Wire Size/Load Question

Neutral load is part of the load calculation but it should also be noted that the neutral in a standard triplex cable is not insulated so it will always look smaller from the ground. The utility side is usually smaller than the customer side too. They use the "free air" column of the ampacity chart and they are on the NESC, not the NEC.

So it's a power dissipation issue and not just circular mils/voltage drop per foot issue? The neutral wire should be twice the circular mils(area) from a voltage drop standpoint if the load is all 120 gadgets, right?

The neutral only has to carry the unbalanced difference between the other phases. So if each hot wire carries 40 amps, the neutral has nothing.

In a more realistic situation of 50 on one and 30 on the other, the neutral only (if you're talking a single phase 120/240 set of circuits) carries 20 amps.

(If you're talking about a neutral as part of a three phase circuit, the math is a bit trickier, but the concept is the same).

Note that this use of smaller neutrals is frowned upon by many current local codes, and may have been dumped in the latest National ones. (I don't have a copy at hand to double check).

Nope. The neutral only has to carry the DIFFERENCE between the currents drawn off the 2 hot wires. When the branch circuits are connected to the panel, the electrician SHOULD distribute the loads equally between the 2 sides of the panel, minimizing neutral current.

Bob Weiss N2IXK

DUH, thanks.

Unless you have connected both hot wires to the same hot coming from the transformer, the neutral wire need be no larger then the larger of the two hot wires ... they should all three be the same. The neutral wire carries only the imbalance between the two sides. Consider putting a 100 watt bulb on one hot-to-neutral, and a 100 watt bulb on the other hot-to-neutral as the only load. Disconnect the neutral in this case and both lights remain lit at the same brightness. The neutral is lame in this condition and is carrying no current. Shut one of those lights off, and now the neutral is carrying the current for the remaining light (unless you left it open).

An open neutral is dangerous because in cases where the load is way out of balance, one side (the one with less load) will get more voltage than the other side. Just having a motor start up can do that. Or as in a case I personally saw, a clock on one side and a stove burner on the other side can heat up the clock more than the burner when the neutral is open.

Personally, I'd rather not have a center tapped neutral system like this. But many devices, such as electric stoves, depend on it by design.

Generally, you have to size the neutral for the maximum unbalanced loads. But keep in mind that utilities aren't always governed by the NEC. Depending on the jurisdiction, they are allowed to use their own 'engineering judgement' in calculating load diversity and equipment sizing to a greater degree than allowed by the NEC.

And I would add, depending on the application, harmonics may require the neutral be increased.

Sincerely,

Donald L. Phillips, Jr., P.E. Worthington Engineering, Inc.

145 Greenglade Avenue Worthington, OH 43085-2264

snipped-for-privacy@worthingtonNSengineering.com (remove NS to use the address)

614.937.0463 voice 208.975.1011 fax

|> Generally, you have to size the neutral for the maximum unbalanced |> loads. But keep in mind that utilities aren't always governed by the |> NEC. Depending on the jurisdiction, they are allowed to use their own |> 'engineering judgement' in calculating load diversity and equipment |> sizing to a greater degree than allowed by the NEC. | | And I would add, depending on the application, harmonics may require the | neutral be increased.

I see a lot of equipment (transfomers, switch panels) rated for 200% neutral current. But I can see scenarios where it could reach as high as 300%. Take, for example, an AC to DC power supply that uses three phases and a full wave rectification. Wouldn't these rectifiers only be conducting when the respective phase is the highest voltage? If this were happening and it were the only (very large) load, you'd have to make sure everything is well overrated for it (3x on neutral for wye, and 2x on each leg for delta). Does anyone even use stuff like that?

I understand the common switching power supply for computers switches on the current when the voltage is high enough to replenish the tank it keeps. So it would see the smaller the computer load relative to the power supply capacity, the narrower the current pulses would be. Once they are smaller than 50%, you could be going past the 200% mark on the neutral, if all your load is these power supplies. For a few computers in an office, maybe that would never happen. But for building a large data center, I suppose it could get serious.

So you maintain that, as the load drawn from the computer power supply DECREASES, the neutral load INCREASES.

Lets go with that. I'm open minded.

What is magic about the 50% current pulses? Why would a

The increased harmonic current can only happen in 3 phase line to neutral loads and not split phase

120/240 line to neutral loads. What happens is that the current is increasing as the voltage is decreasing along the waveform. At some point, the current really drops. If they all drew current proportional to their voltage, then the neutral currents would cancel. But the definition of non-linear load is that the current drawn is not proportional to the voltage at any given time. Therefore, you can get two harmonic waveforms adding current and the third waveform that normally subtracts current happens to be 0 at that time.

From what I read, the worst case possible current overdraw was 1.717 times the rated load current of one phase.

Search for "3 phase harmonic current" and you'll get a better explanation than the one I just gave.

-- Mark Kent, WA

No.

As the computer power supply CAPACITY increases over its ACTUAL LOAD, the period of time needed to refill capacitors back to full charge is a SMALLER portion of each AC cycle.

| Lets go with that. I'm open minded. | | What is magic about the 50% current pulses? Why would a

| The increased harmonic current can only happen in 3 phase line to neutral loads and not split phase | 120/240 line to neutral loads. What happens is that the current is increasing as the voltage is | decreasing along the waveform. At some point, the current really drops. If they all drew current | proportional to their voltage, then the neutral currents would cancel. But the definition of | non-linear load is that the current drawn is not proportional to the voltage at any given time. | Therefore, you can get two harmonic waveforms adding current and the third waveform that normally | subtracts current happens to be 0 at that time. | | From what I read, the worst case possible current overdraw was 1.717 times the rated load current of | one phase.

