Pulling 120 using neutral and a leg of 208 off a double pole breaker.

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This kind of expands on the sticky thread at the top of this forum,
[h=3]Understanding the Neutral Conductor[/h]
So we're building a machine for a customer. Some of the equipment in the rack requires 208 since it pulls enough current that 120 would be a pain. They wanted to use a 120V UPS to run the computer equipment, and use 208 to run the large power supplies.

We normally don't do that, since modern computers all have universal supplies in them and it's much easier to wire with 208 in the entire rack and just use a 208V supply. We told them as much. They're probably going that way. But it got me thinking, can you run a 120V load on one leg of a double pole breaker, and still retain wire protection for the neutral lead using a double pole breaker?

With a split phase (residential) system where L1 and L2 are 180 degrees out of phase, the answer is obviously "yes" since the loads on each leg cancel eachother out:

180phase.png

But what about 208 where the phases are 120 degrees out? Can the current on the neutral leg exceed the breaker rating? The answer is no, it cannot. Here is the worse case scenario, where each leg has a full 20A load on it. The neutral return has the sum of both currents on it, and does not exceed the max of each leg (I'll be honest, I had to graph this before I got my head around it):

120phase.png

So that answers the question, no you cannot overload the neutral conductor by using each leg of the double pole breaker using max load to a single neutral conductor.

That said, I'm curious if anything in the NEC covers this? Can you install a 120V outlet on a double pole breaker?
 

don_resqcapt19

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I am not aware of any code rule that would prohibit the use if a 2 or 3 pole breaker for that application. 210.4(B) requires a simultaneous means of disconnect for multiwire branch circuits. That can be accomplished by using either multi-pole breakers, or single pole breakers with a handle tie.
 

winnie

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Springfield, MA, USA
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Electric motor research
Two points:

1) If the two loads on the two supply 'legs' have different power factor characteristics, then the current on the neutral could exceed the current on the supply legs. This would require almost intentionally building something to try to break the system, eg. putting a highly capacitive load on one leg and a highly inductive load on the other.

2) My understanding is that under the NEC, the neutral needs to be sized for the maximum calculated neutral load.

This can mean that the neutral is smaller than the ungrounded conductors, and thus not protected from overload by the breaker.

There was a discussion on this topic a few years back: http://forums.mikeholt.com/showthread.php?t=129759
The question raised there was the possibility of a grounded conductor even smaller than the EGC, the question raising the possibility that the breaker would provide neither overload nor short circuit protection...but I think the consensus was that you couldn't make the neutral quite that small.

-Jon
 

LarryFine

Master Electrician Electric Contractor Richmond VA
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Henrico County, VA
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This kind of expands on the sticky thread at the top of this forum,
Understanding the Neutral Conductor
Wow! My post became a sticky! :D

That said, I'm curious if anything in the NEC covers this? Can you install a 120V outlet on a double pole breaker?
Yes, you absolutely can. The wiring would be done the same as you would on a 120/240 1ph system.

The neutral wouldn't carry the "sum" of the two legs, just the equivalent of one. Suppose you had an equally-loaded 3ph circuit, say 20a per phase; zero neutral current. Now, subtract, say, 5a from just one phase; neutral current rises by that same 5a. Now, eliminate one phase completely; 20a neutral current, no more. Just don't make the neutral conductor any smaller than the others.
 
Wow! My post became a sticky! :D


Yes, you absolutely can. The wiring would be done the same as you would on a 120/240 1ph system.

The neutral wouldn't carry the "sum" of the two legs, just the equivalent of one. Suppose you had an equally-loaded 3ph circuit, say 20a per phase; zero neutral current. Now, subtract, say, 5a from just one phase; neutral current rises by that same 5a. Now, eliminate one phase completely; 20a neutral current, no more. Just don't make the neutral conductor any smaller than the others.

Well, it does carry the sum of the "instantaneous current". In other words, since it's AC, a snapshot at any moment in time, with a 20A load the wire is carrying anywhere from -20 to 20 amps. That's what I meant by sum, if you look at the two graphs, you can see the current as the voltage swings from + to - and back. The neutral leg carries the sum of those two.

If you're talking DC (as in your sticky post) it's the same as the 240V graph at 90 degrees (IE L1=20A, L2=-20A, Neutral = 0). I basically just expanded that to look at the sine wave, since I was curious if having things 120 degrees out of phase with full loads on two of the three legs to neutral would cause a current of more than 20A at any time in the neutral leg, which it does not.

Of course this assumes a resistive load, not a reactive load as someone else mentioned.

Good discussion on that one thread above with regards to undersized neutral conductors. Never really considered that, but of course you would need the same gauge wire as the hot legs.

Interesting topic nonetheless :)
 
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