Author Topic: Unexpected ballast finds  (Read 2819 times)
Powell
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Unexpected ballast finds « on: March 12, 2012, 04:11:34 PM » Author: Powell
As I was heading up to Full Moon Farm to play with the woofies, I stopped in Hendersonville, NC to a place I kept seeing in a strip shopping center. It just said "SALVAGE" and I went in and found no lamps or fluorescent fixtures. I did find some ballasts. For $4.00 each a single lamp ballast HIGH power factor F40. A dual lamp trigger start HIGH POWER FACTOR  for (2) 14,15 or 20 watt lamps with sockets attached. Neither ballast looked like they had use. I have 2 fixtures these will go in to and covert from LPF to HPF. 

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dor123
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Re: Unexpected ballast finds « Reply #1 on: March 13, 2012, 06:45:10 AM » Author: dor123
Based of the mothed that i learned from Ash, how to calculate the fixture total wattage (Lamps + ballast losses)(Mains voltage * Ballast current for the lamps configuration * Power factor = Total fixture wattage), i think that the higher the power factor, the higher the ballast losses and the least efficient the ballast will be (All this provided that the current of the ballast isn't changes with the power factor as well. Probably I"m wrong).
For example, lets take my new 9W preheat PL fixture. Our mains voltage is 230V, the ballast current is 0.170A to the 9W PL-S and the power factor is 0.34. So 230V*0.170A*0.34PF=13.294W fixture wattage (9W lamp wattage + 4.294W ballast losses). If a PFC cap, would be added, so the PF of the ballast would be high (For example 0.85 of most HPF ballasts), so 230V*0.170A*0.85PF=33.235W total fixture wattage.
Correct me if I"m wrong. Perhaps the ballast current changes with the power factor as well.
« Last Edit: March 13, 2012, 10:21:02 AM by dor123 » Logged

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Please forgive me if my choice of my words looks like offensive, while that isn't my intention.

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Re: Unexpected ballast finds « Reply #2 on: March 13, 2012, 04:18:57 PM » Author: Medved
@dor123: I guess you misunderstand Ash's explanation.
By connecting the capacitor parallel to the mains input the lamp current does not change, as the lamp + ballast "combo" see the same, unchanged 230V, so for your lamp remain 0.17A.

The reasoning "higher power factor => higher real power input, so higher ballast losses when the output to the lamp is the same" is true only when the two setup's have the same mains current, so work only when you compare two uncompensated series choke ballasts.

If you add the PFC capacitor to increase the power factor, the real power input stay unchanged (assume the capacitor have no losses, what is very close to reality compare to the losses in the choke), so in your case (13.5W) the mains current drop from 0.17A (uncompensated ballast) down to 0.06A (assume 98% power factor with the capacitor), the real power input remain the same 13.5W.
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Re: Unexpected ballast finds « Reply #3 on: March 14, 2012, 03:39:36 AM » Author: dor123
Medved: Thanks.
Because i don't know electrical engineering, i only rely on Ash explanations. The case with the high power factor, was my own thinking, not misunderstanding about Ash explanations. Ash only explaned me how to calculate the real fixture wattage.
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Re: Unexpected ballast finds « Reply #4 on: March 14, 2012, 05:10:46 PM » Author: Ash
This may help you out a bit

Suppose that your fixture is 36W T8 on Eltam Mini N ballast (easy calculation, the 40W Mini N were PF 0.5 exactly) The ballast is rated to 230V so lets also assume it is getting 230V (on 240V both lamp power and ballast losses will be higher)

The total power the fixture is taking is 230*0.43*0.5 = 49.45W from which 36W = lamp and 13.45W = ballast losses (well this does actually explain why they were so scorching hot tiny ballasts)

The total apparent power the fixture is taking is same calculation but without the power factor 230*0.43 = 98.9VA

So yes, by the looks of it, the 36W lamp is taking allmost 100W power. But thats why it is called so, this is not real power. There are no real 100 Joules of energy/sec wasted on the lamp

The apparent power is a sum of the real power (49.4W) + some "ghost" power....

The "ghost" power is indeed like a ghost. It does not dissipate energy, it does not consume coal at the power plant, bu it does load all the wiring etc.. So to run 1000 of those fluorescents, you will need 100KVA transformers, 430A main feed and breakers....

What is this power actually ? The ballast causes the current to lag behind voltage. So right after the AC is zero crossing, the voltage is allready negative, but the current is stll going in the previous direction

Wait, this would happen if we connected a generator that powers the line, not a light fixture which is supposed to use power, not generate it..... But thats exactly what we did. The ballast core is charged with magnetic energy, which is now getting emitted into the line. In the next 1/4 cycle or so it will charge back up etc

So there is some energy thats going perpetually back and forth between the line and the ballast without actually being made or wasted. It is actually wasted a little (and replaced accordingly) on wiring losses

This "ghost" power is called Reactive power as it is caused by the ballast's reactance

Lets calculate how much of it we have : sqrt(98.9^2 - 49.45^2) = 85.65VA. And we want this same amount negative (to subtract from the lamp cicuit current, not add even more reactive power to it...) so wa went to supply -85.65VA of power

