High Resistance in the Three-phase Neutral Circuit

Safeguarding Yourself: A Guide to Personal Safety with Electrical Installations This article is for interested utility engineers. Don't make experiments at home. This will lead to tamper and can damage home appliances. Opening of the terminal cover is allowed by authorized utility staff only.
If you work with energy meters, always maintain the general safety rulesSafeguarding Yourself: A Guide to Personal Safety with Electrical Installations.

Since three-phase 3 wire meters don't need a neutral, the following applies for three-phase 4 wire meters only.

The Problem

The energy meter shows for a longer period (> 30 s) a reverse current, though there is no any power generation on load side.

A possible reason can be a high resistance or an interruption in the neutral wiring. This can come from:

poor connection
too small wire cross-section
wrong or poor earthing

The Reason

Now it gets pretty much theoretical. We look at a wiring diagram first.

Three Phase Four Wire Connection, principle diagram

For sure, you remember the formula for the currents in three-phase meters with star connection.

\displaystyle I_{\mathrm{L1}} = \frac{U_{\mathrm{L1\,N}} – U_{\mathrm{N'\,N}}}{Z_{\mathrm{u}}}

\displaystyle I_{\mathrm{L2}} = \frac{U_{\mathrm{L2\,N}} – U_{\mathrm{N'\,N}}}{Z_{\mathrm{v}}}

\displaystyle I_{\mathrm{L3}} = \frac{U_{\mathrm{L3\,N}} – U_{\mathrm{N'\,N}}}{Z_{\mathrm{w}}}

This is simple Ohm's law.

Calculation for balanced load

Z = Z_U = Z_V = Z_W (all resistances are the same)

\displaystyle I_{\mathrm{L1}} = \frac{U_{\mathrm{L1\,N}}}{Z_{\mathrm{u}}}

\displaystyle I_{\mathrm{L2}} = \frac{U_{\mathrm{L2\,N}}}{Z_{\mathrm{v}}}

\displaystyle I_{\mathrm{L3}} = \frac{U_{\mathrm{L3\,N}}}{Z_{\mathrm{w}}}

\displaystyle I_{\mathrm{L1}} + I_{\mathrm{L2}} + I_{\mathrm{L3}} = \frac{ \left( U_{\mathrm{L1\,N}} + U_{\mathrm{L2\,N}} + U_{\mathrm{L3\,N}} \right) }{ Z } = 0

Now we see, what we all have expected. If we have balanced or symmetric load, the sum current is zero.

Calculation for unbalanced Load

Z_U ≠ Z_V ≠ Z_W (different resistances per phase)

If the impedance in Neutral (Z_N) is zero, we can calculate the current in Neutral (I_N) like this:

\displaystyle I_{\mathrm{N}} = I_{\mathrm{L1}} + I_{\mathrm{L2}} + I_{\mathrm{L3}} = \frac{U_{\mathrm{L1\,N}}}{Z_{\mathrm{u}}} + \frac{U_{\mathrm{L2\,N}}}{Z_{\mathrm{v}}} + \frac{U_{\mathrm{L3\,N}}}{Z_{\mathrm{w}}}

If Z_N ≠ 0, we have an impedance in Neutral. In this case the current in Neutral is:

\displaystyle I_{\mathrm{N}} = I_{\mathrm{L1}} + I_{\mathrm{L2}} + I_{\mathrm{L3}} = \frac{U_{\mathrm{N'\,N}}}{Z_{\mathrm{N}}}

The current is particularly returned in the phases with lower load according to the Kirchhoff law. Now we have a reverse current. In case of missing neutral, the current will be distributed to the phases by 100%.

Neutral Point Shift

In addition, we have a Neutral point shift coming from the potential difference between the Neutral at transformer (N) and the Neutral at load (N').

We set up our known equitations first.

\displaystyle \frac{U_{\mathrm{N'\,N}}}{Z_{\mathrm{N}}} = \frac{(U_{\mathrm{L1\,N}} – U_{\mathrm{N'\,N}})}{Z_{\mathrm{u}}} + \frac{(U_{\mathrm{L2\,N}} – U_{\mathrm{N'\,N}})}{Z_{\mathrm{v}}} + \frac{(U_{\mathrm{L3\,N}} – U_{\mathrm{N'\,N}})}{Z_{\mathrm{w}}}

Then we resolve the above equitation by U_N'N.

\displaystyle U_{\mathrm{N'\,N}} = \frac{ \left( \frac{U_{\mathrm{L1\,N}}}{Z_{\mathrm{u}}} + \frac{U_{\mathrm{L2\,N}}}{Z_{\mathrm{v}}} + \frac{U_{\mathrm{L3\,N}}}{Z_{\mathrm{w}}} \right) }{ \left( \frac{1}{Z_{\mathrm{N}}} + \frac{1}{Z_{\mathrm{u}}} + \frac{1}{Z_{\mathrm{v}}} + \frac{1}{Z_{\mathrm{w}}} \right) }

Now, that we have done the Math, we look at the effect.

Green: mains side
Blue: load side
Red: shift

We see that the real problem of high resistance in Neutral is the shift of the Neutral point. This can lead to extreme phase-neutral voltage differences.

How to tackle the problem?

Check the wiring at the meter terminal. Be sure about your electrical power supply systemElectrical Power Supply Systems. Is the Neutral build from earthing only? In most cases, I saw temporary problems during a dry season when the Earth resistance is increasing.
Use an on-site tester with advanced features like analysis of phase angles, monitor the vectorial diagram and run eventually a long time record to limit the causes of the fault.

r Screenshot 20190514 162409 com.clou .rs350 — Long time measurement RS350

Portable Calibrator RS350 — Long time measurement RS350

What about the Energy Meter?

Our CLOU three-phase 4 wire energy meters have wide range power supply modules. They are powering up, even when the Neutral is missing.
Once the meter is powered up, it simply makes its job by samplingWaveform Sampling for Electrical Measurements the incoming values. The advanced meters have registers for bidirectional measurement.The four-quadrant diagram in electricity metering An arrow on the meter display shows flow in reverse directionNegative Phase Current.
I saw in some tender specifications that the meter should operate normal under missing neutral conditions. So, we did extensive tests. The operation is normal, but with error variations.
There is no standard for error limits under these abnormal situations. The general findings are, that the error variations are below 1% for missing neutral, if all three phases are present. If in addition one phase is missing and the power factor is < 0.5, the error can vary up to 3.5% from the regular value in any direction depending on the load-point.

Still, I have to say that missing neutral is a safety risk due to high occurring voltages. It's also utility responsibility to make sure that the resistance in neutral is low. Each country has regulations, in Germany are 10 Ohm impedance allowed for Neutral earthing.

Thank you for reading and stay safe

Editor's note: This article was originally published in January 2022 and has been updated for comprehensiveness.