circuit (Ferranti effect), where such elevated voltage could cause higher load of the capacitor. But if it heats, it will for sure not survive this condition for too long...

The capacitor is dated 1991, and came from a fluorescent troffer that was in use untill about 2007 (and looks like wrong value of capacitor for 4x18/20W fluotescents on 4 inductor ballasts). If it is failing, it only began failing towards the end of the troffer being in use. What would cause it to begin failing ? The troffer worked for long time with all 4 lamps with stuck starters, but the capacitor was not adjacent to the ballasts so not too hot dispite the ballasts severely overheated

Is the failure mode in those capacitors series (film broke and part of the capacitor disconnected) or parallel (capacitor is shorted, resistive loading of the film) ?

Can i test it by charging it and seeing how fast it self-discharges ?

Here is the capacitor in question - http://www.lighting-gallery.net/gallery/displayimage.php?album=2136&pos=4&pid=58163

I would say not good at all, the capacitor is likely to fail soon... But this is not influenced by connected lamp/ballasts, as the capacitor see always the same 230V.
Maybe on long wires it may boost up the voltage of an unloaded circuit (Ferranti effect), where such elevated voltage could cause higher load of the capacitor. But if it heats, it will for sure not survive this condition for too long...

When assuming loss-less linear (no saturation,...) ballast and sinewave arc voltage (both cause some error when neglected, but comparable between lamps)
     L=sqrt(Vmains^2-Varc^2)/(c*Pi*f*Iarc)
MV 125W: Varc = 120V, Iarc = 1.15A, so about 543mH
PL-C10W Varc ~ 70V, Iarc = 0.16A
PL-C13W Varc ~ 96V, Iarc = 0.15A
  In both cases about 4.4H.

What is c in c*Pi*f*I ?

The currents and voltages dont really give the lamp power, is it because of the current waveform not being sine ? (exa. 120V * 1.15A = 138VA not 125VA)



How much underpowering it takes untill the cathodes of the PL are too cold and reduce lamp life ?



If i would put a standard CFL (input rectified to DC right away) in place of the PL adapter, would the mercury ballast harm it in any way ? or would it only be a good PFC ?

At work, a room is lit by 16 fixtures each running a 400 watt MH lamp (M59) on 120V 60Hz circuits. One fixture failed about ten years ago, and was replaced with a new one. The thing is that the old fixtures were remote ballasted, and the new one is not. They did not disconnect the old ballast despite the electricians telling me that the remote ballasts are for other lighting and have been disconnected, and wired the new fitting in its place. This makes two ballasts in series for one 400 W lamp. The lamp warms up to nearly full power and then dims back to half warmed-up level (greenish, no metal halides vapourized). The ballast in the fitting hums loudly during warmup and then falls almost silent when the light output drops. I believe all the ballasts in the ballast bank get warm when the lights are on. When I get the chance, I will check on that.

@sol: Before I was describing the 230/240V circuits, where the "ballast" mean the series connected inductor, nothing more.
In your case (I would guess CWA) is not possible to speak about "series conection", but rather on cascade connection.
There the problem is with the fact, then the first ballast (the remote one) act as a current source on it's output, so feed constant current to load with whatever voltage. That mean, then the actual power delivered to the load rise with the load voltage. As the ballast is no peretual motion, the rising power delivered mean the power input have to rise, on HPF it mean really the input current rise as well.
And here come the effect:
With the lamp being cold, the voltage drop across it is low, so the power input as well, what mean the second ballast consume small current, below that of the first ballast. That cause the first ballast reach it's OCV, what mean overvoltage on the second ballast input, so it saturate on the primary (so it hum; in fact the real voltage between ballasts would be clamped by that saturation).
As the lamp warm up, it takes more power, so the second ballast would need more current.
At some point the current required by the second ballast exceed the output current of the first one, yielding voltage to collapse.
As the voltage between ballasts collapse, the same does the power delivered to the second ballast, so it goes out of regulation and start acting only as voltage x2 and/or current /2 transformer, dividing the current of the first ballast by two.
So as the consequence the lamp current drop to about a half (or less, depend on the second ballast's OCV, the factor 2 would be valid for 240V).
As the voltage on the second ballast input drop (to about 60V; my guess), the magnetic flux drop way below saturation, so the core is then quiet.

(Questions i asked above)

Is the failure mode in those capacitors series (film broke and part of the capacitor disconnected) or parallel (capacitor is shorted, resistive loading of the film) ?

Can i test it by charging it and seeing how fast it self-discharges ?

Here is the capacitor in question - http://www.lighting-gallery.net/gallery/displayimage.php?album=2136&pos=4&pid=58163



What is c in c*Pi*f*I ?

The currents and voltages dont really give the lamp power, is it because of the current waveform not being sine ? (exa. 120V * 1.15A = 138VA not 125VA)



How much underpowering it takes untill the cathodes of the PL are too cold and reduce lamp life ?



If i would put a standard CFL (input rectified to DC right away) in place of the PL adapter, would the mercury ballast harm it in any way ? or would it only be a good PFC ?

Thanks, Medved. You are correct, it is autotransformer ballasts as it is running at 120V line voltage. I omitted that important information in my earlier post.

This is what I suspected. When time permits, I will try to find the ballast in the ballast bank. Then comes the hard part : convincing the electrician that there are two ballasts for this fixture.

Or you could remove the ballast thats in the fixture

I'm sorry, I overlooked it between other responses...

(Questions i asked above)

Is the failure mode in those capacitors series (film broke and part of the capacitor disconnected) or parallel (capacitor is shorted, resistive loading of the film) ?

Can i test it by charging it and seeing how fast it self-discharges ?

