120/208 Shared Neutral Current Additive?

donniet1977 · Dec 9, 2013

I'm remodeling my kitchen and am bringing it up to code. My service is 120/208. I wired up my first multiwire circuit. I decided to read neutral current flow with a clamp on ammeter to make sure everything was cool. I had a 1 amp load and a 2 amp load on the two hots. I expected to see something less than 2 amps on the neutral but was surprised to see the full 3 amps. The hots are fed from opposite busses. Can anyone help me figure this one out?

Thanks,

Don

crutschow · Dec 9, 2013

The only thing I can think of is that you are not really wired to the two opposite phases of the service input. If you measure the voltage between the two hot leads you can determine that. It will either measure 0V if from the same phase or 240V if from the opposite phase.

donniet1977 · Dec 9, 2013

crutschow said:
The only thing I can think of is that you are not really wired to the two opposite phases of the service input. If you measure the voltage between the two hot leads you can determine that. It will either measure 0V if from the same phase or 240V if from the opposite phase.

I read 208 which is expected on a 120/208 service. I would think that since the phases are not 180 degrees out like in 240V service I wouldn't see a 1 to 1 reduction in neutral current but still see a reduction none the less.

richard.cs · Dec 9, 2013

donniet1977 said:
I read 208 which is expected on a 120/208 service. I would think that since the phases are not 180 degrees out like in 240V service I wouldn't see a 1 to 1 reduction in neutral current but still see a reduction none the less.

This is exactly what you should expect but only for linear loads, harmonic currents add in the neutral rather than cancelling. What were your two loads for this test? If they were electronic things which rectify the incoming power that might explain it.

mcgyvr · Dec 9, 2013

This is exactly why you should hire a qualified electrician..

donniet1977 · Dec 9, 2013

mcgyvr said:
This is exactly why you should hire a qualified electrician..

Supposedly a qualified electrician wired the panel in the first place since it is original. Anyhow here is an update:

Quick recap: My service 120/208, hot to hot is 208.

With an 8.5 amp toaster oven on one circuit and 11 amps on the other circuit which feeds the fridge (about 1 amp) and space heater (temp) my neutral current reads 19.2 amps. I'm using an Amprobe ACD-15 Pro clamp on ammeter. This condo has been wired with mostly MWBCs ever since it was built from what I can tell. Have the neutrals been overloaded the whole time?! I'd really like to get on the feed to my panel which is just a subpanel but I just don't have the room.

Lol, someone caught me. I did post this on 2 or 3 other forums. I figure that enough quantity is bound to generate some quality.

Thanks,

Don

donniet1977 · Dec 9, 2013

I'm both embarrassed and happy to say that after further troubleshooting I discovered that the outlet we wired temp for testing was miswired and being fed by the wrong circuit. This led to all loads being on the same leg. I was a little lazy in my initial troubleshooting and made too many assumptions. One assumption was that there was a problem in the first place.

I think the initial problem I saw with the 3 amp neutral was simply due to low inductive loading along with meter creep.

Final Reads Ignoring Fridge (<1 amp) which is always on:

CKT 1 on 2 off - In=9.1A
CKT 1 off 2 on - In=11.9A
Both CKTs on - In=12.0A

Thanks for the replies.

madmantrapper · Dec 10, 2013

With both circuits on, the neutral should only read the imbalance.

donniet1977 · Dec 10, 2013

madmantrapper said:
With both circuits on, the neutral should only read the imbalance.

After reading this thread, its obvious that i'm no expert but the neutral would only carry the imbalance of the phases if they were 180 degrees out. In this case they were 120 degress out hence the 120/208 and not 120/240 service.

I have also learned that inductive loads and their affect on power factor will throw off the simple imbalance calculations.

Don

WBahn · Dec 10, 2013

If the neutral is grounded, which it is in your case, then it will always carry the net imbalance between the phases. It has no choice. In = Iph1 - Iph2.

The problem is that the currents in each phase are, in general, not in phase with each other. If they are purely resistive, you can assume they are in phase with their respective voltages and figure out what the current in the neutral should be. But if either (or both) currents have a power factor, then they are not in phase with their respective voltages and all bets are off unless you have a means of either measuring the phase or measuring the power factor. If you have a wattmeter on each phase and you measure the RMS current in each phase, then you can do this.

For those that might be reading this (perhaps some distant time in the future) and wondering why 208V and not 240V, the key (as touched on by the OP in the prior post) is that the supplies are derived from two phases of a three phase transformer as opposed to being derived from a single-phase.

When tapped off a single phase using a center-tapped transformer, you end up with two "phases" that are 180 degrees out of phase and the line-to-line RMS voltage is the sum of the two line-to-neutral (center tap) RMS voltages.

