Light fixtures

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Big Guns

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I have a three phase service with 208/120 volts we installed 4 foot HO fixtures with electronic ballast in parking garages. I put an amp probe on the branch ckt wire on load side of breaker. leg A-B-N and I had a reading of 16 amps on A 14 amps on B and 23 amps on the N. Is the N high do to reading through the ballast or because my C phase wasn't wired in yet so it was not cancelling out, all phase had same amount of figures at 2.5 amps per
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Something seems wrong. I calculate 15 amps on the neutral. The neutral current should be less than the largest phase current.


They are 120 volt fixtures, right??

Steve
 

infinity

Moderator
Staff member
Location
New Jersey
Occupation
Journeyman Electrician
Sounds like a non-linear load problem causing the high amperage on the neutral. 16 amps on a 20 amp circuit is borderline high for a continuous load.
 

bob

Senior Member
Location
Alabama
Something seems wrong. I calculate 15 amps on the neutral. The neutral current should be less than the largest phase current.
Steve

Did you use this formula in you calculations? If the load Pf is not unity, you will get an error.
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
what type of Ammeter? you may be reading harmonics on N


I was skeptical about the harmonics. But if we assume a 20% THD ballast, that would be about 3.2 amps of harmonics per phase. So 15 amps + 3.2 + 3.2 would be about 21.4 amps. That's still a little less than the 23 amps measured, but not too far off.

Still, most ballasts are 10% THD.

Guns: Can you be more specific about what type of ballasts they have?
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Did you use this formula in you calculations? If the load Pf is not unity, you will get an error.

I didn't directly use that forumla. I kind of did a graphical vector addition. But I think it probably gives the same result as that forumla.

I was assuming that the PF would be off by the same amount on both phases. So I think that formula should still be pretty close to the real result.

Steve
 
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