Understanding 3 phase 480V lighting circuit

I understand that you would use 480V light fixtures as opposed to 277V because the 480V have a lower current draw, resulting in a more efficient system. What I don't understand is how this circuit works. In the following paragraphs, I'm using the term "supply" as current flowing to the load and "return" as current flowing back to the breaker panel.

In a simple single-phase lighting circuit, you would have two wires: say, 120V hot wire and a neutral wire. Current flows from the hot to the light fixture and back through the neutral, assuming conventional direction of current flow. This completes the circuit and you can calculate the wattage by measuring the current pulled and the voltage supplied.

However, I don't see a 3 phase lighting circuit with 480V light fixtures work like that. Consider a 3 ph system A, B and C. If the A and B phase are connected to the fixture in a "480V single phase" manner, which phase supplies and which one returns current? If A supplies and B returns, then what about a load that's connected across B and C? That would mean C supplies and B is the one carrying the return current at all times, or else you will have a conductor with two currents flowing in opposite directions.

OR is the fixture drawing current from both phases? If so, then that's not single phase. Also, wouldn't you use 3 single-pole breakers as opposed to 1 3-pole breaker to supply the 3 phases. Please advise.

• "supply" and "return" aren't really useful terms in multiphase systems. May 12 '18 at 21:39
• Done. Re-formatted the text and provide explanation for the context of "supply" and "return" terms I'm using in my text. May 12 '18 at 21:46
• They're still not useful. All three conductors (four if neutral is present) supply and return current. May 13 '18 at 9:07

I understand that you would use 480V light fixtures as opposed to 277V because the 480V have a lower current draw, resulting in a more efficient system.

The lamps aren't necessarily more efficient. It's just that there are less losses in power transmission.

In a simple single-phase lighting circuit, you would have two wires: say, 120V hot wire and a neutral wire. Current flows from the hot to the light fixture and back through the neutral, assuming conventional direction of current flow.

Correct.

This completes the circuit and you can calculate the wattage by measuring the current pulled and the voltage supplied.

You have to remember that they can be somewhat out of phase giving a power-factor < 1 which needs to be taken into consideration for power (watts) calculations but you are on the right track.

However, I don't see a 3 phase lighting circuit with 480 V light fixtures work like that. Consider a 3 ph system A, B and C. If the A and B phase are connected to the fixture in a "480V single phase" manner, which phase supplies and which one returns current? Figure 1. The voltage difference between two phases is a sinewave.

Since there is no neutral connection you can look at it either way. Current will alternate between A and B.

If A supplies and B returns, then what about a load that's connected across B and C?

Current will also alternate between B and C.

That would mean C supplies and B is the one carrying the return current at all times, or else you will have a conductor with two currents flowing in opposite directions. Figure 2. Two currents flowing in opposite directions cancel out as indicated at point (1) when the B phase is at zero volts (assuming resistive load). At other times B has to carry the sum of the return currents with a peak value when B is at maximum.

OR is the fixture drawing current from both phases? If so, then that's not single phase.

A load connected between two phases has only two wires and sees a sinusoidal voltage applied to it. As far as the load is concerned it has a single-phase supply. (It knows nothing about neutral.)

Also, wouldn't you use 3 single-pole breakers as opposed to 1 3-pole breaker to supply the 3 phases.

Single-phase breakers would leave the other phase connected and the potentially faulty circuit still live. Good practice would be to use a 2-pole breaker on each load or a 3-phase breaker to isolate everything at once.

• Thank you. The detailed explanation helped me clear some of my doubts. I realize part of my confusion is because I'm perceiving an AC circuit the same way as a DC circuit consisting of a battery and a bulb. What I understand from Figure 2 is that at any given time, any one of the three phases could be the "supply" or "return". Now let's assume you have a 10W light fixture connected across AB and another 10W connected across BC. Am I right in saying that load on phases A and C is 10W, whereas phase B is 20W? May 13 '18 at 22:10

Do you know any 3 Phase powered Luminaires? I don't. Neither do any other answers show it. They are only single phase lights, but there may be 3 phase distribution lighting to share load on 3 phases.

The main reason would be cost reduction on power distribution cables in a stadium.

If you had 480W 1A stadium lights on 480V delta 20A breakers each lamp is single delta phase and AB , AC or BC then one might choose to derate the breaker to 80% then,

20 x .8 = 16 amps per line.
16 / 1.732 = 9.2 amps per phase.
9 fixtures per phase x 3 = 27 fixtures.

The margin allows surge power start.

Any questions?

• I wonder if things like stadium lighting do deliberately use multiple phases (i.e. multiple lights on different phases) so as to reduce flicker, and help TV cameras? May 13 '18 at 8:50
• @HenryCrun I have heard that given as a reason for using multiple phases for fluorescent lighting in machine shops, where the stroboscopic effect can make mains-speed motors (e.g. lathes) look stationary. May 13 '18 at 9:10
• I would think the thermal time constant of MH stadium lights attenuates 100/120Hz flicker pretty well and LED stadium lights would need to be DC. We know fluorescent flickers but mainly I think it is for very long wire cost reasons and 3 phase balanced fixtures May 13 '18 at 12:09

You're forgetting this is AC. Even for a single phase system, half the time one line is the supply and the other the return, then it swaps round a fraction of a second later.

The same goes for 3-phase. If you connect a load between phases A and B, half the time A will have a higher voltage than B, and the other half B will have a higher voltage than A.

• Yes, as I mentioned in my comment to Transistor, my confusion stems from perceiving the current flow in an AC circuit as a simple DC current originating from a battery, going through a light bulb, and then back to the battery. I'm still trying to wrap my head around thinking of AC current flow as you and Transistor describe. May 13 '18 at 22:21