But with pulling the lines off the phase voltages of two of the three Y-connected transformer terminals and the neutral of the neutral point between the phases, you basically have the following, relative to the neutral:

Vph1 = Vo @ -60°
Vph2 = Vo @ +60°

Where Vo is the phase-to-neutral voltage (the phase voltage) which, in this case, is 120V.

The line-to-line voltage is simply:

Vl = Vph2 - Vph1
Vl = (Vo @ +60°) - (Vo @ -60°)
Vl = Vo [ (cos(+60°) + jsin(+60°)) - (cos(-60°) + jsin(-60°)) ]
Vl = Vo [ (cos(+60°) + jsin(+60°)) - (cos(60°) - jsin(60°)) ]
Vl = Vo [ cos(+60°) + jsin(+60°) - cos(60°) + jsin(60°) ]
Vl = Vo [ j2sin(+60°) ]
Vl = Vo*2sin(60°) @ 90°

sin(60°) = √3 / 2

Vl = √3 * Vo @ 90°

If Vo=120V, then Vl = 208V

donniet1977 · Dec 11, 2013

WBahn said:
If the neutral is grounded, which it is in your case, then it will always carry the net imbalance between the phases. It has no choice. In = Iph1 - Iph2.

The problem is that the currents in each phase are, in general, not in phase with each other. If they are purely resistive, you can assume they are in phase with their respective voltages and figure out what the current in the neutral should be. But if either (or both) currents have a power factor, then they are not in phase with their respective voltages and all bets are off unless you have a means of either measuring the phase or measuring the power factor. If you have a wattmeter on each phase and you measure the RMS current in each phase, then you can do this.

For those that might be reading this (perhaps some distant time in the future) and wondering why 208V and not 240V, the key (as touched on by the OP in the prior post) is that the supplies are derived from two phases of a three phase transformer as opposed to being derived from a single-phase.

When tapped off a single phase using a center-tapped transformer, you end up with two "phases" that are 180 degrees out of phase and the line-to-line RMS voltage is the sum of the two line-to-neutral (center tap) RMS voltages.

But with pulling the lines off the phase voltages of two of the three Y-connected transformer terminals and the neutral of the neutral point between the phases, you basically have the following, relative to the neutral:

Vph1 = Vo @ -60°
Vph2 = Vo @ +60°

Where Vo is the phase-to-neutral voltage (the phase voltage) which, in this case, is 120V.

The line-to-line voltage is simply:

Vl = Vph2 - Vph1
Vl = (Vo @ +60°) - (Vo @ -60°)
Vl = Vo [ (cos(+60°) + jsin(+60°)) - (cos(-60°) + jsin(-60°)) ]
Vl = Vo [ (cos(+60°) + jsin(+60°)) - (cos(60°) - jsin(60°)) ]
Vl = Vo [ cos(+60°) + jsin(+60°) - cos(60°) + jsin(60°) ]
Vl = Vo [ j2sin(+60°) ]
Vl = Vo*2sin(60°) @ 90°

sin(60°) = √3 / 2

Vl = √3 * Vo @ 90°

If Vo=120V, then Vl = 208V

I thought it would only carry the net imbalance if they were 180° out. Since they are 120° out I would not expect it. This is backed up by my readings.

WBahn · Dec 12, 2013

donniet1977 said:
I thought it would only carry the net imbalance if they were 180° out. Since they are 120° out I would not expect it. This is backed up by my readings.

If they are not balanced, and if the net imbalance is not carried by the neutral, then where does that current imbalance go?

inwo · Dec 12, 2013

WBahn said:
If they are not balanced, and if the net imbalance is not carried by the neutral, then where does that current imbalance go?

The neutral does carry the imbalance.

The neutral carries MORE than the unbalance however.

If neutral current is zero on a 208 two wire multi-circuit.

Then the two 120 volt circuits would essentially be is series across the 208.

This seems impossible as measured voltage can't be 208 and 240.

There can be no conditions where neutral current is zero and phase currents are more than zero.

I might very well be wrong!

WBahn · Dec 12, 2013

inwo said:
The neutral does carry the imbalance.

The neutral carries MORE than the unbalance however.

How could it do that without violating KCL?

If neutral current is zero on a 208 two wire multi-circuit.

Then the two 120 volt circuits would essentially be is series across the 208.

This seems impossible as measured voltage can't be 208 and 240.

There can be no conditions where neutral current is zero and phase currents are more than zero.

I might very well be wrong!

Well, let's walk into it bit by bit.

Let's say I have a socket that has the two line voltages and a neutral. We have 120V across each line to neutral but 208V between line and line.

