In my bedroom, I have a ceiling LED light with three switches. One at the door, and one at each side of the bed.

The LED light contains a constant current supply (simple switching supply), which failed these days. Until the new one arrives, I have replaced the light with a simple fluorescent light bulb in a simple socket.

Now, I hear a quite ticking at 25Hz from the bulb when the light is OFF. Further investigation shows that the neutral line and earth is OK, there is no voltage between them. But the hot gives 40V with the bulb connected, and 70V without. (And yes, the switches act on the hot wire!)

So OK, this clearly indicates some capacitive coupling from other cables in the walls.

My question is:
Can those low voltages harm power supplies or other devices over time? It's already enough to let the fluorescent light tick at 25Hz. I could imagine that this has also caused the death of my LED light, though this could also be just by chance.


2 Answers 2


"Voltage when turned off" effects could come from capacitance or "leakage". If yo assume a 10 megOhm meter impedance, and 110 VAC supply then measuring 70 VAC open circuit with only meter load suggests that
R_leakage_equivalent = (Vmains-Vmeter) / Vmeter * Rmeter
= (110-70)/70 x 10 M ~= 5.7 megOhm. ie say
5 megOhm with a 10 megOhm input resistance meter. 50 megOhm with a 100 megOhm input resistance meter.

Very roughly - if capacitive coupling is causing this then, very arrproximately, about 0.5 nF would cause this degree of leakage with a 10 megOhm input resistance meter.

Death-watch-beetle like lightbulb ticking:

Either capacitive coupling or leakage resistance will cause reservoir capacitors in electronically driven lightbulbs (CFL or LEDs) to charge. This is well known to cause ticking and/or flashing with CFL bulbs as the charged capacitor triggers the bulb circuitry but there is not enough current to maintain it. I have not heard or seen this happening with LED bubs but the same principle applies. Designers my by now have taken precautions to prevent this. eg a small load at voltages below firing point which are enough to drain stray capacitance or leakage charging would be "easily enough" implemented.

Capacitive coupling CAN cause harm in specific situations, but it is usually when purposefully added capacitance is added.

Electronic mains driven supplies often have "Y" capacitors connected phase to ground and neutral to ground. If the ground lead is grounded these serve their intended function as noise filters. If the "ground" lead is not returned to installation ground the device ground will connect to the midpoint of the two Y capacitors across the supply. In the absence of any loading a high impedance meter will measure about Vmains/2 RMS relative to either neutral or phase.

Y capacitors are typically 1 nF (0.001 uF). This is small enough to not do much harm most of the time and to destroy some equipment sometimes. Long long ago I had a brand new printer destroyed the first time it was connected to a PC via a printer cable. The printer had Y capacitors but a 2 wire plug. The Y capacitors were terminated on the printer ground. The PC had a 3 wire cord. Printer death ensued.

The Y capacitor current is theoretically too small to kill you BUT will give a nasty "nip".

Stray capacitance from cabling will generally not be an issue but will cause measurable voltages with high impedance meters. Capacitances are in the order of nFs per km. Most device cables and house wiring runs are << 1 km.

  • \$\begingroup\$ I agree to all details of your answer concerning the use and side effects of Y-caps but I don't understand, how the Y capacitance of a device could bring voltage into a hot wire of a branch which is switched off. \$\endgroup\$
    – Ariser
    Commented Sep 6, 2016 at 8:34
  • \$\begingroup\$ @Ariser Agree. I'd also covered leakage resistance at the end. I moved my leakage resistance comment from end to start and added to explanation. \$\endgroup\$
    – Russell McMahon
    Commented Sep 6, 2016 at 21:32

Capacitive Coupling should not harm devices.

But considering, you can hear some sound from your bulb, there's a slight possibility, that some components will be stressed. What factors can theoretically lead to failure?

Warning: highly speculative, as it depends on the design of the used circuitry

  • triggering a switch on. Every time the hot wire exceeds some triggering voltage level, the electronics try to start the circuit. They eventually fail, because after applying real load to the hot wire, the voltage collapses and the circuitry looses power before all buffering capacitors get properly loaded. You can try to verify that using a good scope. This switch on process may be deteriorating to some of the components.
  • deformed waveforms. A capacitive coupled voltage has a high internal impedance if you want to put it that way. Circuitry with a lot of semiconductors and capacitors as LED bulbs will deform the waveform easily leading to sharp spikes i.e. the derivative of the voltage on the hot wire is partially very high. This can be heard, because it poses additional mechanical stress on capacitors and inductors. Perhaps this mechanical stress also increases the probability of failure.
  • Perhaps you don't deal with capacitive current but with leakage due to damaged wiring. From your statement that you still measure 40 V when a load is applied, I think leakage currents are within the realms of possibility. You should rule that out, soon! Leakage currents may damage the bulb and your health as well.

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