I've seen other figures from 1.5 to 2.0 times the rated load current. But these seem to be making assumptions of full cycle square waves which will have a uniform declining distribution of odd-order harmonics. Many loads are more radical than that, and can have narrow spikes in their current waveform, leading to an uneven distribution of harmonics that will more favor the odd multiple of 3 harmonics (3rd, 9th, 15th, 21st, 27th, 33rd, known as "triplens"). More of the energy will be in those harmonics that don't cancel out than in the non-multiple of 3 harmonics.

| Search for "3 phase harmonic current" and you'll get a better explanation than the one I just gave.

Or for "triplens".

I read in sci.engr.electrical.compliance that snipped-for-privacy@ipal.net wrote (in ) about 'Wire Size/Load Question', on Wed, 25 Feb 2004:

Yes. You have done well to work out these scenarios independently. But the situation isn't quite as bad a pure arithmetic addition of the current pulses in the neutral. The upper limit is about 200 %, not 300%.

But there is a balancing effect; these short pulses are associated with less than maximum loads on the power supplies. In other words, a 300 W supply would do this if loaded to say 100 W, but the currents involved would be related to 100 W, not 300 W.

Yes, it does. Smoke from the neutral cable.

In alt.engineering.electrical John Woodgate wrote: | I read in sci.engr.electrical.compliance that snipped-for-privacy@ipal.net | wrote (in ) about 'Wire Size/Load | Question', on Wed, 25 Feb 2004: | |>I see a lot of equipment (transfomers, switch panels) rated for 200% |>neutral current. But I can see scenarios where it could reach as high |>as 300%. Take, for example, an AC to DC power supply that uses three |>phases and a full wave rectification. Wouldn't these rectifiers only be |>conducting when the respective phase is the highest voltage? If this |>were happening and it were the only (very large) load, you'd have to |>make sure everything is well overrated for it (3x on neutral for wye, |>and 2x on each leg for delta). Does anyone even use stuff like that? | | | Yes. You have done well to work out these scenarios independently. But | the situation isn't quite as bad a pure arithmetic addition of the | current pulses in the neutral. The upper limit is about 200 %, not 300%. |>

|>I understand the common switching power supply for computers switches on |>the current when the voltage is high enough to replenish the tank it |>keeps. So it would see the smaller the computer load relative to the |>power supply capacity, the narrower the current pulses would be. Once |>they are smaller than 50%, you could be going past the 200% mark on the |>neutral, | | But there is a balancing effect; these short pulses are associated with | less than maximum loads on the power supplies. In other words, a 300 W | supply would do this if loaded to say 100 W, but the currents involved | would be related to 100 W, not 300 W.

But circuit sizing is not usually done for computers based on the power supply capacity, but rather, the actual DC load itself. So a computer with a 300 watt p/s and a 100 watt load could be considered a 100 watt computer. Now put 100 of those computers on one leg of a 208Y/120, and

100 more on the 2nd leg, and 100 more on the 3rd leg. If those computers are drawing power with a narrow enough pulse, you could see 250 amps on the neutral, but only 83.34 amps on each leg.

Keep in mind I'm only describing worst case scenarios which include both very narrow current pulses (and I'm not sure that happens very often) and the bulk of the load being this type (generally not an issue even in an office environment). Real life probably doesn't even see 150% very often.

Look at this problem the other way around. Assume that the 100 W per system is fixed. The situation becomes worse as the capacities of the typical system's power supplies are increasingly overrated.

The question is: If one rates the circuit neutral based upon the connected power supply ratings rather than the estimated system loads, will this overrating compensate adequately for the proportional increase (relative to real load) of the neutral harmonics?

| Look at this problem the other way around. Assume that the 100 W per | system is fixed. The situation becomes worse as the capacities of the | typical system's power supplies are increasingly overrated.

Good way to look at it.

| The question is: If one rates the circuit neutral based upon the | connected power supply ratings rather than the estimated system loads, | will this overrating compensate adequately for the proportional increase | (relative to real load) of the neutral harmonics?

If the current approached linear as the DC load approached power supply capacity, that might work. But I doubt that happens. Maybe there is a virtual point where it could. What someone would need to do is measure the characteristics of a wide sampling of power supplies to see if such a thing could be done. I suspect there is a lot of variation in quality and design that could affect this.

Fortunately, only the radical cases (like a data center with hundreds or more computers) would need to worry about it. There is a "safe harbor" that can be used if these measurements cannot be made, and that is to assume the 300% worst case, and derate the stuff with 200% neutral to 66% of rated capacity, and the rest to 33% (rather expensive).

I read in sci.engr.electrical.compliance that Paul Hovnanian P.E. wrote (in ) about 'Wire Size/Load Question', on Wed, 25 Feb 2004:

The same effect occurs to some extent when increased demands are made for 'hold-up' when the supply voltage decreases or is interrupted, even though there is no increase in system capacity (as opposed to 'capacitance'!); the conduction angle of the rectifiers is still decreased.

Yes. AFAIK, there is no condition in which the ratio of the r.m.s. triplen components to the **full load** fundamental current increases with decreasing conduction angle, at least down to 36 degrees, below which the inrush current becomes a big problem and it is, in any case, difficult to get a low enough total resistance in the rectifier circuit economically.

I read in sci.engr.electrical.compliance that snipped-for-privacy@ipal.net wrote (in ) about 'Wire Size/Load Question', on Wed, 25 Feb 2004:

The matter has been thoroughly investigated many times, largely in connection with the provisions of IEC/EN 61000-3-2 on limitation of harmonic current emissions into the public low-voltage supply. Practical ranges of rectifier conduction angle are approximately 36 degrees (below which inrush current becomes a problem) and 72 degrees (above which efficiency is much too poor).

Instead of derating, new neutral cable can be installed. Still not cheap, but at least the outcome of the expense is that the system has been uprated rather than downrated.