Hi there.

If I were to wire a Capacitor in series before the lamp socket on a everyday 175w MV Yardblaster fixture, would this reduce the lamp lumens output?

What if I wired the Cap in series on the 120v line in before the ballast? 

Thanks

Very generally :

You can assume the ballast output as an AC voltage source (the OCV), with resistance R (the coil wire resistance) and reactance Xl (the coil's inductivity as coil) connected in series

R and X both resist the flow of current, so the more R and X, the less current to the lamp. In a good energy-efficient ballast you will want as much of the job as possible to be done by X, and as little as possible by R, so generally X is much more significant than R



The added capacitor has reactance of its own Xc, of opposite sign to that provided by the ballast. The total X will be therefore Xl-Xc. So there may be few things that can happen. Some examples of what will happen along the way from small Xc (large capacity) to large Xc (small capacity) :

Xc << Xl :
The resulting X = Xl-Xc will be just a little less than Xl, so the lamp current a little higher than with the ballast alone

0 < Xc < Xl :
The resulting X = Xl-Xc will be less and the lamp current higher. The closer Xc to Xl, the higher current, possibly by a lot. If it is high enough, you may damage the lamp / ballast / capacitor

Xc = Xl :
They exactly cancel out each other X = Xl-Xc = 0. The only ballasting left is whatever little is done by R. You can well damage the lamp, ballast and capacitor like that since the lamp current will be huge

Xl < Xc < 2Xl :
Same as when 0 < Xc < Xl. The closer they are together, the closer they are to cancelling each other out. The only difference is, now the current (and lamp flicker) will be leading instead of lagging

Xc = 2Xl :
Here Xc cancels Xl, and then builds up the exact same X to the other side. The lamp current will be exactly same as with the ballast alone, and leading

Xc > 2Xl :
Here Xc cancels Xl, and then builds up more X to the other side. The lamp current will be lower than with the ballast alone, and leading

Xc >> 2Xl :
The resulting X will be very high, with very small lamp current, and leading



Capacitors also tend to degrade with use, and as they degrade, their capacity goes down, so Xc goes up. Therefore, all the range where Xc < Xl is unsafe to work in : As the capacitor degrade, Xc will go towards Xl and current go up, increase risk of damage to the lamp, and accelerate the degradation of the capacitor, so Xc will rush towards Xl. This won't end up well.... If you choose Xc > Xl in the 1st place, the curent will only go down if the capacitor degrades, then you are safe



From the 120V input side, the story is somewhat similar (but not the same), generally not as favorable to work with compared to when the capacitor is on the lamp side (OCV changes with the capacitor value too, nastier math respectively)

Hey thanks for the reply but I'm very lost and quite new to this HID stuff.

So if I select the right Cap of the correct rating then Im headed in the right direction?

You might be, depends on what you are trying to get

The important ratings are :

 - Capacity

 - Temperature

 - Voltage and conditions under which the capacitor have to withstand the voltage. For example, a motor run capacitor rated for 400V may be inappropriate for use as PFC capacitor even at 240V

 - Max current, That is not rated on the capacitor but is not less important. You may have to figure it out from max allowed power dissipation and ESR or loss angle data (that is normally provided in the datasheet)

You might be, depends on what you are trying to get

The important ratings are :

 - Capacity

 - Temperature

 - Voltage and conditions under which the capacitor have to withstand the voltage. For example, a motor run capacitor rated for 400V may be inappropriate for use as PFC capacitor even at 240V

 - Max current, That is not rated on the capacitor but is not less important. You may have to figure it out from max allowed power dissipation and ESR or loss angle data (that is normally provided in the datasheet)

Hey thanks Ash but I'm still lost here. I basically want to reduce the ammount of light that is emmited using either a 100w or 175w MV lamp using there matching ballasts. Could I install a heavy duty, lets say 100w/40ohm resistor in series on the hot side between the ballast and lamp to reduce arc voltage? I believe the Maritime Man did this with his 50w MV fixture.

You could, but what would you do with.....

 - Tens watts of power dissipation ? (that will scorch anything next to it and burn out itself, unless you mount it with adequate cooling and where it can heat up without being a hazard

 - Even more power dissipation in case of a fault, but still not high enough currents to blow a fuse or trip a breaker, so stuff may really heat up out of control

Hey there.

I see what you mean, the extra power needs to go somewhere and to have it as a trade off as heat is not an option.

I migt try a rotary dimmer switch and just lower it abit and see what happens!

Not exactly "the rest of the power". It is less

Easier to explain with resistors than with Mercury lamp and its ballast. For this sake imagine you want to drop a 200W incandescent to 100W. And lets imagine that the incandescent behaves as plain resistor, with some constant R (which is VERY wrong for incandescent in reality....)

Incandescent lamp takes 200W on 120V, at current 200/120 = 1.7A

At half the voltage it will take half the current, so quarter of the power
At 1/3 the voltage it will take 1/3 the current, so 1/9 of the power
....

You get the deal. The power goes down with square of the voltage

So to drop the power in half, we want to drop the voltage in 1/sqrt(2). That is about 0.7...
120V * 0.7 = 84V thats the voltage the lamp gets now
1.7 * 0.7 = 1.2A thats the current the lamp gets now
84 * 1.2 = 100W

The resistor gets the voltage difference 120 - 84 = 36V, and same current as the lamp (as they are in series) 1.2A
36 * 1.2 = 43W

So the resistor does not have to drop 100W, it drops just 43W

Resistor value 36V / 1.2A = 30 Ohm



Now assume a Phase wire got damaged in the lantern and shorted to the Earthed lantern body, which is basically the same as shorted to Neutral. Now the resistor alone is getting the full 120V...

120V / 30 Ohm = 4A
120V * 4A = 480W

So now it is overheating by about 10 times and catching fire, but still takes quite low current - 4A aint trip any breaker, and if you try to protect with an 1.5A fuse, depending on the type of fuse it still might take some seconds to blow at 4A - Hard to tell what will happen first : Fuse blow open, resistor blow open, or resistor catch fire....