If we put a component that uses just the same amount of the reactive power but at opposite timing, the current for this power will go in the shortest path between them and not affect the line. We want to supply -85.65 / 230 = -0.37A reactive current by a local component in the fixture. the "reactive resistance" of such component would have to be 230V / -0.37A = -621.6 "ohm" (well not really)

Capacitors charge from the momentary voltage of the sine wave, so their current is changing direction right at the "previous" voltage peak on the line. The capacitor would charge exactly at the time the ballast is emitting current, and it would discharge and supply this current to the ballast in the following 1/4 cycle

The capacitor reactance ("reactive resistance") is -Xc = 1/(2*pi*f*c). We need to get -631.6 "ohm" so -621.6 = 1/(2 * 3.1416 * 50Hz * X uF). Calculate that and get 5.12uF. So if my calculation is right 5uF would be best for this ballast

With 5uF ballast, the line would supply only 49.45W of power, so 0.215A. The ballast is taking 0.43A as allways, but a lot of them are supplied by the capacitor locally in the fixture and not by the line

In lead lag circuits, the different timing of the current in the capacitor's circuit is used for our advantege : The capacitor circuit, instead of containing just capacitor, is put in series with a complete second lamp circuit with ballast and everything. The capacitor is picked up so that to exactly cancel out the ballast's reactance and get the same reactance in the opposite direction. Now the circuit with the "2nd" (lead) lamp, is acting to provide the reactive power to the "1st" (lag) lamp
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Re: Unexpected ballast finds « Reply #5 on: March 27, 2012, 01:59:41 AM » Author: don93s
As I was heading up to Full Moon Farm to play with the woofies, I stopped in Hendersonville, NC to a place I kept seeing in a strip shopping center. It just said "SALVAGE" and I went in and found no lamps or fluorescent fixtures. I did find some ballasts. For $4.00 each a single lamp ballast HIGH power factor F40. A dual lamp trigger start HIGH POWER FACTOR  for (2) 14,15 or 20 watt lamps with sockets attached. Neither ballast looked like they had use. I have 2 fixtures these will go in to and covert from LPF to HPF.  

Powell

One reason that I prefer HPF over NPF (or LPF) is that the larger lamps end up being brighter. The reason, I believe, for this is because a NPF ballast is more sensitive to lamp voltage for lamp current. Since they are made for 14w-20w or 30/40w they have to under-drive 20w or 40w in order not to over-drive 14w or 30w and overheat ballast. A HPF ballast has a cap that not only corrects power factor for a lower line current (as Ash explains), but allows ballast to drive a larger selection of lamps at a more consistent current. They also seem to start a little more reliably for some reason.

 Finally, those single lamp 14w-20w NPF trigger start ballasts are notorious for over-heating when lamp...especially 14/15w...burn out and rectify. The lamp current raises dramatically because a rectifying lamp causes DC current and reduces the ballast reactance. A HPF ballast has a cap to block the DC so EOL lamps won't overheat ballast.
« Last Edit: March 27, 2012, 02:02:11 AM by don93s » Logged
Medved
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Re: Unexpected ballast finds « Reply #6 on: March 29, 2012, 02:03:59 AM » Author: Medved
@don93s: Your brightness observations are not linked directly with NPF vs HPF (both could be made in the same way), but rather a coincidence given by the expected end use:
- Commercial installation are in larger areas run very high hours and so have to be as efficient as possible. That mean running lamps at their rated power. At the same time as these installations are larger, there is a need for high power factor to reduce the mains current. At the same time these installations usually swallow higher purchase costs of the more complex HPF ballast. So this lead to HPF and unity ballast factor.
- Residential installations run relatively few hours, so are way more sensitive for purchase costs and the general availability of lamps, so the simpler NPF ballasts and using cheapest, most common lamps. At the same time home use mean very close distances to the lamp, so lower brightness is a plus. So the F40T12 (most common at the time) lamps operated at only 25W (sufficient for most home applications). So this lead to NPF and rather low ballast factor.
And as the residential fluorescent market is rather small and distribution complex (one user does not buy 100's of pieces, but 1 or 2), there is strong force for "One size fit all" approach - hence common ballasts for all tubes in ~10..22W range. As the ballast should never overdrive the lamp, the current is then designed according the small lamps, so run the bigger ones with low BF. And again, no power factor correction due to cost reasons.
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Re: Unexpected ballast finds « Reply #7 on: March 29, 2012, 05:46:49 AM » Author: Ash
Interesting - over here the inductor ballasts are exactly the same in all applications

As in, if i need a 25W lamp at home, why would i use underpowered 40W in the 1st place and not a full power 20W or 30W ?



"As the ballast should never overdrive the lamp" - is a 1.1x - 1.15x overdrive really causing any harsh effects on the lamp ?
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don93s
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Re: Unexpected ballast finds « Reply #8 on: March 29, 2012, 10:49:54 AM » Author: don93s
@don93s: Your brightness observations are not linked directly with NPF vs HPF ....

Yes, I know about that. They even make "residential" HPF ballasts (or used to) that operate lamps at ~70% B.F. I also have LPF ballasts that operate lamps at B.F. of 100% (full 430 ma). A typical commercial HPF (1x40w) ballast with B.F. of 100% is around .45 amp line current while a LPF of same ballast factor is at least .80 amp line current.
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