# 208v lighting?



## do0f4i (May 19, 2005)

can anybody school me on a 208v lighting system or 208v systems period? 

and why you might need a 2 pole breaker for a 208v if there is only one phase on the panel that has 208v to ground. 
and where would the neut go? or do you even need one 


only reason why im asking is because i did a job a grocery store and the other electrician that was there before our company ran all the lighting in 208v on a single pole breaker with the neuts on the neut buss.which keep on blowing the capisators so our company rewired all the light fixtures to a 240 system. 

the lights are 400w metal helid  

thanx ,
confused


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## mdshunk (Mar 13, 2005)

Well, they could have used a single pole breaker if this was a 120/240 delta high leg service. The phase to phase voltage on a delta high leg service is 240 volts. The "A" phase to neutral is 120 volts. The "C" phase to neutral is 120 volts. The "B" phase to neutral is right around 208 volts (thus the high leg).

Normally, for 208 you think of a 120/208Y 3 phase service. The phase to phase voltage is 208, and any phase to neutral is 120 volts. This would require a double pole breaker to serve anything at 208 volts. This is why I guessed that they might have had a delta high leg service and were able to use a single pole breaker snapped into the B phase buss only. This would also explain how they were able to be served at 240 so easily, since getting 240 from a 120/208Y service would require the installation of a customer transformer. 

Yes, if the lights were designed to be served at 240 volts, and they were only getting 208, that would be hard on the starting components and the capacitor. Normally, commercial high bay lighting has changable taps on the ballast to serve it at any voltage you want. This must have not been the case in your situation.


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## do0f4i (May 19, 2005)

it was a 120\208\240\277v ballest on the light 

and from A phase to B phase i got 240v 
and from A phase to c phase i got 240v
and from B phase to c phase i got 240v

but from nuetrul to A phase or B phase i got 120v
and from nuetrul to C phase i got 208v what kind of transformer was that?


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## mdshunk (Mar 13, 2005)

do0f4i said:


> it was a 120\208\240\277v ballest on the light
> 
> and from A phase to B phase i got 240v
> and from A phase to c phase i got 240v
> ...


Yes, then that's a delta high leg service, often simply called 120/240 delta. You have to be careful with those. The arrangement you measured is actually a code violation. The high leg is supposed to be on the middle most phase (referred to as the 'B' phase). In your case, the high leg is on the C phase, which is a bit unusual but not unheard of. If you mistakenly add a circuit for a regular light or receptacle and snap your breaker into the C phase, you'll be sending 208 out. That makes for trouble. Don't do that. I'm a little puzzled why they refed the lighting that was on the 208 'C' phase when they could have just simply changed the tap? Must have been a reason. More $$'s for the EC, I suppose.


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## kpi (Apr 30, 2005)

My guess on why they refed the lights from a 240v circuit is because the high leg is NOT 208v. I dont recall specifically what the high leg voltage to gound is on a 120/240 delta wired transformer, but 170-190v rings a bell. Kirk


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## mdshunk (Mar 13, 2005)

kpi said:


> My guess on why they refed the lights from a 240v circuit is because the high leg is NOT 208v.


Your bell doesn't ring very true. The high leg of a delta system is the square root of 3 (1.73205081), devided by two, times the normal phase to phase voltage. This is:

Sqrt(3)/2]x[Vph-ph]=the 'high-leg

[1.73205081 / 2] x 240 = 207.846097 volts.

I don't doubt that you may have observed a delta high leg system with low high leg voltage if the transformers were improperly sized or very unevenly loaded. The normal high leg voltage is right at 208. Simple electrical math every electrician knows. Have a nice day.


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## kpi (Apr 30, 2005)

mdshunk you are correct and I was wrong. 
After thinking about it, I would figure the sum of the voltages (neut-A phase and A-B phase) divided by 1,73 as a formula. I am curious how your formula was derived. Please be nice to me as it has been a loooooong time since my last electrical class. 
BTW, you are also correct about me observing voltage much less than 208v on a 120/240v delta connected transformer. (most likely the first time that I ever encountered it and did not give it much thought since then...untill now)
Thans in advance, Kirk


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## mdshunk (Mar 13, 2005)

kpi said:


> After thinking about it, I would figure the sum of the voltages (neut-A phase and A-B phase) divided by 1,73 as a formula. I am curious how your formula was derived.


Phase to neutral x 1.73 is the formula to find phase to phase voltage of a "Y" connected system
Phase to phase / 1.73 is the formula to find the phase to neutral voltage of a "Y" connected system
Phase to phase x (1.73 / 2) is the forumua to find the high leg phase to neutral voltage of a delta connected system
Phase to phase / 2 is the formula to find the normal phase to neutral voltage of a delta connected system


I notice you used a comma instead of a period in "1,73". Are you not in North America? Just wondering.


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