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I saw a schematic for measuring AC voltage on a 3-phase mains supply (R/Y/B are the 3 phases and N is the neutral)

enter image description here

I understand that the 1 MΩ and 2.7 kΩ resistors are for stepping down the voltage (240 V) to a measurable voltage by the microcontroller (the voltage is measured across the two 2.7 kΩ resistors for each phase).

I can't understand the usage of 100 kΩ resistors. The possible explanation I could think of:

  1. Are the 100 kΩ resistors used to somehow avoid voltage fluctuations in case of neutral loss?
  2. Do the 100 kΩ resistors protect the circuit during neutral loss conditions?

Can anyone explain the issues that might arise if the 100 kΩ resistors were removed?

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    \$\begingroup\$ Best guess is artificial neutral. \$\endgroup\$
    – winny
    Jul 21 at 21:27

5 Answers 5

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It's often a bit of speculation as to what a designer had in mind, especially not knowing the measuring instrument or its purpose. Anyway, it appears that a differential measurement is used here, phase to neutral voltage. It seems that these resistors would be useful in detecting a loss of phase, as well as allowing operation in a single phase situation, by pulling the open phase to the neutral voltage.

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    \$\begingroup\$ The 100k resistors seem to be connected to a common point, labeled "MNEU". This may stand for "Manufactured Neutral", which would be at a potential very close to the actual neutral. There is a connection shown to "N" for the neutral, but these resistors may be there in case the actual neutral is not connected, or does not exist, as in a delta configuration. \$\endgroup\$
    – PStechPaul
    Jul 21 at 20:45
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    \$\begingroup\$ This makes a lot of sense particularly in relation to recent standards changes re, for example, EV charging-point safety re PEN failure in TN-C-S installations, etc. The comparing of supply neutral to a manufactured neutral and phase-loss detection are important for safety in some of these setups. My context is the UK, though, and R/Y/B suggests not the UK. But there's probably similar stuff going on in other countries. \$\endgroup\$
    – Dannie
    Jul 22 at 10:23
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I can't understand the usage of 100K resistors.

NB: I would not connect the lowest MNEU to Neutral, otherwise is no real need for these 100k resistors.
Note also that only one serial 4x 1Meg (at the lowest part of the picture) is useful. The voltage at the lowest part of R65, R73, and R81 can be the same.

Thought: These 100k resistors are used for creating a virtual "neutral" of the three phases.
You can then measure the "displacement" voltage of the Neutral wire.
Depending on what you want to measure, one can have the answer at the three differential outputs. But some "modified" (as stated before) wiring should be used.

enter image description here

Example of results ... Unbalanced load, V1 steps.

enter image description here

If somebody wants to enjoy this ... Run Transient Analysis.
Made with microcap v12.

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  • \$\begingroup\$ That's a good point: if it can be connected to Delta or Wye, this could be useful to measure / diagnose neutral currents, harmonic or other common-mode balance, etc. \$\endgroup\$ Jul 21 at 22:17
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The 100k's will add a very slight load to all phases.

If one of the phases were broken, it will then read significantly lower than the others*.

*Note it probably won't read exactly zero, due to AC coupling between conductors. Such would require a substantial load.

How do you choose [these] resistor's value?

Well lets start with Ohm's law: $$ E = I\cdot R $$ $$ I = \frac{E}{R} $$ $$ I = \frac{240V_{RMS}}{200k} = 1.2mA $$ $$ R_{pow} = V\cdot I $$ $$ 240V_{RMS}\cdot 1.2mA = 0.288W$$

So if each resistor is 1/4W (0.25W) and two are in series, the power dissipated by one is 0.144W. This is on-par with resistor wattage rating convention; that is, they should be rated at twice the dissipated power. Otherwise they get too hot and will fail, especially in hot environments or with inadequate air cooling.

So these values were likely chosen as a series of compromises:

  1. Two resistors in series increases the working voltage rating.
  2. Two smaller resistors are cheaper to mass-produce than one larger, higher-power resistor.
  3. 200k (1.2mA) is enough load to significantly "clamp" a lot of the noise and cross-talk coupled between AC conductors. Of course 5mA would work better, but then you'd need more resistors and then more power is wasted as heat.
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    \$\begingroup\$ Note loading is desirable for AC, since AC couples readily into adjacent wiring. Without these resistors, the output would be significantly noisier. \$\endgroup\$
    – rdtsc
    Jul 21 at 16:47
  • \$\begingroup\$ Yes, to be specific, the stray capacitance between phases can cause this effect. \$\endgroup\$ Jul 21 at 22:15
  • \$\begingroup\$ So if the purpose of the 100K resistors is to avoid fluctuations during a phase loss condition (which seems a very valid reason now), how do you choose the resistor's value? Why 100K and not 1M? Do the 2*100K resistance needs to be significantly lower than 4*1M? \$\endgroup\$ Jul 22 at 3:30
  • \$\begingroup\$ @SakshamJain Simple: the value is a function of frequency (including any harmonics of interest), acceptable error (how close to zero should it read?) and expected line-line capacitance (or coupled wire length, etc.). The circuit is a high-pass filter, from one line, series capacitance to the other line, to a shunt resistor to GND/NEU. \$\endgroup\$ Jul 22 at 10:52
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Assuming that schematic is for a measuring instrument, they're probably there to rapidly bleed off any voltage when J8 is disconnected, to prevent a technician from getting zapped if they disconnect then touch the conductors in J8.

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  • \$\begingroup\$ Why would there be any voltage once J8 is disconnected? I would assume the capacitance would be very low in resistors. \$\endgroup\$ Jul 21 at 15:16
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    \$\begingroup\$ There will be capacitance in the wiring, and there's probably some regulatory compliance issues that have to be met. \$\endgroup\$
    – TimWescott
    Jul 21 at 15:18
  • \$\begingroup\$ There may also be some requirement that a non galvanically isolated connection to mains voltages have maximum current well below accepted safety levels (about 5-20 mA for GFCI), and below threshold of sensation (about 1-5 mA). Two or more resistors in series also ensures redundancy and higher voltage rating. 200k will limit current to a peak value of about 2 mA to ground for a 480/277 VAC connection \$\endgroup\$
    – PStechPaul
    Jul 22 at 20:52
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"MNEU" = Mean Neutral voltage could have been an indication of load imbalance from load regulation error without current sensing but then they shorted it to NEU. so they are redundant resistors. Perhaps it helps on an open phase condition. But I would reckon the stray impedance of an open phase fault would be much lower reactance than 2x 100k , so it won't be 0 V.

As always, it's hard to guess the assumptions without specs.

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