I got a string of LED Xmas lights hung behind my TV connected with an extension cord. I want to buy a power bar with a switch to turn it off but for now, I just unplug the strand from the extension cord and to keep the extension plug from falling I plug the lights using only one prong leaving the other open. At first, I thought it was just a reflection but in complete darkness and with my hand cupped, I can see all the LEDs are lit. Not much but you can easily tell that they are on.

For this to work a 120v to 9v DC plug in block adapter is plugged in to the string at the other end. But that doesn't get power either as it is dependent on the power coming from the string of lights.

So why is the string of lights powered with 1 prong of the plug disconnected?

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    \$\begingroup\$ welcome to energy harvesting from electric field coupling and stray leakage. The frequency of the DCDC converter noise makes it more sensitive to smaller capacitance/ \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 14 at 1:53

You have electric fields, at 120 volts RMS or 160 volts peak, at 60Hz(377 radians/second), coupling tiny currents thru the various wire-to-wire capacitances in that part of your house.

Assume a 1 meter wire (1mm diameter) is only 10mm away. Assume parallel-plate capacitance model, because when you have a jumble of wires, there is no sense to use the standard free-space wire-to-wire equations; besides the parallel-plate capacitance is low, and therefor conservative, compared to wire-to-wire maths.

Compute capacitance: C = (E0 * Er) * Area/Distance

C = 9e-12 Farad/meter * (1meter * 1mm) / 10 mm = 0.9pF

The current is computed as I = C * dV/dT

dV/dT is the slewrate of the voltage waveform. We'll assume this is pure sinusoid; the local behavior of any black-bricks switching supplies may greatly boost this slewrate.

The slewrate of 160 volt sin, at 377 radians/second, is approx. 150*400 = 60,000 volts per second.

Now use the I = C * dV/dT

I = 0.9 pF * 60,000 volts/second = 54 * e-12 * e+3 = 54 e-12+3 = 54 e-9

I = 54 nanoAmps

I = 0.054 microAmps, assuming PURE SINUSOIDAL waveforms. And parallel-plate coupling model.

ON THE OTHER HAND, if that black-brick is working, and providing spikes, such as 200 volts in 200 nanoseconds, then the slewrate has increased from 60,000 volts/second to ONE BILLION volts per second.

In that case, all that charging and discharging will increase by 1,000,000,000 / 64,000 or about 16,000X. To the level of 1 milliamp.

Perhaps you can provide a diagram of the wirings.

  • \$\begingroup\$ I can't really provide more with a diagram as it's just a parallel connection of a string of LED lights to the power adapter both plugged to the wall by only one of 2 wires. That's the whole thing. The coil and capacitance and diodes sounds like a crystal radio setup that lights up 30 LED from the 60hz agitation? \$\endgroup\$ – Eric Huelin Jan 14 at 1:46
  • \$\begingroup\$ @EricHuelin it’s a diagram of all the wires around the area where those lights are, in the walls, roof or floor - read what was suggested in the answer. \$\endgroup\$ – Solar Mike Jan 14 at 4:54

AC doesn't need a continuous conductor to transfer energy, current can travel through capacitance. It wouldn't be too difficult to have a capacitance from the string trough the air to other objects in the pf to nf range, which could generate a current in the uA to mA range enough to light the lights.

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    \$\begingroup\$ @ laptop2d Excellent summary "AC doesn't need a continuous conductor to transfer energy; current can flow through capacitance". Thanks for that. \$\endgroup\$ – analogsystemsrf Jan 15 at 12:25

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