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I have a < 10 year old house. And today I wanted to use the 2nd lamp outlet in the living room, and even though it was turned off, I got hit by electricity. I am in 230V, 50Hz country (Measured 227V). Measuring, I can see that I had 72V AC when the other lamp outlet (should be independent) was on, 50V AC when it was off. When I disconnect the mains wire for all my home TV/Stereo etc, the 50V goes up to 55V.

The voltage from the phase was measured both to Neutral and to GND. It is supposed to be air gapped when off according to all electricians I ask. And if it was connected, it should have 230V.

The other lamp outlet and this one, both have correspondance switches (i.e. turn on/off on either side of the wall). So their wires do run parallel for like 7 meters after the on/off switch. The probably shares the same phase.

I have received a 2 theories, one is the voltage is generated by electromagnetic induction.

The other that Switch mode PSUs are sending spikes out on Neutral through capacitors, that might be out of phase and frequency, such that the Neutral is where the power is. This sounds unlikely as there is around 1.2V AC between GND and Neutral.

I can say that I have switch mode PSUs on my wiring. And I have been told even my LED bulbs do have switch mode PSU.

Since a turned off lamp outlet, by all legal designs of correspondance of switches I have seen, should be air-gapped to the phase, I see no explanation where this ghost voltage is coming from.

I do consider shorting it to N so see if a fuse will be triggered (13A fuses). My electrician witht he switch mode PSU theory expects no fuse will blow.

Can I create a poor mans oscilloscope with an Arduino, a 47MOhm + 1 MOhm voltage divider and connect N to 0 and the divided voltage to A/D pin ? Use bluetooth module for data-connection to PC. Hoping to see irregularities/spikes in voltage to explain things - Supposedly I can get up to 3000 samples/sec, which should be more than sufficient for 50Hz. Should I use battery power, or just a 5V PSU ? The PSU probably uses the same Neutral anyway ?

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  • \$\begingroup\$ Both of those explanations are extremely unlikely to do something crazy like give you a shock. SMPS switch in the 10+ KHz range so you will certainly not see it on your proposed oscilloscope . I would suspect a neutral or phase fault on the misbehaving circuit or a miswiring. A neutral fault does not necessarily mean that the neutral wire is energized to 240V, depending on how and where the fault is there could be significant resistance that drops the voltage at the socket. \$\endgroup\$ – crasic Jan 7 '16 at 8:43
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    \$\begingroup\$ 1.2V between ground and neutral is believable. I don't know about where you are but in the US ground and neutral is tied together at one point at the junction box. Meaning if the neutral wire is carying a few amps on it and you are pretty far away from the junction box (say opposite side of the house), you would expect a small voltage, even up to 1V just due to the small resistance of the neutral wire \$\endgroup\$ – crasic Jan 7 '16 at 8:46
  • \$\begingroup\$ DMMs have a very high impedance, usually about 10Mohm. I would not be surprised to see several uA of induced current in premise wiring. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 7 '16 at 8:48
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    \$\begingroup\$ Either ask a question about possible currents and voltages on appliances that are supposedly "off" and forget about the Arduino question or vice versa. \$\endgroup\$ – Andy aka Jan 7 '16 at 10:22
  • \$\begingroup\$ See my answer. What country are you in? I'm in New Zealand. \$\endgroup\$ – Russell McMahon Jan 7 '16 at 10:49
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When you have a load that is controlled by two switches, there is a cable that runs directly between those two switches that contains two conductors. From your description, that cable is at least 10 meters long.

When the power to the load is off, one of those conductors is live and the other conductor is connected to the load. There is a significant amount of capacitive coupling between the two conductors — about 70-100 pF per meter — which will allow a tiny current to flow.

I have seen a situation in which a much shorter run between the switches (about 3 meters) passed enough current to cause several LED bulbs to glow dimly on a 120 VAC circuit even when they were supposed to be off. In this case, I calculated that the capacitance would allow about 13 µA to flow. In your case, with a longer cable and twice the voltage, you could be gettng something on the order of 80 µA.

The voltage you measure will be a function of the load impedance. With no other load than your multimeter, it is not surprising that you measured 72 VAC, and that you felt a significant shock.

The cure is to either disconnect the power from the circuit altogether (at the main distribution box), or to provide a low-impdeance path for the current to flow through. In other words, temporarily shorting the line to neutral would be perfectly fine. Just be sure to remove the short before you try to switch the power back on.

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  • \$\begingroup\$ In Denmark, the 2-switch configuration is legal only in the 2 configurations here: elviden.dk/uploads/1/0/3/6/10366661/3861434_orig.jpg That means 2 wires running between 2 switches, at least one connected to phase. Since the 2 switches are on each side of a wall, I guess they are connected together thru the 10cm concrete wall. Thus only 10 cm of parallel run. From the switch circuit, there is only Neutral + 1 phase wire running to the lamp in legal installation The solution where the bulb can be turned off by having both sides connected to neutral or phase is illegal. \$\endgroup\$ – povlhp Jan 7 '16 at 13:11
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This situation can occur if a device with "Y" filter capacitors is connected to mains AND if the device ground is not connected to Neutral or true system ground.

If you have a device connected that has "Y" EMI filter caps fitted from phase & neutral to ground then IF the ground is not connected to house ground and thence to Neutral then it will float at Vmains/2 with an impedance of Z-Y_cap /2 (1/(2.Pi.f.C)) from the 2 x Y capacitors.

If you measure this voltage with a meter from eg 'floating ground' to true ground or neutral you will get < Vmains/2 due to meter impedance loading the capacitor divider.

Voltage from this source is enough to give you an unpleasant shock but current is below what is meant to be heart-dangerous.

Voltages derived in this way can be enough to destroy equipment connected to other equipment in which the device ground is connected to earth. Long ago I had a new printer connected on 1st connection to a PC printer port due to lack of a ground on the PC and two Y capacitors in the PC floating its ground and associated signals lines at about Vmains/2.

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  • \$\begingroup\$ Seems plausible. I have local danish wall outlets designed for 3-pin plugs. Much equipment comes with Schuko wires, where ground is either a pin in the outlet, or springs in the outlet. Thus I am running some equipment non-grounded. \$\endgroup\$ – povlhp Jan 7 '16 at 11:31
  • \$\begingroup\$ Some people eliminate noise (like buzzing in fingers when touching a device) by turning the plug upside down. I will try to remove most/all non-grounded equipment and re-measure. Not sure why the independent E27 socketed LED bulb would increase voltage when on, E27 sockets allows only for 2 wires. So maybe I should switch wires in the lamp outlet. \$\endgroup\$ – povlhp Jan 7 '16 at 11:38
  • \$\begingroup\$ The shock level I mention is rather more than "buzzing" - it's quite a nasty bite. \$\endgroup\$ – Russell McMahon Jan 7 '16 at 12:01
  • \$\begingroup\$ What I got (the 72V) was also nasty. Let go of nuts and bolts, and pulled my hand. Not sure if I was aware what happened, or if I was out a split second but at least I did not fall down the ladder. Felt it in my fingers a little after. I have had worse and better chocks with 220V. Worst I could feel numbness in my arm for half an hour or so. \$\endgroup\$ – povlhp Jan 7 '16 at 12:11
  • \$\begingroup\$ Now I unplugged everything in my house, and the voltage stayed at 70-72V. So still no explanation. Next step is to trip fuse groups one at a time. \$\endgroup\$ – povlhp Jan 7 '16 at 16:31

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