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.
+____________________________
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.
Useful EDAboard discussion
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
Related
Plenty of cool white (and straight UV, if you like) LEDs are available (and the cool types actually have better lumens-per-watt efficiency than the "warm" types, as a general rule.)
Lumens-per-watt efficiency is all over the map, so you have to check the particular LEDs you are considering using. Some are awful, some are quite good.
Cooling is something you'll have to sort out, especially with high power LEDs, where thermal design is most of the design work. Stock aquarium hoods are terrible for it, so most folks with serious lighting setups end up building something that allows the lamps to cool more effectively (may involve a fan, may involve a barrier between the water and the lamp which will slightly reduce light into the aquarium, but greatly reduce humidity at the lamps, etc.)
One factor you may have missed is that the LEDs drive all the light towards the water, where the florescent has considerable losses from the parts of the lamp where light is reflected before hitting the water. I have 8.9 watt LED ceiling fixtures which are undeniably brighter than 13 watt CFLs for exactly this reason, even though they are "only" 70 lumens per watt.
Best Answer
Looking at the schematic shown here it does seem to have a
built-in[corrected: suggested application] ballast. I'm guessing your teardown wasn't complete enough. Added: Well, you've found the external ballast it seems, so it was not in the tube as I initially suspected.