Now I connect a resistor from line to line. I have nothing connected to the neutral. Haven't I just created a situation in which I have non-zero phase currents but zero neutral current?

inwo · Dec 12, 2013

That's true.

I didn't word that well at all.

What I meant was, any 120 volt load will have neutral current. Even if the 120 volt loads are the same on two phases.
The extra power has to come from some place.
The two loads are using 120v times current. 2 X 120 X I.
The phases are using, if no neutral current, total power = 208 X I.

May be wrong again. I'll keep trying til I understand it.

WBahn · Dec 13, 2013

Nope, you are definitely onto something and asking the right questions and homing in on the key to the apparent descrepancy.

So spend some time with this:

That resistor that I previously placed across the 208V line-to-line supply, make it two equal resistors of size R in series, okay. Now, by symmetry, the voltage at the junction of the resistors has to be half way between the two line voltages, right? This would lead you to believe that this point must therefore be at the neutral potential, right? Seems obvious. If that were true, then we should be able to connect the neutral to that point and have no current flow, right? But if we did that, then we would go from having a single 2R load across 208V to having just the situation you were talking about, which is two 1R loads each across 120V. The only way to explain it is that, as soon as we connect the neutral to the center point that a neutral current DOES flow, which means that we need to look at the voltage at the center point of the two resistors (when not connected to the neutral) a bit more carefully.

GetDeviceInfo · Dec 13, 2013

you should agree on the definition of neutral. Next to ground, it's probably the most confused. Neutrals do not exist in 2 wire circuits, so the statement

What I meant was, any 120 volt load will have neutral current. Even if the 120 volt loads are the same on two phases.

is incorrect. Neutrals, by definition, only carry the imbalanced current.

If the neutral is grounded, which it is in your case, then it will always carry the net imbalance between the phases. It has no choice. In = Iph1 - Iph2.

Grounding has nothing to do with the neutral definition, nor it's calculated currents.

The extra power has to come from some place.
The two loads are using 120v times current. 2 X 120 X I.
The phases are using, if no neutral current, total power = 208 X I.

The 'extra' power comes from the fact that you are employing different voltages. These supplies are typically identified as 120/208Y.

inwo · Dec 13, 2013

WBahn said:
Nope, you are definitely onto something and asking the right questions and homing in on the key to the apparent descrepancy. I don't yet see a discrepancy.

So spend some time with this:

That resistor that I previously placed across the 208V line-to-line supply, make it two equal resistors of size R in series, okay. Now, by symmetry, the voltage at the junction of the resistors has to be half way between the two line voltages, right? ...Yes. Seems to me. Could I be wrong here............This would lead you to believe that this point must therefore be at the neutral potential, right? ...No, it can't be. .......... Seems obvious. If that were true, then we should be able to connect the neutral to that point and have no current flow, right? But if we did that, then we would go from having a single 2R load across 208V to having just the situation you were talking about, which is two 1R loads each across 120V. ...That's what I've been saying.......The only way to explain it is that, as soon as we connect the neutral to the center point that a neutral current DOES flow, .....Isn't that what I'm saying? ...........which means that we need to look at the voltage at the center point of the two resistors (when not connected to the neutral) a bit more carefully.....It has to be equally divided. Half of 208 across each resistor.............

There has to be voltage from center of resistors to neutral.
Current flows!

Are you changing your story now? It seems you're taking my side.
There either will be neutral current in a balanced multi-circuit, as I picture it, or there is not.

GetDeviceInfo · Dec 13, 2013

inwo said:
There has to be voltage from center of resistors to neutral.
Current flows!

Are you changing your story now? It seems you're taking my side.
There either will be neutral current in a balanced multi-circuit, as I picture it, or there is not.

In a three phase system, the neutral point will have a sum of zero, if the phases contribute equally. Remove one of those phase loads and the neutral carries that vectorial difference. Add your vector currents and tell us what the neutral current will be.

donniet1977 · Dec 13, 2013

GetDeviceInfo said:
In a three phase system, the neutral point will have a sum of zero, if the phases contribute equally. Remove one of those phase loads and the neutral carries that vectorial difference. Add your vector currents and tell us what the neutral current will be.

Doesn't this support my readings then? I'm using 2 phases of a 3 phase system. Even if my 2 phases, A and B, are carrying equal current, the neutral will not carry 0 current because the C phase vector (which I'm not using) of 0 amps needs to be included in the calculation. This makes sense because A and B are 120° and not 180 out of phase.

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120/208 Shared Neutral Current Additive?

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120/208 Shared Neutral Current Additive?

donniet1977

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donniet1977

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madmantrapper

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WBahn

inwo

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inwo

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inwo

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