Here is the capacitor in question - http://www.lighting-gallery.net/gallery/displayimage.php?album=2136&pos=4&pid=58163

Actually I'm not sure about the exact mechanism, but there are two possibilities (both related to the "self-recovery" feature):
- Increase in ESR
- Insulation breakdown: The affected area sometimes stay weakened after the "recovery", so the breakdown progress trough large volume, yielding the capacitor's overheat.
Both mechanisms show increased power dissipation short time before the final destructive runaway, so when the capacitor start to heat, it all the time fail soon completely...

(Questions i asked above)

What is c in c*Pi*f*I ?

It is the typo, there should be "2*Pi*f*I"

(Questions i asked above)
The currents and voltages dont really give the lamp power, is it because of the current waveform not being sine ? (exa. 120V * 1.15A = 138VA not 125VA)

It is, because the current does not match the voltage waveform. On an inductive ballast the current is nearly a sinewave, but the voltage is nearly rectangular, what yield power factor about 0.9 (= sqrt(/Pi)

(Questions i asked above)
How much underpowering it takes untill the cathodes of the PL are too cold and reduce lamp life ?

There are two aspects:
- Insufficient preheat current - till it take noticeably longer time to start using the glowbottle starter; Usually the dominant factor, about 0.7*rated current
- Insufficient run current - till the electrode area start to be bluish (= signature of cold-cathode discharge); usually about 50% of rated current or less

(Questions i asked above)
If i would put a standard CFL (input rectified to DC right away) in place of the PL adapter, would the mercury ballast harm it in any way ? or would it only be a good PFC ?

It would cause huge overshoot on the DC bus tank capacitor upon power up (the ballast L form a resonator with the DC bus tank C, causing the voltage to jump up to the double, so about 700V), causing damage mainly to the capacitor itself. Consequence would be short power cycle life of the ballast. Sometimes transistors may fail too...

Actually I'm not sure about the exact mechanism, but there are two possibilities (both related to the "self-recovery" feature):
- Increase in ESR
- Insulation breakdown: The affected area sometimes stay weakened after the "recovery", so the breakdown progress trough large volume, yielding the capacitor's overheat.
Both mechanisms show increased power dissipation short time before the final destructive runaway, so when the capacitor start to heat, it all the time fail soon completely...

How does the self recovery feature work (how is it built) and what it recovers from ?

Are all such capacitors self recovery ?

It is, because the current does not match the voltage waveform. On an inductive ballast the current is nearly a sinewave, but the voltage is nearly rectangular, what yield power factor about 0.9 (= sqrt(/Pi)

I thought it is due to the current being 0 when the lamp goes out around zero crossing

At what voltages does the lamp strike and extingiush ?

Does the ballast alter those times somehow by providing voltage in series with the mains, or it is just the mains voltage to consider ?

It would cause huge overshoot on the DC bus tank capacitor upon power up (the ballast L form a resonator with the DC bus tank C, causing the voltage to jump up to the double, so about 700V), causing damage mainly to the capacitor itself. Consequence would be short power cycle life of the ballast. Sometimes transistors may fail too...

I opened computer power supplies with passive PFC, where it basically is inductor (i dont know the value, but it reminds a 240V 40W fluorescent ballast in core size and winding volume) connected in series qith the AC input. Then it is rectified and charged in 2x 200V electrolytic caps in series

There is a 200V MOV across each of the capacitors, but i assume hey are not in breakdown 100 times/sec to save the capacitors ? (they are tiny, not much heating capacity)

Why doesnt it harm the caps in there, and what is the difference between it and the CFL ?

The self healing is accomplished by using a very thin later of aluminium to make each electrode. It has a quite significant resistance value, but is used in parallel over a large area, so the complete unit has low resistance, but a high resistance for each unit area. If the insulation that makes the dielectric breaks down and starts to conduct, this high current through the film vapourises the film in the area, isolating the faulty part, and disconnecting the faulty area. This can only occur if the voltage is high enough, and there is sufficient current available from the supply to cause the film to vapourise. This self healing reduces the capacitor value slightly, and eventually the capacitor will be out of spec for capacitance as it drops. If the voltage or current is not enough to complete the self healing then there eventually will exist areas where there is increased leakage, and this will generate heat, which will eventually melt the capacitor. This failure mode is expected, and many industrial capacitors have a feature where the increased heat and pressure inside the case blows the cap end off, disconnecting the capacitor from it's terminals. These will generally have a warning either on the case or in the datasheet about having to have clearance by the terminals for this.

Oil filled units ( mostly metal cased ones) use a thicker foil, and only have the terminal disconnect available, as they run at a much higher current in the foil, and have to use a thicker dielectric to ensure that there will be no weak spots.

If the capacitor is rated at 250V and connected to 240V, why is there not enough power to complete the fusing process and stop heating ?

And why does it get broken down (which starts all the process) in the first place ?

If the capacitor is rated at 250V and connected to 240V, why is there not enough power to complete the fusing process and stop heating ?

The fusing itself damage the dielectric foil around and this damage could reach the active region. The "self recovery" does not work in 100% at all, sometimes the breakdown trigger destructive chain "reaction" yielding those melted fried capacitors...
Moreover the mains voltage normally fluctuate, 250VAC rating is in my eyes "just on the border" for 230V nominal, for 240V I would consider it as insufficient...

And why does it get broken down (which starts all the process) in the first place ?

It may have been some overvoltage surge on weakened (mainly due to hot operation and thermal cycling during the life) dielectric material, it is then only statistical question of time... And that's why "selfrecovery" feature is mandatory for such use - to ensure, then at least majority of such surges do not lead to immediate capacitor destruction, so their practical lifetime is sufficient for the application.

In your case the voltage across the capacitor may be higher then the mains voltage (capacitive load on longer wires - the original design didn't count for the capacitor being powered without the inductive load of the lamp circuit)