This is sort of a spin off of this question. Fluorscent have a tiny transformer inside them, and a bunch of capacitors, as well as transistors, and diodes. What exactly happens when a light bulb goes out? Is it the transformers fault?
Electronic – Voltage in a fluorescent light bulb
currentfailurefluorescent-lamp
Related Solutions
I must have missed this when it was asked in January.
This is a well described question and Al's answer to part of his own question was very good. He subsequently deleted it, but hopefully it will get undeleted sometime soon.
I'll address the core questions first and then come back and talk about some clever circuit aspects.
Q: So now I have one old 15uF, and one new 22uF [in series]. ...Will there be problems?
A: Probably not.
When you charge two capacitors in series so that the same current fklows through both capacitors, as happens here, the larger capacitor will experience a smaller voltage rise. This will be very approximately in inverse proportion to their capacitance. The two capacitors are close in nominal value (15/22 =~ 0.7) Electrolytic capacitor values may vary more widely than this (depends on specification). The older capacitor has probably lost some capacitance with age. So, the older small one will probably have a higher voltage to start when charging finishes. This will offset the capacitor voltage midpoint.
However, as you rightly note in your deleted answer (please undelete), when the capacitors discharge they will be electrically in parallel bu=t behind diodes so that the somewhat higher voltage capacitor will start to discharge first and when the output voltage gets down to the voltage of the lower voltage cap the second cap will "join in" seamlessly.This will have some effect on capacitor ripple currents and the higher voltage MAY stress the old cap more, but overall it should work OK. Arguably, a new cap that is not the same as the old one should be at a somewhat LOWER capacitance so that it takes more of the stress. BUT should be OK.
This is Al's picture of the discharge process. Whichever capacitor is at higer voltage will discharge first.
Q: Those caps are surrounded by a lot of diodes. I expect that normally the potentials around and between those caps are -162V, 0V, +162V. When I replaced one of them by a different one, I probably moved the center potential out of ideal zero. Does it matter here?
A: As above. This is the heart of the Valley Fill circuit. The caps charge to ABOUT Vinpeak/2. All should be well enough.
Q: Note that the reason why there are two strange capacitors instead of one 400V one is probably just the space.
A: No. As above. this provides passive power factor correction by very substantially spreading the conduction period of the input diodes. It also provides Vsupply at half Vin peak during the valley period.
Q: The 0R5 resistors on emitters of each transistor are now 0R56. I don't understand ... if it's dangerous change or not.
A: This is OK. The emitter resistors are current sense resistors which provide voltage drive via the diode D1 D2 to trigger SCR1, which terminates the current switching half cycle via D3. I'd have to spend more time on this circuit to get all the nuances and I'm pretty sure it's not 100% correct, but it gives a reasonably good idea of what happens. Increasing the resistors to 5R6 from 5R increases the voltage across them by a factor of 5.6/5 ~= 12% so they will cause the circuit to turn off at very slightly lower currents causing very slightly lower brightness. You would be very unlikely to see the difference visually.
Valley Fill Circuit:
A Valley Fill Circuit is a piece of brilliant black magic from the beginnings of time that allows surprisingly good power factor correction into a resistive load - which a constant brightness high frequency inverter tends to provide.
Rather than continue to sing their praises - here are some references to basic and more clever versions and some discussion. Well worth acquainting oneself with if you have not met them.
IR (amongst market leaders) AN1074 - New valley fill circuit - A new Circuit for Low-Cost Electronic Ballast Passive Valley Fill with additional Control Circuits for Low Total Harmonic Distortion and Low Crest Factor - passive magic refined.
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A very clever circuit that appears to offer substantial gains over the traditional circuits Improved Valley-Fill Passive Current Shape - 1997
- The original valley-fill current shaper permits input current conduction from 30° to 150°, and then from 210° to 330°. Due to the discontinuities from 0° to 30° and from 150° to 210°, substantial amount of harmonics were introduced into the input current waveform. This article presents an improved version of the valley-fill circuit which extends the conduction angle to near 360°, thus lowering unwanted harmonics as well as improving power line current waveform. Improvements are made with passive components. SPICE simulations compare original circuit with different improved versions of the circuit. 98% power factor is achievable with this new circuit.
IEEE abstract - of interest]The circuit with valley switching technique
And again High power factor correction circuit using valley charge-pumping for low cost electronic ballasts
I'm going to go out on a limb and say this question is valuable from the point of view of electronic design, as it pertains to some fundamental understanding on how fluorescent lights work.
Fluorescent lights work by accelerating electrons from the cathode to the anode in an almost-vacuum environment. In this vacuum is mercury vapour, and when the electron hits a mercury atom, that Hg atom goes into an excited state and outputs one or more photons of UV light upon decay. These UV photons then hit the phosphor-based coating on the inside of the glass tube, which converts these UV photons to visible white light.
So, in order to function, it is vitally important for these lights to have a lot of 'free' electrons available to shoot at the mercury. One way to make electrons more mobile and likely to shoot off the cathode is to heat it up, and this is what a so-called 'starter' circuit does: it is essentially nothing more than a high voltage generator and a heating coil. The heating coil heats up the electrode to mobilize the electrons and the high voltage generator (usually just a resonant LC pump) creates enough voltage for the initial 'spark' to ignite the bulb. Once electrons start flowing and the lamp is 'on', the gas inside the lamp looks more like a plasma and is very conductive, so neither the high voltage nor the addition of heat is necessary to keep it working. Hence, it's just a starter, once the bulb is on, it is shut down.
Old-style starters would keep trying to fire the bulb even when the electrodes were entirely spent. This means that that heating coil would be running until its filament would burn out. In a lot of cases this would mean the bulb has a higher power consumption after it's died.
Modern electronic starters 'give up' after a few tries when they detect that the bulb won't start. After that they use up no or almost no energy until power is cycled to the starter.
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Best Answer
The normal aging process results from inert gas contamination from electrode wear which causes current to increase above a shutdown threshold to prevent excess power dissipation.
Most of the aging occurs during cold start if cycled many times per day in 4' linear tubes rated for 30khr which can last 50khr if left on all the time.
Small embedded CFL lamp failures are subject to design variations and process quality and may fail from active ballast failures.