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# How can a fluorescent lamp ballast have a PFC capacitor?

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#### shockoshocko

##### Junior Member level 1
Won't a series capacitor cancel out the function of the ballast and makes it appear as a short circuit ?

The capacitor is operated in a parallel circuit not series.

If it is connected in parallel then they would appear like an open circuit?!

shockoshocko said:
Won't a series capacitor cancel out the function of the ballast and makes it appear as a short circuit ?
Is the ballast you are refreing, a regular choke with winding inside or an electronic ballast?

If it is connected in parallel then they would appear like an open circuit?!
Before putting exclamation marks behind your statement, you may want to think about the basic circuit properties, particularly the real load. I'm sure, you're able to figure it out.

you put series cap so you get zero average dc current..cuzz

1 you dont want to saturate the inductor

and

2 you dont want dc component in lamp as it damages it over time.

also, it give a bit of slew to your voltage so you get ZVS.

"FvM", I still don't get it, if both the parallel inductor and capacitor have the same impedance at line frequency wL=1/wC , then their total impedance would be:

(wL) * (-1/wC) / (wL) + (-1/wC)=infinity

Mathematically, you are correct, but you are analyzing the wrong scenario.

The inductor is only in series with the load (the fluorescent tube) and the capacitor is in parallel with both of these. The capacitor is not connected directly across the inductor.

Without the capacitor, the lamp appears to be inductive as far as the incoming power supply is concerned. The capacitor, wired directly across the incoming power is to redress the current lag.

Brian.

you are analyzing the wrong scenario
Yes, or as I mentioned, you forgot the real load. Actually, the purpose of a PFC is to zero the reactive part of an existing load. Without a real load (the fluorescent lamp) you neither need a ballast nor a PFC capacitor.

Then the total impedance is equal to:

[(R+jwL) * (-j/wC)] / (R+jwL)-(j/wC)

which simplifies to:

[-jR/wC+L/C] / R

No, it's complex in this case. With a L-R series circuit, you need a smaller C value to zero the reactive part.
No one stated, that it would be still C = 1/(ω²L). Also R is transformed in the resulting impedance.

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