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Ballast Experiment: 2x40w PH With EOL 40w

Ballast Experiment: 2x40w PH With EOL 40w

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There was some discussion in another thread about EOL lamps on 40w PH ballasts. I decided to get some line measurements with a Jefferson ballast (shown) and a similar but older GE brick ballast.

Results (With good lamps, control):

2x40w Line: .90a, 98w, PF 90%
1x40w (red wire, cap side) Line: .51a, 49w, PF 80%
1x40w (blue, lag side) Line: 1.0a, 55w, PF 44%
**************************************************

Results with EOL lamp rectifying on blue wire lag side:
(Jefferson)
2x40w Line: .70a, 66w, PF 75%
1x40w (EOL) Line: .62a, 23w, PF 32%

(GE)
2x40w Line: .67a, 66w, 80%
1x40w (EOL) Line: .48a, 22w, PF 37%

All experiments were done with EOL lamp on the "lag" side so it could rectify, both with and without a good lamp on the red wire side. So far, with these two typical ballasts, there is no indication that a rectifying lamp would damage ballast, according to the numbers. Ideally, I suppose ballast should be left running but right now I don't have a test fixture set up to do so. I would like to in the future!

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Powergroove
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Nov 29, 2016 at 08:04 AM Author: Powergroove
Have you experimented with a slimline ballast and a rectifying lamp. This seems to be hard on the ballast i think on the white and red side.

Keep government out of the lighting industry.

don93s
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Nov 29, 2016 at 11:05 AM Author: don93s
I have done experiments on both F48 and F96 slimline ballasts. It appears that a rectifying lamp causes the power factor to go down and the line current to go up. From that it seems it causes the primary coil to over-heat. Having a working lamp only on the red/white wire starting-side seems to have the same effect.
Powergroove
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Nov 29, 2016 at 11:17 AM Author: Powergroove
Im trying to understand this better. Is it the lead side or the lag side that rectified the lamp?

Keep government out of the lighting industry.

don93s
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Nov 29, 2016 at 06:28 PM Author: don93s
It's probably a little confusing the way I wrote the post, lol, but the "lag" side is what rectifies the lamp. The non-capacitor side, usually on the blue wire, an ohm meter will give a steady reading between the blue and common.
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Nov 29, 2016 at 06:31 PM Author: don93s
The series-slimline ballasts seems to behave like a lead-lag because the blue wire lamp also rectifies but it's technically series since both lamps in the same voltage or current phase whereas lead-lag have both lamps 180 degrees out of phase, such as the above tulamp preheat ballast.
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Nov 29, 2016 at 06:37 PM Author: don93s
Also, if i recall, on the "lead" side, with a capacitor, the lamp current waveform 'leads' the lamp voltage waveform and the inductive side, the lamp current waveform 'lags' the lamp voltage waveform. I hope I got that right, lol Might be more of inside the ballast waveforms. Medved, where are you? lol
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Mike McCann


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Nov 29, 2016 at 07:50 PM Author: streetlight98
@ Don: In my circuit theory class I learned that capacitive loads have current lead the ballast and inductive loads have current lag voltage, if that's what you mean. IIRC each are 90 degrees out of phase from a resistive load. IDK... it's all confusing to me lol... One thing I walked in there knowing was that capacitors ca be used to cancel out the effects of an inductive load thanks to LG.

Please check out my newly-updated website! McCann Lighting Company is where my street light collection is displayed in detail.

ace100w120v
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Nov 29, 2016 at 07:52 PM Author: ace100w120v
Mike, you just lost me at the beginning of that paragraph...
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Mike McCann


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Nov 29, 2016 at 09:56 PM Author: streetlight98
LOL. Capacitive loads are capacitors and inductive loads are ballasts, transformers, motors, etc. Capacitors and inductors have opposite effects on the sine wave of a circuit so by using the capacitor parallel to the inductor (the inductor being the ballast in a lighting scenario) you correct the sine wave and that increases power factor and reduces reactive power (which is basically "wasted power" that isn't used for anything but the circuit still "sees" that extra current). Then there's resistive loads, which do not alter the sine wave at all and have a power factor of 1. Incandescent lamps are probably the most commonly known resistive load used in everyday life (well at one time anyway!). I know with a heavily inductive circuit (think a bunch of LPF fluorescent fixtures) the reactive power is high so there's that extra line current. I'm not sure what would happen with a primarily capacitive circuit (or even if such a thing is possible). It wasn't something my professors went into any detail about.

Anyway, that's just the Micky Mouse version of it lol. I'm new to this stuff so I'm giving you my best explanation. Might not be 100% accurate but it should at least give you a vague idea of the complexity of AC circuits lol. With DC non of this stuff happens because there is no sine wave (constant voltage, so just a flat line). I think capacitors and inductors are used in AC to DC converters alongside diodes since they further flatten out the "mountains" and "valleys" of the sine wave to make it closer to a flat line.

Please check out my newly-updated website! McCann Lighting Company is where my street light collection is displayed in detail.

don93s
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Nov 29, 2016 at 09:58 PM Author: don93s
@Mike, yep, pretty much what I meant if not said better, lol. It's been 20 years since I went to college and actually got all A's in those classes, haha. But yeah, a resistive load such as a toaster element, resister, or light bulb have a current flow waveform and voltage waveform across the load exactly in unison. That also translates to 100% power factor. The current and voltage are directly in line which each other. Across the ballast coils and capacitor, things get different. The lamps themselves are, for the most part, a "resistive load", albeit, discharge lamps have different rules, but for this topic, still resistive.

I guess what I should have said is that across the cap, the current waveform "leads" the voltage waveform and across the coil, the current waveform "lags" the voltage waveform. The current waveform is the actual electron flow while the voltage waveform is the actual voltage measurement or potential across the cap or coil. With the cap being 90 degrees on one side of true resistive and the coil being 90 degrees on the other side of resistive, the lamps each end up being 180 degrees apart. But that where trigonometry/algebra math comes in to truly explain it.
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Nov 29, 2016 at 09:59 PM Author: don93s
I see you just added more after I added mine, lol.
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Nov 29, 2016 at 10:03 PM Author: don93s
BTW, as for the huge inductive loads with NPF ballasts, yes you can indeed add parallel caps to lower the line current and raise the PF. I do it to NEMA heads when I want to run a few of them for display without blowing breakers. Still use the same power but less current...or, reactive current.
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